1. ffirs.qxd 11/14/12 10:09 AM Page iv 2. For more information, visit www.wileyplus.com WileyPLUS builds students confidence because it takes the guesswork out of studying by providing students with a clear roadmap: It offers interactive resources along with a complete digital textbook that help students learn more. With WileyPLUS, students take more initiative so youll have greater impact on their achievement in the classroom and beyond. WileyPLUS is a research-based online environment for effective teaching and learning. Now available for what to do how to do it if they did it right ffirs.qxd 11/14/12 10:09 AM Page i 3. ALL THE HELP, RESOURCES, AND PERSONAL SUPPORT YOU AND YOUR STUDENTS NEED! www.wileyplus.com/resources Technical Support 24/7 FAQs, online chat, and phone support www.wileyplus.com/support Student support from an experienced student user Collaborate with your colleagues, find a mentor, attend virtual and live events, and view resources 2-Minute Tutorials and all of the resources you and your students need to get started Your WileyPLUS Account Manager, providing personal training and support www.WhereFacultyConnect.com Pre-loaded, ready-to-use assignments and presentations created by subject matter experts Student Partner Program Quick Start Courtney Keating/ iStockphoto ffirs.qxd 11/14/12 10:09 AM Page ii 4. Algebra and Trigonometry Third Edition ffirs.qxd 11/14/12 10:09 AM Page iii 5. ffirs.qxd 11/14/12 10:09 AM Page iv 6. CYNTHIA Y. YOUNG | Professor of Mathematics UNIVERSITY OF CENTRAL FLORIDA Algebra and Trigonometry Third Edition ffirs.qxd 11/14/12 10:09 AM Page v 7. PUBLISHER Laurie Rosatone ACQUISITIONS EDITOR Joanna Dingle PROJECT EDITOR Jennifer Brady ASSOCIATE CONTENT EDITOR Beth Pearson EDITORIAL ASSISTANT Elizabeth Baird SENIOR PRODUCTION EDITOR Kerry Weinstein/Ken Santor DESIGNER Madelyn Lesure OPERATIONS MANAGER Melissa Edwards ILLUSTRATION EDITOR Sandra Rigby; Electronic illustrations provided by Techsetters, Inc. SENIOR PHOTO EDITOR Jennifer MacMillan COVER DESIGN Madelyn Lesure COVER PHOTO Front Cover: Combined image: Ty Milford/Masterle and Leslie Banks/iStockphoto Back Cover: Rustem Gurler/iStockphoto Modeling Our World image: David Woodfall/Getty Images This book was set in 10/12 Times by MPS Limited, and printed and bound by Quad/Graphics, Inc. The cover was printed by Quad/Graphics, Inc. This book is printed on acid free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulll their aspirations. 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Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. ISBN: 978-0-470-64803-2 BRV ISBN: 978-1-118-12930-2 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ffirs.qxd 11/15/12 8:35 AM Page vi 8. The Wiley Faculty Network (WFN) is a global community of faculty connected by a passion for teaching and a drive to learn, share, and collaborate.Whether youre seeking guidance, training, and resources or simply looking to re-energize your course, youll nd what you need with the WFN. The WFN also partners with institutions to provide customized professional development opportunities. Connect with the Wiley Faculty Network to collaborate with your colleagues, nd a Mentor, attend virtual and live events, and view a wealth of resources all designed to help you grow as an educator. Attend Discover innovative ideas and gain knowledge you can use. Learn from instructors around the world, as well as recognized leaders across disciplines. Join thousands of faculty just like you who participate in virtual and live events each semester. Youll connect with fresh ideas, best practices, and practical tools for a wide range of timely topics. View Explore your resources and development opportunities. See all that is available to you when you connect with the Wiley Faculty Network. From Learning Modules and archived Guest Lectures to faculty-development and peer-reviewed resources, there is a wealth of materials at your ngertips. Collaborate Connect with colleaguesyour greatest resource. Tap into your greatest resourceyour peers. Exchange ideas and teaching tools, while broadening your perspective. Whether you choose to blog, join interest groups, or connect with a Mentoryouve come to the right place! Embrace the art of teaching Great things happen where faculty connect! Web: www.WhereFacultyConnect.com EMAIL: FacultyNetwork@wiley.com PHONE: 1-866-4FACULTY (1-866-432-2858) The Wiley Faculty Network T h e P l a c e W h e r e F a c u l t y C o n n e c t ffirs.qxd 11/14/12 10:09 AM Page vii 9. For Christopher and Caroline ffirs.qxd 11/14/12 10:09 AM Page viii 10. About the Author Cynthia Y. Young is a native of Tampa, Florida. She currently is a Professor of Mathematics at the University of Central Florida (UCF) and the author of College Algebra, Trigonometry, Algebra and Trigonometry, and Precalculus. She holds a B.A. degree in Secondary Mathematics Education from the University of North Carolina (Chapel Hill), an M.S. degree in Mathematical Sciences from UCF, and both an M.S. in Electrical Engineering and a Ph.D in Applied Mathematics from the University of Washington. She has taught high school in North Carolina and Florida, developmental mathematics at Shoreline Community College in Washington, and undergraduate and graduate students at UCF. Dr. Youngs two main research interests are laser propagation through random media and improving student learning in STEM. She has authored or co-authored over 60 books and articles and been involved in over $2.5M in external funding. Her atmospheric propagation research was recognized by the Ofce of Naval ResearchYoung Investigator award, and in 2007 she was selected as a Fellow of the International Society for Optical Engineers. She is currently the co-director of UCFs EXCEL program whose goal is to improve the retention of STEM majors. Although Dr. Young excels in research, she considers teaching her true calling. She has been the recipient of the UCF Excellence in Undergraduate Teaching Award, the UCF Scholarship of Teaching and Learning Award, and a two-time recipient of the UCF Teaching Incentive Program. Dr.Young is committed to improving student learning in mathematics and has shared her techniques and experiences with colleagues around the country through talks at colleges, universities, and conferences. Dr. Young and her husband, Dr. Christopher Parkinson, enjoy spending time outdoors and competing in Field Trials with their Labrador Retrievers. Lairds Cynful Wisdom (call name Wiley) is titled in Canada and currently pursuing her U.S. title. Lairds Cynful Ellegance (call name Ellie) was a nalist in the Canadian National in 2009 and is retired (relaxing at home). Dr.Young is pictured here with Ellies 2011 litter of puppies! ix Bonnie Farris ffirs.qxd 11/14/12 10:09 AM Page ix 11. Preface x As a mathematics professor I would hear my students say, I understand you in class, but when I get home I am lost. When I would probe further, students would continue with I cant read the book. As a mathematician I always found mathematics textbooks quite easy to readand then it dawned on me: dont look at this book through a mathematicians eyes; look at it through the eyes of students who might not view mathematics the same way that I do. What I found was that the books were not at all like my class. Students understood me in class, but when they got home they couldnt understand the book. It was then that the folks at Wiley lured me into writing. My goal was to write a book that is seamless with how we teach and is an ally (not an adversary) to student learning. I wanted to give students a book they could read without sacricing the rigor needed for conceptual understanding. The following quote comes from a reviewer of this third edition when asked about the rigor of the book: I would say that this text comes across as a little less rigorous than other texts, but I think that stems from how easy it is to read and how clear the author is. When one actually looks closely at the material, the level of rigor is high. Distinguishing Features Four key features distinguish this book from others, and they came directly from my classroom. PARALLEL WORDS AND MATH Have you ever looked at your students notes? I found that my students were only scribbling down the mathematics that I would writenever the words that I would say in class. I started passing out handouts that had two columns: one column for math and one column for words. Each Example would have one or the other; either the words were there and students had to ll in the math, or the math was there and students had to ll in the words. If you look at the Examples in this book, you will see that the words (your voice) are on the left and the mathematics is on the right. In most math books, when the author illustrates an Example, the mathematics is usually down the center of the page, and if the students dont know what mathematical operation was performed, they will look to the right for some brief statement of help. Thats not how we teach; we dont write out an Example on the board and then say, Class, guess what I just did! Instead we lead our students, telling them what step is coming and then performing that mathematical step togetherand reading naturally from left to right. Student reviewers have said that the Examples in this book are easy to read; thats because your voice is right there with them, working through problems together. E X AM P LE 1 Graphing a Quadratic Function Given in Standard Form Graph the quadratic function f(x) (x 3)2 1. Solution: STEP 1 The parabola opens up. a 1, so a 0 STEP 2 Determine the vertex. (h, k) (3, 1) STEP 3 Find the y-intercept. f(0) ( 3)2 1 8 FMPreface.qxd 12/28/12 11:00 AM Page x 12. LECTURE VIDEOS BY THE AUTHOR To ensure consistency in the students learning experiences, I authored the videos myself. Throughout the book wherever a student sees the video icon, that indicates a video. These videos provide a mini lecture in that the chapter openers and chapter summaries are more like class discussion and selected Examples. Your Turns throughout the book also have an accompanying video of me working out that exact problem. P R E FAC E xi SKILLS AND CONCEPTS (LEARNING OBJECTIVES AND EXERCISES) In my experience as a mathematics teacher/instructor/professor, I nd skills to be on the micro level and concepts on the macro level of understanding mathematics. I believe that too often skills are emphasized at the expense of conceptual understanding. I have purposely separated learning objectives at the beginning of every section into two categories: skills objectiveswhat students should be able to do; and conceptual objectiveswhat students should understand. At the beginning of every class I discuss the learning objectives for the dayboth skills and concepts. These are reinforced with both skills exercises and conceptual exercises. CATCH THE MISTAKE Have you ever made a mistake (or had a student bring you his or her homework with a mistake) and you go over it and over it and cant nd the mistake? Its often easier to simply take out a new sheet of paper and solve it from scratch again than it is to actually nd the mistake. Finding the mistake demonstrates a higher level of understanding. I include a few Catch the Mistake exercises in each section that demonstrate a common mistake that I have seen in my experience. I use these in class (either as a whole or often in groups), which leads to student discussion and offers an opportunity for formative assessment in real time. C O N C E P TUAL O BJ E CTIVE S Classify lines as rising, falling, horizontal, and vertical. Understand slope as a rate of change. Associate two lines having the same slope with the graph of parallel lines. Associate two lines having negative reciprocal slopes with the graph of perpendicular lines. L I N E S S K I LLS O BJ E CTIVE S Determine x- and y-intercepts of a line. Calculate the slope of a line. Find the equation of a line using slopeintercept form. Find the equation of a line using pointslope form. Find the equation of a line that is parallel or perpendicular to a given line. S E CTI O N 2.3 In Exercises 9598, explain the mistake that is made. 95. Solve the equation: 4ex 9. Solution: Take the natural log of both sides. ln(4ex ) ln 9 Apply the property of inverses. 4x ln 9 Solve for x. This is incorrect. What mistake was made? 96. Solve the equation: log(x) log(3) 1. Solution: x = ln 9 4 L 0.55 97. Solve the equation: log(x) log(x 3) 1 for x. Solution: Apply the product property (5). log(x2 3x) 1 Exponentiate both sides (base 10). Apply the property of inverses. x2 3x 10 Factor. (x 5)(x 2) 0 Solve for x. x 5 and x 2 This is incorrect. What mistake was made? 10log(x2 +3x) = 101 C AT C H T H E M I S TA K E fprep.qxd 11/14/12 10:10 AM Page xi 13. New to the Third Edition The rst edition was my book, the second edition was our book, and this third edition is our even better book. Ive incorporated some specic line-by-line suggestions from reviewers throughout the exposition, added some new Examples, and added over 200 new Exercises. The three main global upgrades to the third edition are a new Chapter Map with Learning Objectives, End-of-chapter Inquiry-Based Learning Projects, and additional Applications Exercises in areas such as Business, Economics, Life Sciences, Health Sciences, and Medicine. A section (2.5*) on Linear Regression was added, as well as some technology exercises on Quadratic, Exponential, and Logarithmic Regression. LEARNING OBJECTIVES xii P R E FAC E L E A R N I N G O B J E C T I V E S Find the domain and range of a function. Sketch the graphs of common functions. Sketch graphs of general functions employing translations of common functions. Perform composition of functions. Find the inverse of a function. Model applications with functions using variation. INQUIRY-BASED LEARNING PROJECTS C HAP TE R 4 I N Q U I RY-BAS E D LE AR N I N G P R OJ E CT 466 Discovering the Connection between the Standard Form of a Quadratic Function and Transformations of the Square Function In Chapter 3, you saw that if you are familiar with the graphs of a small library of common functions, you can sketch the graphs of many related functions using transformation techniques. These ideas will help you here as you discover the relationship between the standard form of a quadratic function and its graph. Let G and H be functions with: G(x) F(x 1) 3 and H(x) F(x 2) 4 where F(x) x2 . 1. For this part, consider the function G. a. List the transformation youd use to sketch the graph of G from the graph of F. b. Write an equation for G(x) in the form G(x) a(x h)2 k. This is called the standard form of a quadratic function. What are the values of a, h, and k? c. The vertex, or turning point, of the graph of F(x) x2 is (0, 0). How can you use the transformations you listed in part (a) to determine the coordinates of the vertex of the graph of G? d. The vertical line that passes through the vertex of a parabola is called its axis of symmetry. The axis of symmetry of the graph of F(x) x2 is the y-axis, or the vertical line with equation x 0. How can you determine the axis of symmetry of the graph of G? Write the equation of this line. e. Sketch graphs of F and G. 2. Next consider the function H given above. a. List the transformations that will produce the graph of H from the graph of F. b. Write an equation for H(x) in standard form. What are the values of a, h, and k? c. What are the coordinates of the vertex of the graph of H? How do the transformations you listed in part (a) help you determine this? d. Determine the equation of the axis of symmetry of the graph of H. e. Sketch graphs of F and H. 3. a. What do you know about the graph of a quadratic function just by looking at its equation in standard form, f(x) a(x h)2 k? b. Shown below are the graphs of F(x) x2 and another quadratic function, y K(x). Write the equation of K in standard form. Hint: Think about the transformations. x y (0, 0) (4, 6) y = F(x) y = K(x) APPLICATIONS TO BUSINESS, ECONOMICS, HEALTH SCIENCES, AND MEDICINE 49. Area. Find the area enclosed by the system of inequalities. 50. Area. Find the area enclosed by the system of inequalities. 51. Area. Find the area enclosed 10 by the system of linear inequalities 0 (assume y 0). 1 52. Area. Find the area enclosed by the system of linear inequalities (assume y 0). x 2 x 1 -5x + y 0 x x 5x + y x 6 3 y 0 x 0 y 6 x y 6 2 y 7 x 53. Hurricanes. After back-to-back-to-back-to-back hurricanes (Charley, Frances, Ivan, and Jeanne) in Florida in the summer of 2004, FEMA sent disaster relief trucks to Florida. Floridians mainly needed drinking water and generators. Each truck could carry no more than 6000 pounds of cargo or 2400 cubic feet of cargo. Each case of bottled water takes up 1 cubic foot of space and weighs 25 pounds. Each generator takes up 20 cubic feet and weighs 150 pounds. Let x represent the number of cases of water and y represent the number of generators, and write a system of linear inequalities that describes the number of generators and cases of water each truck can haul to Florida. 54. Hurricanes. Repeat Exercise 53 with a smaller truck and different supplies. Suppose the smaller trucks that can haul 2000 pounds and 1500 cubic feet of cargo are used to haul plywood and tarps. A case of plywood is 60 cubic feet and weighs 500 pounds. A case of tarps is 10 cubic feet and weighs 50 pounds. Letting x represent the number of cases of plywood and y represent the number of cases of tarps, write a system of linear inequalities that describes the number of cases of tarps and plywood each truck can haul to Florida. Graph the system of linear inequalities. A P P L I C AT I O N S 55. Health. A diet must be designed to provide at least 275 units of calcium, 125 units of iron, and 200 units of Vitamin B. Each ounce of food A contains 10 units of calcium, 15 units of iron, and 20 units of vitamin B. Each ounce of food B contains 20 units of calcium, 10 units of iron, and 15 units of vitamin B. a. Find a system of inequalities to describe the different quantities of food that may be used (let x the number of ounces of food A and y the number of ounces of food B). b. Graph the system of inequalities. c. Using the graph found in part (b), nd two possible solutions (there are innitely many). 56. Health. A diet must be designed to provide at least 350 units of calcium, 175 units of iron, and 225 units of Vitamin B. Each ounce of food A contains 15 units of calcium, 25 units of iron, and 20 units of vitamin B. Each ounce of food B contains 25 units of calcium, 10 units of iron, and 10 units of vitamin B. a. Find a system of inequalities to describe the different quantities of food that may be used (let x the number of ounces of food A and y the number of ounces of food B). b. Graph the system of inequalities. c. Using the graph found in part (b), nd two possible solutions (there are innitely many). 57. Business. A manufacturer produces two types of computer mouse: USB wireless mouse and a Bluetooth mouse. Past sales indicate that it is necessary to produce at least twice as many USB wireless mice than Bluetooth mice. To meet demand, the manufacturer must produce at least 1000 computer mice per hour. a. Find a system of inequalities describing the production levels of computer mice. Let x be the production level for USB wireless mouse and y be the production level for Bluetooth mouse. b. Graph the system of inequalities describing the production levels of computer mice. c. Use your graph in part (b) to nd two possible solutions. 58. Business. A manufacturer produces two types of mechanical pencil lead: 0.5 millimeter and 0.7 millimeter. Past sales indicate that it is necessary to produce at least 50% more 0.5 millimeter lead than 0.7 millimeter lead. To meet demand, the manufacturer must produce at least 10,000 pieces of pencil lead per hour. a. Find a system of inequalities describing the production levels of pencil lead. Let x be the production level for 0.5 millimeter pencil lead and y be the production level for 0.7 millimeter pencil lead. b. Graph the system of inequalities describing the production levels of pencil lead. c. Use your graph in part (b) to nd two possible solutions. fprep.qxd 11/14/12 10:10 AM Page xii 14. P R E FAC E xiii F E ATU R E B E N E F IT TO S TU D E NT Chapter Opening Vignette Chapter Overview, Flowchart, and Learning Objectives Skills and Conceptual Objectives Clear, Concise, and Inviting Writing Style, Tone, and Layout Parallel Words and Math Common Mistakes Color for Pedagogical Reasons Study Tips Author Videos Your Turn Catch the Mistake Exercises Conceptual Exercises Inquiry-Based Learning Project Modeling OUR World Chapter Review Chapter Review Exercises Chapter Practice Test Cumulative Test Piques the students interest with a real-world application of material presented in the chapter. Later in the chapter, the same concept from the vignette is reinforced. Students see the big picture of how topics relate and overarching learning objectives are presented. Skills objectives represent what students should be able to do. Conceptual objectives emphasize a higher level global perspective of concepts. Students are able to read this book, which reduces math anxiety and promotes student success. Increases students ability to read and understand examples with a seamless representation of their instructors class (instructors voice and what they would write on the board). Addresses a different learning style: teaching by counter-example. Demonstrates common mistakes so that students understand why a step is incorrect and reinforces the correct mathematics. Particularly helpful for visual learners when they see a function written in red and then its corresponding graph in red or a function written in blue and then its corresponding graph in blue. Reinforces specic notes that you would want to emphasize in class. Gives students a mini class of several examples worked by the author. Engages students during class, builds student condence, and assists instructor in real-time assessment. Encourages students to assume the role of teacherdemonstrating a higher mastery level. Teaches students to think more globally about a topic. Lets students discover a mathematical identify, formula, etc. that is derived in the book. Engages students in a modeling project of a timely subject: global climate change. Key ideas and formulas are presented section by section in a chart. Improves study skills. Improves study skills. Offers self-assessment and improves study skills. Improves retention. fprep.qxd 11/14/12 10:10 AM Page xiii 15. xiv P R E FAC E Instructor Supplements INSTRUCTORS SOLUTIONS MANUAL (ISBN: 978-1-118-13759-8) Contains worked-out solutions to all exercises in the text. INSTRUCTORS MANUAL Authored by Cynthia Young, the manual provides practical advice on teaching with the text, including: sample lesson plans and homework assignments suggestions for the effective utilization of additional resources and supplements sample syllabi Cynthia Youngs Top 10 Teaching Tips & Tricks online component featuring the author presenting these Tips & Tricks ANNOTATED INSTRUCTORS EDITION (ISBN: 978-1-118-13491-7) Displays answers to all exercise questions, which can be found in the back of the book. Provides additional classroom examples within the standard difculty range of the in-text exercises, as well as challenge problems to assess your students mastery of the material. POWERPOINT SLIDES For each section of the book, a corresponding set of lecture notes and worked-out examples are presented as PowerPoint slides, available on the Book Companion Site (www.wiley.com/college/young) and WileyPLUS. TEST BANK Contains approximately 900 questions and answers from every section of the text. COMPUTERIZED TEST BANK Electonically enhanced version of the Test Bank that contains approximately 900 algorithmically-generated questions. allows instructors to freely edit, randomize, and create questions. allows instructors to create and print different versions of a quiz or exam. recognizes symbolic notation. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG) Contains all instructor supplements listed plus a selection of personal response system questions. WILEYPLUS Features a full-service, digital learning environment, including additional resources for instructors, such as assignable homework exercises, tutorials, gradebook, and integrated links between the online version of the text and supplements. Student Supplements STUDENT SOLUTIONS MANUAL (ISBN: 978-1-118-13758-1) Includes worked-out solutions for all odd problems in the text. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG) Provides additional resources for students to enhance the learning experience. fprep.qxd 11/14/12 10:10 AM Page xiv 16. P R E FAC E xv What Do Students Receive with WileyPLUS? A RESEARCH-BASED DESIGN WileyPLUS provides an online environment that integrates relevant resources, including the entire digital textbook, in an easy-to-navigate framework that helps students study more effectively. WileyPLUS adds structure by organizing textbook content into smaller, more manageable chunks. Related media, examples, and sample practice items reinforce the learning objectives. Innovative features such as visual progress tracking, and self-evaluation tools improve time management and strengthen areas of weakness. ONE-ON-ONE ENGAGEMENT With WileyPLUS, students receive 24/7 access to resources that promote positive learning outcomes. Students engage with related examples (in various media) and sample practice items, including: Self-Study Quizzes Video Quizzes Prociency Exams Guided Online (GO) Tutorial Problems Concept Questions Lecture Videos by Cynthia Young, including chapter introductions, chapter summaries, and selected video examples. MEASURABLE OUTCOMES Throughout each study session, students can assess their progress and gain immediate feedback. WileyPLUS provides precise reporting of strengths and weaknesses, as well as individualized quizzes, so that students are condent they are spending their time on the right things. With WileyPLUS, students always know the exact outcome of their efforts. What Do Instructors Receive with WileyPLUS? WileyPLUS provides reliable, customizable resources that reinforce course goals inside and outside of the classroom, as well as visibility into individual student progress. Pre-created materials and activities help instructors optimize their time. CUSTOMIZABLE COURSE PLAN WileyPLUS comes with a pre-created Course Plan designed by a subject matter expert uniquely for this course. PRE-CREATED ACTIVITY TYPES INCLUDE: Questions Readings and Resources Print Tests COURSE MATERIALS AND ASSESSMENT CONTENT Lecture Notes PowerPoint Slides Instructors Manual Question Assignments (all end-of-chapter problems coded algorithmically with hints, links to text, whiteboard/show work feature, and instructor controlled problem solving help) fprep.qxd 11/14/12 10:10 AM Page xv 17. xvi P R E FAC E GRADEBOOK WileyPLUS provides instant access to reports on trends in class performance, student use of course materials, and progress toward learning objectives, helping inform decisions and drive classroom discussions. Acknowledgments I want to express my sincerest gratitude to the entire Wiley team. Ive said this before, and I will say it again: Wiley is the right partner for me. There is a reason that my dog is named Wileyshes smart, competitive, a team player, and most of all, a joy to be around. There are several people within Wiley to whom I feel the need to express my appreciation: rst and foremost to Laurie Rosatone who convinced Wiley Higher Ed to invest in a young assistant professors vision for a series and who has been unwavering in her commitment to student learning. To my editor Joanna Dingle whose judgment I trust in both editorial and preschool decisions; thank you for surpassing my greatest expectations for an editor. To the rest of the ladies on the math editorial team (Jen Brady, Beth Pearson, and Liz Baird), you are all rst class! This revision was planned and executed exceptionally well thanks to you three. To the math marketing specialists Jonathan Cottrell and Jen Wreyford, thank you for helping reps tell my story: you both are outstanding at your jobs. To Kerry Weinstein, thank you for your attention to detail. To the art and illustration folks (Jennifer MacMillan, Sandra Rigby, and Dennis Ormond), thank you for bringing to life all of the sketches and gures. And nally, Id like to thank all of the Wiley reps: thank you for your commitment to my series and your tremendous efforts to get professors to adopt this book for their students. I would also like to thank all of the contributors who helped us make this our even better book. Id rst like to thank Mark McKibben. He is known as the author of the solutions manuals that accompany this series, but he is much more than that. Mark, thank you for making this series a priority, for being so responsive, and most of all for being my go-to person to think through ideas. Id also like to especially thank Jodi B.A. McKibben who is a statistician and teamed with Mark to develop the new regression material. Id like to thank Steve Davis who was the inspiration for the Inquiry-Based Learning Projects and a huge thanks to Lyn Riverstone who developed all of the IBLPs. Special thanks to Laura Watkins for nding applications that are real and timely and to Ricki Alexander for updating all of the Technology Tips. Id also like to thank Becky Schantz for her environmental problems (I now use AusPens because of Becky). Id also like to thank the following reviewers whose input helped make this book even better. Aaron Anderson, Hillsborough Community College Bernadette Antkoviak, Harrisburg Area Community College Jan Archibald, Ventura College Shari Beck, Navarro College Patricia K. Bezona, Valdosta State University Connie Buller, Metropolitan Community College James Carolan, Wharton County Junior College Diane Cook, Collegiate High School at Northwest Florida State College Doris C. Cowan, Polk State College Jean Davis, Texas State University Nerissa Felder, Polk State College Sunshine Gibbons, Southeast Missouri State University Mehran Hassanpour, South Texas College Tom Hayes, Montana State University Bozeman Celeste Hernandez, Richland College Carolyn Horseman, Polk State College Dianne Marquart, Valdosta State University Maria Luisa Mendez, Laredo Community College Lily Rai, South Texas College Leela Rakesh, Central Michigan University Denise Reid, Valdosta State University Linda Tansil, Southeast Missouri State University And a special thanks to our student reviewer Luis Suarez del Rio. fprep.qxd 11/14/12 10:10 AM Page xvi 18. 0 Prerequisites and Review 2 0.1 Real Numbers 4 0.2 Integer Exponents and Scientic Notation 18 0.3 Polynomials: Basic Operations 28 0.4 Factoring Polynomials 37 0.5 Rational Expressions 48 0.6 Rational Exponents and Radicals 63 0.7 Complex Numbers 73 Inquiry-Based Learning Project 81 | Review 83 | Review Exercises 86 | Practice Test 89 1 Equations and Inequalities 90 1.1 Linear Equations 92 1.2 Applications Involving Linear Equations 102 1.3 Quadratic Equations 116 1.4 Other Types of Equations 130 1.5 Linear Inequalities 140 1.6 Polynomial and Rational Inequalities 151 1.7 Absolute Value Equations and Inequalities 162 Inquiry-Based Learning Project 170 | Modeling Our World 172 | Review 173 | Review Exercises 174 | Practice Test 178 | Cumulative Test 179 2 Graphs 180 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 182 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 190 2.3 Lines 204 2.4 Circles 221 2.5*Linear Regression: Best Fit 230 Inquiry-Based Learning Project 257 | Modeling Our World 258 | Review 260 | Review Exercises 261 | Practice Test 264 | Cumulative Test 265 Table of Contents GarryWade/Taxi/GettyImages GettyImages/BlendImages/GettyImages RadiusImages/CorbisVetta/GettyImages xvii ftoc.qxd 11/28/12 2:37 PM Page xvii 19. xviii TAB LE O F C O NTE NTS 3 Functions and Their Graphs 266 3.1 Functions 268 3.2 Graphs of Functions; Piecewise-Dened Functions; Increasing and Decreasing Functions; Average Rate of Change 287 3.3 Graphing Techniques: Transformations 308 3.4 Operations on Functions and Composition of Functions 323 3.5 One-to-One Functions and Inverse Functions 334 3.6 Modeling Functions Using Variation 350 Inquiry-Based Learning Project 361 | Modeling Our World 363 | Review 364 | Review Exercises 366 | Practice Test 371 | Cumulative Test 373 4 Polynomial and Rational Functions 374 4.1 Quadratic Functions 376 4.2 Polynomial Functions of Higher Degree 394 4.3 Dividing Polynomials: Long Division and Synthetic Division 410 4.4 The Real Zeros of a Polynomial Function 419 4.5 Complex Zeros: The Fundamental Theorem of Algebra 435 4.6 Rational Functions 445 Inquiry-Based Learning Project 466 | Modeling Our World 467 | Review 469 | Review Exercises 472 | Practice Test 476 | Cumulative Test 477 5 Exponential and Logarithmic Functions 478 5.1 Exponential Functions and Their Graphs 480 5.2 Logarithmic Functions and Their Graphs 496 5.3 Properties of Logarithms 512 5.4 Exponential and Logarithmic Equations 521 5.5 Exponential and Logarithmic Models 532 Inquiry-Based Learning Project 544 | Modeling Our World 545 | Review 547 | Review Exercises 549 | Practice Test 552 | Cumulative Test 553 6 Trigonometric Functions 554 6.1 Angles, Degrees, and Triangles 556 6.2 Denition 1 of Trigonometric Functions: Right Triangle Ratios 574 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 589 6.4 Denition 2 of Trigonometric Functions: Cartesian Plane 603 6.5 Trigonometric Functions of Nonacute Angles 619 6.6 Radian Measure and Applications 635 6.7 Denition 3 of Trigonometric Functions: Unit Circle Approach 653 6.8 Graphs of Sine and Cosine Functions 663 6.9 Graphs of Other Trigonometric Functions 693 Inquiry-Based Learning Project 712 | Modeling Our World 713 | Review 714 | Review Exercises 722 | Practice Test 726 | Cumulative Test 727 JohnGiustina/SuperstockFocusonSport/GettyImagesRichardT.Nowitz/PhotoResearchers,Inc.KathyYturralde/Alamy ftoc.qxd 11/28/12 2:37 PM Page xviii 20. 7 Analytic Trigonometry 728 7.1 Basic Trigonometric Identities 730 7.2 Verifying Trigonometric Identities 741 7.3 Sum and Difference Identities 751 7.4 Double-Angle Identities 765 7.5 Half-Angle Identities 773 7.6 Product-to-Sum and Sum-to-Product Identities 784 7.7 Inverse Trigonometric Functions 793 7.8 Trigonometric Equations 815 Inquiry-Based Learning Project 832 | Modeling Our World 833 | Review 834 | Review Exercises 837 | Practice Test 842 | Cumulative Test 843 8 Additional Topics in Trigonometry 844 8.1 Oblique Triangles and the Law of Sines 846 8.2 The Law of Cosines 861 8.3 The Area of a Triangle 872 8.4 Vectors 881 8.5 The Dot Product 896 8.6 Polar (Trigonometric) Form of Complex Numbers 905 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; DeMoivres Theorem 915 8.8 Polar Equations and Graphs 927 Inquiry-Based Learning Project 943 | Modeling Our World 944 | Review 945 | Review Exercises 949 | Practice Test 952 | Cumulative Test 953 9 Systems of Linear Equations and Inequalities 954 9.1 Systems of Linear Equations in Two Variables 956 9.2 Systems of Linear Equations in Three Variables 973 9.3 Partial Fractions 986 9.4 Systems of Linear Inequalities in Two Variables 998 9.5 The Linear Programming Model 1010 Inquiry-Based Learning Project 1018 | Modeling Our World 1019 | Review 1021 | Review Exercises 1023 | Practice Test 1026 | Cumulative Test 1027 10 Matrices 1028 10.1 Matrices and Systems of Linear Equations 1030 10.2 Matrix Algebra 1053 10.3 Matrix Equations; The Inverse of a Square Matrix 1067 10.4 The Determinant of a Square Matrix and Cramers Rule 1081 Inquiry-Based Learning Project 1096 | Modeling Our World 1097 | Review 1099 | Review Exercises 1102 | Practice Test 1105 | Cumulative Test 1107 decodes sends sends decodes visitor web server encrypts receives encrypts receives TAB LE O F C O NTE NTS xix Q.Sakamaki/ReduxPictures Paul J.Richards/AFP/Getty Images Q.Sakamaki/ReduxPictures JasonBrindelCommercial/Alamy Bermuda Miami, Florida San Juan, Puerto Rico Atlantic Ocean ftoc.qxd 11/28/12 2:37 PM Page xix 21. xx TAB LE O F C O NTE NTS 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 1108 11.1 Conic Basics 1110 11.2 The Parabola 1113 11.3 The Ellipse 1127 11.4 The Hyperbola 1140 11.5 Systems of Nonlinear Equations 1153 11.6 Systems of Nonlinear Inequalities 1164 11.7 Rotation of Axes 1172 11.8 Polar Equations of Conics 1183 11.9 Parametric Equations and Graphs 1194 Inquiry-Based Learning Project 1203 | Modeling Our World 1204 | Review 1205 | Review Exercises 1209 | Practice Test 1213 | Cumulative Test 1215 12 Sequences, Series, and Probability 1216 12.1 Sequences and Series 1218 12.2 Arithmetic Sequences and Series 1229 12.3 Geometric Sequences and Series 1238 12.4 Mathematical Induction 1250 12.5 The Binomial Theorem 1255 12.6 Counting, Permutations, and Combinations 1263 12.7 Probability 1273 Inquiry-Based Learning Project 1283 | Modeling Our World 1285 | Review 1286 | Review Exercises 1288 | Practice Test 1292 | Cumulative Test 1293 Answers to Odd Numbered Exercises 1295 Applications Index 1389 Subject Index 1394 2 3 4 5 6 Dealers Up Card A BASIC STRATEGY FOR BLACKJACK Whensurrenderisallowed,surrender9,7or10,6vs9,10,A;9,6or10,5vs10 Whendoublingdownaftersplittingisallowed,split:2s,3s,7svs2-7;4svs5or6;6svs2-6 7 8 9 10 A S S S S S S S S S S S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H H H S S S H H H H H D D D D D D D D D H D D D D D D D D H H H D D D D H H H H H H H H H H H H H H H S S S S S S S S S S S D D D D S S H H H H D D D D H H H H H H H D D D H H H H H H H D D D H H H H H H H H D D H H H H H H H H D D H H H H H SP SP SP SP SP SP SP SP SP SP S S S S S S S S S S SP SP SP SP SP S SP SP S S SP SP SP SP SP SP H H H H H SP SP SP SP H H H H H D D D D D D D D H H H H H H H H H H H H H H SP SP SP SP H H H H H 17+ 16 15 14 13 12 11 10 9 5 - 8 A, 8 - 10 A, 7 A, 6 A, 5 A, 4 A, 3 A, 2 A, A; 8, 8 10, 10 9, 9 7, 7 I II III YourHand IV 6, 6 5, 5 4, 4 3, 3 2, 2 H SP SP SP SP H H H HIT STAND DOUBLE DOWN SPLIT H PaulSouders/TheImageBank/GettyImages ftoc.qxd 11/28/12 2:37 PM Page xx 22. A Note from the Author to the Student xxi Iwrote this text with careful attention to ways in which to make your learning experience more successful. If you take full advantage of the unique features and elements of this textbook, I believe your experience will be fullling and enjoyable. Lets walk through some of the special book features that will help you in your study of algebra and trigonometry. Prerequisites and Review (Chapter 0) A comprehensive review of prerequisite knowledge (intermediate algebra topics) in Chapter 0 provides a brush up on knowledge and skills necessary for success in the course. Clear, Concise, and Inviting Writing Special attention has been made to present an engaging, clear, precise narrative in a layout that is easy to use and designed to reduce any math anxiety you may have. Chapter Introduction, Flow Chart, Section Headings, and Objectives An opening vignette, ow chart, list of chapter sections, and chapter learning objectives give you an overview of the chapter. Skills and Conceptual Objectives For every section, objectives are further divided by skills and concepts so you can see the difference between solving problems and truly understanding concepts. 0 Prerequisites and Review W ould you be able to walk successfully along a tightrope? Most people probably would say no because the foundation is shaky. Would you be able to walk successfully along a beam (4 inches wide)? Most people would probably say yeseven though for some of us it is still challenging. Think of this chapter as the foundation for your walk. The more solid your foundation is now, the more successful your walk through College Algebra will be. The purpose of this chapter is to review concepts and skills that you already have learned in a previous course. Mathematics is a cumulative subject in that it requires a solid foundation to proceed to the next level. Use this chapter to reaffirm your current knowledge base before jumping into the course. GettyImages/BlendImages/GettyImages GettyImages Functions and Their Graphs On a sales rack of clothes at a department store, you see a shirt you like. The original price of the shirt was $100, but it has been discounted 30%. As a preferred shopper, you get an automatic additional 20% off the sale price at the register. How much will you pay for the shirt? Nave shoppers might be lured into thinking this shirt will cost $50 because they add the 20% and 30% to get 50% off, but they will end up paying more than that. Experienced shoppers know that they first take 30% off of $100, which results in a price of $70, and then they take an additional 20% off of the sale price, $70, which results in a final discounted price of $56. Experienced shoppers have already learned composition of functions. A composition of functions can be thought of as a function of a function. One function takes an input (original price, $100) and maps it to an output (sale price, $70), and then another function takes that output as its input (sale price, $70) and maps that to an output (checkout price, $56). JohnGiustina/Superstock/Photolibrary 3 I N TH I S C HAP TE R you will find that functions are part of our everyday thinking: converting from degrees Celsius to degrees Fahrenheit, DNA testing in forensic science, determining stock values, and the sale price of a shirt. We will develop a more complete, thorough understanding of functions. First, we will establish what a relation is, and then we will determine whether a relation is a function. We will discuss common functions, domain and range of functions, and graphs of functions. We will determine whether a function is increasing or decreasing on an interval and calculate the average rate of change of a function. We will perform operations on functions and composition of functions. We will discuss one-to-one functions and inverse functions. Finally, we will model applications with functions using variation. 267 Recognizing and Classifying Functions Increasing and Decreasing Functions Average Rate of Change Piecewise- Defined Functions Relations and Functions Functions Defined by Equations Function Notation Domain of a Function Horizontal and Vertical Shifts Reflection about the Axes Stretching and Compressing Adding, Subtracting, Multiplying, and Dividing Functions Composition of Functions Determine Whether a Function Is One-to-One Inverse Functions Graphical Interpretation of Inverse Functions Finding the Inverse Function Direct Variation Inverse Variation Joint Variation and Combined Variation FUNCTIONS AND THEIR GRAPHS L E A R N I N G O B J E C T I V E S Find the domain and range of a function. Sketch the graphs of common functions. Sketch graphs of general functions employing translations of common functions. Perform composition of functions. Find the inverse of a function. Model applications with functions using variation. 3.1 Functions 3.2 Graphs of Functions; Piecewise- Defined Functions; Increasing and Decreasing Functions; Average Rate of Change 3.3 Graphing Techniques: Transformations 3.4 Operations on Functions and Composition of Functions 3.5 One-To-One Functions and Inverse Functions 3.6 Modeling Functions Using Variation P O LYN O M IAL F U N CTI O N S O F H I G H E R D E G R E E S E CTI O N 4.2 C O N C E P TUAL O BJ E CTIVE S Understand that real zeros of polynomial functions correspond to x-intercepts. Understand the intermediate value theorem and how it assists in graphing polynomial functions. Realize that end behavior is a result of the leading term dominating. Understand that zeros correspond to factors of the polynomial. S K I LLS O BJ E CTIVE S Identify a polynomial function and determine its degree. Graph polynomial functions using transformations. Identify real zeros of a polynomial function and their multiplicities. Determine the end behavior of a polynomial function. Graph polynomial functions. x-intercepts multiplicity (touch/cross) of each zero end behavior ftoc.qxd 11/28/12 2:37 PM Page xxi 23. xxii A N OTE F R O M TH E AUTH O R TO TH E STU D E NT Examples Examples pose a specific problem using concepts already presented and then work through the solution. These serve to enhance your understanding of the subject matter. Your Turn Immediately following many examples, you are given a similar problem to reinforce and check your understanding. This helps build confidence as you progress in the chapter. These are ideal for in-class activity or for preparing for homework later. Answers are provided in the margin for a quick check of your work. Common Mistake/ Correct vs. Incorrect In addition to standard examples, some problems are worked out both correctly and incorrectly to highlight common errors students make. Counter examples like these are often an effective learning approach for many students. Parallel Words and Math This text reverses the common textbook presentation of examples by placing the explanation in words on the left and the mathematics in parallel on the right. This makes it easier for students to read through examples as the material ows more naturally from left to right and as commonly presented in class. Study Tips and Caution Notes These marginal reminders call out important hints or warnings to be aware of related to the topic or problem. Technology Tips These marginal notes provide problem solving instructions and visual examples using graphing calculators. E X AM P LE 9 Evaluating the Difference Quotient For the function nd Solution: Use placeholder notation for the function (x) x2 x. ( ) ( )2 ( ) Calculate (x h). (x h) (x h)2 (x h) Write the difference quotient. Let (x h) (x h)2 (x h) and (x) x2 x. Eliminate the parentheses inside the rst set of brackets. Eliminate the brackets in the numerator. Combine like terms. Factor the numerator. Divide out the common factor, h. 2x h 1 YO U R TU R N Evaluate the difference quotient for (x) x2 1. h Z 0 = h(2x + h - 1) h = 2xh + h2 - h h = x2 + 2xh + h2 - x - h - x2 + x h = [x2 + 2xh + h2 - x - h] - [x2 - x] h h Z 0= C(x + h)2 - (x + h)D - Cx2 - xD h f(x + h) - f(x) h f(x + h) - f(x) h f(x + h) - f(x) h , h Z 0.f(x) = x2 - x, f(x+h) f(x) u r I N C O R R E CT The ERROR is in interpreting the notation as a sum. Z x2 - 3x - 2 f(x + 1) Z f(x) + f(1) C O R R E CT Write the original function. Replace the argument x with a placeholder. f( ) ( )2 3( ) Substitute x 1 for the argument. Eliminate the parentheses. Combine like terms. f(x + 1) = x2 - x - 2 f(x + 1) = x2 + 2x + 1 - 3x - 3 f(x + 1) = (x + 1)2 - 3(x + 1) f(x) = x2 - 3x A common misunderstanding is to interpret the notation (x 1) as a sum: f(x + 1) Z f(x) + f(1). C O M M O N M I S TA K E WORDS MATH Write the interest formula for compounding continuously. A Pert Let A 2P (investment doubles). 2P Pert Divide both sides of the equation by P. 2 ert Take the natural log of both sides of the equation. ln 2 ln ert Simplify the right side by applying the property ln ex x. ln 2 rt Divide both sides by r. Approximate t L 0.7 r ln 2 L 0.7. t = ln 2 r Technology Tip A graphing utility can be used to evaluate P(2). Enter P(x) 4x5 3x4 2x3 7x2 9x 5 as Y1. To evaluate P(2), press VARS Y-VARS 1:Function 1:Y1 ( 2 ) ENTER Study Tip The largest number of zeros a polynomial can have is equal to the degree of the polynomial. C A U T I O N f g Z f # g ftoc.qxd 11/28/12 2:37 PM Page xxii 24. A N OTE F R O M TH E AUTH O R TO TH E STU D E NT xxiii Video icons Video icons appear on all chapter and section introductions, chapter and section reviews, as well as selected examples throughout the chapter to indicate that the author has created a video segment for that element. These video clips help you work through the selected examples with the author as your private tutor. I N TH I S C HAP TE R you will find that functions are part of our everyday thinking: converting from degrees Celsius to degrees Fahrenheit, DNA testing in forensic science, determining stock values, and the sale price of a shirt. We will develop a more complete, thorough understanding of functions. First, we will establish what a relation is, and then we will determine whether a relation is a function. We will discuss common functions, domain and range of functions, and graphs of functions. We will determine whether a function is increasing or decreasing on an interval and calculate the average rate of change of a function. We will perform operations on functions and composition of functions. We will discuss one-to-one functions and inverse functions. Finally, we will model applications with functions using variation. Joint variation occurs when one quantity is directly proportional to two or more quantities. Combined variation occurs when one quantity is directly proportional to one or more quantities and inversely proportional to one or more other quantities. S U M MARY Direct, inverse, joint, and combined variation can be used to model the relationship between two quantities. For two quantities x and y, we say that y is directly proportional to x if y kx. y is inversely proportional to x if .y = k x S E CTI O N 3.6 E X AM P LE 9 Evaluating the Difference Quotient For the function nd Solution: Use placeholder notation for the function (x) x2 x. ( ) ( )2 ( ) Calculate (x h). (x h) (x h)2 (x h) f(x + h) - f(x) h , h Z 0.f(x) = x2 - x, C HAP TE R 3 R EVI EW SECTION CONCEPT KEY IDEAS/FORMULAS 3.1 Functions Relations and functions All functions are relations, but not all relations are functions. Functions dened by equations A vertical line can intersect a function in at most one point. Function notation Placeholder notation: Difference quotient: (x + h) - (x) h ; h Z 0 f(x) = 3x2 - 6x + 2 f(n) = 3(n)2 - 6(n) + 2 Six Different Types of Exercises Every text section ends with Skills, Applications, Catch the Mistake, Conceptual, Challenge, and Technology exercises. The exercises gradually increase in difculty and vary in skill and conceptual emphasis. Catch the Mistake exercises increase the depth of understanding and reinforce what you have learned. Conceptual and Challenge exercises specically focus on assessing conceptual understanding. Technology exercises enhance your understanding and ability using scientic and graphing calculators. E X E R C I S E S S E CTI O N 3.5 In Exercises 116, determine whether the given relation is a function. If it is a function, determine whether it is a one-to-one function. 1. 2. 3 4 Domain Range (202) 555-1212 (307) 123-4567 (878) 799-6504 10-DIGIT PHONE # PERSON Mary Jason Chester Domain Range 78F 68F AVERAGE TEMPERATURE MONTH October January April S K I LL S Security, write a function E(x) that expresses the students take-home pay each week. Find the inverse function E1 (x). What does the inverse function tell you? 70. Salary. A grocery store pays you $8 per hour for the rst 40 hours per week and time and a half for overtime. Write a piecewise-dened function that represents your weekly earnings E(x) as a function of the number of hours worked x. Find the inverse function E1 (x). What does the inverse function tell you? In Exercises 7174, refer to the following: By analyzing available empirical data it was determined that during an illness a patients body temperature uctuated during one 24-hour period according to the function h T h i i d T(t) = 0.0003(t - 24)3 + 101.70 65. Temperature. The equation used to convert from degrees Celsius to degrees Fahrenheit is .f(x) = 9 5 x + 32 Determine the inverse function C1 (x). What does the inverse function represent? 67. Budget. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to gure out how much money to raise. The entry fee is $250 per boat for the rst 10 boats and $175 for each additional boat. Find the cost function C(x) as a function of the number of boats the club enters x. Find the inverse function that will yield how many boats the club can enter Determine the inverse function f 1 (x). What does the inverse function represent? 66. Temperature. The equation used to convert from degrees Fahrenheit to degrees Celsius is .C(x) = 5 9 (x - 32) A P P L I C AT I O N S In Exercises 7578, explain the mistake that is made. 75. Is x y2 a one-to-one function? Solution: Yes, this graph represents a one-to-one function because it passes the horizontal line test. This is incorrect. What mistake was made? 76. A linear one-to-one function is graphed below. Draw its inverse. Solution: Note that the points (3, 3) and (0, 4) lie on the graph of the function. This is incorrect. What mistake was made? By symmetry, the points (3, 3) and x y y (0, 4) C AT C H T H E M I S TA K E x y (3, 3) (0, 4) In Exercises 9194, determine if each statement is true or false. 94. The range of all polynomial functions is ( , ). 95. What is the maximum number of zeros that a polynomial of degree n can have? 96. What is the maximum number of turning points a graph of an nth-degree polynomial can have? 91. The graph of a polynomial function might not have any y-intercepts. 92. The graph of a polynomial function might not have any x-intercepts. 93. The domain of all polynomial functions is ( , ). C O N C E P T U A L 93. For the functions and g(x) x2 a nd and state its domain. 94. For the functions and , nd and state its domain. Assume a 7 1 and b 7 1. g fg(x) = 1 xb f (x) = 1 xa g f f(x) = 1x + a91. For the functions (x) x a and , nd and state its domain. 92. For the functions (x) ax2 bx c and , nd and state its domain.g f g(x) = 1 x - c g f g(x) = 1 x - a C HALLE N G E 95. Using a graphing utility, plot and . Plot y3 y1 y2. What is the domain of y3? 96. Using a graphing utility, plot , , and . What is the domain of y3?y3 = y1 y2 y2 = 1 13 - x y1 = 3 1x + 5 y2 = 19 - xy1 = 1x + 7 97. Using a graphing utility, plot , , and . If y1 represents a functiony3 = 1 y2 1 - 14 y2 = 1 x2 - 14 y1 = 2x2 - 3x - 4 f and y2 represents a function g, then y3 represents the composite function g f. The graph of y3 is only dened for the domain of g f. State the domain of g f. 98. Using a graphing utility, plot ,y1 = 11 - x, y2 = x2 + 2 T E C H N O L O G Y ftoc.qxd 11/28/12 2:37 PM Page xxiii 25. Modeling Our World These unique end-of-chapter exercises provide a fun and interesting way to take what you have learned and model a real world problem. By using climate change as the continuous theme, these exercises can help you to develop more advanced modeling skills with each chapter while seeing how modeling can help you better understand the world around you. xxiv A N OTE F R O M TH E AUTH O R TO TH E STU D E NT Inquiry-Based Learning Projects These end of chapter projects enable you to discover mathematical concepts on your own! Transformations of Functions Being a creature of habit, Dylan usually sets out each morning at 7 AM from his house for a jog. Figure 1 shows the graph of a function, y = d(t), that represents Dylans jog on Friday. a. Use the graph in Figure 1 to ll in the table below. Describe a jogging scenario that ts the graph and table above. b. The graph shown in Figure 2 represents Dylans jog on Saturday. It is a transformation of the function y = d(t) shown in Figure 1. Complete the table of values below for this transformation. You may nd it helpful to refer to the table in part (a). What is the real-world meaning of this transformation? How is Dylans jog on Saturday different from his usual jog? How is it the same? The original function (in Figure 1) is represented by the equation y = d(t). Write an equation, in terms of d(t), that represents the function graphed in Figure 2. Explain. c. The graph shown in Figure 3 represents Dylans jog on Sunday It is a transformation C HAP TE R 3 I N Q U I RY-BA S E D LE AR N I N G P R OJ E CT Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Friday y = d(t) t y = d(t) t y Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Saturday es) 5 Dylan's Jog on Sunday Figure 1 Figure 2 Chapter Review, Review Exercises, Practice Test, Cumulative Test At the end of every chapter, a summary review chart organizes the key learning concepts in an easy to use one or two-page layout. This feature includes key ideas and formulas, as well as indicating relevant pages and review exercises so that you can quickly summarize a chapter and study smarter. Review Exercises, arranged by section heading, are provided for extra study and practice. A Practice Test, without section headings, offers even more self-practice before moving on. A new Cumulative Test feature offers study questions based on all previous chapters content, thus helping you build upon previously learned concepts. The U.S. National Oceanic and Atmospheric Association (NOAA) monitors temperature and carbon emissions at its observatory in Mauna Loa, Hawaii. NOAAs goal is to help foster an informed society that uses a comprehensive understanding of the role of the oceans, coasts, and atmosphere in the global ecosystem to make the best social and economic decisions. The data presented in this chapter is from the Mauna Loa Observatory, where historical atmospheric measurements have been recorded for the last 50 years. You will develop linear models based on this data to predict temperature and carbon emissions in the future. The following table summarizes average yearly temperature in degrees Fahrenheit F and carbon dioxide emissions in parts per million (ppm) for Mauna Loa, Hawaii. 1. Plot the temperature data with time on the horizontal axis and temperature on the vertical axis. Let t 0 correspond to 1960. 2. Find a linear function that models the temperature in Mauna Loa. a. Use data from 1965 and 1995. b. Use data from 1960 and 1990. c. Use linear regression and all data given. 3. Predict what the temperature will be in Mauna Loa in 2020. a. Apply the line found in Exercise 2(a). b. Apply the line found in Exercise 2(b). c. Apply the line found in Exercise 2(c). Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Temperature (F) 44.45 43.29 43.61 43.35 46.66 45.71 45.53 47.53 45.86 46.23 CO2 Emissions 316.9 320.0 325.7 331.1 338.7 345.9 354.2 360.6 369.4 379.7 (ppm) M O D E LI N G O U R W O R LD C HAP TE R 3 R EVI EW SECTION CONCEPT KEY IDEAS/FORMULAS 3.1 Functions Relations and functions All functions are relations, but not all relations are functions. Functions dened by equations A vertical line can intersect a function in at most one point. Function notation Placeholder notation: Difference quotient: Domain of a function Are there any restrictions on x? 3.2 Graphs of functions; piecewise-dened functions; increasing and decreasing functions; average rate of change Recognizing and classifying functions Common functions (x) mx b, (x) x, (x) x2 , Even and odd functions Even: Symmetry about y-axis: (x) (x) Odd: Symmetry about origin: (x) (x) Increasing and decreasing Increasing: rises (left to right) functions Decreasing: falls (left to right) Average rate of change Piecewise-dened functions Points of discontinuity 3.3 Graphing techniques: Shift the graph of (x). Transformations Horizontal and vertical shifts (x c) c units to the left, c 0 (x c) c units to the right, c 0 (x) c c units upward, c 0 (x) c c units downward, c 0 Reection about the axes (x) Reection about the x-axis (x) Reection about the y-axis Stretching and compressing c(x) if c 1; stretch vertically c(x) if 0 c 1; compress vertically (cx) if c 1; compress horizontally (cx) if 0 c 1; stretch horizontally x1 Z x2 (x2) - (x1) x2 - x1 (x) = x , (x) = 1 x f(x) = x3 , f(x) = 1x, f(x) = 3 1x, (x + h) - (x) h ; h Z 0 f(x) = 3x2 - 6x + 2 f(n) = 3(n)2 - 6(n) + 2 CHAPTERREVIEW 364 C HAP TE R 3 R EVI EW E X E R C I S E S 3.1 Functions Determine whether each relation is a function. 1. 2. {(1, 2), (3, 4), (2, 4), (3, 7)} 3. {(2, 3), (1, 3), (0, 4), (2, 6)} 4. {(4, 7), (2, 6), (3, 8), (1, 7)} 5. x2 y2 36 6. x 4 7. y x 2 8. 9. 10. Use the graphs of the functions to nd: 11. 12. x y x y x y x y y = 1x Domain Range NAMES Allie Hannah Danny Ethan Vickie AGES 27 10 4 21 13. 14. Evaluate the given quantities using the following three functions. (x) 4x 7 F(t) t2 4t 3 g(x) x2 2x 4 15. (3) 16. F(4) 17. f(7) g(3) 18. 19. 20. 21. 22. Find the domain of the given function. Express the domain in interval notation. 23. (x) 3x 4 24. g(x) x2 2x 6 25. 26. 27. 28. Challenge 29. If , (4) and (4) are undened, and , nd D. 30. Construct a function that is undened at x 3 and x 2 such that the point (0, 4) lies on the graph of the function. 3.2 Graphs of Functions Determine whether the function is even, odd, or neither. 31. (x) 2x 7 32. g(x) 7x5 4x3 2x 33. h(x) x3 7x 34. (x) x4 3x2 35. (x) x1/4 x 36. 37. 38. f(x) = 1 x2 + 3x4 + xf(x) = 1 x3 + 3x f(x) = 1x + 4 f(5) = 2 f(x) = D x2 - 16 H(x) = 1 12x - 6 G(x) = 1x - 4 F(x) = 7 x2 + 3 h(x) = 1 x + 4 F(t + h) - F(t) h f(3 + h) - f(3) h f(3 + h) f(2) - F(2) g(0) F(0) g(0) x y x y a. (1) b. (1) c. x, where (x) 0 a. (4) b. (0) c. x, where (x) 0 a. (2) b. (4) c. x, where (x) 0 a. (5) b. (0) c. x, where (x) 0 REVIEWEXERCISES 366 1. Graph the parabola y (x 4)2 1. 2. Write the parabola in standard form y x2 4x 1. 3. Find the vertex of the parabola 4. Find a quadratic function whose vertex is (3, 1) and whose graph passes through the point (4, 1). 5. Find a sixth-degree polynomial function with the given zeros: 2 of multiplicity 3 1 of multiplicity 2 0 of multiplicity 1 6. For the polynomial function f(x) x4 6x3 7x: a. List each real zero and its multiplicity. b. Determine whether the graph touches or crosses at each x-intercept. c. Find the y-intercept and a few points on the graph. d. Determine the end behavior. e. Sketch the graph. 7. Divide 4x4 2x3 7x2 5x 2 by 2x2 3x 1. 8. Divide 17x5 4x3 2x 10 by x 2. 9. Is x 3 a factor of x4 x3 13x2 x 12? 10. Determine whether 1 is a zero of P(x) x21 2x18 5x12 7x3 3x2 2. 11. Given that x 7 is a factor of P(x) x3 6x2 9x 14, factor the polynomial in terms of linear factors. 12. Given that 3i is a zero of P(x) x4 3x3 19x2 27x 90, nd all other zeros. 13. Can a polynomial have zeros that are not x-intercepts? Explain. 14. Apply Descartes rule of signs to determine the possible combinations of positive real zeros, negative real zeros, and complex zeros of P(x) 3x5 2x4 3x3 2x2 x 1. 15. From the rational zero test, list all possible rational zeros of P(x) 3x4 7x2 3x 12. In Exercises 1618, determine all zeros of the polynomial function and graph. 16. P(x) x3 4x 17. P(x) 2x3 3x2 8x 12 18. P(x) x4 6x3 10x2 6x 9 19. Sports. A football player shows up in August at 300 pounds. After 2 weeks of practice in the hot sun, he is down to 285 pounds. Ten weeks into the season he is up to 315 pounds because of weight training. In the spring he does not work out, and he is back to 300 pounds by the next August. Plot these points on a graph. What degree polynomial could this be? f(x) = -1 2x2 + 3x - 4. 20. Prot. The prot of a company is governed by the polynomial P(x) x3 13x2 47x 35, where x is the number of units sold in thousands. How many units does the company have to sell to break even? 21. Interest Rate. The interest rate for a 30-year xed mortgage uctuates with the economy. In 1970, the mortgage interest rate was 8%, and in 1988 it peaked at 13%. In 2002, it dipped down to 4%, and in 2005, it was up to 6%. What is the lowest degree polynomial that can represent this function? In Exercises 2225, determine (if any) the: a. x- and y-intercepts b. vertical asymptotes c. horizontal asymptotes d. slant asymptotes e. graph 22. 23. 24. 25. 26. Food. On eating a sugary snack, the average body almost doubles its glucose level. The percentage increase in glucose level y can be approximated by the rational function , where x represents the number of minutes after eating the snack. Graph the function. 27. a. Use the calculator commands STAT QuadReg to model the data using a quadratic function. b. Write the quadratic function in standard form and identify the vertex. c. Find the x-intercepts. d. Plot this quadratic function with a graphing calculator. Do they agree with the given values? y = 25x x2 + 50 F(x) = x - 3 x2 - 2x - 8 h(x) = 3x3 - 3 x2 - 4 g(x) = x x2 - 4 f(x) = 2x - 9 x + 3 C HAP TE R 4 P R ACTI C E TE ST 476 x 3 2.2 7.5 y 10.01 9.75 25.76 28. Find the asymptotes and intercepts of the rational function . Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph conrm what you found? f(x) = x(2x - 3) x2 - 3x + 1 PRACTICETEST C HAP TE R S 14 C U M U LATIVE TE ST 17. Evaluate g( f(1)) for f(x) 7 2x2 and g(x) 2x 10. 18. Find the inverse of the function f(x) (x 4)2 2, where x 4. 19. Find a quadratic function that has the vertex (2, 3) and point (1, 4). 20. Find all of the real zeros and state their multiplicities of the function f(x) 3.7x4 14.8x3 . 21. Use long division to nd the quotient Q(x) and the remainder r(x) of (20x3 8x2 7x 5) (5x 3). 22. Use synthetic division to nd the quotient Q(x) and the remainder r(x) of (2x3 3x2 11x 6) (x 3). 23. List the possible rational zeros, and test to determine all rational zeros for P(x) 12x3 29x2 7x 6. 24. Given the real zero x 5 of the polynomial P(x) 2x3 3x2 32x 15, determine all the other zeros and write the polynomial in terms of a product of linear factors. 25. Find all vertical and horizontal asymptotes of . 26. Graph the function . 27. Find the asymptotes and intercepts of the rational function . Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph conrm what you found? 28. Find the asymptotes and intercepts of the rational function . Note: Combine the twof(x) = 6x 3x + 1 - 6x 4x - 1 f(x) = 5 2x - 3 - 1 x f(x) = 2x3 - x2 - x x2 - 1 f(x) = 3x - 5 x2 - 4 CUMULATIVETEST 1. Simplify and express in terms of positive exponents. (5x-1 y-2 ) 3 (-10x2 y2 ) 2 2. Factor . 3. Multiply and simplify . 4. Solve for x: |2x 5| 3 10. 5. Austin can mow a lawn in 75 minutes. The next week Stacey mows the same lawn in 60 minutes. How long would it take them to mow the lawn working together? 6. Use the discriminant to determine the number and type of roots: 4x2 3x 15 0. 7. Solve and check . 8. Apply algebraic tests to determine whether the equations graph is symmetric with respect to the x-axis, y-axis, or origin for y |x| 3. 9. Write an equation of the line that is parallel to the line x 3y 8 and has the point (4, 1). 10. Find the x-intercept and y-intercept and sketch the graph for 2y 6 0. 11. Write the equation of a circle with center (0, 6) and that passes through the point (1, 5). 12. Express the domain of the function with interval notation. 13. Determine whether the function is even, odd, or neither. 14. For the function y (x 1)2 2, identify all of the transformations of y x2 . 15. Sketch the graph of and identify ally = 1x - 1 + 3 g(x) = 1x + 10 f(x) = 16x - 7 216 + x2 = x + 2 4x2 - 36 x2 # x3 + 6x2 x2 + 9x + 18 2xy - 2x + 3y - 3 ftoc.qxd 11/28/12 2:37 PM Page xxiv 26. Algebra and Trigonometry Third Edition ftoc.qxd 11/28/12 2:37 PM Page 1 27. DE F I N ITIO N S, RU LE S, FO R M U LAS, AN D GR AP H S ARITHMETIC OPERATIONS a a b c b = ac b a a b b c = a bc a Z 0 ab + ac a = b + c, a - b c - d = b - a d - c aa b c b = ab c a a b b a c d b = ad bc a + b c = a c + b c a b + c d = ad + bc bd ab + ac = a(b + c) EXPONENTS AND RADICALS A n a a b b = 2 n a 2 n b 2 n a = a1/n (ab)x = ax bx ax ay = ax+y 2 n ab = 2 n a2 n b2a = a1/ 2 (ax )y = axy a-x = 1 ax 2 n am = am/n = A1 n aB m a a b b x = ax bx ax ay = ax-y a Z 0a0 = 1, ABSOLUTE VALUE SPECIAL FACTORIZATIONS 1. 2. If then or 3. If then 4. If then or (c 7 0)x 7 c.x 6 -cx 7 c, (c 7 0)-c 6 x 6 c.x 6 c, (c 7 0)x = -c.x = cx = c, x = e x if x 0 -x if x 7 0 1. Difference of two squares: 2. Perfect square trinomials: 3. Sum of two cubes: 4. Difference of two cubes: A3 - B3 = (A - B)AA2 + AB + B2 B A3 + B3 = (A + B)AA2 - AB + B2 B A2 - 2AB + B2 = (A - B)2 A2 + 2AB + B2 = (A + B)2 A2 - B2 = (A + B)(A - B) PROPERTIES OF LOGARITHMS 1. 2. 3. 4. 5. 6. x 7 0eln x = xblogb x = x; lnex = xlogb bx = x; logb M = loga M logab = lnM lnb = logM logb logb Mp = plogb M logb a M N b = logb M - logb N logb(MN) = logb M + logb N SYMMETRY x y (x, y) (x, y) x y (x, y) (x, y) y-Axis Symmetry x-Axis Symmetry Origin Symmetry x y (x, y) (x, y) FMEndpapers.qxd 11/27/12 9:59 AM Page F2 28. FORMULAS/EQUATIONS The distance from to is The midpoint of the line segment with endpoints and is The standard equation of a circle of radius r with center at is The slope m of the line containing the points and is m is undened if The equation of a line with slope m and y-intercept (0, b) is The equation of a line with slope m containing the point is The solutions of the equation are If there are two distinct real solutions. If there is a repeated real solution. If there are two complex solutions (complex conjugates).b2 - 4ac 6 0, b2 - 4ac = 0, b2 - 4ac 7 0, x = -b ; 2b2 - 4ac 2a a Z 0,ax2 + bx + c = 0, y - y1 = m(x - x1)(x1, y1) y = mx + b x1 = x2 (x1 Z x2)slope (m) = change in y change in x = y2 - y1 x2 - x1 (x2, y2)(x1, y1) (x - h)2 + (y - k)2 = r2 (h, k) a x1 + x2 2 , y1 + y2 2 b.(x2, y2)(x1, y1) 2(x2 - x1)2 + (y2 - y1)2 .(x2, y2)(x1, y1)Distance Formula Midpoint Formula Standard Equation of a Circle Slope Formula Slope-Intercept Equation of a Line Point-Slope Equation of a Line Quadratic Formula GEOMETRY FORMULAS Circle Triangle Rectangle Rectangular Box Sphere Right Circular Cylinder Height (Altitude), S = 2r2 + 2rhV = r2 h S = Surface areaV = Volume,h = Height,r = Radius, S = 4r2 V = 4 3r3 S = Surface areaV = Volume,r = Radius, S = 2lw + 2lh + 2whV = lwh S = Surface areaV = Volume,h = Height,w = Width,l = Length, P = 2l + 2wA = lw P = perimeterA = area,w = Width,l = Length, A = 1 2bh A = areah =b = Base, C = 2prA = pr2 C = CircumferenceA = Area,r = Radius, CONVERSION TABLE 1 pound L 0.454 kilogramL 0.305 meter1 newton L 0.225 pound 1 ounce L 28.350 grams1 foot L 30.480 centimeters1 liter L 0.264 gallon 1 foot-lb L 1.356 Joules1 inch L 2.540 centimeters1 kilometer L 0.621 mile 1 pound L 4.448 newtons1 kilogram L 2.205 poundsL 3.281 feet 1 gallon L 3.785 liters1 gram L 0.035 ounce1 meter L 39.370 inches 1 mile L 1.609 kilometers1 joule L 0.738 foot-pound1 centimeter L 0.394 inch r h b w l r r h h w l FMEndpapers.qxd 11/27/12 9:59 AM Page F3 29. FUNCTIONS GRAPHS OF COMMON FUNCTIONS Constant Function Linear Function Quadratic Function Polynomial Function Rational Function Exponential Function Logarithmic Function where m is the slope and b is the y-intercept or parabola vertex f (x) = logbx, b 7 0, b Z l b Z 1f(x) = bx , b 7 0, R(x) = n(x) d(x) = anxn + an-1xn-1 + + a1x + a0 bmxm + am-1xm-1 + + b1x + b0 f(x) = anxn + an-1xn-1 + + a1x + a0 (h, k)f (x) = a(x - h)2 + ka Z 0f(x) = ax2 + bx + c, f(x) = mx + b, f(x) = b x y f(x) = |x| x y f (x) = c f(x) = x x y x y f(x) = x (0, 0) 5 3 1 3 421 5 2 3 4 5 3 4 1 2 5 5 3 1 3 421 5 2 3 4 5 3 4 1 2 5 3 5 6 9 10874 3 2 1 4 5 8 9 6 7 10 3 2 1 4 5 8 9 6 7 10 5 3 1 3 421 5 x y x y (0, 1) (1, b) f (x) = x3 f(x) = bx f(x) = logbx f(x) = x2 x y (1, 0) (b, 1) 5 4 3 2 3 421 5 4 2 6 8 10 6 8 2 4 10 x y Constant Function Identity Function Absolute Value Function Square Root Function Square Function Cube Function Exponential Function Logarithmic Function TRANSFORMATIONS In each case, c represents a positive real number. Function Draw the graph of f and: Vertical translations Horizontal translations Reections Shift f upward c units. Shift f downward c units. Shift f to the right c units. Shift f to the left c units. Reect f about the x-axis. Reect f about the y-axis. e y = - f (x) y = f (- x) e y = f (x - c) y = f (x + c) e y = f ( x) + c y = f ( x) - c HERONS FORMULA FOR AREA If the semiperimeter, s, of a triangle is then the area of that triangle is A = 1s(s - a) (s - b)(s - c) s = a + b + c 2 FMEndpapers.qxd 11/27/12 9:59 AM Page F4 30. ffirs.qxd 11/14/12 10:09 AM Page iv 31. 0 Prerequisites and Review W ould you be able to walk successfully along a tightrope? Most people probably would say no because the foundation is shaky. Would you be able to walk successfully along a beam (4 inches wide)? Most people would probably say yeseven though for some of us it is still challenging. Think of this chapter as the foundation for your walk. The more solid your foundation is now, the more successful your walk through College Algebra will be. The purpose of this chapter is to review concepts and skills that you already have learned in a previous course. Mathematics is a cumulative subject in that it requires a solid foundation to proceed to the next level. Use this chapter to reaffirm your current knowledge base before jumping into the course. GettyImages/BlendImages/GettyImages GarryWade/Taxi/GettyImages c00a.qxd 11/14/12 9:43 PM Page 2 32. I N TH I S C HAP TE R real numbers, integer exponents, and scientific notation will be discussed, followed by rational exponents and radicals. Simplification of radicals and rationalization of denominators will be reviewed. Basic operations such as addition, subtraction, and multiplication of polynomials will be discussed followed by a review of how to factor polynomials. Rational expressions will be discussed and a brief overview of solving simple algebraic equations will be given. After reviewing all of these aspects of real numbers, this chapter will conclude with a review of complex numbers. The Set of Real Numbers Approxima- tions: Rounding and Truncation Order of Operations Properties of Real Numbers Integer Exponents Scientific Notation Adding and Subtracting Polynomials Multiplying Polynomials Special Products Greatest Common Factor Factoring Formulas: Special Polynomial Forms Factoring a Trinomial as a Product of Two Binomials Factoring by Grouping A Strategy for Factoring Polynomials Rational Expressions and Domain Restrictions Simplifying Rational Expressions Multiplying and Dividing Rational Expressions Adding and Subtracting Rational Expressions Complex Rational Expressions Square Roots Other (nth) Roots Rational Exponents The Imaginary Unit, i Adding and Subtracting Complex Numbers Multiplying Complex Numbers Dividing Complex Numbers Raising Complex Numbers to Integer Powers PREREQUISITES AND REVIEW 0.1 Real Numbers 0.2 Integer Exponents and Scientific Notation 0.3 Polynomials: Basic Operations 0.4 Factoring Polynomials 0.5 Rational Expressions 0.6 Rational Exponents and Radicals 0.7 Complex Numbers Understand that rational and irrational numbers together constitute the real numbers. Apply properties of exponents. Perform operations on polynomials. Factor polynomials. Simplify expressions that contain rational exponents. Simplify radicals. Write complex numbers in standard form. L E A R N I N G O B J E C T I V E S 3 c00a.qxd 8/7/12 5:34 PM Page 3 33. The Set of Real Numbers A set is a group or collection of objects that are called members or elements of the set. If every member of set B is also a member of set A, then we say B is a subset of A and denote it as . For example, the starting lineup on a baseball team is a subset of the entire team. The set of natural numbers, {1, 2, 3, 4, . . .}, is a subset of the set of whole numbers, {0, 1, 2, 3, 4, . . .}, which is a subset of the set of integers, {. . . , 4, 3, 2, 1, 0, 1, 2, 3, . . .}, which is a subset of the set of rational numbers, which is a subset of the set of real numbers. The three dots, called an ellipsis, indicate that the pattern continues indenitely. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol . The set of real numbers consists of two main subsets: rational and irrational numbers. B ( A C O N C E P TUAL O BJ E CTIVE S Understand that rational and irrational numbers are mutually exclusive and complementary subsets of real numbers. Learn the order of operations for real numbers. R EAL N U M B E R S S K I LLS O BJ E CTIVE S Classify real numbers as rational or irrational. Round or truncate real numbers. Simplify expressions using correct order of operations. Evaluate algebraic expressions. Apply properties of real numbers. S E CTI O N 0.1 A rational number is a number that can be expressed as a quotient (ratio) of two integers, where the integer a is called the numerator and the integer b is called the denominator and where b Z 0. a b , Rational Number Rational numbers include all integers or all fractions that are ratios of integers. Note that any integer can be written as a ratio whose denominator is equal to 1. In decimal form, the rational numbers are those that terminate or are nonterminating with a repeated decimal pattern, which is represented with an overbar. Those decimals that do not repeat and do not terminate are irrational numbers. The numbers are examples of real numbers, where 5, 17, , 1.37, 0, , and are rational numbers, and , and 3.2179. . . are irrational numbers. It is important to note that the ellipsis following the last decimal digit denotes continuing in an irregular fashion, whereas the absence of such dots to the right of the last decimal digit implies the decimal expansion terminates. 12, p 3.666-19 17 1 3 5, -17, 1 3 , 12, p, 1.37, 0, - 19 17 , 3.666, 3.2179. . . D E F I N I T I O N 4 c00a.qxd 8/7/12 5:34 PM Page 4 34. 0.1 Real Numbers 5 Notice that the overbar covers the entire repeating pattern. The following gure and table illustrate the subset relationship and examples of different types of real numbers. RATIONAL NUMBER CALCULATOR DECIMAL (FRACTION) DISPLAY REPRESENTATION DESCRIPTION 3.5 3.5 Terminates 1.25 1.25 Terminates 0.666666666 Repeats 0.09090909 Repeats0.091 11 0.62 3 15 12 7 2 SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers 1, 2, 3, 4, 5, . . . W Whole numbers Natural numbers and zero 0, 1, 2, 3, 4, 5, . . . Z Integers Whole numbers and negative natural numbers . . . , 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, . . . Q Rational numbers Ratios of integers: 17, , 0, , 1.37, , 5 Decimal representation terminates, or Decimal representation repeats I Irrational numbers Numbers whose decimal representation , 1.2179. . . , does not terminate or repeat R Real numbers Rational and irrational numbers , 5, , 17.25, 17- 2 3p p12 3.6661 3-19 7 a b (b Z 0) Real Numbers Irrational Numbers Integers Negative Counting Numbers Zero Rational Numbers Fractions Ratio of two nonzero integers, and Does not reduce to an integer Whole Numbers Natural Numbers Since the set of real numbers can be formed by combining the set of rational numbers and the set of irrational numbers, then every real number is either rational or irrational. The set of rational numbers and the set of irrational numbers are both mutually exclusive (no shared elements) and complementary sets. The real number line is a graph used to represent the set of all real numbers. 3 2 1 10 2 3 19 17 2 Study Tip Every real number is either a rational number or an irrational number. c00a.qxd 8/7/12 5:34 PM Page 5 35. 6 C HAP TE R 0 Prerequisites and Review EXAMPLE 1 Classifying Real Numbers Classify the following real numbers as rational or irrational: Solution: Rational: 3, 0, , 7.51, , , Irrational: YO U R TU R N Classify the following real numbers as rational or irrational: -7 3, 5.9999, 12, 0, -5.27, 15, 2.010010001. . . 13, p-6.66666- 8 5 1 3 1 4 -3, 0, 1 4, 13, p, 7.51, 1 3, - 8 5, 6.66666 Answer: Rational: , , 12, 0, 5.27 Irrational: , 2.010010001 . . .15 5.9999-7 3 Answer: a. Truncation: 23.02 b. Rounding: 23.02 Answer: a. Truncation: 2.3818 b. Rounding: 2.3819 EXAMPLE 2 Approximating Decimals to Two Places Approximate 17.368204 to two decimal places by a. truncation b. rounding Solution: a. To truncate, eliminate all digits to the right of the 6. 17.36 b. To round, look to the right of the 6. Because 8 is greater than 5, round up (add 1 to the 6). 17.37 YO U R TU R N Approximate 23.02492 to two decimal places by a. truncation b. rounding EXAMPLE 3 Approximating Decimals to Four Places Approximate 7.293516 to four decimal places by a. truncation b. rounding Solution: The 5 is in the fourth decimal place. a. To truncate, eliminate all digits to the right of 5. 7.2935 b. To round, look to the right of the 5. Because 1 is less than 5, the 5 remains the same. 7.2935 YO U R TU R N Approximate 2.381865 to four decimal places by a. truncation b. rounding Study Tip When rounding, look to the right of the specied decimal place and use that digit (do not round that digit rst). 5.23491 rounded to two decimal places is 5.23 (do not round the 4 to a 5 rst). Approximations: Rounding and Truncation Every real number can be represented by a decimal. When a real number is in decimal form, it can be approximated by either rounding off or truncating to a given decimal place. Truncation is cutting off or eliminating everything to the right of a certain decimal place. Rounding means looking to the right of the specied decimal place and making a judgment. If the digit to the right is greater than or equal to 5, then the specied digit is rounded up, or increased by one unit. If the digit to the right is less than 5, then the specied digit stays the same. In both of these cases all decimal places to the right of the specied place are removed. c00a.qxd 8/7/12 5:34 PM Page 6 36. 0.1 Real Numbers 7 OPERATION NOTATION RESULT Addition a b Sum Subtraction a b Difference Multiplication a b or ab or (a)(b) Product Division or ab Quotient (Ratio)(b Z 0) a b # Since algebra involves variables such as x, the traditional multiplication sign is not used. Three alternatives are shown in the preceding table. Similarly, the arithmetic sign for division is often represented by vertical or slanted fractions. The symbol is called the equal sign, and is pronounced equals or is, and it implies that the expression on one side of the equal sign is equivalent to (has the same value as) the expression on the other side of the equal sign. WORDS MATH The sum of seven and eleven equals eighteen: 7 11 18 Three times ve is fteen: 3 5 15 Four times six equals twenty-four: 4(6) 24 Eight divided by two is four: Three subtracted from ve is two: 5 3 2 8 2 = 4 # It is important to note that rounding and truncation sometimes yield the same approximation (Example 3), but not always (Example 2). Order of Operations Addition, subtraction, multiplication, and division are called arithmetic operations. The results of these operations are called the sum, difference, product, and quotient, respectively. These four operations are summarized in the following table. When evaluating expressions involving real numbers, it is important to remember the correct order of operations. For example, how do we simplify the expression 3 2 5? Do we multiply rst and then add, or add rst and then multiply? In mathematics, conventional order implies multiplication rst, and then addition: 3 2 5 3 10 13. Parentheses imply grouping of terms, and the necessary operations should always be performed inside them rst. If there are nested parentheses, always start with the innermost parentheses and work your way out. Within parentheses follow the conventional order of operations. Exponents are an important part of order of operations and will be discussed in Section 0.2. # # 1. Start with the innermost parentheses (grouping symbols) and work outward. 2. Perform all indicated multiplications and divisions, working from left to right. 3. Perform all additions and subtractions, working from left to right. ORDER OF OPERATIONS c00a.qxd 8/7/12 5:34 PM Page 7 37. 8 C HAP TE R 0 Prerequisites and Review EXAMPLE 5 Simplifying Expressions That Involve Grouping Signs Using the Correct Order of Operations Simplify the expression 3[5 (4 2) 2 7]. Solution: Simplify the inner parentheses. 3[5 (4 2) 2 7] 3[5 2 2 7] Inside the brackets, perform the multiplication 5 2 10 and 2 7 14. 3[10 14] Inside the brackets, perform the subtraction. 3[4] Multiply. 12 YO U R TU R N Simplify the expression 2[3 (13 5) 4 3].## ## #### ## EXAMPLE 4 Simplifying Expressions Using the Correct Order of Operations Simplify the expressions. a. 4 3 2 7 5 6 b. Solution (a): Perform multiplication rst. Then perform the indicated additions and subtractions. 4 6 35 6 19 Solution (b): The numerator and the denominator are similar to expressions in parentheses. Simplify these separately rst, following the correct order of operations. Perform multiplication in the denominator rst. Then perform subtraction in the numerator and addition in the denominator. YO U R TU R N Simplify the expressions. a. 7 4 5 2 6 9 b. 9 - 6 2 # 5 + 6 ## = 7 - 6 6 + 8 = 1 14 7 - 6 2 # 3 + 8 4 + 3 # 2 - 7 # 5 + 6 7 - 6 2 # 3 + 8 ## u 6 u 6 Answer: a. 10 b. 3 16 Parentheses ( ) and brackets [ ] are the typical notations for grouping and are often used interchangeably. When nesting (groups within groups), use parentheses on the innermost and then brackets on the outermost. u 35 Answer: 24 c00a.qxd 8/7/12 5:34 PM Page 8 38. 0.1 Real Numbers 9 An algebraic expression is the combination of variables and constants using basic operations such as addition, subtraction, multiplication, and division. Each term is separated by addition or subtraction. Algebraic Expression Variable Term Constant Term Coefcient 5x 3 5x 3 5 When we know the value of the variables, we can evaluate an algebraic expression using the substitution principle: Algebraic expression: 5x 3 Value of the variable: x 2 Substitute x 2: 5(2) 3 10 3 13 Algebraic ExpressionD E F I N I T I O N Algebraic Expressions Everything discussed until now has involved real numbers (explicitly). In algebra, however, numbers are often represented by letters (such as x and y), which are called variables. A constant is a xed (known) number such as 5. A coefcient is the constant that is multiplied by a variable. Quantities within the algebraic expression that are separated by addition or subtraction are referred to as terms. EXAMPLE 6 Evaluating Algebraic Expressions Evaluate the algebraic expression 7x 2 for x 3. Solution: Start with the algebraic expression. 7x 2 Substitute x 3. 7(3) 2 Perform the multiplication. 21 2 Perform the addition. 23 YO U R TU R N Evaluate the algebraic expression 6y 4 for y 2. Answer: 16 In Example 6, the value for the variable was specied in order for us to evaluate the algebraic expression. What if the value of the variable is not specied; can we simplify an expression like 3(2x 5y)? In this case, we cannot subtract 5y from 2x. Instead, we rely on the basic properties of real numbers, or the basic rules of algebra. Properties of Real Numbers You probably already know many properties of real numbers. For example, if you add up four numbers, it does not matter in which order you add them. If you multiply ve numbers, it does not matter what order you multiply them. If you add 0 to a real number or multiply a real number by 1, the result yields the original real number. Basic properties of real numbers are summarized in the following table. Because these properties are true for variables and algebraic expressions, these properties are often called the basic rules of algebra. c00a.qxd 8/7/12 5:34 PM Page 9 39. MATH (LET a, b, AND c EACH NAME DESCRIPTION BE ANY REAL NUMBER) EXAMPLE Commutative property Two real numbers can be added a b b a 3x 5 5 3x of addition in any order. Commutative property Two real numbers can be ab ba y 3 3y of multiplication multiplied in any order. Associative property When three real numbers are (a b) c a (b c) (x 5) 7 x (5 7) of addition added, it does not matter which two numbers are added rst. Associative property When three real numbers are (ab)c a(bc) (3x)y 3(xy) of multiplication multiplied, it does not matter which two numbers are multiplied rst. Distributive property Multiplication is distributed a(b c) ab ac 5(x 2) 5x 10 over all the terms of the sums or differences within the a(b c) ab ac 5(x 2) 5x 10 parentheses. Additive identity Adding zero to any real number a 0 a 7y 0 7y property yields the same real number. 0 a a Multiplicative identity Multiplying any real number by a 1 a (8x)(1) 8x property 1 yields the same real number. 1 a a Additive inverse The sum of a real number and its a (a) 0 4x (4x) 0 property additive inverse (opposite) is zero. Multiplicative inverse The product of a nonzero real property number and its multiplicative inverse (reciprocal) is 1. x Z -2 (x + 2) # a 1 x + 2 b = 1a # 1 a = 1 a Z 0 # # # 10 C HAP TE R 0 Prerequisites and Review The properties in the previous table govern addition and multiplication. Subtraction can be dened in terms of addition of the additive inverse, and division can be dened in terms of multiplication by the multiplicative inverse (reciprocal). Let a and b be real numbers. SUBTRACTION AND DIVISION MATH TYPE OF INVERSE WORDS Subtraction a b a (b) b is the additive inverse or Subtracting a real number is equal to opposite of b adding its opposite. Division is the multiplicative inverse or Dividing by a real number is equal to reciprocal of b multiplying by its reciprocal. b Z 0 1 b a , b = a # 1 b PROPERTIES OF REAL NUMBERS (BASIC RULES OF ALGEBRA) c00a.qxd 8/7/12 5:34 PM Page 10 40. 0.1 Real Numbers 11 EXAMPLE 7 Using the Distributive Property Use the distributive property to eliminate the parentheses. a. 3(x 5) b. 2(y 6) Solution (a): Use the distributive property. 3(x 5) 3(x) 3(5) Perform the multiplication. 3x 15 Solution (b): Use the distributive property. 2(y 6) 2(y) 2(6) Perform the multiplication. 2y 12 YO U R TU R N Use the distributive property to eliminate the parentheses. a. 2(x 3) b. 5(y 3) Answer: a. 2x 6 b. 5y 15 MATH (LET a AND b BE DESCRIPTION POSITIVE REAL NUMBERS) EXAMPLE A negative quantity times a positive quantity is a negative quantity. (a)(b) ab (8)(3) 24 A negative quantity divided by a positive quantity is a negative quantity. A positive quantity divided by a negative quantity is a negative quantity. A negative quantity times a negative quantity is a positive quantity. (a)(b) ab (2x)(5) 10x A negative quantity divided by a negative quantity is a positive quantity. The opposite of a negative quantity is a positive quantity (subtracting a (a) a (9) 9 negative quantity is equivalent to adding a positive quantity). A negative sign preceding an expression is distributed throughout (a b) a b 3(x 5) 3x 15 the expression. (a b) a b 3(x 5) 3x 15 -12 -3 = 4 -a -b = a b 15 -3 = -5 a -b = - a b -16 4 = -4 or -a b = - a b or You also probably know the rules that apply when multiplying a negative real number. For example, a negative times a negative is a positive. EXAMPLE 8 Using Properties of Negatives Eliminate the parentheses and perform the operations. a. 5 7 (2) b. (3)(4)(6) Solution: a. Distribute the negative. 2 5 7 2 Combine the three quantities. 4 b. Group the terms. [(3)][(4)(6)] Perform the multiplication inside the [ ]. [3][24] Multiply. 72 -5 + 7 - (-2) u Technology Tip a. b. Here are the calculator keystrokes for (3)(4)(6). or PROPERTIES OF NEGATIVES c00a.qxd 8/7/12 5:34 PM Page 11 41. 12 C HAP TE R 0 Prerequisites and Review We use properties of negatives to dene the absolute value of any real number. The absolute value of a real number a, denoted a, is its magnitude. On a number line this is the distance from the origin, 0, to the point. For example, algebraically, the absolute value of 5 is 5, that is, 5 5; and the absolute value of 5 is 5, or 5 5. Graphically, the distance on the real number line from 0 to either 5 or 5 is 5. Notice that the absolute value does not change a positive real number, but changes a negative real number to a positive number. A negative number becomes a positive number if it is multiplied by 1. IF a IS A a EXAMPLE Positive real number a a 5 5 Negative real number a a 5 (5) 5 Zero a a 0 0 5 0 5 5 5 EXAMPLE 9 Finding the Absolute Value of a Real Number Evaluate the expressions. a. 3 7 b. 2 8 Solution: a. 3 7 4 b. 2 8 6 4 6 Properties of the absolute value will be discussed in Section 1.7. EXAMPLE 10 Using Properties of Negatives and the Distributive Property Eliminate the parentheses (2x 3y). I N C O R R E CT Error: 2x 3y The negative () was not distributed through the second term. C O R R E CT (2x 3y) (2x) (3y) 2x 3y A common mistake is applying a negative only to the rst term. C O M M O N M I S TA K E YO U R TU R N Eliminate the parentheses. a. 2(x 5y) b. (3 2b) Answer: a. 2x 10y b. 3 2b What is the product of any real number and zero? The answer is zero. This property also leads to the zero product property, which is the basis for factoring (one of the methods used to solve quadratic equations, which will be discussed in Section 1.3). c00a.qxd 8/7/12 5:34 PM Page 12 42. 0.1 Real Numbers 13 DESCRIPTION MATH (LET a BE A REAL NUMBER) EXAMPLE A real number multiplied by zero is zero. a 0 0 0 x 0 Zero divided by a nonzero real number is zero. A real number divided by zero is undened. is undened is undened x + 2 0 a 0 0 3 - x = 0 x Z 3 0 a = 0 a Z 0 ## Fractions always seem to intimidate students. In fact, many instructors teach students to eliminate fractions in algebraic equations. It is important to realize that you can never divide by zero. Therefore, in the following table of fractional properties it is assumed that no denominators are zero. DESCRIPTION MATH ZERO CONDITION EXAMPLE Equivalent fractions if and only if ad bc and since 12y 12y Multiplying two fractions and Adding fractions that have the x 3 + 2 3 = x + 2 3 b Z 0 a b + c b = a + c b 3 5 # x 7 = 3x 35 d Z 0b Z 0 a b # c d = ac bd y 2 = 6y 12 d Z 0b Z 0 a b = c d same denominator Subtracting fractions that have 7 3 - 5 3 = 7 - 5 3 = 2 3 b Z 0 a b - c b = a - c b the same denominator Adding fractions with different and denominators using a common denominator Subtracting fractions with and different denominators using a common denominator Dividing by a fraction is , , x 3 , 2 7 = x 3 # 7 2 = 7x 6 c Z 0b Z 0 a b , c d = a b # d c 1 3 - 1 4 = (1)(4) - (1)(3) (3)(4) = 1 12 d Z 0b Z 0 a b - c d = ad bd - cb bd = ad - bc bd 1 2 + 5 3 = (1)(3) + (5)(2) (2)(3) = 13 6 d Z 0b Z 0 a b + c d = ad bd + cb bd = ad + bc bd equivalent to multiplying and by its reciprocal d Z 0 PROPERTIES OF ZERO ZERO PRODUCT PROPERTY DESCRIPTION MATH EXAMPLE If the product of two real numbers is zero, If ab 0, then a 0 or b 0 If x(x 2) 0, then x 0 or x 2 0 then one of those numbers has to be zero. therefore x 0 or x 2 Note: If a and b are both equal to zero, then the product is still zero. FRACTIONAL PROPERTIES c00a.qxd 8/7/12 5:34 PM Page 13 43. Technology Tip To change the decimal number to a fraction, press: a. ENTER Frac1MATH b. 14 C HAP TE R 0 Prerequisites and Review EXAMPLE 11 Performing Operations with Fractions Perform the indicated operations involving fractions and simplify. a. b. c. Solution (a): Determine the LCD. Rewrite fractions applying the LCD. Eliminate the parentheses. Combine terms in the numerator. Solution (b): Rewrite 4 with an understood 1 in the denominator. Dividing by a fraction is equivalent to multiplying by its reciprocal. Multiply numerators and denominators, respectively. Reduce the fraction to simplest form. Solution (c): Determine the LCD. 2 . 5 10 Rewrite fractions in terms of the LCD. Simplify the numerator. YO U R TU R N Perform the indicated operations involving fractions. a. b. c. 2 3 - x 5 1 5 , 3 10 3 5 + 1 2 = 5x + 6 10 x 2 + 3 5 = 5x + 3(2) (2)(5) = 1 6 = 2 12 = 2 3 # 1 4 = 2 3 , 4 1 = 5 12 = 8 - 3 12 = 2(4) - 1(3) 3(4) 2 3 - 1 4 = 2 # 4 3 # 4 - 1 # 3 4 # 3 3 # 4 = 12 x 2 + 3 5 2 3 , 4 2 3 - 1 4 Answer: a. b. c. 10 - 3x 15 2 3 11 10 The least common multiple of two or more integers is the smallest integer that is evenly divisible by each of the integers. For example, the least common multiple (LCM) of 3 and 4 is 12. The LCM of 8 and 6 is 24. The reason the LCM of 8 and 6 is not 48 is that 8 and 6 have a common factor of 2. When adding and subtracting fractions, a common denominator can be found by multiplying the denominators. When there are common factors in the denominators, the LCM is the least common denominator (LCD) of the original denominators. c00a.qxd 8/7/12 5:34 PM Page 14 44. 0.1 Real Numbers 15 In Exercises 18, classify the following real numbers as rational or irrational. 1. 2. 3. 2.07172737. . . 4. 5. 2.776677 6. 7. 8. In Exercises 916, approximate the real number to three decimal places by (a) rounding and (b) truncation. 9. 7.3471 10. 9.2549 11. 2.9949 12. 6.9951 13. 0.234492 14. 1.327491 15. 5.238473 16. 2.118465 117155.2222226677 p22 3 11 3 E X E R C I S E S S E CTI O N 0.1 Subtraction and division can be dened in terms of addition and multiplication. Subtraction: a b a (b) (add the opposite) Division: , where (multiply by the reciprocal) Properties of negatives were reviewed. If a and b are positive real numbers, then: (a)(b) ab (a)(b) ab (a) a (a b) a b and (a b) a b Absolute value of real numbers: a a if a is nonnegative, and a a if a is negative. Properties of zero were reviewed. -a -b = a b -a b = - a b b Z 0a , b = a # 1 b S U M MARY In this section, real numbers were dened as the set of all rational and irrational numbers. Decimals are approximated by either truncating or rounding. Truncating: Eliminate all values after a particular digit. Rounding: Look to the right of a particular digit. If the number is 5 or greater, increase the digit by 1; otherwise, leave it as is and eliminate all digits to the right. The order in which we perform operations is 1. parentheses (grouping); work from inside outward. 2. multiplication/division; work from left to right. 3. addition/subtraction; work from left to right. The properties of real numbers are employed as the basic rules of algebra when dealing with algebraic expressions. Commutative property of addition: a b b a Commutative property of multiplication: ab ba Associative property of addition: (a b) c a (b c) Associative property of multiplication: (ab)c a(bc) Distributive property: a(b c) ab ac or a(b c) ab ac Additive identity: a 0 a Multiplicative identity: a 1 a Additive inverse (opposite): a (a) 0 Multiplicative inverse (reciprocal): a # 1 a = 1 a Z 0 # S E CTI O N 0.1 a 0 0 and a Z 0 0 a = 0# is undened Zero product property: If ab 0, then a 0 or b 0 Properties of fractions were also reviewed. and , , and d Z 0c Z 0b Z 0 a b , c d = a b # d c d Z 0b Z 0 a b ; c d = ad ; bc bd a 0 S K I LL S c00a.qxd 8/7/12 5:34 PM Page 15 45. 16 C HAP TE R 0 Prerequisites and Review In Exercises 1740, perform the indicated operations in the correct order. 17. 5 2 3 7 18. 2 5 4 3 6 19. 2 (5 7 4 20) 20. 3 (2 7) 8 (7 2 1) 21. 2 3[4(2 3 5)] 22. 4 6(5 9) 23. 8 (2) 7 24. 10 (9) 25. 3 (6) 26. 5 2 (3) 27. x (y) z 28. a b (c) 29. (3x y) 30. (4a 2b) 31. 32. 33. 4 6 [(5 8)(4)] 34. 35. (6x 4y) (3x 5y) 36. 37. (3 4x) (4x 7) 38. 2 3[(4x 5) 3x 7] 39. 40. 6(2x 3y) [3x (2 5y)] In Exercises 4156, write as a single fraction and simplify. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. In Exercises 5768, perform the indicated operation and simplify, if possible. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. In Exercises 6972, evaluate the algebraic expression for the specied values. 69. for c 4, d 3 70. 2l 2w for l 5, w 10 71. for m1 3, m2 4, r 10 72. for x 100, m 70, s 15 x - m s m1 # m2 r2 -c 2d 2 1 3 # 7 5 6 y Z 0 6x 7 , 3y 28 14m 2 # 4 7 y Z 0 3x 4 , 9 16y 4 1 5 , 7 1 20 x Z 0 3x 10 , 6x 15 b Z 0 3a 7 , b 21 a Z 0 4b 9 , a 27 4 5 , 7 10 2 7 , 10 3 2 3 # 9 10 2 7 # 14 3 -3 10 - a -7 12 b 3 40 + 7 24 6x 12 - 7x 20 4y 15 - (-3y) 4 y 10 - y 15 x 3 - 2x 7 y 3 - y 6 x 5 + 2x 15 3 7 - (-4) 3 - 5 6 1 9 - 2 27 1 3 + 5 9 3 2 + 5 12 7 3 - 1 6 5 6 - 1 3 1 2 - 1 5 1 3 + 5 4 -4(5) - 5 -5 -4x 6 - (-2) -14 5 - (-2) - 12 (-3)(-4) -3 (5)(-1) ## ######## 73. U.S. National Debt. Round the debt to the nearest million. 74. U.S. Population. Round the number of citizens to the nearest thousand. 75. U.S. Debt. If the debt is distributed evenly to all citizens, what is the national debt per citizen? Round your answer to the nearest dollar. 76. U.S. Debt. If the debt is distributed evenly to all citizens, what is the national debt per citizen? Round your answer to the nearest cent. On December 16, 2007, the United States debt was estimated at $9,176,366,494,947, and at that time the estimated population was 303,818,361 citizens. A P P L I C AT I O N S c00a.qxd 8/7/12 5:34 PM Page 16 46. 0.1 Real Numbers 17 In Exercises 8184, determine whether each of the following statements is true or false. 81. Student athletes are a subset of the students in the honors program. 82. The students who are members of fraternities or sororities are a subset of the entire student population. 83. Every integer is a rational number. 84. A real number can be both rational and irrational. 85. What restrictions are there on x for the following to be true: 86. What restrictions are there on x for the following to be true: x 2 , x 6 = 3 3 x , 5 x = 3 5 88. 2{5(y x) 2[3(2x 5) 7(2) 4] 3} 7 In Exercises 87 and 88, simplify the expressions. 87. 2[3(x 2y) 7] [3(2 5x) 10] 7[2(x 3) 5] 89. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? 90. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? A 144 25 11260 91. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? 92. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? A 882 49 14489 77. Round 13.2749 to two decimal places. Solution: The 9, to the right of the 4, causes the 4 to round to 5. 13.275 The 5, to the right of the 7, causes the 7 to be rounded to 8. 13.28 This is incorrect. What mistake was made? 78. Simplify the expression . Solution: Add the numerators and denominators. Reduce. This is incorrect. What mistake was made? = 1 4 2 + 1 3 + 9 = 3 12 2 3 + 1 9 79. Simplify the expression 3(x 5) 2(4 y). Solution: Eliminate parentheses. 3x 15 8 y Simplify. 3x 7 y This is incorrect. What mistake was made? 80. Simplify the expression 3(x 2) (1 y). Solution: Eliminate parentheses. 3x 6 1 y Simplify. 3x 7 y This is incorrect. What mistake was made? In Exercises 7780, explain the mistake that is made. C AT C H T H E M I S TA K E C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y c00a.qxd 8/7/12 5:34 PM Page 17 47. Let a be a real number and n be a natural number (positive integer); then an is dened as where n is the exponent, or power, and a is the base. (a appears as a factor n times)an = a # a # a p a Natural-Number ExponentD E F I N I T I O N d n factors Answer: a. 216 b. 1 81 EXAMPLE 1 Evaluating Expressions Involving Natural-Number Exponents Evaluate the expressions. a. 43 b. 81 c. 54 d. Solution: a. 43 4 4 4 64 b. 81 8 c. 54 5 5 5 5 625 d. YO U R T U R N Evaluate the expressions. a. 63 b. A1 3 B4 A1 2 B5 = 1 2 # 1 2 # 1 2 # 1 2 # 1 2 = 1 32 ### ## A1 2B5 Technology Tip Integer Exponents Exponents represent repeated multiplication. For example, 2 2 2 2 2 25 . The 2 that is repeatedly multiplied is called the base, and the small number 5 above and to the right of the 2 is called the exponent. #### C O N C E P TUAL O BJ E CTIVE S Visualize negative exponents as reciprocals. Understand that scientic notation is an effective way to represent very large or very small real numbers. I NTE G E R E X P O N E NTS AN D S C I E NTI F I C N OTATI O N S E CTI O N 0.2 S K I LLS O BJ E CTIVE S Evaluate expressions involving integer exponents. Apply properties of exponents. Use scientic notation. Study Tip an : a raised to the nth power a2 : a squared a3 : a cubed We now include exponents in our order of operations: 1. Parentheses 2. Exponents 3. Multiplication/Division 4. Addition/Subtraction 18 c00a.qxd 8/7/12 5:34 PM Page 18 48. EXAMPLE 2 Evaluating Expressions Involving Natural-Number Exponents Evaluate the expressions. a. (3)4 b. 34 c. (2)3 52 Solution: a. b. c. YO U R TU R N Evaluate the expression 43 23 .# (-2)3 # 52 = (-2)(-2)(-2) # 5 # 5 = -200 -34 = -(3 # 3 # 3 # 3) = -81 (-3)4 = (-3)(-3)(-3)(-3) = 81 # 0.2 Integer Exponents and Scientific Notation 19 Answer: 512 Let a be any nonzero real number and n be a natural number (positive integer); then a-n = 1 an a Z 0 NEGATIVE-INTEGER EXPONENT PROPERTY In other words, a base raised to a negative-integer exponent is equivalent to the reciprocal of the base raised to the opposite (positive) integer exponent. Study Tip A negative exponent implies a reciprocal. So far, we have discussed only exponents that are natural numbers (positive integers). When the exponent is a negative integer, we use the following property. EXAMPLE 3 Evaluating Expressions Involving Negative-Integer Exponents Evaluate the expressions. a. 24 b. c. d. Solution: a. b. c. d. YO U R TU R N Evaluate the expressions. a. b. 1 3-2 # 6-2 - 1 5-2 -23 # 1 (-6)-2 = -23 # (-6)2 = (-8)(36) = -288 4-3 # 1 2-4 = 1 43 # 24 = 16 64 = 1 4 1 3-3 = 1 a 1 33 b = 1 , a 1 33 b = 1 # 33 1 = 33 = 27 2-4 = 1 24 = 1 16 -23 # 1 (-6)-2 4-3 # 1 2-4 1 3-3 8 36 u u Answer: a. 25 b. 1 4 Technology Tip Technology Tip 81 25 d 8 d u c00a.qxd 8/7/12 5:34 PM Page 19 49. 20 C HAP TE R 0 Prerequisites and Review MATH (LET a AND b BE NONZERO REAL NUMBERS NAME DESCRIPTION AND m AND n BE INTEGERS) EXAMPLE Product property When multiplying exponentials with the am an amn x2 x5 x25 x7 same base, add exponents. Quotient property When dividing exponentials with the same base, subtract the exponents (numerator denominator). Power property When raising an exponential to a power, multiply exponents. Product to a A product raised to a power is equal to the (ab)n an bn (2x)3 23 x3 8x3 power property product of each factor raised to the power. Quotient to a A quotient raised to a power is equal to the power property quotient of the factors raised to the power. y Z 0a x y b 4 = x4 y4a a b b n = an bn # (x2 ) 4 = x2#4 = x8 (am ) n = amn x Z 0 x5 x3 = x5-3 = x2am an = am-n ## Let a be any nonzero real number; then a0 = 1 a Z 0 ZERO-EXPONENT PROPERTY EXAMPLE 4 Evaluating Expressions Involving Zero Exponents Evaluate the expressions. a. 50 b. c. (3)0 d. 40 Solution: a. 50 1 b. c. (3)0 1 d. 40 1 1 20 = 1 1 = 1 1 20 WORDS MATH When expressions with the same base are multiplied, the exponents are added. When expressions with the same base are divided, the exponents are subtracted. When an expression involving an exponent is (8)2 64 raised to a power, the exponents are multiplied. or (23 ) 2 = 23#2 = 26 = 64 (23 ) 2 or 25-3 = 22 25 23 = 2 # 2 # 2 # 2 # 2 2 # 2 # 2 = 2 # 2 1 = 22 23 # 24 = 2 # 2 # 2 # 2 # 2 # 2 # 2 = 23+4 = 27 u u 23 24 Now we can evaluate expressions involving positive and negative exponents. How do we evaluate an expression with a zero exponent? We dene any nonzero real number raised to the zero power as 1. We now can evaluate expressions involving integer (positive, negative, or zero) exponents. What about when expressions involving integer exponents are multiplied, divided, or raised to a power? { 1 The following table summarizes the properties of integer exponents. PROPERTIES OF INTEGER EXPONENTS c00a.qxd 8/7/12 5:34 PM Page 20 50. 0.2 Integer Exponents and Scientific Notation 21 Common Errors Made Using Properties of Exponents INCORRECT CORRECT ERROR x4 x3 x12 x4 x3 x7 Exponents should be added (not multiplied). Exponents should be subtracted (not divided). Exponents should be multiplied (not raised to a power). (2x)3 2x3 (2x)3 8x3 Both factors (the 2 and the x) should be cubed. 23 24 47 23 24 27 The original common base should be retained. 23 35 68 23 35 The properties of integer exponents require the same base.## ## (x2 ) 3 = x6 (x2 ) 3 = x8 x18 x6 = x12 ; x Z 0 x18 x6 = x3 ## We will now use properties of integer exponents to simplify exponential expressions. EXAMPLE 5 Simplifying Exponential Expressions Simplify the expressions (assume all variables are nonzero). a. b. c. Solution (a): Parentheses imply multiplication. Group the same bases together. Apply the product property. Multiply the constants. 10x5 y4 Solution (b): Apply the product to a power property. Apply the power property. 8x2 # 3 y1 # 3 z3 # 3 Simplify. 8x6 y3 z9 Solution (c): Group the same bases together. Apply the quotient property. (5)x35 y64 5x2 y2 Apply the negative exponent property. YO U R TU R N Simplify the expressions (assume all variables are nonzero). a. b. c. -16x4 y3 4xy7 (-3xy3 z2 ) 3 (-3x3 y2 )(4xy3 ) = -5y2 x2 25x3 y6 -5x5 y4 = a 25 -5 ba x3 x5 ba y6 y4 b (2x2 yz3 ) 3 = (2)3 (x2 ) 3 (y) 3 (z3 ) 3 = (-2)(5)x2 x3 y3 y (-2x2 y3 )(5x3 y) = (-2)(5)x2 x3 y3 y 25x3 y6 -5x5 y4 (2x2 yz3 ) 3 (-2x2 y3 )(5x3 y) x23 y31 u u Study Tip It is customary not to leave negative exponents. Instead we use the negative exponent property to write exponential expressions with only positive exponents. Answer: a. 12x4 y5 b. 27x3 y9 z6 c. -4x3 y4 An exponential expression is simplied when: All parentheses (groupings) have been eliminated. A base appears only once. No powers are raised to other powers. All exponents are positive. c00a.qxd 8/7/12 5:34 PM Page 21 51. 22 C HAP TE R 0 Prerequisites and Review EXAMPLE 6 Simplifying Exponential Expressions Write each expression so that all exponents are positive (assume all variables are nonzero). a. b. c. Solution (a): Apply the product to a power property. Apply the power property. 33 x6 z12 Apply the negative-integer exponent property. Evaluate 33 . Solution (b): Apply the product to a power property. Apply the quotient property. x43 y6(12) Simplify. xy6 Solution (c): Apply the product to a power property on both the numerator and denominator. Apply the power property. Group constant terms and x terms. Apply the quotient property. Simplify. YO U R TU R N Simplify the exponential expression and express it in terms of positive exponents . (tv2 ) -3 (2t4 v3 ) -1 = 2xy6 9z6 = a 8 36 b(x3-2 )a y6 z6 b = a -8 -36 ba x3 x2 ba y6 z6 b = -8x3 y6 -36x2 z6 (-2xy2 )3 -(6xz3 ) 2 = (-2)3 (x)3 (y2 )3 -(6)2 (x)2 (z3 ) 2 (x2 y-3 ) 2 (x-1 y4 ) -3 = x4 y-6 x3 y-12 = z12 27x6 = z12 33 x6 (3x2 z-4 ) -3 = (3) -3 (x2 ) -3 (z-4 ) -3 (-2xy2 ) 3 -(6xz3 ) 2 (x2 y-3 ) 2 (x-1 y4 ) -3(3x2 z-4 ) -3 Answer: 2t v3 Scientific Notation You are already familiar with base 10 raised to positive-integer powers. However, it can be inconvenient to write all the zeros out, so we give certain powers of 10 particular names: thousand, million, billion, trillion, and so on. For example, we say there are 300 million U.S. citizens as opposed to writing out 300,000,000 citizens. Or we say that the national debt is $14 trillion as opposed to writing out $14,000,000,000,000. The following table contains scientic notation for positive exponents and examples of some common prexes and abbreviations. One of the fundamental applications of scientic notation is measurement. c00a.qxd 8/7/12 5:34 PM Page 22 52. 0.2 Integer Exponents and Scientific Notation 23 NUMBER OF ZEROS EXPONENTIAL FOLLOWING FORM REAL NUMBER THE 1 PREFIX ABBREVIATION EXAMPLE 101 10 1 102 100 2 103 1000 3 kilo- k The relay-for-life team ran a total of 80 km (kilometers). (one thousand) 104 10,000 4 105 100,000 5 106 1,000,000 6 mega- M Modern high-powered dieselelectric railroad (one million) locomotives typically have a peak power output of 3 to 5 MW (megawatts). 107 10,000,000 7 108 100,000,000 8 109 1,000,000,000 9 giga- G A ash drive typically has 1 to 4 GB (gigabytes) of storage. (one billion) 1010 10,000,000,000 10 1011 100,000,000,000 11 1012 1,000,000,000,000 12 tera- T Laser systems offer higher frequencies on the order of (one trillion) THz (terahertz). Notice that 108 is a 1 followed by 8 zeros; alternatively, you can start with 1.0 and move the decimal point 8 places to the right (insert zeros). The same type of table can be made for negative-integer powers with base 10. To nd the real number associated with exponential form, start with 1.0 and move the decimal a certain number of places to the left (ll in missing decimal places with zeros). NUMBER OF PLACES DECIMAL EXPONENTIAL (1.0) MOVES FORM REAL NUMBER TO THE LEFT PREFIX ABBREVIATION EXAMPLE 101 0.1 1 102 0.01 2 103 0.001 3 milli- m Excedrin Extra Strength tablets each have 250 mg (one thousandth) (milligrams) of acetaminophen. 104 0.0001 4 105 0.00001 5 106 0.000001 6 micro- m A typical laser has a wavelength of 1.55 mm (micrometers*). (one millionth) 107 0.0000001 7 108 0.00000001 8 109 0.000000001 9 nano- n PSA levels less than 4 ng/ml (nanogram per milliliter of (one billionth) blood) represent low risk for prostate cancer. 1010 0.0000000001 10 1011 0.00000000001 11 1012 0.000000000001 12 pico- p A single yeast cell weighs 44 pg (picograms). (one trillionth) *In optics a micrometer is called a micron. c00a.qxd 8/7/12 5:34 PM Page 23 53. 24 C HAP TE R 0 Prerequisites and Review A positive real number can be written in scientic notation with the form c 10n , where 1 c 10 and n is an integer. SCIENTIFIC NOTATION EXAMPLE 7 Expressing a Positive Real Number in Scientic Notation Express the numbers in scientic notation. a. 3,856,000,000,000,000 b. 0.00000275 Solution: a. Rewrite the number with the implied decimal point. 3,856,000,000,000,000. Move the decimal point to the left 15 places. 3.856 1015 b. Move the decimal point to the right 6 places. 0.00000275 2.75 106 YO U R TU R N Express the numbers in scientic notation. a. 4,520,000,000 b. 0.00000043 Study Tip Scientic notation is a number between 1 and 10 that is multiplied by 10 to a power. Study Tip Real numbers greater than 1 correspond to positive exponents in scientic notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in scientic notation. Answer: a. 4.52 109 b. 4.3 107 Note that c is a real number between 1 and 10. Therefore, 22.5 103 is not in scientic notation, but we can convert it to scientic notation: 2.25 104 . For example, there are approximately 50 trillion cells in the human body. We write 50 trillion as 50 followed by 12 zeros 50,000,000,000,000. An efcient way of writing such a large number is using scientic notation. Notice that 50,000,000,000,000 is 5 followed by 13 zeros, or in scientic notation, 5 1013 . Very small numbers can also be written using scientic notation. For example, in laser communications a pulse width is 2 femtoseconds, or 0.000000000000002 second. Notice that if we start with 2.0 and move the decimal point 15 places to the left (adding zeros in between), the result is 0.000000000000002, or in scientic notation, 2 1015 . EXAMPLE 8 Converting from Scientic Notation to Decimals Write each number as a decimal. a. 2.869 105 b. 1.03 103 Solution: a. Move the decimal point 5 places to the right (add zeros in between). 286,900. or 286,900 b. Move the decimal point 3 places to the left (add zeros in between). 0.00103 YO U R TU R N Write each number as a decimal. a. 8.1 104 b. 3.7 108 Answer: a. 81,000 b. 0.000000037 Technology Tip Technology Tip c00a.qxd 8/7/12 5:34 PM Page 24 54. 0.2 Integer Exponents and Scientific Notation 25 SMHS U M MARY In Exercises 120, evaluate each expression. 1. 44 2. 53 3. (3)5 4. (4)2 5. 52 6. 72 7. 22 4 8. 32 5 9. 90 10. 8x0 11. 101 12. a1 13. 82 14. 34 15. 6 52 16. 2 42 17. 8 23 5 18. 5 24 32 19. 6 32 81 20. 6 42 44######### # ## E X E R C I S E S S E CTI O N 0.2 In this section we discussed properties of exponents. Integer Exponents The following table summarizes integer exponents. Let a be any real number and n be a natural number. Scientic Notation Scientic notation is a convenient way of using exponents to represent either very small or very large numbers. Real numbers greater than 1 correspond to positive exponents in scientic notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in scientic notation. Scientic notation offers the convenience of multiplying and dividing real numbers by applying properties of exponents. NAME DESCRIPTION MATH Natural-number exponent Multiply n factors of a. Negative-integer exponent A negative exponent implies a reciprocal. property Zero-exponent property Any nonzero real number raised to the zero power is equal to one. a0 1 a Z 0 a Z 0a-n = 1 an an = a # a # a p a NAME DESCRIPTION MATH Product property When multiplying exponentials with the same base, add exponents. am an amn Quotient property When dividing exponentials with the same base, subtract the exponents (numerator denominator). Power property When raising an exponential to a power, multiply exponents. Product to a power property A product raised to a power is equal to the product of each factor raised (ab)n an bn to the power. Quotient to a power property A quotient raised to a power is equal to the quotient of the factors raised to the power. a a b b n = an bn (am ) n = amn am an = am-n # REAL NUMBER (DECIMAL FORM) PROCESS SCIENTIFIC NOTATION 2,357,000,000 Move the implied decimal point to the left 9 places 2.357 109 0.00000465 Move the decimal point to the right 6 places 4.65 106e n factors Properties of Integer Exponents The following table summarizes properties of integer exponents. Let a and b be nonzero real numbers and m and n be integers. S E CTI O N 0.2 S K I LL S c00a.qxd 8/7/12 5:34 PM Page 25 55. 26 C HAP TE R 0 Prerequisites and Review In Exercises 2150, simplify and write the resulting expression with only positive exponents. 21. x2 x3 22. y3 y5 23. x2 x3 24. y3 y7 25. (x2 ) 3 26. (y3 )2 27. (4a)3 28. (4x2 ) 3 29. (2t)3 30. (3b)4 31. (5xy2 ) 2 (3x3 y) 32. (4x2 y)(2xy3 ) 2 33. 34. 35. 36. 37. 38. 39. (9a2 b3 ) 2 40. (9x3 y2 ) 4 41. 42. 43. 44. 45. 46. 47. 48. 2x2 (2x3 ) 5 49. 50. 51. Write 28 163 (64) as 52. Write 39 815 (9) as a power of 2 : 2? a power of 3 : 3? In Exercises 5360, express the given number in scientic notation. 53. 27,600,000 54. 144,000,000,000 55. 93,000,000 56. 1,234,500,000 57. 0.0000000567 58. 0.00000828 59. 0.000000123 60. 0.000000005 In Exercises 6166, write the number as a decimal. 61. 4.7 107 62. 3.9 105 63. 2.3 104 64. 7.8 103 65. 4.1 105 66. 9.2 108 ####s b-3 (-x3 y2 ) 4 y2 (-b2 x5 ) 3 t 5 s a2 (- xy4 ) 3 x4 (- a3 y2 ) 2 t 3 (x-4 y5 ) -2 [-2(x3 ) 2 y-4 ] 5 (-4x-2 ) 2 y3 z (2x3 ) -2 (y-1 z) 4 3(x2 y) 3 12(x-2 y) 4 (x3 y-2 ) 2 (x4 y3 ) -3 (x3 y-1 ) 2 (xy2 ) -2 x-3 y2 y-4 x5 a-2 b3 a4 b5 a c 3 b -2 a b 2 b -4 (-3x3 y) -4(x2 y3 ) 3 (2xy)2 (-2xy)3 y5 x2 y-2 x-5 x5 y3 x7 y ### 69. Astronomy. The distance from Earth to Mars on a particular day can be 200 million miles. Express this distance in scientic notation. 70. Astronomy. The distance from Mars to the Sun on a particular day can be 142 million miles. Express this distance in scientic notation. 71. Lasers. The wavelength of a typical laser used for communication systems is 1.55 microns (or 1.55 106 meters). Express the wavelength in decimal representation in terms of meters. 72. Lasers. A ruby-red laser has a wavelength of 694 nanometers (or 6.93 107 meters). Express the wavelength in decimal representation in terms of meters. In Exercises 67 and 68, refer to the following: It is estimated that there are currently cell phones being used worldwide. Assume that the average cell phone measures 5 inches in length and there are 5280 feet in a mile. 67. Cell Phones Spanning the Earth. a. If all of the cell phones currently in use were to be lined up next to each other tip to tip, how many feet would the line of cell phones span? Write the answer in scientic notation. b. The circumference of the Earth (measured at the equator) is approximately 25,000 miles. If the cell phones in part (a) were to be wrapped around the Earth at the equator, would they circle the Earth completely? If so, approximately how many times? 68. Cell Phones Reaching the Moon. a. If all of the cell phones currently in use were to be lined up next to each other tip to tip, how many miles would the line of cell phones span? Write the answer in scientic notation. b. The Moon traces an elliptical path around the Earth, with the average distance between them being approximately 239,000 miles. Would the line of cell phones in part (a) reach the Moon? 5.0 * 109 A P P L I C AT I O N S c00a.qxd 8/7/12 5:34 PM Page 26 56. 0.2 Integer Exponents and Scientific Notation 27 73. Simplify (2y3 )(3x2 y2 ). Group like factors together. (2)(3)x2 y3 y2 Use the product property. 6x2 y6 This is incorrect. What mistake was made? 75. Simplify (2xy3 ) 2 (5x2 y) 2 . Apply the product to a power property. (2)2 x2 (y3 ) 2 (5)2 (x2 ) 2 y2 Apply the power rule. 4x2 y9 25x4 y2 Group like factors. (4)(25)x2 x4 y9 y2 Apply the product property. 100x6 y11 This is incorrect. What mistake was made? 74. Simplify (2xy2 ) 3 . Eliminate the parentheses. (2xy2 ) 3 2x3 y6 This is incorrect. What mistake was made? 76. Simplify . Group like factors. Use the quotient property. This is incorrect. What mistake was made? = - 1 2 x8 y3 = q -4 8 rq x16 x2 rq y9 y3 r -4x16 y9 8x2 y3 In Exercises 7376, explain the mistake that is made. C AT C H T H E M I S TA K E In Exercises 7780, determine whether each of the following statements is true or false. 77. 2n (2)n , if n is an integer. 78. Any nonzero real number raised to the zero power is one. 79. for x any real number. 80. x1 x2 x3 81. Simplify 82. Simplify A(a-m ) -n B -k . A(am ) n B k . xn+1 xn = x In Exercises 8386, evaluate the expression for the given value. 83. a2 2ab for a 2, b 3 84. 2a3 7a2 for a 4 85. 16t2 100t for t 3 86. for a 2 a3 - 27 a - 4 C O N C E P T U A L 88. The population of the United States is approximately 3.0 108 people, and there are approximately 3.79 106 square miles of land in the United States. If one square mile is approximately 640 acres, how many acres per person are there in the United States? Round to the nearest tenth of an acre. 90. Evaluate: . Express your answer in both scientic and decimal notation. (2 * 10-17 )(5 * 1013 ) (1 * 10-6 ) 87. The Earths population is approximately 6.6 109 people, and there are approximately 1.5 108 square kilometers of land on the surface of the Earth. If one square kilometer is approximately 247 acres, how many acres per person are there on Earth? Round to the nearest tenth of an acre. 89. Evaluate: . Express your answer in both scientic and decimal notation. (4 * 10-23 )(3 * 1012 ) (6 * 10-10 ) C H A L L E N G E Scientic calculators have an EXP button that is used for scientic notation. For example, 2.5 103 can be input into the calculator by pressing 2.5 EXP 3. 91. Repeat Exercise 87 and conrm your answer with a calculator. 92. Repeat Exercise 88 and conrm your answer with a calculator. In Exercises 93 and 94, use a graphing utility or scientic calculator to evaluate the expression. Express your answer in scientic notation. 93. 94. (1.6849 * 1032 ) (8.12 * 1016 )(3.32 * 10-9 ) (7.35 * 10-26 )(2.19 * 1019 ) (3.15 * 10-21 ) T E C H N O L O G Y c00a.qxd 8/7/12 5:34 PM Page 27 57. A polynomial in x is an algebraic expression of the form where a0, a1, a2, . . . , an are real numbers, with , and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefcient, and a0 is the constant term. an Z 0 anxn + an-1xn-1 + an-2xn-2 + + a2x2 + a1x + a0 PolynomialD E F I N I T I O N C O N C E P TUAL O BJ E CTIVE S Recognize like terms. Learn formulas for special products. P O LYN O M IALS: BAS I C O P E R ATI O N S S K I LLS O BJ E CTIVE S Add and subtract polynomials. Multiply polynomials. Recognize special products. S E CTI O N 0.3 Adding and Subtracting Polynomials Polynomials in Standard Form The expressions 3x2 7x 1 4y3 y 5z are all examples of polynomials in one variable. A monomial in one variable, axk , is the product of a constant and a variable raised to a nonnegative-integer power. The constant a is called the coefcient of the monomial, and k is called the degree of the monomial. A polynomial is the sum of monomials. The monomials that are part of a polynomial are called terms. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. Polynomials are typically written in standard form in order of decreasing degrees, and the degree of the polynomial is determined by the highest degree (exponent) of any single term. POLYNOMIAL STANDARD FORM SPECIAL NAME DEGREE DESCRIPTION 4x3 5x7 2x 6 5x7 4x3 2x 6 Polynomial 7 A seventh-degree polynomial in x 5 2y3 4y 2y3 4y 5 Trinomial 3 A third-degree polynomial in y 7z2 2 7z2 2 Binomial 2 A second-degree polynomial in z 17x5 17x5 Monomial 5 A fth-degree monomial in x 28 c00b.qxd 8/3/12 11:16 AM Page 28 58. 0.3 Polynomials: Basic Operations 29 Adding and Subtracting Polynomials Polynomials are added and subtracted by combining like terms. Like terms are terms having the same variables and exponents. Like terms can be combined by adding their coefcients. WORDS MATH Identify like terms. Add coefcients of like terms. 7x2 2x 5 Note: The 2x and 5 could not be combined because they are not like terms. 3x2 + 2x + 4x2 + 5 EXAMPLE 1 Writing Polynomials in Standard Form Write the polynomials in standard form and state their degree, leading coefcient, and constant term. a. 4x 9x5 2 b. 3 x2 c. 3x2 8 14x3 20x8 x d. 7x3 25x Solution: Standard Form Degree Leading Coefcient Constant Term a. 9x5 4x 2 5 9 2 b. x2 3 2 1 3 c. 20x8 14x3 3x2 x 8 8 20 8 d. 7x3 25x 3 7 0 YO U R TU R N Write the polynomial in standard form and state its degree, leading coefcient, and constant term. 17x2 - 4x3 + 5 - x Answer: 4x3 17x2 x 5 Degree: 3 Leading coefcient: 4 Constant term: 5 Answer: 2x5 6x3 2x2 5x 11 EXAMPLE 2 Adding Polynomials Find the sum and simplify . Solution: Eliminate parentheses. 5x2 2x 3 3x3 4x2 7 Identify like terms. Combine like terms. x2 2x 10 3x3 Write in standard form. 3x3 x2 2x 10 YO U R TU R N Find the sum and simplify: (3x2 + 5x - 2x5 ) + (6x3 - x2 + 11) 5x2 - 2x + 3 + 3x3 - 4x2 + 7 (5x2 - 2x + 3) + (3x3 - 4x2 + 7) c00b.qxd 8/3/12 11:16 AM Page 29 59. 30 C HAP TE R 0 Prerequisites and Review Multiplying Polynomials The product of two monomials is found by using the properties of exponents (Section 0.2). For example, To multiply a monomial and a polynomial we use the distributive property (Section 0.1). (-5x3 )(9x2 ) = (-5)(9)x3+2 = -45x5 Answer: 12x4 6x3 3x2 EXAMPLE 4 Multiplying a Monomial and a Polynomial Find the product and simplify 5x2 . Solution: Use the distributive property. 5x2 5x2 5x2 5x2 (7x) 5x2 (4) Multiply each set of monomials. 15x7 5x5 35x3 20x2 YO U R TU R N Find the product and simplify 3x2 (4x2 2x 1). (x3 )(3x5 ) (3x5 - x3 + 7x - 4) (3x5 - x3 + 7x - 4) How do we multiply two polynomials if neither one is a monomial? For example, how do we nd the product of a binomial and a trinomial such as (2x 5)(x2 2x 3)? Notice that the binomial is a combination of two monomials. Therefore, we treat each monomial, 2x and 5, separatelyandthencombineourresults.Inotherwords,usethedistributivepropertyrepeatedly. EXAMPLE 3 Subtracting Polynomials Find the difference and simplify (3x3 - 2x + 1) - (x2 + 5x - 9). Study Tip When subtracting polynomials, it is important to distribute the negative through all of the terms in the second polynomial. I N C O R R E CT ERROR: 3x3 2x 1 x2 5x 9 Dont forget to distribute the negative through the entire second polynomial. C O R R E CT Eliminate the parentheses. 3x3 2x 1 x2 5x 9 Identify like terms. Combine like terms. 3x3 x2 7x 10 3x3 - 2x + 1 - x2 - 5x + 9 Distributing the negative to only the rst term in the second polynomial. C O M M O N M I S TA K E YO U R TU R N Find the difference and simplify: (-7x2 - x + 5) - (2 - x3 + 3x) Answer: x3 7x2 4x 3 WORDS MATH Apply the distributive property. Apply the distributive property. 5(2x) 5(3)(2x)(x2 ) + (2x)(-2x) + (2x)(3) - 5(x2 ) (2x - 5)(x2 - 2x + 3) = 2x(x2 - 2x + 3) - 5(x2 - 2x + 3) Multiply the monomials. 2x3 4x2 6x 5x2 10x 15 Combine like terms. 2x3 9x2 16x 15 c00b.qxd 8/3/12 11:16 AM Page 30 60. 0.3 Polynomials: Basic Operations 31 First Inner Product of First Terms Product of Last Terms Outer Last Product of Outer Terms Product of Inner Terms Special Products The method outlined for multiplying polynomials works for all products of polynomials. For the special case when both polynomials are binomials, the FOIL method can also be used. WORDS MATH Apply the distributive property. (5x 1)(2x 3) 5x(2x 3) 1(2x 3) Apply the distributive property. 5x(2x) 5x(3) 1(2x) 1(3) Multiply each set of monomials. 10x2 15x 2x 3 Combine like terms. 10x2 13x 3 The FOIL method nds the products of the First terms, Outer terms, Inner terms, and (5x 1)(2x 3) 10x2 15x 2x 3 Last terms. EXAMPLE 6 Multiplying Binomials Using the FOIL Method Multiply (3x 1)(2x 5) using the FOIL method. Solution: Multiply the rst terms. (3x)(2x) 6x2 Multiply the outer terms. (3x)(5) 15x Multiply the inner terms. (1)(2x) 2x Multiply the last terms. (1)(5) 5 Add the rst, outer, inner, and last terms, and identify the like terms. Combine like terms. 6x2 13x 5 YO U R TU R N Multiply (2x 3)(5x 2). (3x + 1)(2x - 5) = 6x2 - 15x + 2x - 5 Answer: 10x2 19x 6 Study Tip When the binomials are of the form (ax b)(cx d), the outer and inner terms will be like terms and can be combined. EXAMPLE 5 Multiplying Two Polynomials Multiply and simplify . Solution: Multiply each term of the rst trinomial by the entire second trinomial. Identify like terms. Combine like terms. 2x4 13x3 30x2 26x 7 YO U R TU R N Multiply and simplify (-x3 + 2x - 4)(3x2 - x + 5). = 2x4 - 10x3 + 14x2 - 3x3 + 15x2 - 21x + x2 - 5x + 7 = 2x2 (x2 - 5x + 7) - 3x(x2 - 5x + 7) + 1(x2 - 5x + 7) (x2 - 5x + 7)(2x2 - 3x + 1) Answer: 3x5 x4 x3 14x2 14x 20 c00b.qxd 8/3/12 11:16 AM Page 31 61. EXAMPLE 7 Multiplying Binomials Resulting in Special Products Find the following: a. (x 5)(x 5) b. (x 5)2 c. (x 5)2 Solution: a. (x - 5)(x + 5) = x2 + 5x - 5x - 52 = x2 - 52 = x2 - 25 32 C HAP TE R 0 Prerequisites and Review Square of a binomial sum: (a b)2 (a b)(a b) a2 2ab b2 Square of a binomial difference: (a b)2 (a b)(a b) a2 2ab b2 PERFECT SQUARES (a + b)(a - b) = a2 - b2 DIFFERENCE OF TWO SQUARES EXAMPLE 8 Finding the Square of a Binomial Sum Find (x 3)2 . I N C O R R E CT ERROR: Dont forget the middle term, which is twice the product of the two terms in the binomial. (x + 3)2 Z x2 + 9 C O R R E CT (x 3)2 (x 3)(x 3) x2 3x 3x 9 x2 6x 9 Forgetting the middle term, which is twice the product of the two terms in the binomial. C O M M O N M I S TA K E Let a and b be any real number, variable, or algebraic expression in the following special products. Some products of binomials occur frequently in algebra and are given special names. Example 7 illustrates the difference of two squares and perfect squares. r First r Inner s Difference of two squares r Outer r Last r First r Inner r Outer r First r Inner r Outer r Last r Last b. c. (x - 5)2 = (x - 5)(x - 5) = x2 - 5x - 5x + 52 = x2 - 2(5x) + 52 = x2 - 10x + 25 (x + 5)2 = (x + 5)(x + 5) = x2 + 5x + 5x + 52 = x2 + 2(5x) + 52 = x2 + 10x + 25 Study Tip (a b)(a b) a2 ab ab b2 a2 b2 c00b.qxd 8/3/12 11:17 AM Page 32 62. 0.3 Polynomials: Basic Operations 33 Answer: a. 9x2 6x 1 b. 9y2 6y 1 c. 9x2 4 EXAMPLE 9 Using Special Product Formulas Find the following: a. (2x 1)2 b. (3 2y)2 c. (4x 3)(4x 3) Solution (a): Write the square of a binomial difference formula. (a b)2 a2 2ab b2 Let a 2x and b 1. (2x 1)2 (2x)2 2(2x)(1) 12 Simplify. 4x2 4x 1 Solution (b): Write the square of a binomial sum formula. (a b)2 a2 2ab b2 Let a 3 and b 2y. (3 2y)2 (3)2 2(3)(2y) (2y)2 Simplify. 9 12y 4y2 Write in standard form. 4y2 12y 9 Solution (c): Write the difference of two squares formula. (a b)(a b) a2 b2 Let a 4x and b 3. (4x 3)(4x 3) (4x)2 32 Simplify. 16x2 9 YO U R TU R N Find the following: a. (3x 1)2 b. (1 3y)2 c. (3x 2)(3x 2) EXAMPLE 10 Cubing a Binomial Find the following: a. (x 2)3 b. (x 2)3 Solution (a): Write the cube as a product of three binomials. Apply the perfect square formula. Apply the distributive property. Apply the distributive property. x3 4x2 4x 2x2 8x 8 Combine like terms. x3 6x2 12x 8 Solution (b): Write the cube as a product of three binomials. Apply the perfect square formula. (x 2)(x2 4x 4) Apply the distributive property. x(x2 4x 4) 2(x2 4x 4) Apply the distributive property. x3 4x2 4x 2x2 8x 8 Combine like terms. x3 6x2 12x 8 (x - 2)2 (x - 2)3 = (x - 2)(x - 2)(x - 2) x(x2 + 4x + 4) + 2(x2 + 4x + 4) (x + 2)(x2 + 4x + 4) (x + 2)2 (x + 2)3 = (x + 2)(x + 2)(x + 2) ff C A U T I O N (x - 2)3 Z x3 - 8 (x + 2)3 Z x3 + 8 c00b.qxd 8/3/12 11:17 AM Page 33 63. 34 C HAP TE R 0 Prerequisites and Review EXAMPLE 11 Applying the Special Product Formulas Find the following: a. (2x 1)3 b. (2x 5)3 Solution (a): Write the cube of a binomial sum formula. (a b)3 a3 3a2 b 3ab2 b3 Let a 2x and b 1. (2x 1)3 (2x)3 3(2x)2 (1) 3(2x)(1)2 13 Simplify. 8x3 12x2 6x 1 Solution (b): Write the cube of a binomial difference formula. (a b)3 a3 3a2 b 3ab2 b3 Let a 2x and b 5. (2x 5)3 (2x)3 3(2x)2 (5) 3(2x)(5)2 53 Simplify. 8x3 60x2 150x 125 YO U R TU R N Find (3x 4)3 . Answer: 27x3 108x2 144x 64 EXAMPLE 12 Applying the Special Product Formulas for Binomials in Two Variables Find (2x 3y)2 . Solution: Write the square of a binomial difference formula. (a b)2 (a b)(a b) a2 2ab b2 Let a 2x and b 3y. (2x 3y)2 (2x)2 2(2x)(3y) (3y)2 Simplify. 4x2 12xy 9y2 YO U R TU R N Find (3x 2y)2 . Answer: 9x2 12xy 4y2 Cube of a binomial sum: (a b)3 a3 3a2 b 3ab2 b3 Cube of a binomial difference: (a b)3 a3 3a2 b 3ab2 b3 PERFECT CUBES c00b.qxd 8/3/12 11:17 AM Page 34 64. 0.3 Polynomials: Basic Operations 35 In Exercises 18, write the polynomial in standard form and state the degree of the polynomial. 1. 5x2 2x3 16 7x4 2. 7x3 9x2 5x 4 3. 4x 3 6x3 4. 5x5 7x3 8x4 x2 10 5. 15 6. 14 7. y 2 8. x 5 In Exercises 924, add or subtract the polynomials, gather like terms, and write the simplied expression in standard form. 9. (2x2 x 7) (3x2 6x 2) 10. (3x2 5x 2) (2x2 4x 9) 11. (7x2 5x 8) (4x 9x2 10) 12. (8x3 7x2 10) (7x3 8x2 9x) 13. (2x4 7x2 8) (3x2 2x4 9) 14. (4x2 9x 2) (5 3x 5x2 ) 15. (7z2 2) (5z2 2z 1) 16. (25y3 7y2 9y) (14y2 7y 2) 17. (3y3 7y2 8y 4) (14y3 8y 9y2 ) 18. (2x2 3xy) (x2 8xy 7y2 ) 19. (6x 2y) 2(5x 7y) 20. 3a [2a2 (5a 4a2 3)] 21. (2x2 2) (x 1) (x2 5) 22. (3x3 1) (3x2 1) (5x 3) 23. (4t t2 t3 ) (3t2 2t 2t3 ) (3t3 1) 24. (z3 2z2 ) (z2 7z 1) (4z3 3z2 3z 2) In Exercises 2564, multiply the polynomials and write the expressions in standard form. 25. 5xy2 (7xy) 26. 6z(4z3 ) 27. 2x3 (1 x x2 ) 28. 4z2 (2 z z2 ) 29. 2x2 (5 x 5x2 ) 30. 31. (x2 x 2)2x3 32. (x2 x 2)3x3 33. 2ab2 (a2 2ab 3b2 ) 34. bc3 d2 (b2 c cd3 b2 d4 ) 35. (2x 1)(3x 4) 36. (3z 1)(4z 7) 37. (x 2)(x 2) 38. (y 5)(y 5) 39. (2x 3)(2x 3) 40. (5y 1)(5y 1) 41. (2x 1)(1 2x) 42. (4b 5y)(4b 5y) 43. (2x2 3)(2x2 3) 44. (4xy 9)(4xy 9) 45. (7y 2y2 )(y y2 1) 46. (4 t2 )(6t 1 t2 ) 47. (x 1)(x2 2x 3) 48. (x 3)(x2 3x 9) -1 2 z(2z + 4z2 - 10) E X E R C I S E S S E CTI O N 0.3 Perfect Squares Square of a binomial sum. (a b)2 (a b)(a b) a2 2ab b2 Square of a binomial difference. (a b)2 (a b)(a b) a2 2ab b2 Perfect Cubes Cube of a binomial sum. (a b)3 a3 3a2 b 3ab2 b3 Cube of a binomial difference. (a b)3 a3 3a2 b 3ab2 b3 In this section, polynomials were dened. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. Polynomials are added and subtracted by combining like terms. Polynomials are multiplied by distributing the monomials in the rst polynomial throughout the second polynomial. In the special case of the product of two binomials, the FOIL method can also be used. The following are special products of binomials. Difference of Two Squares (a b)(a b) a2 b2 S U M MARY S E CTI O N 0.3 S K I LL S c00b.qxd 8/3/12 11:17 AM Page 35 65. 36 C HAP TE R 0 Prerequisites and Review 65. Prot. Donna decides to sell fabric cord covers on eBay for $20 a piece. The material for each cord cover costs $9, and it costs her $100 a month to advertise on eBay. Let x be the number of cord covers sold. Write a polynomial representing her monthly prot. 66. Prot. Calculators are sold for $25 each. Advertising costs are $75 per month. Let x be the number of calculators sold. Write a polynomial representing the monthly prot earned by selling x calculators. 67. Prot. If the revenue associated with selling x units of a product is R x2 100x, and the cost associated with producing x units of the product is C 100x 7500, nd the polynomial that represents the prot of making and selling x units. 68. Prot. A business sells a certain quantity x of items. The revenue generated by selling x items is given by the equation . The costs are given by C 8000 150x. Find a polynomial representing the net prot of this business when x items are sold. 69. Volume of a Box. A rectangular sheet of cardboard is to be used in the construction of a box by cutting out squares of side length x from each corner and turning up the sides. Suppose the dimensions of the original rectangle are 15 inches by 8 inches. Determine a polynomial in x that would give the volume of the box. 70. Volume of a Box. Suppose a box is to be constructed from a square piece of material of side length x by cutting out a 2-inch square from each corner and turning up the sides. Express the volume of the box as a polynomial in the variable x. 71. Geometry. Suppose a running track is constructed of a rectangular portion that measures 2x feet wide by 2x 5 feet long. Each end of the rectangular portion consists of a semicircle whose diameter is 2x. Write a polynomial that determines the a. perimeter of the track in terms of the variable x. b. area of the track in terms of x. 2x + 5 2x 2x R = -1 2x2 + 50x 72. Geometry. A right circular cylinder whose radius is r and whose height is 2r is surmounted by a hemisphere of radius r. a. Find a polynomial in the variable r that represents the volume of the silo shape. b. Find a polynomial in r that represents the total surface area of the silo. 73. Engineering. The force of an electrical eld is given by the equation . Suppose q1 x, q2 3x, and r 10x. Find a polynomial representing the force of the electrical eld in terms of the variable x. 74. Engineering. If a football (or other projectile) is thrown upward, its height above the ground is given by the equation s 16t2 v0t s0, where v0 and s0 are the initial velocity and initial height of the football, respectively, and t is the time in seconds. Suppose the football is thrown from the top of a building that is 192 feet tall, with an initial speed of 96 feet per second. a. Write the polynomial that gives the height of the football in terms of the variable t (time). b. What is the height of the football after 2 seconds have elapsed? Will the football hit the ground after 2 seconds? F = k q1q2 r2 r r 2r In Exercises 6568, prot is equal to revenue minus cost: P R C. 49. (t 2)2 50. (t 3)2 51. (z 2)2 52. (z 3)2 53. [(x y) 3]2 54. (2x2 3y) 2 55. (5x 2)2 56. (x 1)(x2 x 1) 57. y(3y 4)(2y 1) 58. p2 (p 1)(p 2) 59. (x2 1)(x2 1) 60. (t 5)2 (t 5)2 61. (b 3a)(a 2b)(b 3a) 62. (x 2y)(x2 2xy 4y2 ) 63. (x y z)(2x 3y 5z) 64. (5b2 2b 1)(3b b2 2) A P P L I C AT I O N S c00b.qxd 8/3/12 11:17 AM Page 36 66. 0.4 Factoring Polynomials 37 In Exercises 7780, determine whether each of the following statements is true or false. 77. All binomials are polynomials. 78. The product of two monomials is a binomial. 79. (x y)3 x3 y3 80. (x y)2 x2 y2 In Exercises 81 and 82, let m and n be real numbers and . 81. What degree is the product of a polynomial of degree n and a polynomial of degree m? 82. What degree is the sum of a polynomial of degree n and a polynomial of degree m? m>n In Exercises 8386, perform the indicated operations and simplify. 83. (7x 4y2 ) 2 (7x 4y2 ) 2 84. (3x 5y2 ) 2 (3x 5y2 ) 2 85. (x a)(x2 ax a2 ) 86. (x a)(x2 ax a2 ) 87. Use a graphing utility to plot the graphs of the three expressions , and Which two graphs agree with each other?2x2 - 5x - 12. (2x + 3)(x - 4), 2x2 + 5x - 12 88. Use a graphing utility to plot the graphs of the three expressions Which two graphs agree with each other? (x + 5)2 , x2 + 25, and x2 + 10x + 25. C O N C E P TUAL O BJ E CTIVE S Understand that factoring has its basis in the distributive property. Identify prime (irreducible) polynomials. Develop a general strategy for factoring polynomials. S K I LLS O BJ E CTIVE S Factor out the greatest common factor. Factor the difference of two squares. Factor perfect squares. Factor the sum or difference of two cubes. Factor a trinomial as a product of binomials. Factor by grouping. S E CTI O N 0.4 In Section 0.3 we discussed multiplying polynomials. In this section we examine the reverse of that process, which is called factoring. Consider the following product: (x + 3)(x + 1) = x2 + 4x + 3 FACTO R I N G P O LYN O M IALS 75. Subtract and simplify Solution: Eliminate the parentheses. 2x2 5 3x x2 1 Collect like terms. x2 3x 4 This is incorrect. What mistake was made? (2x2 - 5) - (3x - x2 + 1). 76. Simplify (2 x)2 . Solution: Write the square of the binomial as the sum of the squares. (2 x)2 22 x2 Simplify. x2 4 This is incorrect. What mistake was made? In Exercises 75 and 76, explain the mistake that is made. C AT C H T H E M I S TA K E C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y c00b.qxd 8/3/12 11:17 AM Page 37 67. To factor the resulting polynomial, you reverse the process to undo the multiplication: The polynomials (x 3) and (x 1) are called factors of the polynomial x2 4x 3. The process of writing a polynomial as a product is called factoring. In Chapter 1 we will solve quadratic equations by factoring. In this section we will restrict our discussion to factoring polynomials with integer coefcients, which is called factoring over the integers. If a polynomial cannot be factored using integer coefcients, then it is prime or irreducible over the integers. When a polynomial is written as a product of prime polynomials, then the polynomial is said to be factored completely. Greatest Common Factor The simplest type of factoring of polynomials occurs when there is a factor common to every term of the polynomial. This common factor is a monomial that can be factored out by applying the distributive property in reverse: For example, 4x2 6x can be written as 2x(x) 2x(3). Notice that 2x is a common factor to both terms, so the distributive property tells us we can factor this polynomial to yield 2x(x 3). Although 2 is a common factor and x is a common factor, the monomial 2x is called the greatest common factor. ab + ac = a(b + c) x2 + 4x + 3 = (x + 3)(x + 1) The monomial axk is called the greatest common factor (GCF) of a polynomial in x with integer coefcients if both of the following are true: a is the greatest integer factor common to all of the polynomial coefcients. k is the smallest exponent on x found in all of the terms of the polynomial. GREATEST COMMON FACTOR WRITE EACH TERM AS A PRODUCT OF GCF AND POLYNOMIAL GCF REMAINING FACTOR FACTORED FORM 7x 21 7 7(x) 7(3) 7(x 3) 3x2 12x 3x 3x(x) 3x(4) 3x(x 4) 4x3 2x 6 2 2(2x3 ) 2x 2(3) 2(2x3 x 3) 6x4 9x3 12x2 3x2 3x2 (2x2 ) 3x2 (3x) 3x2 (4) 3x2 (2x2 3x 4) 5x4 25x3 20x2 5x2 5x2 (x2 ) 5x2 (5x) 5x2 (4) 5x2 (x2 5x 4) EXAMPLE 1 Factoring Polynomials by Extracting the Greatest Common Factor Factor: a. 6x5 18x4 b. 6x5 10x4 8x3 12x2 Solution (a): Identify the greatest common factor. 6x4 Write each term as a product with the GCF as a factor. 6x5 18x4 6x4 (x) 6x4 (3) Factor out the GCF. 6x4 (x 3) 38 C HAP TE R 0 Prerequisites and Review c00b.qxd 8/3/12 11:17 AM Page 38 68. 0.4 Factoring Polynomials 39 Answer: a. 4x(3x2 1) Difference of two squares a2 b2 (a b)(a b) Perfect squares a2 2ab b2 (a b)2 a2 2ab b2 (a b)2 Sum of two cubes a3 b3 (a b)(a2 ab b2 ) Difference of two cubes a3 b3 (a b)(a2 ab b2 ) Solution (b): Identify the greatest common factor. 2x2 Write each term as a product with the GCF as a factor. 6x5 10x4 8x3 12x2 2x2 (3x3 ) 2x2 (5x2 ) 2x2 (4x) 2x2 (6) Factor out the GCF. 2x2 (3x3 5x2 4x 6) YO U R TU R N Factor: a. 12x3 4x b. 3x5 9x4 12x3 6x2 Factoring Formulas: Special Polynomial Forms The rst step in factoring polynomials is to look for a common factor. If there is no common factor, then we look for special polynomial forms that we learned were special products in Section 0.3 and reverse the process. EXAMPLE 2 Factoring the Difference of Two Squares Factor: a. x2 9 b. 4x2 25 c. x4 16 Solution (a): Rewrite as the difference of two squares. x2 9 x2 32 Let a x and b 3 in a2 b2 (a b)(a b). (x 3)(x 3) Solution (b): Rewrite as the difference of two squares. 4x2 25 (2x)2 52 Let a 2x and b 5 in a2 b2 (a b)(a b). (2x 5)(2x 5) Solution (c): Rewrite as the difference of two squares. x4 16 42 Let a x2 and b 4 in a2 b2 (a b)(a b). Note that x2 4 is also a difference of two squares (part a of the following Your Turn). YO U R TU R N Factor: a. x2 4 b. 9x2 16 c. x4 81 (x + 2)(x - 2)(x2 + 4) (x2 + 4)(x2 - 4) (x2 ) 2 Answer: a. (x 2)(x 2) b. (3x 4)(3x 4) c. (x2 9)(x 3)(x 3) b. 3x2 (x3 3x2 4x 2) c00b.qxd 8/3/12 11:17 AM Page 39 69. 40 C HAP TE R 0 Prerequisites and Review A trinomial is a perfect square if it has the form a2 ; 2ab b2 . Notice that: The rst term and third term are perfect squares. The middle term is twice the product of the bases of these two perfect squares. The sign of the middle term determines the sign of the factored form: a2 ; 2ab + b2 = (a ; b)2 EXAMPLE 3 Factoring Trinomials That Are Perfect Squares Factor: a. x2 6x 9 b. x2 10x 25 c. 9x2 12x 4 Solution (a): Rewrite the trinomial so that the rst and third terms are perfect squares. x2 6x 9 x2 6x 32 Notice that if we let a x and b 3 in a2 2ab b2 (a b)2 , then the middle term 6x is 2ab. x2 6x 9 x2 2(3x) 32 (x 3)2 Solution (b): Rewrite the trinomial so that the rst and third terms are perfect squares. x2 10x 25 x2 10x 52 Notice that if we let a x and b 5 in a2 2ab b2 (a b)2 , then the middle term 10x is 2ab. x2 10x 25 x2 2(5x) 52 (x 5)2 Solution (c): Rewrite the trinomial so that the rst and third terms are perfect squares. 9x2 12x 4 (3x)2 12x 22 Notice that if we let a 3x and b 2 in a2 2ab b2 (a b)2 , then the middle term 12x is 2ab. 9x2 12x 4 (3x)2 2(3x)(2) 22 (3x 2)2 YO U R TU R N Factor: a. x2 8x 16 b. x2 4x 4 c. 25x2 20x 4 Answer: a. (x 4)2 b. (x 2)2 c. (5x 2)2 EXAMPLE 4 Factoring the Sum of Two Cubes Factor x3 27. Solution: Rewrite as the sum of two cubes. x3 27 x3 33 Write the sum of two cubes formula. a3 b3 (a b)(a2 ab b2 ) Let a x and b 3. x3 27 x3 33 (x 3)(x2 3x 9) c00b.qxd 8/3/12 11:17 AM Page 40 70. 0.4 Factoring Polynomials 41 Answer: a. (x 2)(x2 2x 4) b. (x 4)(x2 4x 16) Factoring a Trinomial as a Product of Two Binomials The rst step in factoring is to look for a common factor. If there is no common factor, look to see whether the polynomial is a special form for which we know the factoring formula. If it is not of such a special form and if it is a trinomial, then we proceed with a general factoring strategy. We know that (x 3)(x 2) x2 5x 6, so we say the factors of x2 5x 6 are (x 3) and (x 2). In factored form we have x2 5x 6 (x 3)(x 2). Recall the FOIL method from Section 0.3. The product of the last terms (3 and 2) is 6, and the sum of the products of the inner terms (3x) and the outer terms (2x) is 5x. Lets pretend for a minute that we didnt know this factored form but had to work with the general form: The goal is to nd a and b. We start by multiplying the two binomials on the right. Compare the expression we started with on the left with the expression on the far right x2 5x 6 x2 (a b)x ab. We see that ab 6 and (a b) 5. Start with the possible combinations of a and b whose product is 6, and then look among those for the combination whose sum is 5. ab 6 a, b: 1, 6 1, 6 2, 3 2, 3 a b 7 7 5 5 All of the possible a, b combinations in the rst row have a product equal to 6, but only one of those has a sum equal to 5. Therefore the factored form is x2 + 5x + 6 = (x + a)(x + b) = (x + 2)(x + 3) x2 + 5x + 6 = (x + a)(x + b) = x2 + ax + bx + ab = x2 + (a + b)x + ab x2 + 5x + 6 = (x + a)(x + b) EXAMPLE 5 Factoring the Difference of Two Cubes Factor x3 125. Solution: Rewrite as the difference of two cubes. x3 125 x3 53 Write the difference of two cubes formula. a3 b3 (a b)(a2 ab b2 ) Let a x and b 5. x3 125 x3 53 (x 5)(x2 5x 25) YO U R TU R N Factor: a. x3 8 b. x3 64 c00b.qxd 8/3/12 11:17 AM Page 41 71. 42 C HAP TE R 0 Prerequisites and Review Answer: (x 4)(x 5) Answer: (x 6)(x 3) In Example 6, all terms in the trinomial are positive. When the constant term is negative, then (regardless of whether the middle term is positive or negative) the factors will be opposite in sign, as illustrated in Example 7. EXAMPLE 7 Factoring a Trinomial Factor x2 3x 28. Solution: Write the trinomial as a product of two binomials in general form. x2 3x 28 (x )(x ) Write all of the integers whose product is 28. Integers whose product is 28 1, 28 1, 28 2, 14 2, 14 4, 7 4, 7 Determine the sum of the integers. Integers whose product is 28 1, 28 1, 28 2, 14 2, 14 4, 7 4, 7 Sum 27 27 12 12 3 3 Select 4, 7 because the product is 28 (last term of the trinomial) and the sum is 3 (middle term coefcient of the trinomial). x2 3x 28 (x 4)(x 7) Check: (x 4)(x 7) x2 7x 4x 28 x2 3x 28 YO U R TU R N Factor x2 3x 18. nn EXAMPLE 6 Factoring a Trinomial Factor x2 10x 9. Solution: Write the trinomial as a product of two binomials in general form. x2 10x 9 (x n)(x n) Write all of the integers whose product is 9. Integers whose product is 9 1, 9 1, 9 3, 3 3, 3 Determine the sum of the integers. Integers whose product is 9 1, 9 1, 9 3, 3 3, 3 Sum 10 10 6 6 Select 1, 9 because the product is 9 (last term of the trinomial) and the sum is 10 (middle term coefcient of the trinomial). x2 10x 9 (x 9)(x 1) Check: (x 9)(x 1) x2 1x 9x 9 x2 10x 9 YO U R TU R N Factor x2 9x 20. c00b.qxd 8/3/12 11:17 AM Page 42 72. 0.4 Factoring Polynomials 43 Answer: (2t 3)(t 1) Factors of a ax2 bx c (nx n)(nx n) Factors of c EXAMPLE 8 Factoring a Trinomial Whose Leading Coefcient Is Not 1 Factor 5x2 9x 2. Solution: STEP 1 Start with the rst term. Note that 5x x 5x2 . (5x ; n)(x ; n) STEP 2 The product of the last terms should yield 2. 1, 2 or 1, 2 STEP 3 Consider all possible factors based on Steps 1 and 2. (5x 1)(x 2) (5x 1)(x 2) (5x 2)(x 1) (5x 2)(x 1) Since the outer and inner products must sum to 9x, the factored form must be: 5x2 9x 2 (5x 1)(x 2) Check: (5x 1)(x 2) 5x2 10x 1x 2 5x2 9x 2 YO U R TU R N Factor 2t2 t 3. When the leading coefcient of the trinomial is not equal to 1, then we consider all possible factors using the following procedure, which is based on the FOIL method in reverse. EXAMPLE 9 Factoring a Trinomial Whose Leading Coefcient Is Not 1 Factor 15x2 x 6. Solution: STEP 1 Start with the rst term. (5x ; n)(3x ; n) or (15x ; n)(x ; n) STEP 2 The product of the last terms should yield 6. 1, 6 or 1, 6 or 2, 3 or 2, 3 Step 1: Find two First terms whose product is the rst term of the trinomial. Step 2: Find two Last terms whose product is the last term of the trinomial. Step 3: Consider all possible combinations found in Steps 1 and 2 until the sum of the Outer and Inner products are equal to the middle term of the trinomial. FACTORING A TRINOMIAL WHOSE LEADING COEFFICIENT IS NOT 1 c00b.qxd 8/3/12 11:17 AM Page 43 73. 44 C HAP TE R 0 Prerequisites and Review STEP 3 Consider all possible factors (5x 1)(3x 6) (15x 1)(x 6) based on Steps 1 and 2. (5x 6)(3x 1) (15x 6)(x 1) (5x 1)(3x 6) (15x 1)(x 6) (5x 6)(3x 1) (15x 6)(x 1) (5x 2)(3x 3) (15x 2)(x 3) (5x 3)(3x 2) (15x 3)(x 2) (5x 2)(3x 3) (15x 2)(x 3) (5x 3)(3x 2) (15x 3)(x 2) Since the outer and inner products must sum to x, the factored form must be: 15x2 x 6 (5x 3)(3x 2) Check: (5x 3)(3x 2) 15x2 10x 9x 6 15x2 x 6 YO U R TU R N Factor 6x2 x 12. Answer: (3x 4)(2x 3) EXAMPLE 11 Factoring a Polynomial by Grouping Factor x3 x2 2x 2. Solution: Group the terms that have a common factor. (x3 x2 ) (2x 2) Factor out the common factor in each pair of parentheses. x2 (x 1) 2(x 1) Use the distributive property. (x2 2)(x 1) EXAMPLE 10 Identifying Prime (Irreducible) Polynomials Factor x2 x 8. Solution: Write the trinomial as a product of two binomials in general form. x2 x 8 (x n)(x n) Write all of the integers whose product is 8. Factoring by Grouping Much of our attention in this section has been on factoring trinomials. For polynomials with more than three terms we rst look for a common factor to all terms. If there is no common factor to all terms of the polynomial, we look for a group of terms that have a common factor. This strategy is called factoring by grouping. The middle term of the trinomial is x, so we look for the sum of the integers that equals 1. Since no sum exists for the given combinations, we say that this polynomial is prime (irreducible) over the integers. Determine the sum of the integers. Integers whose product is 8 1, 8 1, 8 4, 2 4, 2 Integers whose product is 8 1, 8 1, 8 4, 2 4, 2 Sum 7 7 2 2 Study Tip In Example 9, Step 3, we can eliminate any factors that have a common factor since there is no common factor to the terms in the trinomial. c00b.qxd 8/3/12 11:17 AM Page 44 74. 0.4 Factoring Polynomials 45 1. Factor out the greatest common factor (monomial). 2. Identify any special polynomial forms and apply factoring formulas. 3. Factor a trinomial into a product of two binomials: (ax b)(cx d). 4. Factor by grouping. STRATEGY FOR FACTORING POLYNOMIALS Study Tip When factoring, always start by factoring out the GCF. A Strategy for Factoring Polynomials The rst step in factoring a polynomial is to look for the greatest common factor. When specically factoring trinomials, look for special known forms: a perfect square or a difference of two squares. A general approach to factoring a trinomial uses the FOIL method in reverse. Finally, we look for factoring by grouping. The following strategy for factoring polynomials is based on the techniques discussed in this section. EXAMPLE 12 Factoring a Polynomial by Grouping Factor 2x2 2x x 1. Solution: Group the terms that have a common factor. (2x2 2x) (x 1) Factor out the common factor in each pair of parentheses. 2x(x 1) 1(x 1) Use the distributive property. (2x 1)(x 1) YO U R TU R N Factor x3 x2 3x 3. EXAMPLE 13 Factoring Polynomials Factor: a. 3x2 6x 3 b. 4x3 2x2 6x c. 15x2 7x 2 d. x3 x 2x2 2 Solution (a): Factor out the greatest common factor. 3x2 6x 3 3(x2 2x 1) The trinomial is a perfect square. 3(x 1)2 Solution (b): Factor out the greatest common factor. 4x3 2x2 6x 2x(2x2 x 3) Use the FOIL method in reverse to factor the trinomial. 2x(2x 3)(x 1) Solution (c): There is no common factor. 15x2 7x 2 Use the FOIL method in reverse to factor the trinomial. (3x 2)(5x 1) Solution (d): Factor by grouping. x3 x 2x2 2 (x3 x) (2x2 2) x(x2 1) 2(x2 1) (x 2)(x2 1) Factor the difference of two squares. (x 2)(x 1)(x 1) Answer: (x 1)(x2 3) c00b.qxd 8/3/12 11:17 AM Page 45 75. 46 C HAP TE R 0 Prerequisites and Review Factoring a Trinomial as a Product of Two Binomials x2 bx c (x ?)(x ?) 1. Find all possible combinations of factors whose product is c. 2. Of the combinations in Step 1, look for the sum of factors that equals b. ax2 bx c (?x ?)(?x ?) 1. Find all possible combinations of the rst terms whose product is ax2 . 2. Find all possible combinations of the last terms whose product is c. 3. Consider all possible factors based on Steps 1 and 2. Factoring by Grouping Group terms that have a common factor. Use the distributive property. In this section, we discussed factoring polynomials, which is the reverse process of multiplying polynomials. Four main techniques were discussed. Greatest Common Factor: axk a is the greatest common factor for all coefcients of the polynomial. k is the smallest exponent found on all of the terms in the polynomial. Factoring Formulas: Special Polynomial Forms Difference of two squares: a2 b2 (a b)(a b) Perfect squares: a2 2ab b2 (a b)2 a2 2ab b2 (a b)2 Sum of two cubes: a3 b3 (a b)(a2 ab b2 ) Difference of two cubes: a3 b3 (a b)(a2 ab b2 ) S E CTI O N 0.4 S E CTI O N 0.4 In Exercises 112, factor each expression. Start by nding the greatest common factor (GCF). 1. 5x 25 2. x2 2x 3. 4t2 2 4. 16z2 20z 5. 2x3 50x 6. 4x2 y 8xy2 16x2 y2 7. 3x3 9x2 12x 8. 14x4 7x2 21x 9. x3 3x2 40x 10. 9y2 45y 11. 4x2 y3 6xy 12. 3z3 6z2 18z In Exercises 1320, factor the difference of two squares. 13. x2 9 14. x2 25 15. 4x2 9 16. 1 x4 17. 2x2 98 18. 144 81y2 19. 225x2 169y2 20. 121y2 49x2 In Exercises 2132, factor the perfect squares. 21. x2 8x 16 22. y2 10y 25 23. x4 4x2 4 24. 1 6y 9y2 25. 4x2 12xy 9y2 26. x2 6xy 9y2 27. 9 6x x2 28. 25x2 20xy 4y2 29. x4 2x2 1 30. x6 6x3 9 31. p2 2pq q2 32. p2 2pq q2 In Exercises 3342, factor the sum or difference of two cubes. 33. t3 27 34. z3 64 35. y3 64 36. x3 1 37. 8 x3 38. 27 y3 39. y3 125 40. 64x x4 41. 27 x3 42. 216x3 y3 S U M MARY E X E R C I S E S S K I LL S c00b.qxd 8/3/12 11:17 AM Page 46 76. 0.4 Factoring Polynomials 47 In Exercises 4352, factor each trinomial into a product of two binomials. 43. x2 6x 5 44. t2 5t 6 45. y2 2y 3 46. y2 3y 10 47. 2y2 5y 3 48. 2z2 4z 6 49. 3t2 7t 2 50. 4x2 2x 12 51. 6t2 t 2 52. 6x2 17x 10 In Exercises 5360, factor by grouping. 53. x3 3x2 2x 6 54. x5 5x3 3x2 15 55. a4 2a3 8a 16 56. x4 3x3 x 3 57. 3xy 5rx 10rs 6sy 58. 6x2 10x 3x 5 59. 20x2 8xy 5xy 2y2 60. 9x5 a2 x3 9x2 a2 In Exercises 6192, factor each of the polynomials completely, if possible. If the polynomial cannot be factored, state that it is prime. 61. x2 4y2 62. a2 5a 6 63. 3a2 a 14 64. ax b bx a 65. x2 16 66. x2 49 67. 4z2 25 68. 69. 6x2 10x 4 70. x2 7x 5 71. 6x2 13xy 5y2 72. 15x 15xy 73. 36s2 9t2 74. 3x3 108x 75. a2 b2 25c2 76. 2x3 54 77. 4x2 3x 10 78. 10x 25 x2 79. 3x3 5x2 2x 80. 2y3 3y2 2y 81. x3 9x 82. w3 25w 83. xy x y 1 84. a b ab b2 85. x4 5x2 6 86. x6 7x3 8 87. x2 2x 24 88. 25x2 30x 9 89. x4 125x 90. x4 1 91. x4 81 92. 10x2 31x 15 1 16 - b4 96. Business. The break-even point for a company is given by solving the equation 3x2 9x 4x 12 0. Factor the polynomial on the left side of the equation. 97. Engineering. The height of a projectile is given by the equation s 16t2 78t 10. Factor the expression on the right side of the equal sign. 98. Engineering. The electrical eld at a point P between two charges is given by . Factor the numerator of this expression. k = 10x - x2 100 93. Geometry. A rectangle has a length of 2x 4 and a width of x. Express the perimeter of the rectangle as a factored polynomial in x. 94. Geometry. The volume of a box is given by the expression x3 7x2 12x. Express the volume as a factored polynomial in the variable x. 95. Business. The prot of a business is given by the expression P 2x2 15x 4x 30. Express the prot as a factored polynomial in the variable x. 100. Factor 4x2 12x 40. Solution: Factor the trinomial into a product of binomials. (2x 4)(2x 10) Factor out a 2. 2(x 2)(x 5) This is incorrect. What mistake was made? 99. Factor x3 x2 9x 9. Solution: Group terms with common factors. (x3 x2 ) (9x 9) Factor out common factors. x2 (x 1) 9(x 1) Distributive property. (x 1)(x2 9) Factor x2 9. (x 1)(x 3)2 This is incorrect. What mistake was made? In Exercises 99 and 100, explain the mistake that is made. A P P L I C AT I O N S C AT C H T H E M I S TA K E c00b.qxd 8/3/12 11:17 AM Page 47 77. 48 C HAP TE R 0 Prerequisites and Review In Exercises 101104, determine whether each of the following statements is true or false. 101. All trinomials can be factored into a product of two binomials. 102. All polynomials can be factored into prime factors with respect to the integers. 103. x2 y2 (x y)(x y) 104. x2 y2 (x y)2 107. Use a graphing utility to plot the graphs of the three expressions 8x3 1, (2x 1)(4x2 2x 1), and (2x 1)(4x2 2x 1). Which two graphs agree with each other? 105. Factor a2n b2n completely, assuming a, b, and n are positive integers. 106. Find all the values of c such that the trinomial x2 cx 14 can be factored. C O N C E P TUAL O BJ E CTIVE S Understand why rational expressions have domain restrictions. Understand the least common denominator method for rational expressions. S K I LLS O BJ E CTIVE S Find the domain of an algebraic expression. Reduce a rational expression to lowest terms. Multiply and divide rational expressions. Add and subtract rational expressions. Simplify complex rational expressions. R ATI O NAL E X P R E S S I O N S S E CTI O N 0.5 Rational Expressions and Domain Restrictions Recall that a rational number is the ratio of two integers with the denominator not equal to zero. Similarly, the ratio, or quotient, of two polynomials is a rational expression. Rational Numbers: Rational Expressions: 9 3x - 2 5x2 x2 + 1 3x + 2 x - 5 9 11 5 9 3 7 108. Use a graphing utility to plot the graphs of the three expressions 27x3 1, (3x 1)3 , and (3x 1)(9x2 3x 1). Which two graphs agree with each other? C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y c00b.qxd 8/3/12 11:17 AM Page 48 78. As with rational numbers, the denominators of rational expressions are never equal to zero. In the rst and third rational expressions, there are values of the variable that would correspond to a denominator equal to zero; these values are not permitted: In the second rational expression, , there are no real numbers that will correspond to a zero denominator. The set of real numbers for which an algebraic expression is dened is called the domain. Since a rational expression is not dened if its denominator is zero, we must eliminate from the domain those values of the variable that would result in a zero denominator. To nd the domain of an algebraic expression we ask the question, What can x (the variable) be? For rational expressions the answer in general is any values except those that make the denominator equal to zero. 5x2 x2 + 1 9 3x - 2 x Z 2 3 3x + 2 x - 5 x Z 5 0.5 Rational Expressions 49 EXAMPLE 1 Finding the Domain of an Algebraic Expression Find the domain of the expressions. a. 2x2 5x 3 b. c. d. Solution: 3x + 1 x x x2 + 1 2x + 1 x - 4 YO U R TU R N Find the domain of the expressions. a. b. c. 3x2 2x 7 d. 2x + 5 x2 + 4 5x - 1 x 3x - 1 x + 1 Answer: a. b. c. all real numbers d. all real numbers x Z 0 x Z -1 ALGEBRAIC EXPRESSION TYPE DOMAIN NOTE a. 2x2 5x 3 Polynomial All real numbers The domain of all polynomials is the set of all real numbers. Rational All real numbers When x 4, the rational b. expression except x 4 expression is undened. Rational All real numbers There are no real numbers c. expression that will result in the denominator being equal to zero. d. Rational All real numbers When x 0, the rational expression except x 0 expression is undened. 2x + 1 x - 4 x x2 + 1 3x + 1 x In this text, it will be assumed that the domain is the set of all real numbers except the real numbers shown to be excluded. c00c.qxd 8/7/12 4:30 PM Page 49 79. 50 C HAP TE R 0 Prerequisites and Review 1. Factor the numerator and denominator completely. 2. State any domain restrictions. 3. Cancel (divide out) the common factors in the numerator and denominator. REDUCING A RATIONAL EXPRESSION TO LOWEST TERMS (SIMPLIFYING) Technology Tip Compare the graphs of and . To show the graph and table of Y1, unhighlight for Y2 and press GRAPH . Y2 = x - 2 2 , x Z -1 Y1 = x2 - x - 2 2x + 2 In this section, we will simplify rational expressions and perform operations on rational expressions such as multiplication, division, addition, and subtraction. The resulting expressions may not have explicit domain restrictions, but it is important to note that there are implicit domain restrictions, because the domain restrictions on the original rational expression still apply. Simplifying Rational Expressions Recall that a fraction is reduced when it is written with no common factors. Similarly, rational expressions are reduced to lowest terms, or simplied, if the numerator and denominator have no common factors other than . As with real numbers, the ability to write fractions in reduced form is dependent upon your ability to factor. ;1 16 12 = 4 # 4 4 # 3 = a 4 4 b # a 4 3 b = (1) # a 4 3 b = 4 3 or 16 12 = 4 # 4 4 # 3 = 4 3 Note that x 1 is not dened. EXAMPLE 2 Excluding Values from the Domain of Rational Expressions Determine what real numbers must be excluded from the domain of the following rational expressions. a. b. Solution (a): Factor the denominator. Determine the values of x that will and must be excluded make the denominator equal to zero. from the domain. Solution (b): Factor the denominator. Determine the values of x that will and must be excluded make the denominator equal to zero. from the domain. x = 5x = 0 3x + 2 x2 - 5x = 3x + 2 x(x - 5) x = 2x = -2 7x + 5 x2 - 4 = 7x + 5 (x + 2)(x - 2) 3x + 2 x2 - 5x 7x + 5 x2 - 4 c00c.qxd 8/7/12 4:30 PM Page 50 80. 0.5 Rational Expressions 51 EXAMPLE 3 Reducing a Rational Expression to Lowest Terms Simplify and state any domain restrictions. Solution: Factor the numerator and denominator. State any domain restrictions. Cancel (divide out) the common factor, x 1. The rational expression is now in lowest terms (simplied). YO U R TU R N Simplify and state any domain restrictions. x2 + x - 2 2x - 2 x - 2 2 x Z -1 = (x - 2)(x + 1) 2(x + 1) x Z -1 x2 - x - 2 2x + 2 = (x - 2)(x + 1) 2(x + 1) x2 - x - 2 2x + 2 The following table summarizes two common mistakes made with rational expressions. CORRECT INCORRECT COMMENT is already simplied. = 1 x + 1 x Z 0, x Z -1 x Z 0, x Z -1= x x(x + 1) x x2 + x x + 5 y + 5 Error: Error: Note: Missing x Z 0. x x2 + x = 1 x + 1 x Z -1 x + 5 y + 5 = x y Factors can be divided out (canceled). Terms or parts of terms cannot be divided out. Remember to factor the numerator and denominator rst, and then divide out common factors. Determine the domain restrictions before dividing out common factors. Study Tip Factors can be divided out (canceled). Factor the numerator and denominator rst, and then divide out common factors. Study Tip Determine domain restrictions of a rational expression before dividing out (canceling) common factors. Answer: x Z 1x Z -3, x - 2 x - 1 EXAMPLE 4 Simplifying Rational Expressions Reduce to lowest terms and state any domain restrictions. Solution: Factor the numerator and denominator. State domain restrictions. Divide out the common factor, x 2. Simplify. YO U R TU R N Reduce to lowest terms and state any domain restrictions. x2 + x - 6 x2 + 2x - 3 x - 3 x - 1 x Z -2, x Z 1 = (x - 3)(x + 2) (x - 1)(x + 2) x Z -2, x Z 1 x2 - x - 6 x2 + x - 2 = (x - 3)(x + 2) (x - 1)(x + 2) x2 - x - 6 x2 + x - 2 Answer: x + 2 2 x Z 1 c00c.qxd 8/7/12 4:30 PM Page 51 81. 52 C HAP TE R 0 Prerequisites and Review Answer: (x 5) x Z 5 1. Factor all numerators and denominators completely. 2. State any domain restrictions. 3. Divide the numerators and denominators by any common factors. 4. Multiply the remaining numerators and denominators, respectively. Dividing Rational Expressions 1. Factor all numerators and denominators completely. 2. State any domain restrictions. 3. Rewrite division as multiplication by a reciprocal. 4. State any additional domain restrictions. 5. Divide the numerators and denominators by any common factors. 6. Multiply the remaining numerators and denominators, respectively. Multiplying Rational Expressions PROPERTY RESTRICTION DESCRIPTION Multiply numerators and denominators, respectively. Dividing is equivalent to multiplying by a reciprocal.b Z 0, d Z 0, c Z 0 a b , c d = a b # d c b Z 0, d Z 0 a b # c d = ac bd Multiplying and Dividing Rational Expressions The same rules that apply to multiplying and dividing rational numbers also apply to rational expressions. EXAMPLE 5 Simplifying Rational Expressions Reduce to lowest terms and state any domain restrictions. Solution: Factor the numerator and denominator. State domain restrictions. Factor out a negative in the denominator. Cancel (divide out) the common factor, x 2. Simplify. YO U R TU R N Reduce to lowest terms and state any domain restrictions. x2 - 25 5 - x -(x + 2) x Z 2 = (x - 2)(x + 2) -(x - 2) = (x - 2)(x + 2) -(x - 2) x Z 2 x2 - 4 2 - x = (x - 2)(x + 2) (2 - x) x2 - 4 2 - x c00c.qxd 8/7/12 4:30 PM Page 52 82. 0.5 Rational Expressions 53 EXAMPLE 6 Multiplying Rational Expressions Multiply and simplify . Solution: Factor the numerators and denominators. State any domain restrictions. , , Divide the numerators and denominators by common factors. Simplify. YO U R TU R N Multiply and simplify . 2x + 1 3x2 - 3x # x3 + 2x2 - 3x 8x + 4 x + 2 12 x Z 0, x Z -1, x Z - 1 3 = (3x + 1) 4x(x + 1) # x(x + 1) (x + 2) 3(3x + 1) x Z - 1 3 x Z -1x Z 0 = (3x + 1) 4x(x + 1) # x(x + 1)(x + 2) 3(3x + 1) 3x + 1 4x2 + 4x # x3 + 3x2 + 2x 9x + 3 Answer: x + 3 12 x Z 0, x Z 1, x Z - 1 2 Answer: 7x2 2 x Z -3, x Z 0, x Z 3 EXAMPLE 7 Dividing Rational Expressions Divide and simplify . Solution: Factor numerators and denominators. State any domain restrictions. Write the quotient as a product. State any additional domain restrictions. Divide out the common factors. Simplify. YO U R TU R N Divide and simplify . x2 - 9 x , 2x3 - 18x 7x4 5x 3 x Z -2, x Z 0, x Z 2 = (x - 2)(x + 2) x # 5x3 3x (x - 2)(x + 2) x Z -2, x Z 2 = (x - 2)(x + 2) x # 5x3 3x(x - 2)(x + 2) x Z 0 (x - 2)(x + 2) x , 3x(x - 2)(x + 2) 5x3 x2 - 4 x , 3x3 - 12x 5x3 Adding and Subtracting Rational Expressions The same rules that apply to adding and subtracting rational numbers also apply to rational expressions. PROPERTY RESTRICTION DESCRIPTION Adding or subtracting rational expressions when the denominators are the same and Adding or subtracting rational expressions when the denominators are different d Z 0b Z 0 a b ; c d = ad ; bc bd b Z 0 a b ; c b = a ; c b c00c.qxd 8/7/12 4:30 PM Page 53 83. 54 C HAP TE R 0 Prerequisites and Review Study Tip When subtracting a rational expression, distribute the negative of the quantity to be subtracted over all terms in the numerator. EXAMPLE 9 Adding and Subtracting Rational Expressions: No Common Factors in Denominators Perform the indicated operation and simplify. a. b. Solution (a): The common denominator is the product of the denominators. Eliminate parentheses in the numerator. Combine like terms in the numerator. = x2 + 5x - 3 (2x + 1)(x - 1) x Z - 1 2 , x Z 1 = 3x - 3 - x2 + x + 2x2 + x (2x + 1)(x - 1) = (3 - x)(x - 1) + x(2x + 1) (2x + 1)(x - 1) 3 - x 2x + 1 + x x - 1 = (3 - x)(x 1) (2x + 1)(x 1) + x(2x 1) (x - 1)(2x 1) 1 x2 - 2 x + 1 3 - x 2x + 1 + x x - 1 EXAMPLE 8 Adding and Subtracting Rational Expressions: Equal Denominators Perform the indicated operation and simplify. a. b. Solution (a): Write as a single expression. State any domain restrictions. Combine like terms in the numerator. Factor out the common factor in the numerator. Cancel (divide out) the common factor, x 2. Solution (b): Write as a single expression. Use parentheses around the second numerator to ensure that the negative will be distributed throughout all terms. State any domain restrictions. Eliminate parentheses. Distribute the negative. Combine like terms in the numerator. Factor out the common factor in the numerator. Divide out (cancel) the common factor, 2x 1. 2 x Z 1 2 = 2(2x - 1) 2x - 1 = 4x - 2 2x - 1 = 6x + 7 - 2x - 9 2x - 1 x Z 1 2 = 6x + 7 - (2x + 9) 2x - 1 4 x + 2 x Z -2 = 4(x + 2) (x + 2)2 = 4x + 8 (x + 2)2 x Z -2 = x + 7 + 3x + 1 (x + 2)2 6x + 7 2x - 1 - 2x + 9 2x - 1 x + 7 (x + 2)2 + 3x + 1 (x + 2)2 c00c.qxd 8/7/12 4:30 PM Page 54 84. 0.5 Rational Expressions 55 Solution (b): The common denominator is the product of the denominators. Eliminate parentheses in the numerator. Write the numerator in factored form to ensure no further simplication is possible. YO U R TU R N Perform the indicated operation and simplify. a. b. 2 x2 + 1 - 1 x 2x - 1 x + 3 + x 2x + 1 = -(2x + 1)(x - 1) x2 (x + 1) x Z -1, x Z 0 = x + 1 - 2x2 x2 (x + 1) = -2x2 + x + 1 x2 (x + 1) = -(2x2 - x - 1) x2 (x + 1) 1 x2 - 2 x + 1 = (1)(x + 1) - 2(x2 ) x2 (x + 1) Study Tip When adding or subtracting rational expressions whose denominators have no common factors, the least common denominator is the product of the two denominators. Answer: a. b. -(x - 1)2 x(x2 + 1) x Z 0 x Z -3, x Z - 1 2 5x2 + 3x - 1 (x + 3)(2x + 1) 2 3 + 1 6 - 4 9 = 12 + 3 - 8 18 = 7 18 When combining two or more fractions through addition or subtraction, recall that the least common multiple, or least common denominator (LCD), is the smallest real number that all of the denominators divide into evenly (that is, the smallest of which all are factors). For example, To nd the LCD of these three fractions, factor the denominators into prime factors: Rational expressions follow this same procedure, only now variables are also considered: 3 # 2 # 3 = 18 9 = 3 # 3 6 = 3 # 2 3 = 3 2 3 + 1 6 - 4 9 1. Factor each of the denominators completely. 2. The LCD is the product of each of these distinct factors raised to the highest power to which that factor appears in any of the denominators. 3. Write each rational expression using the LCD for each denominator. 4. Add or subtract the resulting numerators. 5. Factor the resulting numerator to check for common factors. THE LCD METHOD FOR ADDING AND SUBTRACTING RATIONAL EXPRESSIONS LCD 2x3 LCD (x 1)2 (2x 1) 2 x + 1 - x 2x + 1 + 3 - x (x + 1)2 = 2 (x + 1) - x (2x + 1) + 3 - x (x + 1)(x + 1) = 1 2x + 1 x # x # x 1 2x + 1 x3 The following box summarizes the LCD method for adding and subtracting rational expressions whose denominators have common factors. c00c.qxd 8/7/12 4:30 PM Page 55 85. 56 C HAP TE R 0 Prerequisites and Review EXAMPLE 10 Subtracting Rational Expressions: Common Factors in Denominators (LCD) Perform the indicated operation and write in simplied form. Solution: Factor the denominators. Identify the LCD. LCD 2(x 3)(x 2) Write each expression using the LCD as the denominator. Combine into one expression. Distribute the negative through the entire second numerator. Simplify. YO U R TU R N Perform the indicated operation and write in simplied form. 2x 3x - 6 - 5x + 1 x2 + 2x - 8 = 5x2 - 4x + 4 2(x - 3)(x + 2) x Z -2, x Z 3 = 5x2 + 10x - 14x + 4 2(x - 3)(x + 2) = 5x(x 2) 2(x - 3)(x 2) - 2(7x - 2) 2(x - 3)(x + 2) = 5x 2(x - 3) - 7x - 2 (x - 3)(x + 2) 5x 2x - 6 - 7x - 2 x2 - x - 6 Answer: 2x2 - 7x - 3 3(x - 2)(x + 4) x Z -4, x Z 2 Complex Rational Expressions A rational expression that contains another rational expression in either its numerator or denominator is called a complex rational expression. The following are examples of com- plex rational expressions. 1 x - 5 2 + x 2 - x 4 + 3 x - 1 3 x - 7 6 2x - 5 - 1 Procedure 1: Write a sum or difference of rational expressions that appear in either the numerator or denominator as a single rational expression. Once the complex rational expression contains a single rational expression in the numerator and one in the denominator, then rewrite the division as multiplication by the reciprocal. OR Procedure 2: Find the LCD of all rational expressions contained in both the numerator and denominator. Multiply the numerator and denominator by this LCD and simplify. TWO METHODS FOR SIMPLIFYING COMPLEX RATIONAL EXPRESSIONS c00c.qxd 8/7/12 4:30 PM Page 56 86. Technology Tip Compare the graphs and tables of values of and Y2 = x + 1 x , x Z 0. x Z -2, x Z -1, x Z 0 Y1 = 2/x + 1 1 + 1/(x + 1) , Note that , and are not dened.x = -2 x = 0,x = -1 Note that is not dened. Be careful not to forget that the original domain restrictions still hold. (x Z -2, -1, 0) x = 0 EXAMPLE 11 Simplifying a Complex Rational Expression Write the rational expression in simplied form. Solution: State the domain restrictions. , , and Procedure 1: Add the expressions in both the numerator and denominator. Simplify. Express the quotient as a product. Divide out the common factors. Write in simplied form. Procedure 2: Find the LCD of the numerator and denominator. Identify the LCDs. Numerator LCD: x Denominator LCD: x 1 Combined LCD: x(x 1) Multiply both numerator and denominator by their combined LCD. Multiply the numerators and denominators, respectively, applying the distributive property. Divide out common factors. Simplify. = 2(x + 1) + x(x + 1) x(x + 1) + x = 2 x # x(x + 1) + 1x(x + 1) x(x + 1) + 1 x + 1 # x(x + 1) = 2 x # x(x + 1) + 1x(x + 1) 1 # x(x + 1) + 1 x + 1 # x(x + 1) = 2 x + 1 1 + 1 x + 1 # x(x + 1) x(x + 1) 2 x + 1 1 + 1 x + 1 = x + 1 x x Z -2, x Z -1, x Z 0 = 2 + x x # x + 1 x + 2 = 2 + x x # x + 1 x + 2 = 2 + x x x + 2 x + 1 = 2 x + x x x + 1 x + 1 + 1 x + 1 = 2 + x x (x + 1) + 1 x + 1 1 + 1 x + 1 x Z -2x Z -1x Z 0 2 x + 1 2 x + 1 1 + 1 x + 1 0.5 Rational Expressions 57 c00c.qxd 8/7/12 4:31 PM Page 57 87. 58 C HAP TE R 0 Prerequisites and Review EXAMPLE 12 Simplifying a Complex Rational Expression Write the rational expression in simplied form. Solution: Using Procedure 1 Factor the respective denominators. Identify the LCDs. Numerator LCD: (x 3)(x 3) Denominator LCD: 2(x 3) Combined LCD: 2(x 3)(x 3) Multiply both the numerator and the denominator by the combined LCD. Multiply the numerators and denominators, respectively, applying the distributive property. Simplify. Eliminate the parentheses. Combine like terms. Factor the numerator and denominator to make sure there are no common factors. = 2(3x2 - 26) (x + 6)(x - 3) x Z -6, x Z -3, x Z 3 = 6x2 - 52 x2 + 3x - 18 = 2 + 6x2 - 54 2x2 - 18 - x2 + 3x = 2 + 6(x + 3)(x - 3) 2(x + 3)(x - 3) - x(x - 3) = 2(x + 3)(x - 3) (x - 3)(x + 3) + 3 # 2(x + 3)(x - 3) 2(x + 3)(x - 3) - x # 2(x + 3)(x - 3) 2(x + 3) = 1 (x - 3)(x + 3) + 3 1 - x 2(x + 3) # 2(x + 3)(x - 3) 2(x + 3)(x - 3) 1 (x - 3)(x + 3) + 3 1 - x 2(x + 3) 1 x2 - 9 + 3 1 - x 2x + 6 x Z -6, -3, 3 Apply the distributive property. Combine like terms. Factor the numerator and denominator. Divide out the common factor. Write in simplied form. = x + 1 x x Z -2, x Z -1, x Z 0 = (x + 2)(x + 1) x(x + 2) = (x + 2)(x + 1) x(x + 2) = x2 + 3x + 2 x2 + 2x = 2x + 2 + x2 + x x2 + x + x c00c.qxd 8/7/12 4:31 PM Page 58 88. 0.5 Rational Expressions 59 S U M MARY In this section, rational expressions were dened as quotients of polynomials. The domain of any polynomial is the set of all real numbers. Since rational expressions are ratios of polynomials, the domain of rational expressions is the set of all real numbers except those values that make the denominator equal to zero. In this section, rational expressions were simplied (written with no common factors), multiplied, divided, added, and subtracted. S E CTI O N 0.5 OPERATION Multiplying rational expressions Dividing rational expressions Adding/subtracting rational expressions with no common factors Adding/subtracting rational expressions with common factors EXAMPLE = x + 4 x(x + 1)(x + 2) x Z -2, -1, 0 = 3x + 6 - 2x - 2 x(x + 1)(x + 2) = 3(x + 2) - 2(x + 1) x(x + 1)(x + 2) LCD = x(x + 1)(x + 2) 3 x(x + 1) - 2 x(x + 2) x Z -2, -1, 0 = (3x - 1)(x + 2) (x + 1)(x - 1) x Z ;1 = 3x2 + 5x - 2 (x + 1)(x - 1) = 2x - 2 + 3x2 + 3x (x + 1)(x - 1) = 2(x - 1) + 3x(x + 1) (x + 1)(x - 1) LCD = (x + 1)(x - 1) 2 x + 1 + 3x x - 1 x Z ;1 = 2(x - 1) 3x(x + 1) x Z 0, x Z ;1 = 2 x + 1 # x - 1 3x x Z 0 2 x + 1 , 3x x - 1 x Z ;1 2 x + 1 # 3x x - 1 = 6x x2 - 1 x Z ;1 NOTE State domain restrictions. When dividing rational expressions, remember to check for additional domain restrictions once the division is rewritten as multiplication by a reciprocal. The least common denominator (LCD) is the product of the two denominators. The LCD is the product of each of these distinct factors raised to the highest power that appears in any of the denominators. Complex rational expressions are simplied in one of two ways: 1. Combine the sum or difference of rational expressions in a numerator or denominator as a single rational expression. The result is a rational expression in the numerator and a rational expression in the denominator. Then write the division as multiplication by the reciprocal. 2. Multiply the numerator and denominator by the overall LCD (LCD for all rational expressions that appear). The result is a single rational expression. Then simplify, if possible. c00c.qxd 8/7/12 4:31 PM Page 59 89. 60 C HAP TE R 0 Prerequisites and Review In Exercises 110, state any real numbers that must be excluded from the domain of each rational expression. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. In Exercises 1130, reduce the rational expression to lowest terms and state any real numbers that must be excluded from the domain. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. In Exercises 3148, multiply the rational expressions and simplify. State any real numbers that must be excluded from the domain. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 3x2 - 2x 12x3 - 8x2 # x2 - 7x - 18 2x2 - 162 6x2 - 11x - 35 8x2 - 22x - 21 # 4x2 - 49 9x2 - 25 5t - 1 4t # 4t2 + 3t 16t2 - 9 3x2 - 15x 2x3 - 50x # 2x2 - 7x - 15 3x2 + 15x t2 + t - 6 t2 - 4 # 8t 2t2 y2 - 4 y - 3 # 3y y + 2 7a2 + 21a 14(a2 - 9) # a + 3 7 t2 + 4 t - 3 # 3t t + 2 y + 3 3y + 9 # y2 - 10y + 25 y2 + 3y - 40 t + 2 3t - 9 # t2 - 6t + 9 t2 + 4t + 4 4x2 - 32x x # x2 + 3x x2 - 5x - 24 3x2 - 12 x # x2 + 5x x2 + 3x - 10 5x - 5 10x # x2 + x x2 - 1 2x - 2 3x # x2 + x x2 - 1 4(x - 2)(x + 5) 8x # 16x (x - 5)(x + 5) 5x + 6 x # 2x 5x - 6 4x + 5 x - 2 # 3x + 4 4x + 5 x - 2 x + 1 # 3x + 5 x - 2 15x2 - x - 2 5x2 + 13x - 6 6x2 - x - 1 2x2 + 9x - 5 x2 + 19x + 60 x2 + 8x + 16 x2 + 5x + 6 x2 - 3x - 10 x2 + 4 2x + 4 x2 + 9 2x + 9 2y + 9 2y + 9 x + 7 x + 7 t3 - t t - 1 x2 - 4 x - 2 (t - 7)(2t + 5) 3t - 21 (3x + 7)(x - 4) 4x - 16 (2t - 1)(t + 2) 4t + 8 (5y - 1)(y + 1) 25y - 5 7(2y + 1)(3y - 1) 5(3y - 1)(2y) 2(3y + 1)(2y - 1) 3(2y - 1)(3y) (2x + 1)(x - 3) 3(x - 3) (x - 3)(x + 1) 2(x + 1) 4y(y - 8)(y + 7) 8y(y + 7)(y + 8) (x + 3)(x - 9) 2(x + 3)(x + 9) 2t - 2 t2 + 4 3p - 1 p2 + 1 3t t2 - 9 2p2 p2 - 1 2x 3 - x 5x - 1 x + 1 6 y - 1 3 x - 1 5 x 3 x E X E R C I S E S S E CTI O N 0.5 S K I LL S c00c.qxd 8/7/12 4:31 PM Page 60 90. 0.5 Rational Expressions 61 In Exercises 4966, divide the rational expressions and simplify. State any real numbers that must be excluded from the domain. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. In Exercises 6782, add or subtract the rational expression and simplify. State any real numbers that must be excluded from the domain. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. In Exercises 8390, simplify the complex rational expressions. State any real numbers that must be excluded from the domain. 83. 84. 85. 86. 87. 88. 89. 90. 3 x + 1 - 3 x - 1 5 x2 - 1 1 x - 1 + 1 1 x + 1 + 1 7 y + 7 1 y + 7 - 1 y 1 x - 1 + 1 1 - 1 x + 1 1 x + 2 x2 9 x - 5 x2 3 + 1 x 9 - 1 x2 3 y - 5 4 - 2 y 1 x - 1 1 - 2 x 3 5y + 6 - 4 y - 2 + y2 - y 5y2 - 4y - 12 7 + 1 x - 3 1 y + 4 y2 - 4 - 2 y2 - 2y 5a a2 - b2 - 7 b - a 2 y - 3 + 7 y + 2 x - 1 x - 2 + x - 6 x2 - 4 x - 1 4 - x2 - x + 1 2 + x 3x x2 - 4 + 3 + x x + 2 3 1 - x + 4 x - 1 3y2 y + 1 + 1 - 2y y - 1 7 2x - 1 - 5 1 - 2x 2x + 1 5x - 1 - 3 - 2x 1 - 5x 4 9 + x - 5x x - 2 3 p - 2 + 5p p + 1 5 7x - 3 x 3 x - 2 5x x2 - 6x - 27 2x2 + 13x - 7 , 2x2 - 15x - 27 2x2 + 9x - 5 20x2 - 3x - 2 25x2 - 4 , 12x2 + 23x + 5 3x2 + 5x 2y2 - 5y - 3 2y2 - 9y - 5 , 3y - 9 y2 - 5y x2 + 4x - 21 x2 + 3x - 10 , x2 - 2x - 63 x2 + x - 20 y2 - 3y 2y , y3 - 3y2 8y w2 - w w , w3 - w 5w3 x3 + 8x2 + 12x 5x2 - 10x , 4x + 8 x2 - 4 3t3 - 6t2 - 9t 5t - 10 , 6 + 6t 4t - 8 49 - y2 y2 - 25 , 7 + y 2y + 10 36 - n2 n2 - 9 , n + 6 n + 3 4 - x x2 - 16 , 12 - 3x x - 4 2 - p p2 - 1 , 2p - 4 p + 1 5 3x - 4 , 10 9x2 - 16 1 x - 1 , 5 x2 - 1 5(x + 6) 10(x - 6) , 20(x + 6) 8 6 x - 2 , 12 (x - 2)(x + 2) 5 x2 , 10 x3 3 x , 12 x2 93. Circuits. If two resistors are connected in parallel, the combined resistance is given by the formula where R1 and R2 are the individual resistances. Simplify the formula. 94. Optics. The focal length of a lens can be calculated by applying the formula , where p is the distance that the object is from the lens and q is the distance that the image is from the lens. Simplify the formula. f = 1 1/p + 1/q R = 1 1/R1 + 1/R2 , 91. Finance. The amount of payment made on a loan is given by the formula , where p is the principal (amount borrowed), and , where r is the interest rate expressed as a decimal and n is the number of payments per year. Suppose n 5. Simplify the formula as much as possible. 92. Finance. Use the formula to calculate the amount your monthly payment will be on a loan of $150,000 at an interest rate of 6.5% for 30 years (nt 360). A = pi 1 - 1 (1 + i)nt i = r n A = pi 1 - 1/(1 + i)n A P P L I C AT I O N S c00c.qxd 8/7/12 4:31 PM Page 61 91. 62 C HAP TE R 0 Prerequisites and Review 95. Simplify . Solution: Factor the numerator. Cancel the common factor, x 1. Write in simplied form. This is incorrect. What mistake was made? = x + 1 = (x + 1)(x + 1) (x + 1) x2 + 2x + 1 x + 1 = (x + 1)(x + 1) (x + 1) x2 + 2x + 1 x + 1 96. Simplify . Solution: Cancel the common 1s. Factor the denominator. Cancel the common x. Write in simplied form. , This is incorrect. What mistake was made? x Z 0x Z -2= 1 x + 2 = x x(x + 2) = x x(x + 2) x + 1 x2 + 2x + 1 x + 1 x2 + 2x + 1 In Exercises 97100, determine whether each of the statements is true or false. 97. 98. x Z -9, 9 x - 9 x2 - 81 = 1 x + 9 x2 - 81 x - 9 = x + 9 99. When adding or subtracting rational expressions, the LCD is always the product of all the denominators. 100. for all values of x. x - c c - x = -1 101. Perform the operation and simplify (remember to state domain restrictions). x + a x + b , x + c x + d 102. Write the numerator as the product of two binomials. Divide out any common factors of the numerator and denominator. a2n - b2n an - bn 103. Utilizing a graphing technology, plot the expression . Zoom in near x 7. Does this agree with what you found in Exercise 23? 104. Utilizing a graphing technology, plot the expression . Zoom in near x 2. Does this agree with what you found in Exercise 21? y = x2 - 4 x - 2 y = x + 7 x + 7 In Exercises 105 and 106, for each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other. 105. 106. 1 - 2 x + 3 1 + 1 x + 4 1 + 1 x - 2 1 - 1 x + 2 In Exercises 95 and 96, explain the mistake that is made. C AT C H T H E M I S TA K E C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y c00c.qxd 8/7/12 4:31 PM Page 62 92. C O N C E P TUAL O BJ E CTIVE S Understand that radicals are equivalent to rational exponents. Understand that a radical implies one number (the principal root), not two ( the principal root).; R ATI O NAL E X P O N E NTS AN D R AD I CALS S K I LLS O BJ E CTIVE S Simplify square roots. Simplify radicals. Add and subtract radicals. Rationalize denominators containing radicals. Apply properties of rational exponents. S E CTI O N 0.6 In Section 0.2, we discussed integer exponents and their properties. For example, 42 16 and x2 x3 x5 . In this section we expand our discussion of exponents to include any rational numbers. For example, 161/2 ? and . We will rst start with a more familiar notation (roots) and discuss operations on radicals, and then rational exponents will be discussed. Square Roots (x1/2 )3/4 = ? Let a be any nonnegative real number; then the nonnegative real number b is called the principal square root of a, denoted , if b2 a. The symbol is called a radical sign, and a is called the radicand. 1b = 1a Principal Square RootD E F I N I T I O N It is important to note that the principal square root b is nonnegative. The principal square root of 16 is 4 implies 42 16. Although it is also true that (4)2 16, the principal square root is dened to be nonnegative. It is also important to note that negative real numbers do not have real square roots. For example, is not a real number because there are no real numbers that when squared yield . Since principal square roots are dened to be nonnegative, this means they must be zero or positive. The square root of zero is equal to zero: . All other nonnegative principal square roots are positive. 10 = 0 -9 1-9 EXAMPLE 1 Evaluating Square Roots Evaluate the square roots, if possible. a. b. c. Solution: a. What positive real number squared results in 169? Check: 132 169 b. What positive real number squared results in ? Check: c. What positive real number squared results in 36? No real number A2 3 B 2 = 4 9 A 4 9 = 2 3 4 9 1169 = 13 1-36 A 4 9 1169 Study Tip Incorrect: Correct: The principal square root, , is dened as a nonnegative real number. 1 125 = 5 125 = ;5 63 c00c.qxd 8/7/12 4:31 PM Page 63 93. 64 C HAP TE R 0 Prerequisites and Review EXAMPLE 2 Finding Square Roots of Perfect Squares Evaluate the following: a. b. c. Solution: a. b. c. 2x2 = x2(-7)2 = 149 = 7262 = 136 = 6 2x2 2(-7)2 262 Let a be any real number; then: 1a2 = a SQUARE ROOTS OF PERFECT SQUARES Let a and b be nonnegative real numbers, then: PROPERTIES OF SQUARE ROOTS Property A a b = 1a 2b b Z 0 1a # b = 1a # 1b Example A 40 49 = 140 149 = 14 # 110 7 = 2110 7 120 = 14 # 15 = 215 Description The square root of a product is the product of the square roots. The square root of a quotient is the quotient of the square roots. Simplifying Square Roots So far only square roots of perfect squares have been discussed. Now we consider how to simplify square roots such as . We rely on the following properties.112 EXAMPLE 3 Simplifying Square Roots Simplify: a. b. c. d. Solution: a. b. since x 0= 2x17x= 2x 17x= 2x 17 1x = 14 # 17 # 2x2 # 1x228x3 = 128 # 2x3 = 14 # 7 # 2x2 # x 248x2 = 148 # 2x2 = 116 # 3 # 2x2 = 116 # 13 # 2x2 = 4 x 13 245x3 25x 112x # 16x228x3 248x2 2 s x s x s 4 s Study Tip If you have a single odd power under the square root, like , the variable is forced to be nonnegative because a negative cubed results in a negative inside the radical; therefore, the absolute value is not necessary. 2x3 c. since d. Note: , YO U R TU R N Simplify: a. b. 2125x5 225x3 260x3 x 7 0x Z 0 = A 45x3 5x = 29x2 = 19 # 2x2 = 3 x = 3x since x 7 0 245x3 15x x 06x12 112x # 16x = 272x2 = 236 # 2 # x2 = 136 # 12 # 2x2 = 6 x 12 6 s x s 3 s x s Answer: a. b. x152x115x c00c.qxd 8/7/12 4:31 PM Page 64 94. 0.6 Rational Exponents and Radicals 65 Let a be a real number and n be a positive integer. Then the real number b is called the principal nth root of a, denoted , if . If n is even, then a and b are nonnegative real numbers. The positive integer n is called the index. The square root corresponds to n 2, and the cube root corresponds to n 3. bn = ab = 1 n a Principal nth RootD E F I N I T I O N Other (nth) Roots We now expand our discussion from square roots to other nth roots. A radical sign, , combined with a radicand is called a radical.2 n a b EXAMPLE Even Positive Positive because 24 16 Even Negative Not a real number is not a real number Odd Positive Positive because Odd Negative Negative because (-5)3 = -1251 3 -125 = -5 33 = 271 3 27 = 3 14 -16 1 4 16 = 2 Let a and b be real numbers, then PROPERTY DESCRIPTION EXAMPLE PROPERTIES OF RADICALS if and both exist if and both exist n is odd n is even2 n an = a 2 n an = a 2 n am = A1 n a B m 2 n b2 n a A n a b = 2 n a 2 n b b Z 0 2 n b2 n a 2 n ab = 2 n a # 2 n b The nth root of a product is the product of the nth roots. The nth root of a quotient is the quotient of the nth roots. The nth root of a power is the power of the nth root. When n is odd, the nth root of a raised to the nth power is a. When n is even, the nth root of a raised to the nth power is the absolute value of a. 2 4 x4 = x 2 3 x3 = x 2 3 82 = A1 3 8B 2 = (2)2 = 4 A 4 81 16 = 1 4 81 1 4 16 = 3 2 1 3 16 = 1 3 8 # 1 3 2 = 21 3 2 c00c.qxd 8/7/12 4:31 PM Page 65 95. 66 C HAP TE R 0 Prerequisites and Review EXAMPLE 4 Simplifying Radicals Simplify: a. b. Solution: a. b. since x 02 4 32x5 = 2 4 16 # 2 # x4 # x = 1 4 16 # 1 4 2 # 2 4 x4 # 1 4 x = 2x 1 4 2x = 2x1 4 2x 2 3 -24x5 = 2 3 (-8)(3)x3 x2 = 1 3 -8 # 1 3 3 # 2 3 x3 # 2 3 x2 = -2x2 3 3x2 2 4 32x5 2 3 -24x5 -2 s x s 2 s x s Combining Like Radicals We have already discussed properties for multiplying and dividing radicals. Now we focus on combining (adding or subtracting) radicals. Radicals with the same index and radicand are called like radicals. Only like radicals can be added or subtracted. EXAMPLE 5 Combining Like Radicals Combine the radicals if possible. a. b. c. d. Solution (a): Use the distributive property. Eliminate the parentheses. Solution (b): None of these radicals are alike. The expression is in simplied form. Solution (c): Write the radicands as products with a factor of 5. The square root of a product is the product of square roots. Simplify the square roots of perfect squares. All three radicals are now like radicals. Use the distributive property. Simplify. Solution (d): None of these radicals are alike because they have different indices. The expression is in simplied form. YO U R TU R N Combine the radicals. a. b. 5124 - 215441 3 7 - 61 3 7 + 91 3 7 1 4 10 - 21 3 10 + 3110 - 15 (3 + 2 - 6)15 315 + 215 - 615 315 + 215 - 2(3)15 315 + 14 # 15 - 219 # 15 315 + 120 - 2145 = 315 + 14 # 5 - 219 # 5 215 - 317 + 613 = 513 413 - 613 + 713 = (4 - 6 + 7)13 1 4 10 - 21 3 10 + 3110315 + 120 - 2145 215 - 317 + 613413 - 613 + 713 6 s Answer: a. b. 41671 3 7 c00c.qxd 8/7/12 4:31 PM Page 66 96. 0.6 Rational Exponents and Radicals 67 s 1 Rationalizing Denominators When radicals appear in a quotient, it is customary to write the quotient with no radicals in the denominator. This process is called rationalizing the denominator and involves multiplying by an expression that will eliminate the radical in the denominator. For example, the expression contains a single radical in the denominator. In a case like this, multiply the numerator and denominator by an appropriate radical expression, so that the resulting denominator will be radical free: If the denominator contains a sum of the form , multiply both the numerator and the denominator by the conjugate of the denominator, , which uses the difference of two squares to eliminate the radical term. Similarly, if the denominator contains a difference of the form , multiply both the numerator and the denominator by the conjugate of the denominator, . For example, to rationalize , take the conjugate of the denominator, which is : In general we apply the difference of two squares: Notice that the product does not contain a radical. Therefore, to simplify the expression multiply the numerator and denominator by : The denominator now contains no radicals: A1a - 1bB (a - b) 1 A1a + 1bB # A1a - 1bB A1a - 1bB A1a - 1bB 1 A1a + 1bB A1a + 1bB A1a - 1bB = A1aB 2 - A1bB 2 = a - b 1 A3 - 15B # A3 + 15B A3 + 15B = 3 + 15 32 + 315 - 315 - A15B 2 = 3 + 15 9 - 5 = 3 + 15 4 3 + 15 1 3 - 15 a + 1b a - 1b a - 1b a + 1b 1 13 # A13B A13B = 13 13 # 13 = 13 3 1 13 e like terms EXAMPLE 6 Rationalizing Denominators Rationalize the denominators and simplify. a. b. c. Solution (a): Multiply the numerator and denominator by . Simplify. Divide out the common 2 in the numerator and denominator. = 110 15 = 2110 3A110B2 = 2110 3(10) = 2110 30 110 = 2 3110 # 110 110 15 12 - 17 5 3 - 12 2 3110 c00c.qxd 8/7/12 4:31 PM Page 67 97. 68 C HAP TE R 0 Prerequisites and Review Solution (b): Multiply the numerator and denominator by the conjugate, . The denominator now contains no radicals. Simplify. Solution (c): Multiply the numerator and denominator by the conjugate, . Multiply the numerators and denominators, respectively. The denominator now contains no radicals. Simplify. YO U R TU R N Write the expression in simplied form. 7 1 - 13 = - 110 + 135 5 = 110 + 135 2 - 7 = 15A 12 + 17B A 12 - 17B A 12 + 17B 12 + 17 = 15 A 12 - 17B # A 12 + 17B A 12 + 17B = 15 + 512 7 = 15 + 512 9 - 2 = 5A3 + 12B A3 - 12B A3 + 12B 3 + 12 = 5 A3 - 12B # A3 + 12B A3 + 12B Answer: - 7A1 + 13B 2 A radical expression is in simplied form if No factor in the radicand is raised to a power greater than or equal to the index. The power of the radicand does not share a common factor with the index. The denominator does not contain a radical. The radical does not contain a fraction. SIMPLIFIED FORM OF A RADICAL EXPRESSION EXAMPLE 7 Expressing a Radical Expression in Simplied Form Express the radical expression in simplied form: Solution: Rewrite the expression so that the radical does not contain a fraction. Let Factors in both radicands are raised to powers greater than the index (3). Rewrite the expression so that each power in the radicand is less than the index. = 2x2 3 2x2 3y2 2 3 3y = 2 3 24 # x5 2 3 34 # y7 16 = 24 and 81 = 34 . B 3 16x5 81y7 = 2 3 16x5 2 3 81y7 A 3 16x5 81y7 x 0, y 7 0 c00c.qxd 8/7/12 4:31 PM Page 68 98. 0.6 Rational Exponents and Radicals 69 The denominator contains a radical. In order to eliminate the radical in the denominator, we multiply the numerator and denominator by . The radical expression now satises the conditions for simplied form. = 2x2 3 18x2 y2 9y3 = 2x2 3 18x2 y2 9y3 = 2x2 3 18x2 y2 3y2 2 3 27y3 2 3 9y2 = 2x2 3 2x2 3y2 2 3 3y # 3 29y2 3 29y2 Rational Exponents We now use radicals to dene rational exponents. Let a be any real number and n be a positive integer, then where is the rational exponent of a. When n is even and a is negative, then and are not real numbers. Furthermore, if m is a positive integer with m and n having no common factors, then Note: Any of the four notations can be used. am/n = (a1/n ) m = (am ) 1/n = 2 n am 1 n aa1/n 1 n a1/n = 1 n a RATIONAL EXPONENTS: 1 EXAMPLE 8 Simplifying Expressions with Rational Exponents Simplify: a. b. Solution: a. b. YO U R TU R N Simplify 272/3 . (-8)2/3 = [(-8)1/3 ]2 = (-2)2 = 4 163/2 = (161/2 ) 3 = A 116B 3 = 43 = 64 (-8)2/3 163/2 Answer: 9 The properties of exponents that hold for integers also hold for rational numbers: and a Z 0a-m/n = 1 am/n a-1/n = 1 a1/n Technology Tip n c00c.qxd 8/7/12 4:31 PM Page 69 99. 70 C HAP TE R 0 Prerequisites and Review Answer:18x2 EXAMPLE 10 Simplifying Algebraic Expressions with Rational Exponents Simplify Solution: Write in terms of positive exponents. YO U R TU R N Simplify and write your answer with only positive exponents. (16x3 y)1/2 (27x2 y3 )1/3 = - 2x1/6 3y5/3 (-8x2 y)1/3 (9xy4 )1/2 = (-8)1/3 (x2 )1/3 y1/3 91/2 x1/2 (y4 )1/2 = (-8) 1/3 x 2/3 y 1/3 91/2 x1/2 y2 = a -2 3 bx2/3-1/2 y1/3-2 = - 2 3 x1/6 y-5/3 x 7 0, y 7 0. (-8x2 y) 1/3 (9xy4 ) 1/2 Answer: 4x5/6 3y1/2 EXAMPLE 11 Factoring Expressions with Rational Exponents Factor completely . Solution: Factor out the greatest common factor . Factor the trinomial. YO U R TU R N Factor completely .x7/3 - x4/3 - 2x1/3 = x2/3 (x - 6)(x + 1) x2/3 x8/3 - 5x5/3 - 6x2/3 = x2/3 (x2 - 5x - 6) x8/3 - 5x5/3 - 6x2/3 EXAMPLE 9 Simplifying Expressions with Negative Rational Exponents Simplify Solution: Negative exponents correspond to positive exponents in the reciprocal. Eliminate the parentheses. Apply the quotient property on x. Simplify. YO U R TU R N Simplify x 7 0. 9x3/2 (4x)-1/2 = x 12 = x1 4 # 3 = x3/2-1/2 4 # 91/2 = x3/2 4 # 91/2 x1/2 (9x)-1/2 4x-3/2 = x3/2 4 # (9x)1/2 x 7 0. (9x)-1/2 4x-3/2 Answer: x1/3 (x - 2)(x + 1) s 3 s x2/3 x6/3 s x2/3 x3/3 c00c.qxd 8/7/12 4:31 PM Page 70 100. 0.6 Rational Exponents and Radicals 71 S U M MARY S E CTI O N 0.6 In this section, we dened radicals as means for a and b positive real numbers when n is a positive even integer, and a and b any real numbers when n is a positive odd integer. Properties of Radicals PROPERTY EXAMPLE 2 4 x6 = 2 4 x4 # 2 4 x2 = x 2 4 x2 1 n an = a n is even 2 3 x5 = 2 3 x3 # 2 3 x2 = x2 3 x21 n an = a n is odd 2 3 82 = A1 3 8B 2 = (2)2 = 41 n am = A1 n aB m A 4 81 16 = 1 4 81 1 4 16 = 3 2A n a b = 1 n a 1 n b b Z 0 1 3 16 = 1 3 8 # 1 3 2 = 21 3 21 n ab = 1 n a # 1 n b a = bn b = 1 n a Radicals can be combined only if they are like radicals (same radicand and index). Quotients with radicals in the denominator are usually rewritten with no radicals in the denominator. Rational exponents were dened in terms of radicals: The properties for integer exponents we learned in Section 0.2 also hold true for rational exponents: Negative rational exponents: , for m and n positive integers with no common factors, a 0.Z a-m/n = 1 am/n am/n = (a1/n ) m = A1 n aB m and am/n = (am )1/n = 1 n am 1 n a.a1/n = s x In Exercises 124, evaluate each expression or state that it is not a real number. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. In Exercises 2540, simplify (if possible) the radical expressions. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. In Exercises 4156, rationalize the denominators. 41. 42. 43. 44. 45. 46. 47. 48. 3 - 15 3 + 15 1 + 12 1 - 12 2 1 + 13 3 1 - 15 5 312 2 3111B 2 5B 1 3 2 5 -32x10 y8 2 3 -81x6 y8 216x3 y24x2 y 16236y4 8225x2 15121317 1 4 8 # 1 4 41 3 12 # 1 3 4215 # 3140112 # 12 617 + 717 - 1017315 - 215 + 715315 - 71512 - 512 (27)2/3 93/2 15/2 (-1)1/3 (-243)1/3 (-32)1/5 (-64)2/3 82/3 (-64)1/3 (-27)1/3 1 5 -11-16 1 5 01 3 01 7 -11 8 1- 1 3 -271 3 343 1 3 -1251 3 -2161-169- 114411211100 E X E R C I S E S S E CTI O N 0.6 S K I LL S c00c.qxd 8/7/12 4:31 PM Page 71 101. 72 C HAP TE R 0 Prerequisites and Review 49. 50. 51. 52. 53. 54. 55. 56. In Exercises 5764, simplify by applying the properties of rational exponents. Express your answers in terms of positive exponents. 57. 58. 59. 60. 61. 62. 63. 64. In Exercises 6568, factor each expression completely. 65. x7/3 x4/3 2x1/3 66. 8x1/4 4x5/4 67. 7x3/7 14x6/7 21x10/7 68. 7x1/3 70x (2x-2/3 ) 3 (4x-4/3 ) 2 (2x2/3 ) 3 (4x-1/3 ) 2 (y-3/4 x-2/3 ) 12 (y1/4 x7/3 ) 24 x1/2 y1/5 x-2/3 y-9/5 (x-2/3 y-3/4 ) -2 (x1/3 y1/4 ) 4 (x1/3 y1/2 ) -3 (x-1/2 y1/4 ) 2(y2/3 y1/4 ) 12 (x1/2 y2/3 ) 6 1y 1x - 1y 17 + 3 12 - 15 6 312 + 4 4 + 15 3 + 215 7 213 + 312 4 312 + 213 5 12 + 15 3 12 - 13 69. Gravity. If a penny is dropped off a building, the time it takes (seconds) to fall d feet is given by . If a penny is dropped off a 1280-foot-tall building, how long will it take until it hits the ground? Round to the nearest second. 70. Gravity. If a ball is dropped off a building, the time it takes (seconds) to fall d meters is approximately given by . If a ball is dropped off a 600-meter-tall building, how long will it take until it hits the ground? Round to the nearest second. B d 5 B d 16 71. Keplers Law. The square of the period p (in years) of a planets orbit around the Sun is equal to the cube of the planets maximum distance from the Sun, d (in astronomical units or AU). This relationship can be expressed mathematically as p2 d3 . If this formula is solved for d, the resulting equation is d p2/3 . If Saturn has an orbital period of 29.46 Earth years, calculate Saturns maximum distance from the Sun to the nearest hundredth of anAU. 72. Period of a Pendulum. The period (in seconds) of a pendulum of length L (in meters) is given by . If a certain pendulum has a length of 19.6 meters, determine the period P of this pendulum to the nearest tenth of a second. P = 2 # p # a L 9.8 b 1/2 In Exercises 73 and 74, explain the mistake that is made. 74. Simplify . Solution: Multiply numerator and denominator by . Multiply numerators and denominators. Simplify. This is incorrect. What mistake was made? 2A5 - 111B 14 = 5 - 111 7 2A5 - 111B 25 - 11 2 5 - 111 # A5 - 111B A5 - 111B5 - 111 2 5 - 111 73. Simplify Solution: Use properties of exponents. Simplify. 4xy1/2 This is incorrect. What mistake was made? 4(x1/2 ) 2 (y1/4 ) 2 (4x1/2 y1/4 ) 2 A P P L I C AT I O N S C AT C H T H E M I S TA K E c00c.qxd 8/7/12 4:31 PM Page 72 102. 0.7 Complex Numbers 73 In Exercises 7578, determine whether each statement is true or false. 75. 76. , where x is any real number. 77. 78. 1-4 = -2 2a2 + b2 = 2a + 2b 2x2 = x 1121 = ;11 In Exercises 79 and 80, a, m, n, and k are any positive real numbers. 79. Simplify . 80. Simplify . In Exercises 81 and 82, evaluate each algebraic expression for the specied values. 81. for a 1, b 7, c 12 82. for a 1, b 7, c 122b2 - 4ac 2b2 - 4ac 2a (a-k ) -1/k ((am ) n ) k 85. Use a calculator to approximate to three decimal places. 86. Use a calculator to approximate to three decimal places. 87. Given a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 4 512 + 413 1 3 7 111 88. Given a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 2 415 - 316 C O N C E P TUAL O BJ E CTIVE S Understand that real numbers and imaginary numbers are subsets of complex numbers. Understand how to eliminate imaginary numbers in denominators. S K I LLS O BJ E CTIVE S Write radicals with negative radicands as imaginary numbers. Add and subtract complex numbers. Multiply complex numbers. Divide complex numbers. Raise complex numbers to powers. C O M P LE X N U M B E R S S E CTI O N 0.7 The Imaginary Unit, i In Section 1.3, we will be studying equations whose solutions sometimes involve the square roots of negative numbers. In Section 0.6, when asked to evaluate the square root of a negative number, like , we said it is not a real number, because there is no real number such that . To include such roots in the number system, mathematicians created a new expanded set of numbers, called the complex numbers. The foundation of this new set of numbers is the imaginary unit i. x2 = -16 1-16 C O N C E P T U A L 83. Rationalize the denominator and simplify: . 1 A 1a + 1bB 2 84. Rationalize the denominator and simplify: . 1a + b - 1a 1a + b + 1a C H A L L E N G E T E C H N O L O G Y c00c.qxd 8/7/12 4:31 PM Page 73 103. 74 C HAP TE R 0 Prerequisites and Review The imaginary unit is denoted by the letter i and is dened as where .i2 = -1 i = 1-1 The Imaginary Unit iD E F I N I T I O N Recall that for positive real numbers a and b we dened the principal square root as which means Similarly, we dene the principal square root of a negative number as , since .Ai1aB 2 = i2 a = -a 1-a = i1a b2 = ab = 1a If a is a negative real number, then the principal square root of a is where i is the imaginary unit and i2 1. 1-a = i1a We write instead of to avoid any confusion as to what is included in the radical.1a ii1a EXAMPLE 1 Using Imaginary Numbers to Simplify Radicals Simplify using imaginary numbers. a. b. Solution: a. b. YO U R TU R N Simplify .1-144 1-8 = i18 = i # 212 = 2i121-9 = i19 = 3i 1-81-9 Technology Tip Be sure to put the graphing calculator in mode. a. b. 1-81-9 a + bi Answer: 12i A complex number in standard form is dened as where a and b are real numbers and i is the imaginary unit. We denote a as the real part of the complex number and b as the imaginary part of the complex number. a + bi Complex NumberD E F I N I T I O N A complex number written as a bi is said to be in standard form. If a 0 and , then the resulting complex number bi is called a pure imaginary number. If b 0, then a bi is a real number. The set of all real numbers and the set of all imaginary numbers are both subsets of the set of complex numbers. b Z 0 Study Tip = i 1a 1-a = 1-1 # 1a c00d.qxd 8/3/12 11:11 AM Page 74 104. 0.7 Complex Numbers 75 Complex Numbers a bi Real Numbers a (b 0) Imaginary Numbers bi (a 0) The following are examples of complex numbers. 17 -9i3 - i111-5 + i2 - 3i The complex numbers a bi and c di are equal if and only if a c and b d. In other words, two complex numbers are equal if and only if both real parts are equal and both imaginary parts are equal. Equality of Complex NumbersD E F I N I T I O N Adding and Subtracting Complex Numbers Complex numbers in the standard form a bi are treated in much the same way as binomials of the form a bx. We can add, subtract, and multiply complex numbers the same way we performed these operations on binomials. When adding or subtracting complex numbers, combine real parts with real parts and combine imaginary parts with imaginary parts. EXAMPLE 2 Adding and Subtracting Complex Numbers Perform the indicated operation and simplify. a. (3 2i) (1 i) b. (2 i) (3 4i) Solution (a): Eliminate the parentheses. (3 2i) (1 i) 3 2i 1 i Group real and imaginary numbers, respectively. (3 1) (2i i) Simplify. 2 i Solution (b): Eliminate the parentheses (distribute the negative). (2 i) (3 4i) 2 i 3 4i Group real and imaginary numbers, respectively. (2 3) (i 4i) Simplify. 1 3i YO U R TU R N Perform the indicated operation and simplify: (4 i) (3 5i). Technology Tip Be sure to put the graphing calculator in mode. a. b. (2 - i) - (3 - 4i) (3 - 2i) + (-1 + i) a + bi Answer: 1 6i c00d.qxd 8/3/12 11:11 AM Page 75 105. 76 C HAP TE R 0 Prerequisites and Review WORDS MATH Multiply the complex numbers. (5 i)(3 4i) Multiply using the distributive property. 5(3) 5(4i) i(3) (i)(4i) Eliminate the parentheses. 15 20i 3i 4i2 Let i2 1. 15 20i 3i 4(1) Simplify. 15 20i 3i 4 Combine real parts and imaginary parts, respectively. 11 23i EXAMPLE 3 Multiplying Complex Numbers Multiply the complex numbers and express the result in standard form, a bi. a. (3 i)(2 i) b. i(3 i) Solution (a): Use the distributive property. (3 i)(2 i) 3(2) 3(i) i(2) i(i) Eliminate the parentheses. 6 3i 2i i2 Substitute i2 1. 6 3i 2i (1) Group like terms. (6 1) (3i 2i) Simplify. 7 i Solution (b): Use the distributive property. i(3 i) 3i i2 Substitute i2 1. 3i 1 Write in standard form. 1 3i YO U R TU R N Multiply the complex numbers and express the result in standard form, a ; bi: .(4 - 3i)(-1 + 2i) Answer: 2 11i Technology Tip Be sure to put the graphing calculator in mode. a. b. i(-3 + i) (3 - i)(2 + i) a + bi Dividing Complex Numbers Recall the special product that produces a difference of two squares, (a b)(a b) a2 b2 . This special product has only rst and last terms because the outer and inner terms cancel each other out. Similarly, if we multiply complex numbers in the same manner, the result is a real number because the imaginary terms cancel each other out. The product of a complex number, z a bi, and its complex conjugate, , is a real number. zz = (a + bi)(a - bi) = a2 - b2 i2 = a2 - b2 (-1) = a2 + b2 z = a - bi COMPLEX CONJUGATE Multiplying Complex Numbers When multiplying complex numbers, you apply all the same methods as you did when multiplying binomials. It is important to remember that i2 1. Study Tip When multiplying complex numbers, remember that i2 1. c00d.qxd 8/3/12 11:11 AM Page 76 106. 0.7 Complex Numbers 77 In order to write a quotient of complex numbers in standard form, a bi, multiply the numerator and the denominator by the complex conjugate of the denominator. It is important to note that if i is present in the denominator, then the complex number is not in standard form. EXAMPLE 4 Dividing Complex Numbers Write the quotient in standard form: . Solution: Multiply numerator and denominator by the complex conjugate of the denominator, 1 3i. Multiply the numerators and denominators, respectively. Use the FOIL method (or distributive property). Combine imaginary parts. Substitute i2 1. Simplify the numerator and denominator. Write in standard form. Recall that . YO U R TU R N Write the quotient in standard form: . 3 + 2i 4 - i = - 1 10 - 7 10 i a + b c = a c + b c = -1 - 7i 10 = 2 - 7i - 3 1 - 9(-1) = 2 - 7i + 3i2 1 - 9i2 = 2 - 6i - i + 3i2 1 - 3i + 3i - 9i2 = (2 - i)(1 - 3i) (1 + 3i)(1 - 3i) a 2 - i 1 + 3i ba 1 - 3i 1 - 3i b 2 - i 1 + 3i Answer: 10 17 + 11 17 i Technology Tip Be sure to put the graphing calculator in mode. To change the answer to the fraction form, press , highlightMATH 2 - i 1 + 3i a + bi Raising Complex Numbers to Integer Powers Note that i raised to the fourth power is 1. In simplifying imaginary numbers, we factor out i raised to the largest multiple of 4. i8 = (i4 )2 = 1 i7 = i4 # i3 = (1)(-i) = -i i6 = i4 # i2 = (1)(-1) = -1 i5 = i4 # i = (1)(i) = i i4 = i2 # i2 = (-1)(-1) = 1 i3 = i2 # i = (-1)i = -i i2 = -1 i = 1-1 , press and .ENTER ENTER1: Frac c00d.qxd 8/3/12 11:11 AM Page 77 107. This is due to rounding off error in the programming. Since can be approximated as 0, .i7 = -i -3 * 10-13 EXAMPLE 5 Raising the Imaginary Unit to Integer Powers Simplify: a. i7 b. i13 c. i100 Solution: a. i7 i4 i3 (1)(i) i b. i13 i12 i (i4 )3 i 13 i i c. i100 (i4 )25 125 1 YO U R TU R N Simplify i27 . ### # Answer: i EXAMPLE 6 Raising a Complex Number to an Integer Power Write (2 i)3 in standard form. Solution: Recall the formula for cubing a binomial. (a b)3 a3 3a2 b 3ab2 b3 Let a 2 and b i. (2 i)3 23 3(2)2 (i) 3(2)(i)2 i3 Let i2 1 and i3 i. 23 3(2)2 (i) (3)(2)(1) (i) Eliminate parentheses. 8 6 12i i Combine the real parts and imaginary parts, respectively. 2 11i YO U R TU R N Write (2 i)3 in standard form. Answer: 2 11i Technology Tip Technology Tip Be sure to put the graphing calculator in mode. (2 - i)3 a + bi Multiplying Complex Numbers (a bi)(c di) (ac bd) (ad bc)i Apply the same methods as for multiplying binomials. It is important to remember that i2 1. Dividing Complex Numbers Complex conjugate of a bi is a bi. In order to write a quotient of complex numbers in standard form, multiply the numerator and the denominator by the complex conjugate of the denominator: a + bi c + di # (c - di) (c - di) S U M MARY The Imaginary Unit i Complex Numbers Standard Form: a bi, where a is the real part and b is the imaginary part. The set of real numbers and the set of pure imaginary numbers are subsets of the set of complex numbers. Adding and Subtracting Complex Numbers (a bi) (c di) (a c) (b d)i (a bi) (c di) (a c) (b d)i To add or subtract complex numbers, add or subtract the real parts and imaginary parts, respectively. i2 = -1 i = 1-1 S E CTI O N 0.7 78 C HAP TE R 0 Prerequisites and Review c00d.qxd 8/3/12 11:11 AM Page 78 108. 0.7 Complex Numbers 79 E X E R C I S E S S E CTI O N 0.7 In Exercises 112, write each expression as a complex number in standard form. Some expressions simplify to either a real number or a pure imaginary number. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. In Exercises 1340, perform the indicated operation, simplify, and express in standard form. 13. (3 7i) (1 2i) 14. (1 i) (9 3i) 15. (3 4i) (7 10i) 16. (5 7i) (10 2i) 17. (4 5i) (2 3i) 18. (2 i) (1 i) 19. (3 i) (2 i) 20. (4 7i) (5 3i) 21. 3(4 2i) 22. 4(7 6i) 23. 12(8 5i) 24. 3(16 4i) 25. 3(16 9i) 26. 5(6i 3) 27. 6(17 5i) 28. 12(8 3i) 29. (1 i)(3 2i) 30. (3 2i)(1 3i) 31. (5 7i)(3 4i) 32. (16 5i)(2 i) 33. (7 5i)(6 9i) 34. (3 2i)(7 4i) 35. (12 18i)(2 i) 36. (4 3i)(4 3i) 37. 38. 39. (i 17)(2 3i) 40. (3i 2)(2 3i) In Exercises 4148, for each complex number z, write the complex conjugate and nd . 41. z 4 7i 42. z 2 5i 43. z 2 3i 44. z 5 3i 45. z 6 4i 46. z 2 7i 47. z 2 6i 48. z 3 9i In Exercises 4964, write each quotient in standard form. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. In Exercises 6576, simplify. 65. i15 66. i99 67. i40 68. i18 69. (5 2i)2 70. (3 5i)2 71. (2 3i)2 72. (4 9i)2 73. (3 i)3 74. (2 i)3 75. (1 i)3 76. (4 3i)3 10 - i 12 + 5i 8 + 3i 9 - 2i 7 + 4i 9 - 3i 4 - 5i 7 + 2i 2 + i 3 - i 2 + 3i 3 - 5i 3 - i 3 + i 1 - i 1 + i 8 1 + 6i 2 7 + 2i 1 4 - 3i 1 3 + 2i 2 7 - i 1 3 - i 3 i 2 i zzz A-3 4 + 9 16 iB A2 3 + 4 9 iBA1 2 + 2iB A4 9 - 3iB 7 - 1 3 -125-10 - 1-1444 - 1-1213 - 1-100 1-271-641 3 -271 3 -64 1-241-201-1001-16 Electrical impedance is the ratio of voltage to current in ac circuits. Let Z represent the total impedance of an electrical circuit. If there are two resistors in a circuit, let Z1 3 6i ohms and Z2 5 4i ohms. 78. Electrical Circuits in Parallel. When the resistors in the circuit are placed in parallel, the total impedance is given by . Find the total impedance of the electrical circuit in parallel. 1 Z = 1 Z1 + 1 Z2 77. Electrical Circuits in Series. When the resistors in the circuit are placed in series, the total impedance is the sum of the two impedances . Find the total impedance of the electrical circuit in series. Z = Z1 + Z2 S K I LL S A P P L I C AT I O N c00d.qxd 8/3/12 11:11 AM Page 79 109. 80 C HAP TE R 0 Prerequisites and Review In Exercises 79 and 80, explain the mistake that is made. 79. Write the quotient in standard form: . Solution: Multiply the numerator and the denominator by 4 i. Multiply the numerator using the distributive property and the denominator using the FOIL method. Simplify. Write in standard form. This is incorrect. What mistake was made? 8 15 - 2 15 i 8 - 2i 15 8 - 2i 16 - 1 2 4 - i # 4 - i 4 - i 2 4 - i 80. Write the product in standard form: (2 3i)(5 4i). Solution: Use the FOIL method to multiply the complex numbers. 10 7i 12i2 Simplify. 2 7i This is incorrect. What mistake was made? In Exercises 8184, determine whether each statement is true or false. 83. Real numbers are a subset of the complex numbers. 84. There is no complex number that equals its conjugate. 81. The product is a real number: (a bi)(a bi). 82. Imaginary numbers are a subset of the complex numbers. 85. Factor completely over the complex numbers: x4 2x2 1. In Exercises 8790, use a graphing utility to simplify the expression. Write your answer in standard form. 87. 88. 89. 90. 1 (4 + 3i)2 1 (2 - i)3 (3 - i)6 (1 + 2i)5 86. Factor completely over the complex numbers: x4 18x2 81. C AT C H T H E M I S TA K E C O N C E P T U A L C HALLE N G E T E C H N O L O G Y c00d.qxd 8/3/12 11:11 AM Page 80 110. Understanding Rules through Patterns and Examples In this chapter, you reviewed many denitions, rules, and properties from your previous algebra classes. It can be a lot to remember, and sometimes its easy to misremember something. Observing patterns can help you see that rules are not arbitrary; rather they make sense in the context of other mathematics you already know. And by looking at examples, you can test whether youve correctly remembered a rule. You will explore both of these strategies here. 1. In this part, you will observe patterns to try to discover some of the rules for exponents given in this chapter. a. For instance, to discover a rule for zero as an exponent, rst complete the next four steps. a5 = a a a a a a4 = ___________ a3 = ___________ a2 = ___________ a1 = ___________ As the power decreases by one at each step, what pattern do you notice? Extend this pattern to nd the next step. Now complete this rule: Notice in the statement of the above rule, the base a is required to be nonzero. To see why, consider these two patterns: 40 = 1 04 = 0 30 = 1 03 = 0 20 = 1 02 = 0 10 = 1 01 = 0 According to this pattern, According to this pattern, 00 should be _____. 00 should be _____. Since these two patterns are not consistent, we say that 00 is undened. b. Look again at the pattern in part (a) that shows why a0 ought to be dened as 1, for nonzero a. Write the next several steps after a0 , following the same pattern. To be consistent with the pattern, how should we dene an ? Observing patterns can be helpful for making sense of rules, but an equally useful tool is looking at examples. Trying several examples can help you better understand given rules, or decide whether an equation is in fact a rule, as you will see in the next several parts. #### C HAP TE R 0 I N Q U I RY- BAS E D LE AR N I N G P R OJ E CT Let a be any nonzero whole number. Then a0 = Let a be any nonzero whole number. Then an = 81 c00d.qxd 8/3/12 11:11 AM Page 81 111. 2. A property of radicals in your text is stated as follows: when n is odd, and when n is even. Investigate several examples with various values of a and n, to try to discover the reasons for the two different rules for odd and even n. For instance, a positive value of a and an even value for n are given in the following chart. Based on your examples in this chart, explain why the property is when n is odd, but when n is even. 3. In this chapter you practiced using the distributive property of multiplication over addition: a(bc)abac. Suppose a fellow student wonders, Does it work the same way if there is a multiplication in the parentheses? Investigate whether the equation a(bc)abac is a property of whole-number multiplication, and then explain to this student what you did to decide. 4. Another student says, is a rule, because the bs cancel out. Do you agree with the student? Explain your answer by looking at some examples. a + b b = a n 1an = a n 1an = a n 1an = a n 1an = a a n an n 1an 3 2 82 c00d.qxd 8/3/12 11:11 AM Page 82 112. SECTION CONCEPT KEY IDEAS/FORMULAS 0.1 Real numbers The set of real numbers Rational: , where a and b are integers or a decimal that terminates or repeats. Irrational: Nonrepeating/nonterminating decimal. Approximations: Rounding Rounding: Examine the digit to the right of the and truncation last desired digit. Digit 5: Keep last desired digit as is. Digit 5: Round the last desired digit up 1. Truncating: Eliminate all digits to the right of the desired digit. Order of operations 1. Parentheses 2. Multiplication and Division 3. Addition and Subtraction Properties of real numbers a(b c) ab ac If xy 0, then x 0 or y 0 and , , and 0.2 Integer exponents and scientic notation Integer exponents Scientic notation c 10n where c is a positive real number 1 c 10 and n is an integer. 0.3 Polynomials: Basic operations Adding and subtracting polynomials Combine like terms. Multiplying polynomials Distributive property Special products (x a)(x b) x2 (a b)x ab Perfect Squares (a b)2 (a b)(a b) a2 2ab b2 (a b)2 (a b)(a b) a2 2ab b2 Difference of Two Squares (a b)(a b) a2 b2 Perfect Cubes (a b)3 a3 3a2 b 3ab2 b3 (a b)3 a3 3a2 b 3ab2 b3 (am )n = amn a-n = 1 an = a 1 a b n a Z 0 am an = am-n a0 = 1am # an = am+n an = a # a # a . . . a d Z 0c Z 0 a b , c d = a b # d c b Z 0 d Z 0 a b ; c d = ad ; bc bd b Z 0 6 a b CHAPTERREVIEW C HAP TE R 0 R EVI EW d n factors 83 c00d.qxd 8/3/12 11:11 AM Page 83 113. 0.4 Factoring polynomials Greatest common factor Factor out using distributive property: axk Factoring formulas: Special Difference of Two Squares polynomial forms a2 b2 (a b)(a b) Perfect Squares a2 2ab b2 (a b)2 a2 2ab b2 (a b)2 Sum of Two Cubes a3 b3 (a b)(a2 ab b2 ) Difference of Two Cubes a3 b3 (a b)(a2 ab b2 ) Factoring a trinomial as x2 bx c (x ?)(x ?) a product of two binomials ax2 bx c (?x ?)(?x ?) Factoring by grouping Group terms with common factors. A strategy for factoring polynomials 1. Factor out any common factors. 2. Recognize any special products. 3. Use the foil method in reverse for trinomials. 4. Look for factoring by grouping. 0.5 Rational expressions Rational expressions and Note domain restrictions when denominator is domain restrictions equal to zero. Simplifying rational expressions Multiplying and dividing rational Use properties of rational numbers. expressions State additional domain restrictions once division is rewritten as multiplication of a reciprocal. Adding and subtracting Least common denominator (LCD) rational expressions Complex rational Two strategies: expressions 1. Write sum/difference in numerator/ denominator as a rational expression. 2. Multiply by the LCD of the numerator and denominator. 0.6 Rational exponents and radicals Square roots Other (nth) roots means a bn for a and b positive real numbers and n a positive even integer, or for a and b any real numbers and n a positive odd integer. 1 n an = a n is even 1 n an = a n is odd 1 n am = A1 n aB m 1 n ab = 1 n a # 1 n b A n a b = 1 n a 1 n b b Z 0 b = 1 n a 125 = 5 SECTION CONCEPT KEY IDEAS/FORMULAS CHAPTERREVIEW 84 c00d.qxd 8/3/12 11:11 AM Page 84 114. Rational exponents for m and n positive integers with no common factors, . 0.7 Complex numbers The Imaginary Unit, i Adding and subtracting Complex Numbers: where a and b are real numbers. complex numbers Combine real parts with real parts and imaginary parts with imaginary parts. Multiplying complex numbers Use the FOIL method and i2 1 to simplify. Dividing complex numbers If a bi is in the denominator, then multiply the numerator and the denominator by a bi. The result is a real number in the denominator. Raising complex numbers to integer powers i4 = 1i3 = -ii2 = -1i = 1-1 a + bi i = 1-1 a Z 0 a-mn = 1 amn am/n = (a1/n ) m = A1 n aB m a1/n = 1 n a SECTION CONCEPT KEY IDEAS/FORMULAS CHAPTERREVIEW 85 c00d.qxd 8/3/12 11:11 AM Page 85 115. 0.1 Real Numbers Approximate to two decimal places by (a) rounding and (b) truncating. 1. 5.21597 2. 7.3623 Simplify. 3. 7 2 5 4 3 5 4. 2(5 3) 7(3 2 5) 5. 6. 3(x y) 4(3x 2y) Perform the indicated operation and simplify. 7. 8. 9. 10. 0.2 Integer Exponents and Scientic Notation Simplify using properties of exponents. 11. (2z)3 12. 13. 14. 15. Express 0.00000215 in scientic notation. 16. Express 7.2 109 as a real number. 0.3 Polynomials: Basic Operations Perform the indicated operation and write the results in standard form. 17. (3z 4) 18. 19. 20. 21. 5xy2 (3x 4y) 22. 2st2 (t s 2st) 23. (x 7)(x 9) 24. (2x 1)(3x 2) 25. (2x 3)2 26. (5x 7)(5x 7) 27. (x2 1) 2 28. (1 x2 ) 2 [2x - (4x2 - 7x)] - [3x - (2x2 + 5x - 4)] (6x - 9x2 + 10)(36x2 - 4x - 5) (y2 + 3y - 7)(27y2 - 6y + 2) (14z2 + 2) (2x2 y3 ) 2 (4xy) 3 (3x3 y2 ) 2 2(x2 y) 4 (-4z2 ) 3 a2 b3 , 2a b2 12 7 # 21 4 y 3 + y 5 - y 6 x 4 - x 3 - 16 (-2)(-4) # ## 0.4 Factoring Polynomials Factor out the common factor. 29. 14x2 y2 10xy3 30. 30x4 20x3 10x2 Factor the trinomial into a product of two binomials. 31. 2x2 9x 5 32. 6x2 19x 7 33. 16x2 25 34. 9x2 30x 25 Factor the sum or difference of two cubes. 35. x3 125 36. 1 8x3 Factor into a product of three polynomials. 37. 2x3 4x2 30x 38. 6x3 5x2 x Factor into a product of two binomials by grouping. 39. x3 x2 2x 2 40. 2x3 x2 6x 3 0.5 Rational Expressions State the domain restrictions on each of the rational expressions. 41. 42. Simplify. 43. 44. 45. 46. Perform the indicated operation and simplify. 47. 48. 49. 50. 1 x - 1 x + 1 + 1 x + 2 1 x + 1 - 1 x + 3 x2 - x - 2 x3 + 3x2 , x + 1 x2 + 2x x2 + 3x - 10 x2 + 2x - 3 # x2 + x - 2 x2 + x - 6 z3 - z z2 + z t2 + t - 6 t2 - t - 2 x - 5 x - 5 x2 - 4 x - 2 1 x2 + 1 4x2 - 3 x2 - 9 C HAP TE R 0 R EVI EW E X E R C I S E S REVIEWEXERCISES 86 c00d.qxd 8/3/12 11:11 AM Page 86 116. Review Exercises 87 Simplify. 51. 52. 0.6 Rational Exponents and Radicals Simplify. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 0.7 Complex Numbers Simplify. 67. 68. 69. i19 70. i9 Perform the indicated operation, simplify, and express in standard form. 71. (3 2i) (5 4i) 72. (4 7i) (2 3i) 73. (12 i) (2 5i) 74. (9 8i) (4 2i) 75. (2 2i)(3 3i) 76. (1 6i)(1 5i) 77. (4 7i)2 78. (7 i)2 Express the quotient in standard form. 79. 80. 81. 7 + 2i 4 + 5i 1 3 + i 1 2 - i 1-321-169 (x-23 y14 ) 12512 513 (4x34 ) 2 (2x-13 ) 2 (3x23 ) 2 (4x13 ) 2 1 3 - 1x 1 2 - 13 A3 + 1xB A4 - 1xBA2 + 15B A1 - 15B 4127x - 8112x3120 + 5180 24 32x4 y5 2 3 -125x5 y4 180120 1 x + 2 x2 3 - 1 x2 2 + 1 x - 3 1 5x - 15 + 4 82. 83. 84. Technology Section 0.1 85. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? 86. Use your calculator to evaluate . Does the answer appear to be a rational or an irrational number? Why? Section 0.2 Use a graphing utility to evaluate the expression. Express your answer in scientic notation. 87. 88. Section 0.3 89. Use a graphing utility to plot the graphs of the three expressions and Which two graphs agree with each other? 90. Use a graphing utility to plot the graphs of the three expressions and Which two graphs agree with each other? Section 0.4 91. Use a graphing utility to plot the graphs of the three expressions and Which two graphs agree with each other? 92. Use a graphing utility to plot the graphs of the three expressions and Which two graphs agree with each other? Section 0.5 For each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other. 93. 94. 1 - (3/x) 1 + (9/x2 ) 1 - (4/x) 1 - (4/x2 ) (x - 4)2 .x2 - 8x + 16, (x + 4)2 , (x - 6)(x + 3).x2 - 3x + 18, (x + 6)(x - 3), x2 - 6x + 9.(x - 3)2 , 8x2 + 9, 8x3 + 36x2 + 54x + 27.(2x + 3)3 , 8x3 + 27, (1.4805 * 1021 ) (5.64 * 1026 )(1.68 * 10-9 ) (8.2 * 1011 )(1.167 * 10-35 ) (4.92 * 10-18 ) A 1053 81 2272.25 7 2i 10 3i 6 - 5i 3 - 2i REVIEWEXERCISES c00d.qxd 8/3/12 11:11 AM Page 87 117. Section 0.6 95. Given a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 96. Given a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 11 216 + 113 6 15 - 12 Section 0.7 In Exercises 97 and 98, use a graphing utility to simplify the expression. Write your answer in standard form. 97. (3 5i)5 98. 99. Apply a graphing utility to simplify the expression and write your answer in standard form. 1 (6 + 2i)4 1 (1 + 3i)4 REVIEWEXERCISES 88 C HAP TE R 0 Prerequisites and Review c00d.qxd 8/3/12 11:11 AM Page 88 118. Perform the indicated operations and simplify. 25. 26. 27. 28. 29. 30. Write the resulting expression in standard form. 31. (1 3i)(7 5i) 32. 33. Rationalize the denominator: . 34. Represent 0.0000155 in scientic notation. 35. Simplify and state any domain restrictions. 36. For the given expression: a. Simplify the expression. b. Use a graphing utility to plot the expression and the answer in (a) in the same viewing window. c. Determine the domain restriction(s) where the graphs will agree with each other. 37. Apply a graphing utility to evaluate the expression. Round your answer to three decimal places. 15 113 - 17 1 + 5 x 1 - 25 x2 1 x - 2 x + 1 x - 1 7 - 213 4 - 513 2 - 11i 4 + i 1 - t 3t + 1 , t2 - 2t + 1 7t + 21t2 x - 3 2x - 5 , x2 - 9 5 - 2x 4x2 - 9 x2 - 11x - 60 # x2 - 16 2x + 3 x - 1 x2 - 1 # x2 + x + 1 x3 - 1 5x x2 - 7x + 10 - 4 x2 - 25 2 x + 3 x - 1 Simplify. 1. 2. 3. 4. 5. 6. i17 7. 8. 9. 10. Perform the indicated operation and simplify. 11. 12. (2x 3)(5x 7) Factor. 13. x2 16 14. 3x2 15x 18 15. 4x2 12xy 9y2 16. x4 2x2 1 17. 2x2 x 1 18. 6y2 y 1 19. 2t3 t2 3t 20. 2x3 5x2 3x 21. x2 3yx 4yx 12y2 22. x4 5x2 3x2 15 23. 81 3x3 24. 27x x4 (3y2 - 5y + 7) - (y2 + 7y - 13) A516 - 212B A 16 + 312B 3118 - 4132 31x - 41x + 51x (x2 y-3 z-1 ) -2 (x-1 y2 z3 ) 12 2-12x2 25 -32 -3(2 + 52 ) + 2(3 - 7) - (32 - 1) 2 3 54x6 116 C HAP TE R 0 P R ACTI C E TE ST PRACTICETEST 89 c00d.qxd 8/3/12 11:11 AM Page 89 119. Equations and Inequalities G olf courses usually charge both greens fees (cost of playing the course) and cart fees (cost of renting a golf cart). Two friends who enjoy playing golf decide to investigate becoming members at a golf course. The course they enjoy playing the most charges $40 for greens fees and $15 for cart rental (per person), so it currently costs each of them $55 every time they play. The membership offered at that course costs $160 per month with no greens fees, but there is still the per person cart rental fee. How many times a month would they have to play golf in order for the membership option to be the better deal?* This is just one example of how the real world can be modeled with equations and inequalities. RadiusImages/Corbis *See Section 1.5, Exercises 109 and 110. 1 c01a.qxd 11/23/11 5:33 PM Page 90 120. I N TH I S C HAP TE R you will solve linear and quadratic equations. You will then solve more complicated equations (polynomial, rational, radical, and absolute value) by first transforming them into linear or quadratic equations. Then you will solve linear, quadratic, polynomial, rational, and absolute value inequalities. Throughout this chapter you will solve applications of equations and inequalities. Solving Linear Equations in One Variable Solving Rational Equations That Are Reducible to Linear Equations Solving Application Problems Using Mathemati- cal Models Geometry Problems Interest Problems Mixture Problems Distance RateTime Problems Factoring Square Root Method Completing the Square Quadratic Formula Radical Equations Equations Quadratic in Form: u-Substitu- tion Factorable Equations Graphing Inequalities and Interval Notation Solving Linear Inequalities Polynomial Inequalities Rational Inequalities Equations Involving Absolute Value Inequalities Involving Absolute Value EQUATIONS AND INEQUALITIES 1.1 Linear Equations 1.2 Applications Involving Linear Equations 1.3 Quadratic Equations 1.4 Other Types of Equations 1.5 Linear Inequalities 1.6 Polynomial and Rational Inequalities 1.7 Absolute Value Equations and Inequalities Solve linear equations. Solve application problems involving linear equations. Solve quadratic equations. Solve rational, polynomial, and radical equations. Solve linear inequalities. Solve polynomial and rational inequalities. Solve absolute value equations and inequalities. L E A R N I N G O B J E C T I V E S 91 c01a.qxd 11/23/11 5:33 PM Page 91 121. Solving Linear Equations in One Variable An algebraic expression (see Chapter 0) consists of one or more terms that are combined through basic operations such as addition, subtraction, multiplication, or division; for example: An equation is a statement that says two expressions are equal. For example, the following are all equations in one variable, x: To solve an equation means to nd all the values of x that make the equation true. These values are called solutions, or roots, of the equation. The rst of these statements shown above, x 7 11, is true when x 4 and false for any other values of x. We say that x 4 is the solution to the equation. Sometimes an equation can have more than one solution, as in x2 9. In this case, there are actually two values of x that make this equation true, x 3 and x 3. We say the solution set of this equation is {3, 3}. In the third equation, 7 3x 2 3x, no values of x make the statement true. Therefore, we say this equation has no solution. And the fourth equation, 4x 7 x 2 3x 5, is true for any values of x. An equation that is true for any value of the variable x is called an identity. In this case, we say the solution set is the set of all real numbers. Two equations that have the same solution set are called equivalent equations. For example, are all equivalent equations because each of them has the solution set {2}. Note that x2 4 is not equivalent to these three equations because it has the solution set {2, 2}. When solving equations it helps to nd a simpler equivalent equation in which the variable is isolated (alone). The following table summarizes the procedures for generating equivalent equations. x = 23x = 63x + 7 = 13 4x + 7 = x + 2 + 3x + 57 - 3x = 2 - 3xx2 = 9x + 7 = 11 x + y5 - 2y3x + 2 C O N C E P TUAL O BJ E CTIVE Eliminate values that result in a denominator being equal to zero. LI N EAR E Q UATI O N S S K I LLS O BJ E CTIVE S Solve linear equations in one variable. Solve rational equations that are reducible to linear equations. S E CTI O N Generating Equivalent Equations ORIGINAL EQUATION DESCRIPTION EQUIVALENT EQUATION 3(x 6) 6x x Eliminate parentheses. 3x 18 5x Combine like terms on one or both sides of an equation. 7x 8 29 Add (or subtract) the same quantity to 7x 21 (from) both sides of an equation. 7x 8 8 29 8 5x 15 Multiply (or divide) both sides of an equation x 3 by the same nonzero quantity: . 7 x Interchange the two sides of the equation. x 7 5x 5 = 15 5 1.1 92 c01a.qxd 11/23/11 5:33 PM Page 92 122. 1.1 Linear Equations 93 You probably already know how to solve simple linear equations. Solving a linear equation in one variable is done by nding an equivalent equation. In generating an equivalent equation, remember that whatever operation is performed on one side of an equation must also be performed on the other side of the equation. Technology Tip Use a graphing utility to display graphs of y1 3x 4 and y2 16. The x-coordinate of the point of intersection is the solution to the equation 3x 4 16. Example 1 illustrates solving linear equations in one variable. What is a linear equation in one variable? A linear equation in one variable, x, can be written in the form ax b 0 where a and b are real numbers and .a Z 0 Linear EquationD E F I N I T I O N Answer: The solution is x 3. The solution set is {3}. What makes this equation linear is that x is raised to the rst power. We can also classify a linear equation as a rst-degree equation. Equation Degree General Name x 7 0 First Linear x2 6x 9 0 Second Quadratic x3 3x2 8 0 Third Cubic E X AM P LE 1 Solving a Linear Equation Solve the equation 3x 4 16. Solution: Subtract 4 from both sides of the equation. Divide both sides by 3. The solution is x 4. x 4 The solution set is {4}. YO U R TU R N Solve the equation 2x 3 9. 3x 3 = 12 3 3x + 4 = 16 -4 -4 3x = 12 c01a.qxd 12/22/11 6:46 PM Page 93 123. 94 C HAP TE R 1 Equations and Inequalities Technology Tip Use a graphing utility to display graphs of and y2 = 3 4 p.y1 = 1 2 p - 5 Answer: The solution is m 18. The solution set is {18}. Study Tip Prime Factors 2 2 6 2 3 5 5 LCD 2 3 5 30 Answer: The solution is x 2. The solution set is {2}. To solve a linear equation involving fractions, nd the least common denominator (LCD) of all terms and multiply both sides of the equation by the LCD. We will rst review how to nd the LCD. To add the fractions , we must rst nd a common denominator. Some people are taught to nd the lowest number that 2, 6, and 5 all divide evenly into. Others prefer a more systematic approach in terms of prime factors. 1 2 + 1 6 + 2 5 The x-coordinate of the point of intersection is the solution. EXAMPLE 2 Solving a Linear Equation Solve the equation 5x (7x 4) 2 5 (3x 2). Solution: Eliminate the parentheses. 5x (7x 4) 2 5 (3x 2) Dont forget to distribute the negative sign through both terms inside the parentheses. 5x 7x 4 2 5 3x 2 Combine x terms on the left, constants on the right. Add 3x to both sides. Subtract 2 from both sides. Check to verify that x 1 is a 5 1 (7 1 4) 2 5 (3 1 2) solution to the original equation. 5 (7 4) 2 5 (3 2) 5 (3) 2 5 (5) 0 0 Since the solution x 1 makes the equation true, the solution set is {1}. YO U R TU R N Solve the equation 4(x 1) 2 x 3(x 2). ### - 2 - 2 x = 1 -2x + 2 = 3 - 3x +3x + 3x x + 2 = 3 Technology Tip Use a graphing utility to display graphs of y1 5x (7x 4) 2 and y2 5 (3x 2). The x-coordinate of the point of intersection is the solution to the equation 5x (7x 4) 2 5 (3x 2). E X AM P LE 3 Solving a Linear Equation Involving Fractions Solve the equation . Solution: Write the equation. Multiply each term in the equation by the LCD, 4. The result is a linear equation with no fractions. Subtract 2p from both sides. p 20 Since p 20 satises the original equation, the solution set is {20}. YO U R TU R N Solve the equation .1 4 m = 1 12 m - 3 -2p -2p -20 = p 2p - 20 = 3p (4) 1 2 p - (4)5 = (4) 3 4 p 1 2 p - 5 = 3 4 p 1 2 p - 5 = 3 4 p c01a.qxd 11/23/11 5:33 PM Page 94 124. 1.1 Linear Equations 95 Solving Rational Equations That Are Reducible to Linear Equations Arational equation is an equation that contains one or more rational expressions (Chapter 0). Some rational equations can be transformed into linear equations that you can then solve, but as you will see momentarily, you must be certain that the solution to the linear equation also satises the original rational equation. Technology Tip Use a graphing utility to display graphs of and .y2 = 4 x + 4 3 y1 = 2 3x + 1 2 E X AM P LE 4 Solving a Rational Equation That Can Be Reduced to a Linear Equation Solve the equation Solution: State the excluded values (those that make any denominator equal 0). Eliminate fractions by multiplying each term by the LCD, 6x. Simplify both sides. 4 3x 24 8x Subtract 4. 4 4 Subtract 8x. Divide by 5. x 4 Since x 4 satises the original equation, the solution set is {4}. 6xa 2 3x b + 6xa 1 2 b = 6xa 4 x b + 6xa 4 3 b 2 3x + 1 2 = 4 x + 4 3 x Z 0 2 3x + 1 2 = 4 x + 4 3 . STEP DESCRIPTION EXAMPLE 1 Simplify the algebraic expressions on both sides of the equation. 2 Gather all variable terms on one side of the equation and all constant terms on the other side. 3 Isolate the variable. x = 4 10x = 40 40 = 10x +29 +29 11 = 10x - 29 +3x +3x -3x + 11 = 7x - 29 -3x + 11 = 7x - 29 -3x + 6 + 5 = 7x - 28 - 1 -3(x - 2) + 5 = 7(x - 4) - 1 Solving a Linear Equation in One Variable YO U R TU R N Solve the equation 3 y + 2 = 7 2y . Study Tip Since dividing by 0 is not dened, we exclude values of the variable that correspond to a denominator equaling 0. Answer: The solution is . The solution set is E1 4 F. y = 1 4 The x-coordinate of the point of intersection is the solution to the equation 2 3x + 1 2 = 4 x + 4 3 . 3x 20 8x 8x 8x 5x 20 c01a.qxd 11/23/11 5:33 PM Page 95 125. 96 C HAP TE R 1 Equations and Inequalities Answer: no solution Extraneous solutions are solutions that satisfy a transformed equation but do not satisfy the original equation. It is important to rst state any values of the variable that must be eliminated based on the original rational equation. Once the rational equation is transformed to a linear equation and solved, remove any excluded values of the variable. Study Tip When a variable is in the denominator of a fraction, the LCD will contain the variable. This sometimes results in an extraneous solution. We have reviewed nding the least common denominator (LCD) for real numbers. Now let us consider nding the LCD for rational equations that have different denominators. We multiply the denominators in order to get a common denominator. Rational expression: LCD: x(x 1) In order to nd a least common denominator, it is useful to rst factor the denominators to identify common multiples. Rational equation: Factor the denominators: LCD: 6x(x 1) 1 3(x - 1) + 1 2(x - 1) = 1 x(x - 1) 1 3x - 3 + 1 2x - 2 = 1 x2 - x 1 x + 2 x - 1 Technology Tip Use a graphing utility to display graphs of and .y2 = 3 x - 1 y1 = 3x x - 1 + 2 The x-coordinate of the point of intersection is the solution to the equation 3x x - 1 + 2 = 3 x - 1 . E X AM P LE 5 Solving Rational Equations That Can Be Reduced to Linear Equations Solve the equation Solution: State the excluded values (those that make any denominator equal 0). Eliminate the fractions by multiplying each term by the LCD, x 1. Simplify. 3x 2(x 1) 3 Distribute the 2. 3x 2x 2 3 Combine x terms on the left. 5x 2 3 Add 2 to both sides. 5x 5 Divide both sides by 5. x 1 It may seem that x 1 is the solution. However, the original equation had the restriction . Therefore, x 1 is an extraneous solution and must be eliminated as a possible solution. Thus, the equation has no solution . YO U R TU R N Solve the equation . 2x x - 2 - 3 = 4 x - 2 3x x - 1 + 2 = 3 x - 1 x Z 1 3x x - 1 # (x - 1) + 2 # (x - 1) = 3 x - 1 # (x - 1) 3x x - 1 # (x - 1) + 2 # (x - 1) = 3 x - 1 # (x - 1) 3x x - 1 + 2 = 3 x - 1 x Z 1 3x x - 1 + 2 = 3 x - 1 . No intersection implies no solution. c01a.qxd 11/23/11 5:33 PM Page 96 126. 1.1 Linear Equations 97 E X AM P LE 6 Solving Rational Equations Solve the equation . Solution: Factor the denominators. State the excluded values. Multiply the equation by the LCD, 6x(x 6). Divide out the common factors. Simplify. 2x 3x 6 Solve the linear equation. x 6 Since one of the excluded values is , we say that x 6 is an extraneous solution. Therefore, this rational equation has no solution . x Z -6 6x(x + 6) # 1 3(x + 6) - 6x(x + 6) # 1 2(x + 6) = 6x(x + 6) # 1 x(x + 6) 6x(x + 6) # 1 3(x + 6) - 6x(x + 6) # 1 2(x + 6) = 6x(x + 6) # 1 x(x + 6) x Z 0, -6 1 3(x + 6) - 1 2(x + 6) = 1 x(x + 6) 1 3x + 18 - 1 2x + 12 = 1 x2 + 6x YO U R TU R N Solve the equation 2 x + 1 x + 1 = - 1 x (x + 1) . E X AM P LE 7 Solving Rational Equations Solve the equation Solution: What values make either denominator equal to zero? The values x 2 and x 3 must be excluded from possible solutions to the equation. Multiply the equation by the LCD, (x 3)(2 x). Divide out the common factors. Eliminate the parentheses. Collect x terms on the left, constants on the right. Since x 5 satises the original equation, the solution set is {5}. YO U R TU R N Solve the equation . -4 x + 8 = 3 x - 6 x = 5 4 - 2x = -3x + 9 2(2 - x) = -3(x - 3) 2 x - 3 (x - 3)(2 - x) = -3 2 - x (x - 3)(2 - x) 2 x - 3 = -3 2 - x x Z 2, x Z 3 2 x - 3 = -3 2 - x . Answer: no solution Answer: The solution is x 0. The solution set is {0}. Technology Tip Use a graphing utility to display graphs of and .y2 = -3 2 - x y1 = 2 x - 3 The x-coordinate of the point of intersection is the solution to the equation . 2 x - 3 = -3 2 - x c01a.qxd 11/23/11 5:34 PM Page 97 127. 98 C HAP TE R 1 Equations and Inequalities E X AM P LE 8 Automotive Service A car dealership charges for parts and an hourly rate for labor. If parts cost $273, labor is $53 per hour, and the total bill is $458.50, how many hours did the dealership spend working on your car? Solution: Let x equal the number of hours the dealership worked on your car. Write the cost equation. Subtract 273 from both sides of the equation. 53x 185.50 Divide both sides of the equation by 53. x 3.5 The dealership charged for 3.5 hours of labor. 53x + 273 = 458.50 E X AM P LE 9 Grades Dante currently has the following three test scores: 82, 79, and 90. If the score on the nal exam is worth two test scores and his goal is to earn an 85 for his class average, what score on the nal exam does Dante need to achieve his course goal? Solution: Let x equal nal exam grade. Write the equation that determines the course grade. Simplify the numerator. Multiply the equation by 5 (or cross multiply). 251 2x 425 Solve the linear equation. x 87 Dante needs to score at least an 87 on the nal exam. 251 + 2x 5 = 85 82 + 79 + 90 + 2x 5 = 85 labor parts total cost total of ve test scores average 2## SMH S U M MARY S E CTI O N 1.1 Rational equations are solved by: 1. Determining any excluded values (denominator equals 0). 2. Multiplying the equation by the LCD. 3. Solving the resulting equation. 4. Eliminating any extraneous solutions. Linear equations, are solved by: 1. Simplifying the algebraic expressions on both sides of the equation. 2. Gathering all variable terms on one side of the equation and all constant terms on the other side. 3. Isolating the variable. ax + b = 0, scores 1, 2, and 3 nal is worth two test scores # # # c01a.qxd 12/22/11 6:06 PM Page 98 128. 1.1 Linear Equations 99 In Exercises 136, solve for the indicated variable. 1. 5x 35 2. 4t 32 3. 3 n 12 4. 4 5 y 5. 24 3x 6. 50 5t 7. 8. 9. 3x 5 7 10. 4p 5 9 11. 9m 7 11 12. 2x 4 5 13. 5t 11 18 14. 7x 4 21 24x 15. 3x 5 25 6x 16. 5x 10 25 2x 17. 20n 30 20 5n 18. 14c 15 43 7c 19. 4(x 3) 2(x 6) 20. 5(2y 1) 2(4y 3) 21. 3(4t 5) 5(6 2t) 22. 2(3n 4) (n 2) 23. 2(x 1) 3 x 3(x 1) 24. 4(y 6) 8 2y 4(y 2) 25. 5p 6(p 7) 3(p 2) 26. 3(z 5) 5 4z 7(z 2) 27. 28. 29. 30. 31. 2a 9(a 6) 6(a 3) 4a 32. 25 [2 5y 3(y 2)] 3(2y 5) [5(y 1) 3y 3] 33. 32 [4 6x 5(x 4)] 4(3x 4) [6(3x 4) 7 4x] 34. 12 [3 4m 6(3m 2)] 7(2m 8) 3[(m 2) 3m 5] 35. 20 4[c 3 6(2c 3)] 5(3c 2) [2(7c 8) 4c 7] 36. 46 [7 8y 9(6y 2)] 7(4y 7) 2[6(2y 3) 4 6y] Exercises 3748 involve fractions. Clear the fractions by rst multiplying by the least common denominator, and then solve the resulting linear equation. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. In Exercises 4970, specify any values that must be excluded from the solution set and then solve the equation. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 2 - x x - 2 = 3 4 t - 1 1 - t = 3 2 3 4n - 1 = 2 2n - 5 2 5x + 1 = 1 2x - 1 5 m + 3 m - 2 = 6 m(m - 2) n - 5 6n - 6 = 1 9 - n - 3 4n - 4 1 c - 2 + 1 c = 2 c(c - 2) 3 a - 2 a + 3 = 9 a(a + 3) 1 x + 1 x - 1 = 1 x(x - 1) 1 n + 1 n + 1 = -1 n(n + 1) 5y 2y - 1 - 3 = 12 2y - 1 3x x + 2 - 4 = 2 x + 2 4t t + 2 = 3 - 8 t + 2 2p p - 1 = 3 + 2 p - 1 n n - 5 + 2 = n n - 5 x x - 2 + 5 = 2 x - 2 4 x - 2 = 5 2x 2 a - 4 = 4 3a 7 6t = 2 + 5 3t 7 - 1 6x = 10 3x 4 x + 10 = 2 3x 4 y - 5 = 5 2y 1 - x - 5 3 = x + 2 5 - 6x - 1 15 x - 3 3 - x - 4 2 = 1 - x - 6 6 c 4 - 2c = 5 4 - c 2 p + p 4 = 5 2 2m - 5m 8 = 3m 72 + 4 3 5y 3 - 2y = 2y 84 + 5 7 3x 5 - x = x 10 - 5 2 1 3 p = 3 - 1 24 p a 11 = a 22 + 9 x 7 = 2x 63 + 4 1 12 z = 1 24 z + 3 1 5 m = 1 60 m + 1 5 - (2x - 3) = 7 - (3x + 5)2 - (4x + 1) = 3 - (2x - 1) 3x - (4x + 2) = x - 57x - (2x + 3) = x - 2 6 = 1 3 p1 5 n = 3 S K I LL S E X E R C I S E S S E CTI O N 1.1 c01a.qxd 11/23/11 5:34 PM Page 99 129. 100 C HAP TE R 1 Equations and Inequalities In Exercises 79 and 80 refer to the following: Medications are often packaged in liquid form (known as a suspension) so that a precise dose of a drug is delivered within a volume of inert liquid; for example, 250 milligrams amoxicillin in 5 milliliters of a liquid suspension. If a patient is prescribed a dose of a drug, medical personnel must compute the volume of the liquid with a known concentration to administer. The formula denes the relationship between the dose of the drug prescribed d, the concentration of the liquid suspension c, and the amount of the liquid administered a. 79. Medicine. A physician has ordered a 600-milligram dose of amoxicillin. The pharmacy has a suspension of amoxicillin with a concentration of 125 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 80. Medicine. A physician has ordered a 600-milligram dose of carbamazepine. The pharmacy has a suspension of carbamazepine with a concentration of 100 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 81. Speed of Light. The frequency f of an optical signal in hertz (Hz) is related to the wavelength in meters (m) of a laser through the equation where c is the speed of light in a vacuum and is typically taken to be c 3.0 108 meters per second (m/s). What values must be eliminated from the wavelengths? 82. Optics. For an object placed near a lens, an image forms on the other side of the lens at a distinct position determined by the distance from the lens to the object. The position of the image is found using the thin lens equation: where do is the distance from the object to the lens, di is the distance from the lens to the image, and f is the focal length of the lens. Solve for the object distance do in terms of the focal length and image distance. 1 f = 1 do + 1 di f = c l , a = d c 71. Temperature. To calculate temperature in degrees Fahrenheit we use the formula , where F is degrees Fahrenheit and C is degrees Celsius. Find the formula to convert from Fahrenheit to Celsius. 72. Geometry. The perimeter P of a rectangle is related to the length L and width W of the rectangle through the equation P 2L 2W. Determine the width in terms of the perimeter and length. 73. Costs: Cellular Phone. Your cell phone plan charges $15 a month plus 12 cents per minute. If your monthly bill is $25.08, how many minutes did you use? 74. Costs: Rental Car. Becky rented a car on her Ft. Lauderdale vacation. The car was $25 a day plus 10 cents per mile. She kept the car for 5 days and her bill was $185. How many miles did she drive the car? 75. Costs: Internet. When traveling in London, Charlotte decided to check her e-mail at an Internet caf. There was a at charge of $2 plus a charge of 10 cents a minute. How many minutes was she logged on if her bill was $3.70? 76. Sales: Income. For a summer job, Dwayne decides to sell magazine subscriptions. He will be paid $20 a day plus $1 for each subscription he sells. If he works for 25 days and makes $645, how many subscriptions did he sell? 77. Business. The operating costs for a local business are a xed amount of $15,000 and $2500 per day. a. Find C that represents operating costs for the company which depends on the number of days open, x. b. If the business accrues $5,515,000 in annual operating costs, how many days did the business operate during the year? 78. Business. Negotiated contracts for a technical support provider produce monthly revenue of $5000 and $0.75 per minute per phone call. a. Find R that represents the revenue for the technical support provider which depends on the number of minutes of phone calls x. b. In one month the provider received $98,750 in revenue. How many minutes of technical support were provided? F = 9 5 C + 32 A P P L I C AT I O N S In Exercises 8386, explain the mistake that is made. 83. Solve the equation 4x 3 6x 7. Solution: Subtract 4x and add 7 to the equation. 3 6x Divide by 3. x 2 This is incorrect. What mistake was made? C AT C H T H E M I S TA K E 2f f f fdo di 2f Image Object 84. Solve the equation 3(x 1) 2 x 3(x 1). Solution: 3x 3 2 x 3x 3 3x 5 2x 3 5x 8 This is incorrect. What mistake was made? x = - 8 5 c01a.qxd 11/23/11 5:34 PM Page 100 130. 1.1 Linear Equations 101 85. Solve the equation . Solution: (p 3)2 4(5p) Cross multiply. 2p 6 20p 6 18p This is incorrect. What mistake was made? p = - 1 3 p = - 6 18 4 p - 3 = 2 5p 90. x 1 is a solution to the equation 91. Solve for x, given that a, b, and c are real numbers and : ax b c 92. Solve for x, given that a, b, and c are real numbers and : a x - b x = c c Z 0 a Z 0 x2 - 1 x - 1 = x + 1. In Exercises 8790, determine whether each of the statements is true or false. C O N C E P T U A L C H A L L E N G E 93. Solve the equation for . Are there any restrictions given that 94. Solve the equation for . Does y have any restrictions? 95. Solve for x: . 1 - 1/x 1 + 1/x = 1 y: 1 y - a + 1 y + a = 2 y - 1 a Z 0, x Z 0? x: b + c x + a = b - c x - a 96. Solve for t: . 97. Solve the equation for x in terms of y: 98. Find the number a for which y 2 is a solution of the equation y a y 5 3ay. y = a 1 + b/x + c t + 1/t 1/t - 1 = 1 T E C H N O L O G Y In Exercises 99106, graph the function represented by each side of the equation in the same viewing rectangle and solve for x. 104. 105. 0.035x 0.029(8706 x) 285.03 106. 1 0.75x - 0.45 x = 1 9 2x(x + 3) x2 = 2 87. The solution to the equation is the set of all real numbers. 88. The solution to the equation is the set of all real numbers. 89. x 1 is a solution to the equation . x2 - 1 x - 1 = x + 1 1 (x - 1)(x + 2) = 1 x2 + x - 2 x = 1 1/x 99. 3(x 2) 5x 3x 4 100. 5(x 1) 7 10 9x 101. 2x 6 4x 2x 8 2 102. 10 20x 10x 30x 20 10 103. x(x - 1) x2 = 1 86. Solve the equation . Solution: Multiply by the LCD, x(x 1). Simplify. (x 1) x 1 x 1 x 1 2x 2 x 1 This is incorrect. What mistake was made? x(x - 1) x + x(x - 1) x - 1 = x(x - 1) x(x - 1) 1 x + 1 x - 1 = 1 x(x - 1) c01a.qxd 11/23/11 5:34 PM Page 101 131. Solving Application Problems Using Mathematical Models In this section, we will use algebra to solve problems that occur in our day-to-day lives. You typically will read the problem in words, develop a mathematical model (equation) for the problem, solve the equation, and write the answer in words. C O N C E P TUAL O BJ E CTIVE Understand the mathematical modeling process. AP P LI CATI O N S I NVO LVI N G LI N EAR E Q UATI O N S S K I LLS O BJ E CTIVE S Solve application problems involving common formulas. Solve geometry problems. Solve interest problems. Solve mixture problems. Solve distanceratetime problems. S E CTI O N 1.2 You will have to come up with a unique formula to solve each kind of word problem, but there is a universal procedure for approaching all word problems. Step 1: Identify the question. Read the problem one time and note what you are asked to nd. Step 2: Make notes. Read until you can note something (an amount, a picture, anything). Continue reading and making notes until you have read the problem a second* time. Step 3: Assign a variable to whatever is being asked for (if there are two choices, then let it be the smaller of the two). Step 4: Set up an equation. Step 5: Solve the equation. Step 6: Check the solution. Run the solution past the common sense department using estimation. *Step 2 often requires multiple readings of the problem. PROCEDURE FOR SOLVING WORD PROBLEMS Real-World Problem Mathematical Model Solve Mathematical Problem Solution to Real-World Problem Translate from Math to Words Translate Words to Math Solve Using Standard Methods 102 c01a.qxd 11/23/11 5:34 PM Page 102 132. 1.2 Applications Involving Linear Equations 103 E X AM P LE 1 How Long Was the Trip? During a camping trip in North Bay, Ontario, a couple went one-third of the way by boat, 10 miles by foot, and one-sixth of the way by horse. How long was the trip? Solution: STEP 1 Identify the question. How many miles was the trip? STEP 2 Make notes. Read Write ... one-third of the way by boat BOAT: of the trip ...10 miles by foot FOOT: 10 miles ... one-sixth of the way by horse HORSE: of the trip STEP 3 Assign a variable. Distance of total trip in miles x STEP 4 Set up an equation. The total distance of the trip is the sum of all the distances by boat, foot, and horse. Distance by boat Distance by foot Distance by horse Total distance of trip Distance by foot 10 miles Distance by horse STEP 5 Solve the equation. Multiply by the LCD, 6. 2x 60 x 6x Collect x terms on the right. 60 3x Divide by 3. 20 x The trip was 20 miles. x 20 STEP 6 Check the solution. Estimate: The boating distance, of 20 miles, is approximately 7 miles; the riding distance on horse, of 20 miles, is approximately 3 miles. Adding these two distances to the 10 miles by foot gives a trip distance of 20 miles. 1 6 1 3 1 3 x + 10 + 1 6 x = x 1 6x Distance by boat = 1 3x 1 6 1 3 YO U R TU R N A family arrives at the Walt Disney World parking lot. To get from their car in the parking lot to the gate at the Magic Kingdom they walk mile, take a tram for of their total distance, and take a monorail for of their total distance. How far is it from their car to the gate of Magic Kingdom? 1 2 1 3 1 4 Answer: The distance from their car to the gate is 1.5 miles. boat foot horse total # # # # 1 3 x + 10 + 1 6 x = x c01a.qxd 11/23/11 5:34 PM Page 103 133. 104 C HAP TE R 1 Equations and Inequalities E X AM P LE 2 Find the Numbers Find three consecutive even integers so that the sum of the three numbers is 2 more than twice the third. Solution: STEP 1 Identify the question. What are the three consecutive even integers? STEP 2 Make notes. Examples of three consecutive even integers are 14, 16, 18 or 8, 6, 4 or 2, 4, 6. STEP 3 Assign a variable. Let n represent the rst even integer. The next consecutive even integer is n 2 and the next consecutive even integer after that is n 4. n 1st integer n 2 2nd consecutive even integer n 4 3rd consecutive even integer STEP 4 Set up an equation. Read Write ... sum of the three numbers n (n 2) (n 4) ... is ... two more than 2 ... twice the third 2(n 4) STEP 5 Solve the equation. n (n 2) (n 4) 2 2(n 4) Eliminate the parentheses. n n 2 n 4 2 2n 8 Simplify both sides. 3n 6 2n 10 Collect n terms on the left and constants on the right. n 4 The three consecutive even integers are 4, 6, and 8. STEP 6 Check the solution. Substitute the solution into the problem to see whether it makes sense. The sum of the three integers (4 6 8) is 18. Twice the third is 16. Since 2 more than twice the third is 18, the solution checks. n + (n + 2) + (n + 4) = 2 + 2(n + 4) Geometry Problems Some problems require geometric formulas in order to be solved. The following geometric formulas may be useful. sum of the three numbers is 2 more twice the than third YO U R TU R N Find three consecutive odd integers so that the sum of the three integers is 5 less than 4 times the rst. Answer: The three consecutive odd integers are 11, 13, and 15. { c01a.qxd 11/23/11 5:34 PM Page 104 134. 1.2 Applications Involving Linear Equations 105 E X AM P LE 3 Geometry A rectangle 24 meters long has the same area as a square with 12-meter sides. What are the dimensions of the rectangle? Solution: STEP 1 Identify the question. What are the dimensions (length and width) of the rectangle? STEP 2 Make notes. Read Write/Draw A rectangle 24 meters long w l 24 area of rectangle l w 24w ... a square with 12-meter sides area of square 12 12 144 STEP 3 Assign a variable. Let w width of the rectangle. STEP 4 Set up an equation. The area of the rectangle is equal to the area of the square. rectangle area square area Substitute in known quantities. 24w 144 STEP 5 Solve the equation. Divide by 24. The rectangle is 24 meters long and 6 meters wide. STEP 6 Check the solution. A 24 meter by 6 meter rectangle has an area of 144 square meters. w = 144 24 = 6 Rectangle Perimeter Area w l A = l # wP = 2l + 2w Circle Circumference Area r Triangle Perimeter Area a h height c b base A = 1 2 bhP = a + b + c A = pr2 C = 2pr YO U R TU R N A rectangle 3 inches wide has the same area as a square with 9-inch sides. What are the dimensions of the rectangle? Answer: The rectangle is 27 inches long and 3 inches wide. Geometric Formulas 12 m 12 m c01a.qxd 11/23/11 5:34 PM Page 105 135. 106 C HAP TE R 1 Equations and Inequalities Interest Problems In our personal or business nancial planning, a particular concern we have is interest. Interest is money paid for the use of money; it is the cost of borrowing money. The total amount borrowed is called the principal. The principal can be the price of our new car; we pay the bank interest for loaning us money. The principal can also be the amount we keep in a CD or money market account; the bank uses this money and pays us interest. Typically interest rate, expressed as a percentage, is the amount charged for the use of the principal for a given time, usually in years. Simple interest is interest that is paid only on principal during a period of time. Later we will discuss compound interest, which is interest paid on both principal and the interest accrued over a period of time. If a principal of P dollars is borrowed for a period of t years at an annual interest rate r (expressed in decimal form), the interest I charged is This is the formula for simple interest. I = Prt Simple InterestD E F I N I T I O N E X AM P LE 4 Simple Interest Through a summer job Morgan is able to save $2500. If she puts that money into a 6-month certicate of deposit (CD) that pays a simple interest rate of 3% a year, how much money will she have in her CD at the end of the 6 months? Solution: STEP 1 Identify the question. How much money does Morgan have after 6 months? STEP 2 Make notes. The principal is $2500. The annual interest rate is 3%, which in decimal form is 0.03. The time the money spends accruing interest is 6 months, or of a year. STEP 3 Assign a variable. Label the known quantities. P 2500, r 0.03, and t 0.5 STEP 4 Set up an equation. Write the simple interest formula. I Prt STEP 5 Solve the equation. I Prt I (2500)(0.03)(0.5) 37.5 The interest paid on the CD is $37.50. Adding this to the principal gives a total of $2500 $37.50 $2537.50 STEP 6 Check the solution. This answer agrees with our intuition. Had we made a mistake, say, of moving one decimal place to the right, then the interest would have been $375, which is much larger than we would expect on a principal of only $2500. 1 2 c01a.qxd 11/23/11 5:34 PM Page 106 136. 1.2 Applications Involving Linear Equations 107 E X AM P LE 5 Multiple Investments Theresa earns a full athletic scholarship for college, and her parents have given her the $20,000 they had saved to pay for her college tuition. She decides to invest that money with an overall goal of earning 11% interest. She wants to put some of the money in a low-risk investment that has been earning 8% a year and the rest of the money in a medium-risk investment that typically earns 12% a year. How much money should she put in each investment to reach her goal? Solution: STEP 1 Identify the question. How much money is invested in each (the 8% and the 12%) account? STEP 2 Make notes. Read Write/Draw Theresa has $20,000 to invest. $20,000 If part is invested at 8% and the rest at 12%, how much should be invested at each rate to yield 11% Some at 8% Some at 12% on the total amount invested? $20,000 at 11% STEP 3 Assign a variable. If we let x represent the amount Theresa puts into the 8% investment, how much of the $20,000 is left for her to put in the 12% investment? Amount in the 8% investment: x Amount in the 12% investment: 20,000 x STEP 4 Set up an equation. Simple interest formula: I Prt INVESTMENT PRINCIPAL RATE TIME (YR) INTEREST 8% Account x 0.08 1 0.08x 12% Account 20,000 x 0.12 1 0.12(20,000 x) Total 20,000 0.11 1 0.11(20,000) Adding the interest earned in the 8% investment to the interest earned in the 12% investment should earn an average of 11% on the total investment. 0.08x 0.12(20,000 x) 0.11(20,000) STEP 5 Solve the equation. Eliminate the parentheses. 0.08x 2400 0.12x 2200 Collect x terms on the left, constants on the right. 0.04x 200 Divide by 0.04. x 5000 Calculate the amount at 12%. 20,000 5000 15,000 Theresa should invest $5000 at 8% and $15,000 at 12% to reach her goal. c01a.qxd 11/23/11 5:34 PM Page 107 137. 108 C HAP TE R 1 Equations and Inequalities STEP 6 Check the solution. If money is invested at 8% and 12% with a goal of averaging 11%, our intuition tells us that more should be invested at 12% than 8%, which is what we found. The exact check is as follows: 0.08(5000) 0.12(15,000) 0.11(20,000) 400 1800 2200 2200 2200 Mixture Problems Mixtures are something we come across every day. Different candies that sell for different prices may make up a movie snack. New blends of coffees are developed by coffee connoisseurs. Chemists mix different concentrations of acids in their labs. Whenever two or more distinct ingredients are combined the result is a mixture. Our choice at a gas station is typically 87, 89, and 93 octane. The octane number is the number that represents the percentage of iso-octane in fuel; 89 octane is signicantly overpriced. Therefore, if your car requires 89 octane, it would be more cost effective to mix 87 and 93 octane. YO U R TU R N You win $24,000 and you decide to invest the money in two different investments: one paying 18% and the other paying 12%. A year later you have $27,480 total. How much did you originally invest in each account? 89 octane + 87 octane + 93 octane = 89 octane [1 gallon] [? gallons] [? gallons] [16 gallons] Answer: $10,000 is invested at 18% and $14,000 is invested at 12%. E X AM P LE 6 Mixture Problem The manual for your new car suggests using gasoline that is 89 octane. In order to save money, you decide to use some 87 octane and some 93 octane in combination with the 89 octane currently in your tank in order to have an approximate 89 octane mixture. Assuming you have 1 gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many gallons of 87 and 93 octane should be used to ll up your tank to achieve a mixture of 89 octane? Solution: STEP 1 Identify the question. How many gallons of 87 octane and how many gallons of 93 octane should be used? STEP 2 Make notes. Read Write/Draw Assuming you have one gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many gallons of 87 and 93 octane should you add? c01a.qxd 11/23/11 5:34 PM Page 108 138. 1.2 Applications Involving Linear Equations 109 Answer: 40 milliliters of 5% HCl and 60 milliliters of 15% HCl YO U R TU R N For a certain experiment, a student requires 100 milliliters of a solution that is 11% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 11% HCl? DistanceRateTime Problems The next example deals with distance, rate, and time. On a road trip, you see a sign that says your destination is 90 miles away, and your speedometer reads 60 miles per hour. Dividing 90 miles by 60 miles per hour tells you that your arrival will be in 1.5 hours. Here is how you know. If the rate, or speed, is assumed to be constant, then the equation that relates distance (d), rate (r), and time (t) is given by d r t. In the above driving example, Substituting these into d r t, we arrive at Solving for t, we get t = 90 miles 60 miles hour = 1.5 hours 90 miles = c60 miles hour d # t# r = 60 miles hour d = 90 miles STEP 3 Assign a variable. x gallons of 87 octane gasoline added at the pump 15 x gallons of 93 octane gasoline added at the pump 1 gallons of 89 octane gasoline already in the tank STEP 4 Set up an equation. 0.89(1) 0.87x 0.93(15 x) 0.89(16) STEP 5 Solve the equation. 0.89(1) 0.87x 0.93(15 x) 0.89(16) Eliminate the parentheses. 0.89 0.87x 13.95 0.93x 14.24 Collect x terms on the left side. 0.06x 14.84 14.24 Subtract 14.84 from both sides of the equation. 0.06x 0.6 Divide both sides by 0.06. x 10 Calculate the amount of 93 octane. 15 10 5 Add 10 gallons of 87 octane and 5 gallons of 93 octane. STEP 6 Check the solution. Estimate: Our intuition tells us that if the desired mixture is 89 octane, then we should add approximately one part 93 octane and two parts 87 octane. The solution we found, 10 gallons of 87 octane and 5 gallons of 93 octane, agrees with this. c01a.qxd 11/23/11 5:34 PM Page 109 139. 110 C HAP TE R 1 Equations and Inequalities YO U R TU R N A Cessna 150 averages 150 miles per hour in still air. With a tailwind it is able to make a trip in hours. Because of the headwind, it is only able to make the return trip in hours. What is the average wind speed?31 2 21 3 Answer: The wind is blowing 30 mph. E X AM P LE 7 DistanceRateTime It takes 8 hours to y from Orlando to London and 9.5 hours to return. If an airplane averages 550 miles per hour in still air, what is the average rate of the wind blowing in the direction from Orlando to London? Assume the speed of the wind (jet stream) is constant and the same for both legs of the trip. Round your answer to the nearest miles per hour. Solution: STEP 1 Identify the question. At what rate in mph is the wind blowing? STEP 2 Make notes. Read Write/Draw It takes 8 hours to y 8 hours from Orlando to London Orlando London and 9.5 hours to return. 9.5 hours If the airplane averages 550 mph wind 550 miles per hour in still air... Orlando London 550 mph wind STEP 3 Assign a variable. w wind speed STEP 4 Set up an equation. The formula relating distance, rate, and time is d r t. The distance d of each ight is the same. On the Orlando to London ight the time is 8 hours due to an increased speed from a tailwind. On the London to Orlando ight the time is 9.5 hours, and the speed is decreased due to the headwind. Let w represent the wind speed. Orlando to London: d (550 w)8 London to Orlando: d (550 w)9.5 These distances are the same, so set them equal to each other: (550 w)8 (550 w)9.5 STEP 5 Solve the equation. Eliminate the parentheses. 4400 8w 5225 9.5w Collect w terms on the left, constants on the right. 17.5w 825 Divide by 17.5. The wind is blowing approximately 47 miles per hour in the direction from Orlando to London. STEP 6 Check the solution. Estimate: Going from Orlando to London, the tailwind is approximately 50 miles per hour, which added to the planes 550 miles per hour speed yields a ground speed of 600 miles per hour. The Orlando to London route took 8 hours. The distance of that ight is (600 mph)(8 hr), which is 4800 miles. The return trip experienced a headwind of approximately 50 miles per hour, so subtracting the 50 from 550 gives an average speed of 500 miles per hour. That route took 9.5 hours, so the distance of the London to Orlando ight was (500 mph)(9.5 hr), which is 4750 miles. Note that the estimates of 4800 and 4750 miles are close. w = 47.1429 L 47 c01a.qxd 11/23/11 5:34 PM Page 110 140. 1.2 Applications Involving Linear Equations 111 AMOUNT OF TIME AMOUNT OF JOB DONE TO DO ONE JOB PER UNIT OF TIME Connie 2 Alvaro x Together 4 5 5 4 1 x 1 2 ## # 1. Identify the quantity you are to determine. 2. Make notes on any clues that will help you set up an equation. 3. Assign a variable. 4. Set up the equation. 5. Solve the equation. 6. Check the solution against your intuition. S U M MARY In the real world many kinds of application problems can be solved through modeling with linear equations. The following six-step procedure will help you develop the model. Some problems require development of a mathematical model, while others rely on common formulas. S E CTI O N 1.2 E X AM P LE 8 Work Connie can clean her house in 2 hours. If Alvaro helps her, they can clean the house in 1 hour and 15 minutes together. How long would it take Alvaro to clean the house by himself? Solution: STEP 1 Identify the question. How long would it take Alvaro to clean the house by himself? STEP 2 Make notes. Connie can clean her house in 2 hours, so Connie can clean of the house per hour. Together Connie and Alvaro can clean the house in 1 hour and 15 minutes, or of an hour. Therefore together they can clean of the house per hour. STEP 3 Assign a variable. Let x number of hours it takes Alvaro to clean the house by himself. So Alvaro can clean of the house per hour. STEP 4 Set up an equation. Amount of house Connie can Amount of house Alvaro can Amount of house they can clean per hour clean per hour clean per hour if they work together STEP 5 Solve the equation. Multiply by the LCD, 10x. 5x 10 8x Solve for x. It takes Alvaro 3 hours and 20 minutes to clean the house by himself. STEP 6 Check the solution. Connie cleans the house in 2 hours. If Alvaro could clean it in 2 hours, then together it would take them 1 hour. Since together it takes them 1 hour and 15 minutes, we expect that it takes Alvaro more than 2 hours by himself. x = 10 3 = 3 1 3 1 2 + 1 x = 4 5 4 5 = 1 x + 1 2 1 x 1 5/4 = 4 5 5 4 1 2 c01a.qxd 12/23/11 5:14 PM Page 111 141. 112 C HAP TE R 1 Equations and Inequalities E X E R C I S E S S E CTI O N 1.2 11. Budget. A company has a total of $20,000 allocated for monthly costs. Fixed costs are $15,000 per month and variable costs are $18.50 per unit. How many units can be manufactured in a month? 12. Budget. A woman decides to start a small business making monogrammed cocktail napkins. She can set aside $1870 for monthly costs. Fixed costs are $1329.50 per month and variable costs are $3.70 per set of napkins. How many sets of napkins can she afford to make per month? 13. Numbers. Find a number such that 10 less than the number is the number. 14. Numbers. Find a positive number such that 10 times the number is 16 more than twice the number. 15. Numbers. Find two consecutive even integers such that 4 times the smaller number is 2 more than 3 times the larger number. 16. Numbers. Find three consecutive integers such that the sum of the three is equal to 2 times the sum of the rst two integers. 17. Geometry. Find the perimeter of a triangle if one side is 11 inches, another side is the perimeter, and the third side is the perimeter. 18. Geometry. Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 20 feet. 19. Geometry. An NFL playing eld is a rectangle. The length of the eld (excluding the end zones) is 40 more yards than twice the width. The perimeter of the playing eld is 260 yards. What are the dimensions of the eld in yards? 20. Geometry. The length of a rectangle is 2 more than 3 times the width, and the perimeter is 28 inches. What are the dimensions of the rectangle? 21. Geometry. Consider two circles, a smaller one and a larger one. If the larger one has a radius that is 3 feet larger than that of the smaller circle and the ratio of the circumferences is 2:1, what are the radii of the two circles? 22. Geometry. The perimeter of a semicircle is doubled when the radius is increased by 1. Find the radius of the semicircle. 23. Home Improvement. A man wants to remove a tall pine tree from his yard. Before he goes to Home Depot, he needs to know how tall an extension ladder he needs to purchase. He measures the shadow of the tree to be 225 feet long. At the same time he measures the shadow of a 4-foot stick to be 3 feet. Approximately how tall is the pine tree? 24. Home Improvement. The same man in Exercise 23 realizes he also wants to remove a dead oak tree. Later in the day he measures the shadow of the oak tree to be 880 feet long, and the 4-foot stick now has a shadow of 10 feet. Approximately how tall is the oak tree? 1 4 1 5 1 4 2 3 1. Discount Price. Donna redeems a 10% off coupon at her local nursery. After buying azaleas, bougainvillea, and bags of potting soil, her checkout price before tax is $217.95. How much would she have paid without the coupon? 2. Discount Price. The original price of a pair of binoculars is $74. The sale price is $51.80. How much was the markdown? 3. Cost: Fair Share. Jeff, Tom, and Chelsea order a large pizza. They decide to split the cost according to how much they will eat. Tom pays $5.16, Chelsea eats of the pizza, and Jeff eats of the pizza. How much did the pizza cost? 4. Event Planning. A couple decide to analyze their monthly spending habits. The monthly bills are 50% of their take-home pay, and they invest 20% of their take-home pay. They spend $560 on groceries, and 23% goes to miscellaneous. How much is their take-home pay per month? 5. Discounts. A builder of tract homes reduced the price of a model by 15%. If the new price is $125,000, what was its original price? How much can be saved by purchasing the model? 6. Markups. A college bookstore marks up the price it pays the publisher for a book by 25%. If the selling price of a book is $79, how much did the bookstore pay for the book? 7. Puzzle. Angela is on her way from home in Jersey City into New York City for dinner. She walks 1 mile to the train station, takes the train of the way, and takes a taxi of the way to the restaurant. How far does Angela live from the restaurant? 8. Puzzle. An employee at Kennedy Space Center (KSC) lives in Daytona Beach and works in the vehicle assembly building (VAB). She carpools to work with a colleague. She drives 7 miles from her house to the park-and-ride. Then she rides with her colleague from the park-and-ride in Daytona Beach to the KSC headquarters building, and then takes the KSC shuttle from the headquarters building to the VAB. The drive from the park-and-ride to the headquarters building is of her total trip, and the shuttle ride is of her total trip. How many miles does she travel from her house to the VAB on days when her colleague drives? 9. Puzzle. A typical college student spends of her waking time in class, of her waking time eating, of her waking time working out, 3 hours studying, and 2 hours doing other things. How many hours of sleep does the typical college student get? 10. Diet. A particular 1550-calories-per-day diet suggests eating breakfast, lunch, dinner, and two snacks. Dinner is twice the calories of breakfast. Lunch is 100 calories more than breakfast. The two snacks are 100 and 150 calories. How many calories are each meal? 1 2 1 10 1 5 1 3 1 20 5 6 1 6 3 4 1 2 1 8 A P P L I C AT I O N S c01a.qxd 1/2/12 3:27 PM Page 112 142. 1.2 Applications Involving Linear Equations 113 25. Biology: Alligators. It is common to see alligators in ponds, lakes, and rivers in Florida. The ratio of head size (back of the head to the end of the snout) to the full body length of an alligator is typically constant. If a -foot alligator has a head length of 6 inches, how long would you expect an alligator to be whose head length is 9 inches? 26. Biology: Snakes. In the African rainforest there is a snake called a Gaboon viper. The fang size of this snake is proportional to the length of the snake. A 3-foot snake typically has 2-inch fangs. If a herpetologist nds Gaboon viper fangs that are 2.6-inches long, how long a snake would she expect to nd? 27. Investing. Ashley has $120,000 to invest and decides to put some in a CD that earns 4% interest per year and the rest in a low-risk stock that earns 7%. How much did she invest in each to earn $7800 interest in the rst year? 28. Investing. You inherit $13,000 and you decide to invest the money in two different investments: one paying 10% and the other paying 14%. A year later your investments are worth $14,580. How much did you originally invest in each account? 29. Investing. Wendy was awarded a volleyball scholarship to the University of Michigan, so on graduation her parents gave her the $14,000 they had saved for her college tuition. She opted to invest some money in a privately held company that pays 10% per year and evenly split the remaining money between a money market account yielding 2% and a high-risk stock that yielded 40%. At the end of the rst year she had $16,610 total. How much did she invest in each of the three? 30. Interest. A high school student was able to save $5000 by working a part-time job every summer. He invested half the money in a money market account and half the money in a stock that paid three times as much interest as the money market account. After a year he earned $150 in interest. What were the interest rates of the money market account and the stock? 31. Budget: Home Improvement. When landscaping their yard, a couple budgeted $4200. The irrigation system costs $2400 and the sod costs $1500. The rest they will spend on trees and shrubs. Trees each cost $32 and shrubs each cost $4. They plant a total of 33 trees and shrubs. How many of each did they plant in their yard? 32. Budget: Shopping. At the deli Jennifer bought spicy turkey and provolone cheese. The turkey costs $6.32 per pound and the cheese costs $4.27 per pound. In total, she bought 3.2 pounds and the price was $17.56. How many pounds of each did she buy? 33. Chemistry. For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 8% HCl? 34. Chemistry. How many gallons of pure alcohol must be mixed with 5 gallons of a solution that is 20% alcohol to make a solution that is 50% alcohol? 31 2 35. Automobiles. A mechanic has tested the amount of antifreeze in your radiator. He says it is only 40% antifreeze and the remainder is water. How many gallons must be drained from your 5 gallon radiator and replaced with pure antifreeze to make the mixture in your radiator 80% antifreeze? 36. Costs: Overhead. A professor is awarded two research grants, each having different overhead rates. The research project conducted on campus has a rate of 42.5% overhead, and the project conducted in the eld, off campus, has a rate of 26% overhead. If she was awarded $1,170,000 total for the two projects with an average overhead rate of 39%, how much was the research project on campus and how much was the research project off campus? 37. Theater. On the way to the movies a family picks up a custom-made bag of candies. The parents like caramels ($1.50/lb) and the children like gummy bears ($2.00/lb). They bought a 1.25-pound bag of combined candies that cost $2.50. How much of each candy did they buy? 38. Coffee. Joy is an instructional assistant in one of the college labs. She is on a very tight budget. She loves Jamaican Blue Mountain coffee, but it costs $12 a pound. She decides to blend this with regular coffee beans that cost $4.20 a pound. If she spends $14.25 on 2 pounds of coffee, how many pounds of each did she purchase? 39. Communications. The speed of light is approximately 3.0 108 meters per second (670,616,629 mph). The distance from Earth to Mars varies because of the orbits of the planets around the Sun. On average, Mars is 100 million miles from Earth. If we use laser communication systems, what will be the delay between Houston and NASA astronauts on Mars? 40. Speed of Sound. The speed of sound is approximately 760 miles per hour in air. If a gun is red mile away, how long will it take the sound to reach you? 41. Business. During the month of February 2011, the average price of gasoline rose 4.7% in the United States. If the average price of gasoline at the end of February 2011 was $3.21 per gallon, what was the price of gasoline at the beginning of February? 42. Business. During the Christmas shopping season of 2010, the average price of a at screen television fell by 40%. A shopper purchased a 42-inch at screen television for $299 in late November 2010. How much would the shopper have paid, to the nearest dollar, for the same television if it was purchased in September 2010? 43. Medicine. A patient requires an IV of 0.9% saline solution, also known as normal saline solution. How much distilled water, to the nearest milliliter, must be added to 100 milliliters of a 3% saline solution to produce normal saline? 44. Medicine. A patient requires an IV of D5W, a 5% solution of Dextrose (sugar) in water. To the nearest milliliter, how much D20W, a 20% solution of Dextrose in water, must be added to 100 milliliters of distilled water to produce a D5W solution? 1 2 c01a.qxd 11/23/11 5:34 PM Page 113 143. 114 C HAP TE R 1 Equations and Inequalities 45. Boating. A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat makes a trip upstream to a marina in 20 minutes. The return trip takes 15 minutes. What is the speed of the current? 46. Aviation. A Cessna 175 can average 130 miles per hour. If a trip takes 2 hours one way and the return takes 1 hour and 15 minutes, nd the wind speed, assuming it is constant. 47. Exercise. A jogger and a walker cover the same distance. The jogger nishes in 40 minutes. The walker takes an hour. How fast is each exerciser moving if the jogger runs 2 miles per hour faster than the walker? 48. Travel. A high school student in Seattle, Washington, attended the University of Central Florida. On the way to UCF he took a southern route. After graduation he returned to Seattle via a northern trip. On both trips he had the same average speed. If the southern trek took 45 hours and the northern trek took 50 hours, and the northern trek was 300 miles longer, how long was each trip? 49. DistanceRateTime. College roommates leave for their rst class in the same building. One walks at 2 miles per hour and the other rides his bike at a slow 6 miles per hour pace. How long will it take each to get to class if the walker takes 12 minutes longer to get to class and they travel on the same path? 50. DistanceRateTime. A long-distance delivery service sends out a truck with a package at 7 A.M. At 7:30 A.M., the manager realizes there was another package going to the same location. He sends out a car to catch the truck. If the truck travels at an average speed of 50 miles per hour and the car travels at 70 miles per hour, how long will it take the car to catch the truck? 51. Work. Christopher can paint the interior of his house in 15 hours. If he hires Cynthia to help him, they can do the same job together in 9 hours. If he lets Cynthia work alone, how long will it take her to paint the interior of his house? 52. Work. Jay and Morgan work in the summer for a landscaper. It takes Jay 3 hours to complete the companys largest yard alone. If Morgan helps him, it takes only 1 hour. How much time would it take Morgan alone? 53. Work. Tracey and Robin deliver Coke products to local convenience stores. Tracey can complete the deliveries in 4 hours alone. Robin can do it in 6 hours alone. If they decide to work together on a Saturday, how long will it take? 54. Work. Joshua can deliver his newspapers in 30 minutes. It takes Amber 20 minutes to do the same route. How long would it take them to deliver the newspapers if they worked together? 55. Music. A major chord in music is composed of notes whose frequencies are in the ratio 4:5:6. If the rst note of a chord has a frequency of 264 hertz (middle C on the piano), nd the frequencies of the other two notes. Hint: Set up two proportions using 4:5 and 4:6. 56. Music. A minor chord in music is composed of notes whose frequencies are in the ratio 10:12:15. If the rst note of a minor chord is A, with a frequency of 220 hertz, what are the frequencies of the other two notes? 57. Grades. Danielles test scores are 86, 80, 84, and 90. The nal exam will count as of the nal grade. What score does Danielle need on the nal in order to earn a B, which requires an average score of 80? What score does she need to earn an A, which requires an average of 90? 58. Grades. Sams nal exam will count as two tests. His test scores are 80, 83, 71, 61, and 95. What score does Sam need on the nal in order to have an average score of 80? 59. Sports. In Super Bowl XXXVII, the Tampa Bay Buccaneers scored a total of 48 points. All of their points came from eld goals and touchdowns. Field goals are worth 3 points and each touchdown was worth 7 points (Martin Gramatica was successful in every extra point attempt). They scored a total of 8 times. How many eld goals and touchdowns were scored? 60. Sports. A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20 yard line with the defensive back at the 15 yard line. If no other players are nearby, at what yard line will the defensive back catch up to the tight end? 61. Recreation. How do two children of different weights balance on a seesaw? The heavier child sits closer to the center and the lighter child sits farther away. When the product of the weight of the child and the distance from the center is equal on both sides, the seesaw should be horizontal to the ground. Suppose Max weighs 42 pounds and Maria weighs 60 pounds. If Max sits 5 feet from the center, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 62. Recreation. Refer to Exercise 61. Suppose Martin, who weighs 33 pounds, sits on the side of the seesaw with Max. If their average distance to the center is 4 feet, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 63. Recreation. If a seesaw has an adjustable bench, then the board can slide along the fulcrum. Maria and Max in Exercise 61 decide to sit on the very edge of the board on each side. Where should the fulcrum be placed along the board in order to balance the seesaw horizontally to the ground? Give the answer in terms of the distance from each childs end. 64. Recreation. Add Martin (Exercise 62) to Maxs side of the seesaw and recalculate Exercise 63. 2 3 c01a.qxd 11/23/11 5:34 PM Page 114 144. C H A L L E N G E 77. Tricia and Janine are roommates and leave Houston on Interstate 10 at the same time to visit their families for a long weekend. Tricia travels west and Janine travels east. If Tricias average speed is 12 miles per hour faster than Janines, nd the speed of each if they are 320 miles apart in 2 hours and 30 minutes. 79. Suppose you bought a house for $132,500 and sold it 3 years later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you waited 2 additional years? 80. Suppose you bought a house for $132,500 and sold it 3 years later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you sold it 1 year after buying it? 81. A golf club membership has two options. Option A is a $300 monthly fee plus $15 cart fee every time you play. Option B has a $150 monthly fee and a $42 fee every time you play. Write a mathematical model for monthly costs for each plan and graph both in the same viewing rectangle using a graphing utility. Explain when Option A is the better deal and when Option B is the better deal. 82. A phone provider offers two calling plans. Plan A has a $30 monthly charge and a $0.10 per minute charge on every call. Plan B has a $50 monthly charge and a $0.03 per minute charge on every call. Explain when Plan A is the better deal and when Plan B is the better deal. T E C H N O L O G Y In Exercises 6976, solve each formula for the specied variable. C O N C E P T U A L In Exercises 6568, refer to this lens law. (See Exercise 82 in Section 1.1.) The position of the image is found using the thin lens equation: , where do is the distance from the object to the lens, di is the distance from the lens to the image, and f is the focal length of the lens. 2f f f fdo di 2f Image Object 1 f = 1 do + 1 di 65. Optics. If the focal length of a lens is 3 centimeters and the image distance is 5 centimeters from the lens, what is the distance from the object to the lens? 66. Optics. If the focal length of the lens is 8 centimeters and the image distance is 2 centimeters from the lens, what is the distance from the object to the lens? 67. Optics. The focal length of a lens is 2 centimeters. If the image distance from the lens is half the distance from the object to the lens, nd the object distance. 68. Optics. The focal length of a lens is 8 centimeters. If the image distance from the lens is half the distance from the object to the lens, nd the object distance. 78. Rick and Mike are roommates and leave Gainesville on Interstate 75 at the same time to visit their girlfriends for a long weekend. Rick travels north and Mike travels south. If Mikes average speed is 8 miles per hour faster than Ricks, nd the speed of each if they are 210 miles apart in 1 hour and 30 minutes. 69. P 2l 2w for w 70. P 2l 2w for l 71. for h 72. C 2pr for r A = 1 2 bh 73. A lw for w 74. d rt for t 75. V lwh for h 76. V pr2 h for h 1.2 Applications Involving Linear Equations 115 c01a.qxd 11/23/11 5:34 PM Page 115 145. A quadratic equation in x is an equation that can be written in the standard form ax2 bx c 0 where a, b, and c are real numbers and a Z 0. Quadratic EquationD E F I N I T I O N FACTORING METHOD There are several methods for solving quadratic equations: factoring, the square root method, completing the square, and the Quadratic Formula. The factoring method applies the zero product property: WORDS MATH If a product is zero, then at least If B C 0, then B 0 or C 0 or both. one of its factors has to be zero. Consider (x 3)(x 2) 0. The zero product property says that x 3 0 or x 2 0, which leads to x 2 or x 3. The solution set is {2, 3}. When a quadratic equation is written in the standard form ax2 bx c 0 it may be possible to factor the left side of the equation as a product of two rst-degree polynomials. We use the zero product property and set each linear factor equal to zero. We solve the resulting two linear equations to obtain the solutions of the quadratic equation. 1.1 Linear Equations 116 C O N C E P TUAL O BJ E CTIVE S Choose appropriate methods for solving quadratic equations. Interpret different types of solution sets (real, imaginary, complex conjugates, repeated roots). Derive the Quadratic Formula. Q UAD R ATI C E Q UATI O N S S E CTI O N 1.3 S K I LLS O BJ E CTIVE S Solve quadratic equations by factoring. Use the square root method to solve quadratic equations. Solve quadratic equations by completing the square. Use the quadratic formula to solve quadratic equations. Factoring In a linear equation, the variable is raised only to the rst power in any term where it occurs. In a quadratic equation, the variable is raised to the second power in at least one term. Examples of quadratic equations, also called second-degree equations, are: x2 - 3 = 05x2 + 4x - 7 = 0x2 + 3 = 7 116 c01a.qxd 11/23/11 5:34 PM Page 116 146. 1.3 Quadratic Equations 117 E X AM P LE 2 Solving a Quadratic Equation by Factoring Solve the equation x2 6x 5 4. I N C O R R E CT Factor the left side. The error occurs here. x - 5 = -4 or x - 1 = -4 (x - 5)(x - 1) = -4 C O R R E CT Write the original equation. Write the equation in standard form by adding 4 to both sides. Factor the left side. Use the zero product property and set each factor equal to zero. Solve each linear equation. x = 3 x - 3 = 0 or x - 3 = 0 (x - 3)(x - 3) = 0 x2 - 6x + 9 = 0 x2 - 6x + 5 = -4 Dont forget to put the quadratic equation in standard form rst. Note: The equation has one solution, or root, which is 3. The solution set is {3}. Since the linear factors were the same, or repeated, we say that 3 is a double root, or repeated root. A common mistake is to forget to put the equation in standard form rst and then use the zero product property incorrectly. C O M M O N M I S TA K E C A U T I O N Dont forget to put the quadratic equation in standard form rst. YO U R TU R N Solve the quadratic equation 9p2 24p 16 by factoring. Answer: The solution is , which is a double root. The solution set is .E4 3 F p = 4 3 Technology Tip Use a graphing utility to display graphs of y1 x2 6x 5 and y2 4. The point of intersection is the solution to this equation. E X AM P LE 1 Solving a Quadratic Equation by Factoring Solve the equation x2 6x 16 0. Solution: The quadratic equation is already in standard form. x2 6x 16 0 Factor the left side into a product of two linear factors. (x 8)(x 2) 0 If a product equals zero, one of its factors has to be equal to zero. x 8 0 or x 2 0 Solve both linear equations. x 8 or x 2 The solution set is {2, 8} . YO U R TU R N Solve the quadratic equation x2 x 20 0 by factoring. Answer: The solution is x 5, 4. The solution set is {5, 4}. c01a.qxd 11/24/11 6:16 PM Page 117 147. 118 C HAP TE R 1 Equations and Inequalities In Example 3, the root x 0 is lost when the original quadratic equation is divided by x. Remember to put the equation in standard form rst and then factor. Square Root Method The square root of 16, , is 4, not 4. In the review (Chapter 0) the principal square root was discussed. The solutions to x2 16, however, are x 4 and x 4. Let us now investigate quadratic equations that do not have a rst-degree term. They have the form The method we use to solve such equations uses the square root property. ax2 + c = 0 a Z 0 ;116 Technology Tip Use a graphing utility to display graphs of y1 2x2 and y2 3x. Use the zero product property and set each factor equal to zero. Solve each linear equation. The solution set is .E0, 3 2 F x = 0 or x = 3 2 x = 0 or 2x - 3 = 0 WORDS MATH If an expression squared is equal If x2 P, then . to a constant, then that expression is equal to the positive or negative square root of the constant. Note: The variable squared must be isolated rst (coefcient equal to 1). x = ; 1P SQUARE ROOT PROPERTY The points of intersection are the solutions to this equation. E X AM P LE 3 Solving a Quadratic Equation by Factoring Solve the equation 2x2 3x. C A U T I O N Do not divide by a variable (because the value of that variable may be zero). Bring all terms to one side rst and then factor. I N C O R R E C T Write the original equation. The error occurs here when both sides are divided by x. 2x = 3 2x2 = 3x C O R R E CT Write the equation in standard form by subtracting 3x. Factor the left side. x(2x - 3) = 0 2x2 - 3x = 0 The common mistake here is dividing both sides by x, which is not allowed because x might be zero. C O M M O N M I S TA K E c01a.qxd 11/23/11 5:34 PM Page 118 148. 1.3 Quadratic Equations 119 E X AM P LE 4 Using the Square Root Property Solve the equation 3x2 27 0. Solution: Add 27 to both sides. 3x2 27 Divide both sides by 3. x2 9 Apply the square root property. The solution set is {3, 3} . If we alter Example 4 by changing subtraction to addition, we see in Example 5 that we get imaginary roots (as opposed to real roots), which we discussed in Chapter 0. x = ; 19 = ;3 E X AM P LE 5 Using the Square Root Property Solve the equation 3x2 27 0. Solution: Subtract 27 from both sides. 3x2 27 Divide by 3. x2 9 Apply the square root property. Simplify. The solution set is {3i, 3i} . x = ;i19 = ;3i x = ; 1-9 YO U R TU R N Solve the equations y2 147 0 and v2 64 0. E X AM P LE 6 Using the Square Root Property Solve the equation (x 2)2 16. Solution: Approach 1: If an expression squared is 16, then the expression equals Separate into two equations. or x 2 4 x 2 4 x 6 x 2 The solution set is {2, 6} . Approach 2: It is acceptable notation to keep the (x - 2) = ; 116 x - 2 = - 116x - 2 = 116 (x - 2) = ; 116; 116. Answer: The solution is . The solution set is The solution is . The solution set is {-8i, 8i}. v = ;8i E -713, 713F. y = ;713 equations together. x 2 ; 4 x 2 ; 4 x 2, 6 c01a.qxd 11/23/11 5:34 PM Page 119 149. 120 C HAP TE R 1 Equations and Inequalities WORDS MATH Express the quadratic equation in the following form. x2 bx c Divide b by 2 and square the result, then add the square to both sides. Write the left side of the equation as a perfect square. Solve using the square root method. ax + b 2 b 2 = c + a b 2 b 2 x2 + bx + a b 2 b 2 = c + a b 2 b 2 SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE WORDS MATH Write the original equation. x2 10x 3 0 Add 3 to both sides. x2 10x 3 Add 25 to both sides.* x2 10x 25 3 25 The left side can be written as a perfect square. (x 5)2 28 Apply the square root method. Add 5 to both sides. *Why did we add 25 to both sides? Recall that (x c)2 x2 2xc c2 . In this case c 5 in order for 2xc 10x. Therefore the desired perfect square (x 5)2 results in x2 10x 25. Applying this product we see that 25 is needed. A systematic approach is to take the coefcient of the rst-degree term x2 10x 3 0, which is 10. Take half of (10), which is (5), and then square it (5)2 25. x = 5 ; 217 x - 5 = ; 128 Completing the Square Factoring and the square root method are two efcient, quick procedures for solving many quadratic equations. However, some equations, such as x2 10x 3 0, cannot be solved directly by these methods. A more general procedure to solve this kind of equation is called completing the square. The idea behind completing the square is to transform any standard quadratic equation ax2 bx c 0 into the form (x A)2 B, where A and B are constants and the left side, (x A)2 , has the form of a perfect square. This last equation can then be solved by the square root method. How do we transform the rst equation into the second equation? Note that the above-mentioned example, x2 10x 3 0, cannot be factored into expressions in which all numbers are integers (or even rational numbers). We can, however, transform this quadratic equation into a form that contains a perfect square. c01a.qxd 11/23/11 5:34 PM Page 120 150. 1.3 Quadratic Equations 121 E X AM P LE 8 Completing the Square When the Leading Coefcient Is Not Equal to 1 Solve the equation 3x2 12x 13 0 by completing the square. Solution: Divide by the leading coefcient, 3. Collect variables to one side of the equation and constants to the other side. Add to both sides. Write the left side of the equation as a perfect square and simplify the right side. Solve using the square root method. Simplify. Rationalize the denominator (Chapter 0). Simplify. , The solution set is . YO U R TU R N Solve the equation 2x2 4x 3 0 by completing the square. e2 - i13 3 , 2 + i13 3 f x = 2 + i13 3 x = 2 - i13 3 x = 2 ; i 13 # 13 13 x = 2 ; i A 1 3 x - 2 = ; A - 1 3 (x - 2)2 = - 1 3 x2 - 4x + 4 = - 13 3 + 4A- 4 2 B 2 = 4 x2 - 4x = - 13 3 x2 - 4x + 13 3 = 0 E X AM P LE 7 Completing the Square Solve the quadratic equation x2 8x 3 0 by completing the square. Solution: Add 3 to both sides. x2 8x 3 Add to both sides. x2 8x 42 3 42 Write the left side as a perfect square and simplify the right side. (x 4)2 19 Apply the square root method to solve. Subtract 4 from both sides. The solution set is 4 , 4 . In Example 7, the leading coefcient (the coefcient of the x2 term) is 1. When the leading coefcient is not 1, start by rst dividing the equation by that leading coefcient. F119119E x = -4 ; 119 x + 4 = ; 119 A1 2 # 8B2 = 42 Technology Tip Graph y1 x2 8x 3. Study Tip When the leading coefcient is not 1, start by rst dividing the equation by that leading coefcient. Technology Tip Graph y1 3x2 12x 13. Answer: The solution is . The solution set is e1 - i12 2 , 1 + i12 2 f. x = 1 ; i22 2 The x-intercepts are the solutions to this equation. The graph does not cross the x-axis, so there is no real solution to this equation. c01a.qxd 11/23/11 5:34 PM Page 121 151. 122 C HAP TE R 1 Equations and Inequalities Study Tip Read as negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a. x = -b ; 2b2 - 4ac 2a If ax2 bx c 0, , then the solution is Note: The quadratic equation must be in standard form (ax2 bx c 0) in order to identify the parameters: acoefcient of x2 bcoefcient of x cconstant x = -b ; 2b2 - 4ac 2a a Z 0 QUADRATIC FORMULA We read this formula as negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a. It is important to note that negative b could be positive (if b is negative). For this reason, an alternate form is opposite b. . . The Quadratic Formula should be memorized and used when simpler methods (factoring and the square root method) cannot be used. The Quadratic Formula works for any quadratic equation. Study Tip The Quadratic Formula works for any quadratic equation. Quadratic Formula Let us now consider the most general quadratic equation: We can solve this equation by completing the square. WORDS MATH Divide the equation by the leading coefcient a. Subtract from both sides. Square half of and add the result to both sides. Write the left side of the equation as a perfect square and the right side Solve using the square root method. Subtract from both sides and simplify the radical. Write as a single fraction. We have derived the Quadratic Formula. x = -b ; 2b2 - 4ac 2a x = - b 2a ; 2b2 - 4ac 2a b 2a x + b 2a = ; A b2 - 4ac 4a2 ax + b 2a b 2 = b2 - 4ac 4a2 x2 + b ax + a b 2a b 2 = a b 2a b 2 - c aa b 2a b 2 b a x2 + b ax = - c a c a x2 + b ax + c a = 0 ax2 + bx + c = 0 a Z 0 as a single fraction. c01a.qxd 11/23/11 5:34 PM Page 122 152. 1.3 Quadratic Equations 123 E X AM P LE 9 Using the Quadratic Formula and Finding Two Distinct Real Roots Use the Quadratic Formula to solve the quadratic equation x2 4x 1 0. Solution: For this problem a 1, b 4, and c 1. Write the Quadratic Formula. Use parentheses to avoid losing a minus sign. Substitute values for a, b, and c into the parentheses. Simplify. The solution set contains two distinct real numbers.E2 - 15, 2 + 15F x = 4 ; 216 + 4 2 = 4 ; 220 2 = 4 ; 225 2 = 4 2 ; 225 2 = 2 ; 25 x = -(-4) ; 2(-4)2 - 4(1)(-1) 2(1) x = -(n) ; 2(n)2 - 4(n)(n) 2(n) x = -b ; 2b2 - 4ac 2a E X AM P LE 10 Using the Quadratic Formula and Finding Two Complex Roots Use the Quadratic Formula to solve the quadratic equation x2 8 4x. Solution: Write this equation in standard form x2 4x 8 0 in order to identify a 1, b 4, and c 8. Write the Quadratic Formula. Use parentheses to avoid overlooking a minus sign. Substitute the values for a, b, and c into the parentheses. Simplify. The solution set {2 2i, 2 2i} contains two complex numbers. Note that they are complex conjugates of each other. x = 4 ; 116 - 32 2 = 4 ; 1-16 2 = 4 ; 4i 2 = 4 2 ; 4i 2 = 2 ; 2i x = -(-4) ; 2(-4)2 - 4(1)(8) 2(1) x = -(n) ; 2(n)2 - 4(n)(n) 2(n) x = -b ; 2b2 - 4ac 2a YO U R TU R N Use the Quadratic Formula to solve the quadratic equation x2 6x 2 0. Study Tip Using parentheses as place holders helps avoid errors. x = -(n) ; 2(n)2 - 4(n)(n) 2(n) x = -b ; 2b2 - 4ac 2a ; Answer: The solution is The solution set is .E -3 - 211, -3 + 211F -3 ; 111.x = YO U R TU R N Use the Quadratic Formula to solve the quadratic equation x2 2 2x. Answer: The solution set is {1 i, 1 i}. Technology Tip Set the graphing utility in complex number mode. Keystrokes: MODE NORMAL a bi c01a.qxd 11/23/11 5:34 PM Page 123 153. 124 C HAP TE R 1 Equations and Inequalities The term inside the radical, b2 4ac, is called the discriminant. The discriminant gives important information about the corresponding solutions or roots of ax2 bx c 0, where a, b, and c are real numbers. TYPES OF SOLUTIONS b2 4ac SOLUTIONS (ROOTS) Positive Two distinct real roots 0 One real root (a double or repeated root) Negative Two complex roots (complex conjugates) In Example 9, the discriminant is positive and the solution has two distinct real roots. In Example 10, the discriminant is negative and the solution has two complex (conjugate) roots. In Example 11, the discriminant is zero and the solution has one repeated real root. Applications Involving Quadratic Equations In Section 1.2, we developed a procedure for solving word problems involving linear equations. The procedure is the same for applications involving quadratic equations. The only difference is that the mathematical equations will be quadratic, as opposed to linear. E X AM P LE 11 Using the Quadratic Formula and Finding One Repeated Real Root Use the Quadratic Formula to solve the quadratic equation 4x2 4x 1 0. Solution: Identify a, b, and c. a 4, b 4, c 1 Write the Quadratic Formula. Use parentheses to avoid losing a minus sign. Substitute values a 4, b 4, Simplify. The solution set is a repeated real root . Note: This quadratic equation also could have been solved by factoring: (2x 1)2 0. e 1 2 f x = 4 ; 216 - 16 8 = 4 ; 0 8 = 1 2 x = -(-4) ; 2(-4)2 - 4(4)(1) 2(4) x = -(n) ; 2(n)2 - 4(n)(n) 2(n) x = -b ; 2b2 - 4ac 2a YO U R TU R N Use the Quadratic Formula to solve the quadratic equation 9x2 6x 1 0. Answer: E1 3 F and c 1. c01a.qxd 11/23/11 5:34 PM Page 124 154. 1.3 Quadratic Equations 125 E X AM P LE 13 Pythagorean Theorem Hitachi makes a 60-inch HDTV that has a 60-inch diagonal. If the width of the screen is approximately 52-inches, what is the approximate height of the screen? Solution: STEP 1 Identify the question. What is the approximate height of the HDTV screen? STEP 2 Make notes. 60 inches 52 inches ? STEP 3 Set up an equation. Recall the Pythagorean theorem. a2 b2 c2 Substitute in the known values. h2 522 602 STEP 4 Solve the equation. Simplify the constants. h2 2704 3600 Subtract 2704 from both sides. h2 896 Solve using the square root method. Distance is positive, so the negative value is eliminated. The height is approximately 30 inches . STEP 5 Check the solution. 3604 L 3600 900 + 2704 3600 302 + 522 602 h = ; 2896 L ;30 Study Tip Dimensions such as length and width are distances, which are dened as positive quantities. Although the mathematics may yield both positive and negative values, the negative values are excluded. Solution: STEP 1 Identify the question. When is the price of the stock equal to $30? STEP 2 Make notes. Stock price: P 0.2t2 5.6t 50.2 P 30 STEP 3 Set up an equation. 0.2t2 5.6t 50.2 30 STEP 4 Solve the equation. Subtract 30 from both sides. 0.2t2 5.6t 20.2 0 Solve for t using the Quadratic Formula. Simplify. Rounding these two numbers, we nd that and Since corresponds to January 1999, these two solutions correspond to April 1999 and December 2000 . STEP 5 Check the solution. Look at the gure. The horizontal axis represents the year (2000 corresponds to January 2000), and the vertical axis represents the stock price. Estimating when the stock price is approximately $30, we nd April 1999 and December 2000. t = 1t L 24.t L 4 t L 5.6 ; 3.9 0.4 L 4.25, 23.75 t = -(-5.6) ; 2(-5.6)2 - 4(0.2)(20.2) 2(0.2) E X AM P LE 12 Stock Value From 1999 to 2001 the price of Abercrombie & Fitchs (ANF) stock was approximately given by P 0.2t2 5.6t 50.2, where P is the price of stock in dollars, t is in months, and t 1 corresponds to January 1999. When was the value of the stock worth $30? Technology Tip The graphing utility screen for -(-5.6) ; 2(-5.6)2 - 4(0.2)(20.2) 2(0.2) ANF Quarterly 2000 2001 2002 2003 2004 10 20 30 40 50 5/26/04 c01a.qxd 11/23/11 5:34 PM Page 125 155. 126 C HAP TE R 1 Equations and Inequalities In Exercises 122, solve by factoring. 1. x2 5x 6 0 2. v2 7v 6 0 3. p2 8p 15 0 4. u2 2u 24 0 5. x2 12 x 6. 11x 2x2 12 7. 16x2 8x 1 8. 3x2 10x 8 0 9. 9y2 1 6y 10. 4x 4x2 1 11. 8y2 16y 12. 3A2 12A 13. 9p2 12p 4 14. 4u2 20u 25 15. x2 9 0 16. 16v2 25 0 17. x(x 4) 12 18. 3t2 48 0 19. 2p2 50 0 20. 5y2 45 0 21. 3x2 12 22. 7v2 28 In Exercises 2334, solve using the square root method. 23. p2 8 0 24. y2 72 0 25. x2 9 0 26. v2 16 0 27. (x 3)2 36 28. (x 1)2 25 29. (2x 3)2 4 30. (4x 1)2 16 31. (5x 2)2 27 32. (3x 8)2 12 33. (1 x)2 9 34. (1 x)2 9 In Exercises 3544, what number should be added to complete the square of each expression? 35. x2 6x 36. x2 8x 37. x2 12x 38. x2 20x 39. 40. 41. 42. 43. x2 2.4x 44. x2 1.6x In Exercises 4556, solve by completing the square. 45. x2 2x 3 46. y2 8y 2 0 47. t2 6t 5 48. x2 10x 21 49. y2 4y 3 0 50. x2 7x 12 0 51. 2p2 8p 3 52. 2x2 4x 3 0 53. 2x2 7x 3 0 54. 3x2 5x 10 0 55. 56. In Exercises 5768, solve using the Quadratic Formula. 57. t2 3t 1 0 58. t2 2t 1 59. s2 s 1 0 60. 2s2 5s 2 61. 3x2 3x 4 0 62. 4x2 2x 7 63. x2 2x 17 0 64. 4m2 7m 8 0 65. 5x2 7x 3 66. 3x2 5x 11 67. 68. 1 4 x2 - 2 3 x - 1 3 = 01 4 x2 + 2 3 x - 1 2 = 0 t2 3 + 2t 3 + 5 6 = 0 x2 2 - 2x = 1 4 x2 + 4 5xx2 + 2 5xx2 - 1 3x x2 - 1 2x S K I LL S E X E R C I S E S S E CTI O N 1.3 equations and can yield three types of solutions: two distinct real roots, one real root (repeated), or two complex roots (conjugates of each other). Quadratic Formula: x = -b ; 2b2 - 4ac 2a S U M MARY The four methods for solving quadratic equations are factoring, the square root method, completing the square, and the Quadratic Formula. Factoring and the square root method are the quickest and easiest but cannot always be used. The quadratic formula and completing the square work for all quadratic ax2 + bx + c = 0 a Z 0 S E CTI O N 1.3 c01a.qxd 11/23/11 5:34 PM Page 126 156. 1.3 Quadratic Equations 127 In Exercises 6974, determine whether the discriminant is positive, negative, or zero, and indicate the number and type of root to expect. Do not solve. 69. x2 22x 121 0 70. x2 28x 196 0 71. 2y2 30y 68 0 72. 3y2 27y 66 0 73. 9x2 7x 8 0 74. 3x2 5x 7 0 In Exercises 7594, solve using any method. 75. v2 8v 20 76. v2 8v 20 77. t2 5t 6 0 78. t2 5t 6 0 79. (x 3)2 16 80. (x 3)2 16 81. (p 2)2 4p 82. (u 5)2 16u 83. 8w2 2w 21 0 84. 8w2 2w 21 0 85. 3p2 9p 1 0 86. 3p2 9p 1 0 87. 88. 89. 90. 91. 92. 93. x2 0.1x 0.12 94. y2 0.5y 0.06 5 y + 4 = 4 + 3 y - 2 4(x - 2) x - 3 + 3 x = -3 x(x - 3) x - 10 x = -3x + 12 x = 7 1 2 x2 + 2 3 x = 2 5 2 3 t2 + 4 3 t = 1 5 97. Business. Find the break-even point for a month to the nearest unit. 98. Business. Find the production level that produces a monthly prot of $40 million. In Exercises 99 and 100 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly prot. The function models the effect that a price increase of x dollars on a bottle of wine will have on the prot P measured in dollars. 99. Business/Economics. What is the smallest price increase that will produce a weekly prot of $460? 100. Business/Economics. What is the smallest price increase that will produce a weekly prot of $630? In Exercises 101 and 102 refer to the following: An epidemiological study of the spread of the u in a small city nds that the total number P of people who contracted the u t days into an outbreak is modeled by the function 101. Health/Medicine. After approximately how many days will 160 people have contracted the u? 102. Health/Medicine. After approximately how many days will 172 people have contracted the u? P = -t2 + 13t + 130 1 t 6 P = -5(x + 3)(x - 24) 95. Stock Value. From June 2003 until April 2004 JetBlue airlines stock (JBLU) was approximately worth P 4t2 80t 360, where P denotes the price of the stock in dollars and t corresponds to months, with t 1 corresponding to January 2003. During what months was the stock equal to $24? A P P L I C AT I O N S JBLU Daily J J A S O N D 04 F M A M 20 30 40 50 5/27/04 96. Stock Value. From November 2003 until March 2004, Wal-Mart stock (WMT) was approximately worth P 2t2 12t 70, where P denotes the price of the stock in dollars and t corresponds to months, with t 1 corresponding to November 2003. During what months was the stock equal to $60? In Exercises 97 and 98 refer to the following: Research indicates that monthly prot for Widgets R Us is modeled by the function where P is prot measured in millions of dollars and q is the quantity of widgets produced measured in thousands. P = -100 + (0.2q - 3)q WMT Daily J J A S O N D 04 F M A M 50 55 60 65 5/27/04 c01a.qxd 11/23/11 5:34 PM Page 127 157. 128 C HAP TE R 1 Equations and Inequalities 103. Environment: Reduce Your Margins, Save a Tree. Lets dene the usable area of an 8.5-inch by 11-inch piece of paper as the rectangular space between the margins of that piece of paper. Assume the default margins in a word processor in a colleges computer lab are set up to be 1.25 inches wide (top and bottom) and 1 inch wide (left and right). Answer the following questions using this information. a. Determine the amount of usable space, in square inches, on one side of an 8.5-inch by 11-inch piece of paper with the default margins of 1.25-inch and 1-inch. b. The Green Falcons, a campus environmental club, has convinced their colleges computer lab to reduce the default margins in their word-processing software by x inches. Create and simplify the quadratic expression that represents the new usable area, in square inches, of one side of an 8.5-inch by 11-inch piece of paper if the default margins at the computer lab are each reduced by x inches. c. Subtract the usable space in part (a) from the expression in part (b). Explain what this difference represents. d. If 10 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch. 104. Environment: Reduce Your Margins, Save a Tree. Repeat Exercise 103 assuming the computer labs default margins are 1 inch all the way around (left, right, top, and bottom). If 15 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch. 105. Television. A standard 32-inch television has a 32-inch diagonal and a 25-inch width. What is the height of the 32-inch television? 106. Television. A 42-inch LCD television has a 42-inch diagonal and a 20-inch height. What is the width of the 42-inch LCD television? 107. Numbers. Find two consecutive numbers such that their sum is 35 and their product is 306. 108. Numbers. Find two consecutive odd integers such that their sum is 24 and their product is 143. 109. Geometry. The area of a rectangle is 135 square feet. The width is 6 feet less than the length. Find the dimensions of the rectangle. 110. Geometry. A rectangle has an area of 31.5 square meters. If the length is 2 more than twice the width, nd the dimensions of the rectangle. 111. Geometry. A triangle has a height that is 2 more than 3 times the base and an area of 60 square units. Find the base and height. 112. Geometry. A squares side is increased by 3 yards, which corresponds to an increase in the area by 69 square yards. How many yards is the side of the initial square? 113. Falling Objects. If a person drops a water balloon off the rooftop of a 100-foot building, the height of the water balloon is given by the equation h 16t2 100, where t is in seconds. When will the water balloon hit the ground? 114. Falling Objects. If the person in Exercise 113 throws the water balloon downward with a speed of 5 feet per second, the height of the water balloon is given by the equation h 16t2 5t 100, where t is in seconds. When will the water balloon hit the ground? 115. Gardening. A square garden has an area of 900 square feet. If a sprinkler (with a circular pattern) is placed in the center of the garden, what is the minimum radius of spray the sprinkler would need in order to water all of the garden? 116. Sports. A baseball diamond is a square. The distance from base to base is 90 feet. What is the distance from home plate to second base? 117. Volume. A at square piece of cardboard is used to construct an open box. Cutting a 1-foot by 1-foot square off of each corner and folding up the edges will yield an open box (assuming these edges are taped together). If the desired volume of the box is 9 cubic feet, what are the dimensions of the original square piece of cardboard? 118. Volume. A rectangular piece of cardboard whose length is twice its width is used to construct an open box. Cutting a 1-foot by 1-foot square off of each corner and folding up the edges will yield an open box. If the desired volume is 12 cubic feet, what are the dimensions of the original rectangular piece of cardboard? 119. Gardening. A landscaper has planted a rectangular garden that measures 8 feet by 5 feet. He has ordered 1 cubic yard (27 cubic feet) of stones for a border along the outside of the garden. If the border needs to be 4 inches deep and he wants to use all of the stones, how wide should the border be? 120. Gardening. A gardener has planted a semicircular rose garden with a radius of 6 feet, and 2 cubic yards of mulch (1 cubic yard 27 cubic feet) are being delivered. Assuming she uses all of the mulch, how deep will the layer of mulch be? 121. Work. Lindsay and Kimmie, working together, can balance the nancials for the Kappa Kappa Gamma sorority in 6 days. Lindsay by herself can complete the job in 5 days less than Kimmie. How long will it take Lindsay to complete the job by herself? 122. Work. When Jack cleans the house, it takes him 4 hours. When Ryan cleans the house, it takes him 6 hours. How long would it take both of them if they worked together? c01a.qxd 11/23/11 5:34 PM Page 128 158. In Exercises 123126, explain the mistake that is made. 125. 16a2 9 0 16a2 9 126. x = 1 ; 12 x - 1 = ; 22 (x - 1)2 = 2 2(x - 1)2 = 4 2(x2 - 2x + 1) = 3 + 1 2(x2 - 2x) = 3 2x2 - 4x = 3 a = ;3 4 a = ; 2 9 16 a2 = - 9 16 C AT C H T H E M I S TA K E 123. t2 5t 6 0 (t 3)(t 2) 0 t 2, 3 124. (2y 3)2 25 2y 3 5 2y 8 y 4 y = 5 4 In Exercises 135138, solve for the indicated variable in terms of other variables. 135. Solve for t. 136. Solve A P(1 r)2 for r. 137. Solve a2 b2 c2 for c. 138. Solve P EI RI2 for I. 139. Solve the equation by factoring: x4 4x2 0. 140. Solve the equation by factoring: 3x 6x2 0. 141. Solve the equation using factoring by grouping: x3 x2 4x 4 0. 142. Solve the equation using factoring by grouping: x3 2x2 x 2 0. s = 1 2gt2 In Exercises 127130, determine whether the following statements are true or false. 127. The equation (3x 1)2 16 has the same solution set as the equation 3x 1 4. 128. The quadratic equation ax2 bx c 0 can be solved by the square root method only if b 0. 129. All quadratic equations can be solved exactly. 130. The Quadratic Formula can be used to solve any quadratic equation. 131. Write a quadratic equation in general form that has x a as a repeated real root. 132. Write a quadratic equation in general form that has x bi as a root. 133. Write a quadratic equation in general form that has the solution set {2, 5}. 134. Write a quadratic equation in general form that has the solution set {3, 0}. C O N C E P T U A L C H A L L E N G E 147. Aviation. An airplane takes 1 hour longer to go a distance of 600 miles ying against a headwind than on the return trip with a tailwind. If the speed of the wind is 50 miles per hour, nd the speed of the plane in still air. 148. Boating. A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current? 149. Find a quadratic equation whose two distinct real roots are the negatives of the two distinct real roots of the equation ax2 bx c 0. 143. Show that the sum of the roots of a quadratic equation is equal to . 144. Show that the product of the roots of a quadratic equation is equal to . 145. Write a quadratic equation in general form whose solution set is 146. Write a quadratic equation in general form whose solution set is {2 i, 2 i}. E3 + 15, 3 - 15F. c a - b a 1.3 Quadratic Equations 129 c01a.qxd 11/23/11 5:34 PM Page 129 159. 130 C HAP TE R 1 Equations and Inequalities 153. Solve the equation x2 x 2 by rst writing it in standard form and then factoring. Now plot both sides of the equation in the same viewing screen (y1 x2 x and y2 2). At what x-values do these two graphs intersect? Do those points agree with the solution set you found? 154. Solve the equation x2 2x 2 by rst writing it in standard form and then using the quadratic formula. Now plot both sides of the equation in the same viewing screen ( y1 x2 2x and y2 2). Do these graphs intersect? Does this agree with the solution set you found? 155. a. Solve the equation x2 2x b, b 8 by rst writing it in standard form. Now plot both sides of the equation in the same viewing screen (y1 x2 2x and y2 b). At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 3, 1, 0, and 5. 156. a. Solve the equation x2 2x b, b 8 by rst writing it in standard form. Now plot both sides of the equation in the same viewing screen (y1 x2 2x and y2 b). At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 3, 1, 0, and 5. T E C H N O L O G Y 150. Find a quadratic equation whose two distinct real roots are the reciprocals of the two distinct real roots of the equation ax2 bx c 0. 151. A small jet and a 757 leave Atlanta at 1 P.M. The small jet is traveling due west. The 757 is traveling due south. The speed of the 757 is 100 miles per hour faster than the small jet. At 3 P.M. the planes are 1000 miles apart. Find the average speed of each plane. 152. Two boats leave Key West at noon. The smaller boat is traveling due west. The larger boat is traveling due south. The speed of the larger boat is 10 miles per hour faster than the speed of the smaller boat. At 3 P.M. the boats are 150 miles apart. Find the average speed of each boat. C O N C E P TUAL O BJ E CTIVE S Transform a difcult equation into a simpler linear or quadratic equation. Recognize the need to check solutions when the transformation process may produce extraneous solutions. Realize that not all polynomial equations are factorable. OTH E R T YP E S O F E Q UATI O N S S E CTI O N 1.4 S K I LLS O BJ E CTIVE S Solve radical equations. Solve equations that are quadratic in form. Solve equations that are factorable. Radical Equations Radical equations are equations in which the variable is inside a radical (that is, under a square root, cube root, or higher root). Examples of radical equations follow. Until now your experience has been with linear and quadratic equations. Often you can transform a radical equation into a simple linear or quadratic equation. Sometimes the transformation process yields extraneous solutions, or apparent solutions that may solve the transformed problem but are not solutions of the original radical equation. Therefore, it is very important to check your answers. 1x + 2 + 17x + 2 = 612x + 3 = x1x - 3 = 2 c01a.qxd 12/22/11 6:06 PM Page 130 160. 1.4 Other Types of Equations 131 E X AM P LE 1 Solving an Equation Involving a Radical Solve the equation Solution: Square both sides of the equation. Simplify. x 3 4 Solve the resulting linear equation. x 7 The solution set is {7}. Check: 17 - 3 = 14 = 2 Q1x - 3R 2 = 22 1x - 3 = 2. YO U R TU R N Solve the equation 13p + 4 = 5. When both sides of an equation are squared, extraneous solutions can arise. For example, take the equation x 2 If we square both sides of this equation, then the resulting equation, x2 4, has two solutions: x 2 and x 2. Notice that the value x 2 is not in the solution set of the original equation x 2. Therefore, we say that x 2 is an extraneous solution. In solving a radical equation we square both sides of the equation and then solve the resulting equation. The solutions to the resulting equation can sometimes be extraneous in that they do not satisfy the original radical equation. E X AM P LE 2 Solving an Equation Involving a Radical Solve the equation Solution: Square both sides of the equation. Simplify. 2x 3 x2 Write the quadratic equation in standard form. x2 2x 3 0 Factor. (x 3)(x 1) 0 Use the zero product property. x 3 or x 1 Check these values to see whether they both make the original equation statement true. X The solution is x 3 . The solution set is {3}. x = -1: 12(-1) + 3 = -1 1 1-2 + 3 = -1 1 11 = -1 1 1 Z -1 x = 3: 12(3) + 3 = 3 1 16 + 3 = 3 1 19 = 3 1 3 = 3 A12x + 3 B 2 = x2 12x + 3 = x. YO U R TU R N Solve the equation 112 + t = t. Technology Tip Use a graphing utility to display graphs of and y2 x. y1 = 12x + 3 Answer: p 7 or {7} Study Tip Extraneous solutions are common when we deal with radical equations, so remember to check your answers. The x-coordinate of the point of intersection is the solution to the equation 12x + 3 = x. Answer: t 4 or {4} Answer: x 1 and x 3 or {3, 1} YO U R TU R N Solve the equation 12x + 6 = x + 3. c01a.qxd 11/23/11 5:34 PM Page 131 161. 132 C HAP TE R 1 Equations and Inequalities What happened in Example 2? When we transformed the radical equation into a quadratic equation, we created an extraneous solution, x 1, a solution that appears to solve the original equation but does not. When solving radical equations, answers must be checked to avoid including extraneous solutions in the solution set. E X AM P LE 3 Solving an Equation That Involves a Radical Solve the equation . Solution: Subtract 4x from both sides. Divide both sides by 2. Square both sides. x 3 (2x 5)(2x 5) (2x 5)2 Eliminate the parentheses. x 3 4x2 20x 25 Rewrite the quadratic equation in standard form. 4x2 19x 22 0 Factor. (4x 11)(x 2) 0 Solve. The apparent solutions are and 2. Note that does not satisfy the original equation; therefore it is extraneous. The solution is The solution set is {2}.x = -2 . -11 4-11 4 x = - 11 4 and x = -2 1x + 3 = 2x + 5 -21x + 3 = -10 - 4x 4x - 21x + 3 = -10 YO U R TU R N Solve the equation 2x - 41x + 2 = -6. Answer: x 1 or {1} In Examples 1 through 3 each equation only contained one radical each. The next example contains two radicals. Our technique will be to isolate one radical on one side of the equation with the other radical on the other side of the equation. I N C O R R E CT Square the expression. The error occurs here when only individual terms are squared. 9 (x 2)Z A3 + 1x + 2B 2 C O R R E CT Square the expression. Write the square as a product of two factors. Use the FOIL method. 9 + 61x + 2 + (x + 2) A3 + 1x + 2B A3 + 1x + 2B A3 + 1x + 2B 2 C O M M O N M I S TA K E c01a.qxd 11/23/11 5:34 PM Page 132 162. 1.4 Other Types of Equations 133 E X AM P LE 4 Solving an Equation with More Than One Radical Solve the equation Solution: Subtract from both sides. Square both sides. Simplify. Multiply the expressions on the right side of the equation. Isolate the term with the radical on the left side. Combine like terms on the right side. Divide by 6. Square both sides. 4(x 2) (6 x)2 21x + 2 = 6 - x 121x + 2 = 36 - 6x 121x + 2 = 36 + x + 2 - 7x - 2 7x + 2 = 36 - 121x + 2 + (x + 2) 7x + 2 = A6 - 1x + 2B A6 - 1x + 2B A17x + 2B 2 = A6 - 1x + 2B 2 17x + 2 = 6 - 1x + 2 1x + 2 1x + 2 + 17x + 2 = 6. Technology Tip Use a graphing utility to display graphs of and y2 6. y1 = 1x + 2 + 17x + 2 The x-coordinate of the point of intersection is the solution to the equation .2x + 2 + 27x + 2 = 6 YO U R TU R N Solve the equation .1x - 4 = 5 - 1x + 1 Simplify. 4x 8 36 12x x2 Rewrite the quadratic equation in standard form. x2 16x 28 0 Factor. (x 14)(x 2) 0 Solve. x 14 and x 2 The apparent solutions are 2 and 14. Note that x 14 does not satisfy the original equation; therefore it is extraneous. The solution is x 2 . The solution set is {2}. Answer: x 8 or {8} Study Tip Remember to check both solutions. Step 1: Isolate the term with a radical on one side. Step 2: Raise both (entire) sides of the equation to the power that will eliminate this radical, and simplify the equation. Step 3: If a radical remains, repeat Steps 1 and 2. Step 4: Solve the resulting linear or quadratic equation. Step 5: Check the solutions and eliminate any extraneous solutions. Note: If there is more than one radical in the equation, it does not matter which radical is isolated rst. PROCEDURE FOR SOLVING RADICAL EQUATIONS c01a.qxd 11/23/11 5:34 PM Page 133 163. 134 C HAP TE R 1 Equations and Inequalities Equations Quadratic in Form: u-Substitution Equations that are higher order or that have fractional powers often can be transformed into a quadratic equation by introducing a u-substitution. When this is the case, we say that equations are quadratic in form. In the following table, the two original equations are quadratic in form because they can be transformed into a quadratic equation given the correct substitution. ORIGINAL EQUATION SUBSTITUTION NEW EQUATION x4 3x2 4 0 u x2 u2 3u 4 0 t2/3 2t1/3 1 0 u t1/3 u2 2u 1 0 u y1/2 2u2 u 1 0 2 y - 1 1y + 1 = 0 For example, the equation x4 3x2 4 0 is a fourth-degree equation in x. How did we know that u x2 would transform the original equation into a quadratic equation? If we rewrite the original equation as (x2 )2 3(x2 ) 4 0, the expression in parentheses is the u-substitution. Let us introduce the substitution u x2 . Note that squaring both sides implies u2 x4 . We then replace x2 in the original equation with u, and x4 in the original equation with u2 , which leads to a quadratic equation in u: u2 3u 4 0. WORDS MATH Solve for x. x4 3x2 4 0 Introduce u-substitution. u x2 [Note that u2 x4 .] Write the quadratic equation in u. u2 3u 4 0 Factor. (u 4)(u 1) 0 Solve for u. u 4 or u 1 Transform back to x, u x2 . x2 4 or x2 1 Solve for x. x ; 2 or x ;i The solution set is {;2, ;i}. It is important to correctly determine the appropriate substitution in order to arrive at an equation quadratic in form. For example, t2/3 2t1/3 1 0 is an original equation given in the above table. If we rewrite this equation as (t1/3 )2 2(t1/3 ) 1 0, then it becomes apparent that the correct substitution is u t1/3 , which transforms the equation in t into a quadratic equation in u: u2 2u 1 0. Step 1: Identify the substitution. Step 2: Transform the equation into a quadratic equation. Step 3: Solve the quadratic equation. Step 4: Apply the substitution to rewrite the solution in terms of the original variable. Step 5: Solve the resulting equation. Step 6: Check the solutions in the original equation. PROCEDURE FOR SOLVING EQUATIONS QUADRATIC IN FORM c01a.qxd 11/23/11 5:34 PM Page 134 164. 1.4 Other Types of Equations 135 E X AM P LE 5 Solving an Equation Quadratic in Form with Negative Exponents Find the solutions to the equation x2 x1 12 0. Solution: Rewrite the original equation. (x1 )2 (x1 ) 12 0 Determine the u-substitution. u x1 [Note that u2 x2 .] The original equation in x corresponds to a quadratic equation in u. u2 u 12 0 Factor. (u 4)(u 3) 0 Solve for u. u 4 or u 3 The most common mistake is forgetting to transform back to x. Transform back to x. Let u x1 . x1 4 or x1 3 Write . Solve for x. or The solution set is .E- 1 3, 1 4 F x = - 1 3 x = 1 4 1 x = 4 or 1 x = -3x-1 as 1 x Technology Tip Use a graphing utility to graph y1 x2 x1 12. The x-intercepts are the solutions to this equation. YO U R TU R N Find the solutions to the equation x2 x1 6 0. Study Tip Remember to transform back to the original variable. Answer: The solution is or . The solution set is .E- 1 2, 1 3 Fx = 1 3 x = - 1 2 E X AM P LE 6 Solving an Equation Quadratic in Form with Fractional Exponents Find the solutions to the equation x2/3 3x1/3 10 0. Solution: Rewrite the original equation. 3x1/3 10 0 Identify the substitution as u x1/3 . u2 3u 10 0 (x1/3 ) 2 Factor. (u 5)(u 2) 0 Solve for u. u 5 or u 2 Let u x1/3 again. x1/3 5 x1/3 2 Cube both sides of the equations. (5)3 (2)3 Simplify. x 125 x 8 The solution set is {8, 125} , which a check will conrm. (x1/3 ) 3 (x1/3 ) 3 YO U R TU R N Find the solution to the equation 2t 5t1/2 3 0. Answer: t 9 or {9}. c01a.qxd 11/23/11 5:34 PM Page 135 165. 136 C HAP TE R 1 Equations and Inequalities Technology Tip Use a graphing utility to graph .y1 = x7/3 - 3x4/3 - 4x1/3 The x-intercepts are the solutions to this equation. YO U R TU R N Solve the equation x3 x2 4x 4 0. Answer: x 1 or x 2 or {2, 1, 2} ; Factorable Equations Some equations (both polynomial and with rational exponents) that are factorable can be solved using the zero product property. Equations Quadratic in Form: Identify the u-substitution that transforms the equation into a quadratic equation. Solve the quadratic equation and then remember to transform back to the original variable. Factorable Equations: Look for a factor common to all terms or factor by grouping. Radical equations, equations quadratic in form, and factorable equations can often be solved by transforming them into simpler linear or quadratic equations. Radical Equations: Isolate the term containing a radical and raise it to the appropriate power that will eliminate the radical. If there is more than one radical, it does not matter which radical is isolated rst. Raising radical equations to powers may cause extraneous solutions, so check each solution. S E CTI O N 1.4 S U M MARY E X AM P LE 7 Solving an Equation with Rational Exponents by Factoring Solve the equation x7/3 3x4/3 4x1/3 0. Solution: Factor the left side of the equation. x1/3 (x2 3x 4) 0 Factor the quadratic expression. x1/3 (x 4)(x 1) 0 Apply the zero product property. x1/3 0 or x 4 0 or x 1 0 Solve for x. x 0 or x 4 or x 1 The solution set is {1, 0, 4}. E X AM P LE 8 Solving a Polynomial Equation Using Factoring by Grouping Solve the equation x3 2x2 x 2 0. Solution: Factor by grouping (Chapter 0). 0 Identify the common factors. 2 0 Factor. (x 2) 0 Factor the quadratic expression. (x 2)(x 1)(x 1) 0 Apply the zero product property. x 2 0 or x 1 0 or x 1 0 Solve for x. x 2 or x 1 or x 1 The solution set is {2, 1, 1}. (x2 - 1) (x2 - 1)x(x2 - 1) (2x2 - 2)(x3 - x) c01a.qxd 11/23/11 5:34 PM Page 136 166. 1.4 Other Types of Equations 137 In Exercises 140, solve the radical equation for the given variable. 1. 2. 3. (4p 7)1/2 5 4. 11 (21 p)1/2 5. 6. 7. 8. 9. (4y 1)1/3 1 10. (5x 1)1/3 4 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. In Exercises 4170, solve the equations by introducing a substitution that transforms these equations to quadratic form. 41. x2/3 2x1/3 0 42. x1/2 2x1/4 0 43. x4 3x2 2 0 44. x4 8x2 16 0 45. 2x4 7x2 6 0 46. x8 17x4 16 0 47. (2x 1)2 5(2x 1) 4 0 48. (x 3)2 6(x 3) 8 0 49. 4(t 1)2 9(t 1) 2 50. 2(1 y)2 5(1 y) 12 0 51. x8 17x4 16 0 52. 2u2 5u1 12 0 53. 3y2 y1 4 0 54. 5a2 11a1 2 0 55. z2/5 2z1/5 1 0 56. 2x1/2 x1/4 1 0 57. (x 3)5/3 32 58. (x 2)4/3 16 59. (x 1)2/3 4 60. (x 7)4/3 81 61. 6t2/3 t1/3 1 0 62. t2/3 t1/3 6 0 63. 64. 65. 66. 67. u4/3 5u2/3 4 68. u4/3 5u2/3 4 69. 70. In Exercises 7186, solve by factoring. 71. x3 x2 12x 0 72. 2y3 11y2 12y 0 73. 4p3 9p 0 74. 25x3 4x 75. u5 16u 0 76. t5 81t 0 77. x3 5x2 9x 45 0 78. 2p3 3p2 8p 12 0 79. y(y 5)3 14(y 5)2 0 80. v(v 3)3 40(v 3)2 0 81. x9/4 2x5/4 3x1/4 0 82. u7/3 u4/3 20u1/3 0 83. t5/3 25t1/3 0 84. 4x9/5 9x1/5 0 85. y3/2 5y1/2 6y1/2 0 86. 4p5/3 5p2/3 6p1/3 0 u = 4 2-2u2 - 1t = 4 2t2 + 6 5 (2x + 1)2 - 3 (2x + 1) = 2a 1 2x - 1 b 2 + a 1 2x - 1 b - 12 = 0 1 (x + 1)2 + 4 (x + 1) + 4 = 03 = 1 (x + 1)2 + 2 (x + 1) 32 - 2x = 2x32 + 2x = 2x 2x + 5 = 1 + 2x - 223x - 5 = 7 - 2x + 2 28 - x = 2 + 22x + 322x - 1 - 2x - 1 = 1 15 - x + 13x + 1 = 41x + 12 + 18 - x = 6 12 - x + 16 - 5x = 613x + 1 - 16x - 5 = 1 22x2 - 8x + 1 = x - 32x2 - 2x - 5 = x + 1 225 - x2 = x + 12x2 - 4 = x - 121x + 1 - 3x = -531x + 4 - 2x = 9 2x - 41x + 1 = 43x - 61x + 2 = 35x - 101x + 2 = -103x - 61x - 1 = 3 22 - x = x - 221 - 3x = x + 128 - 2x = 2x - 222x + 6 = x + 3 -2s = 23 - ss = 32s - 21y = y 4 y = 51y x = 256 - x212 + x = x 3 21 - x = -23 25x + 2 = 3- 23 - 2u = 92u + 1 = -4 22t - 7 = 32t - 5 = 2 S K I LL S E X E R C I S E S S E CTI O N 1.4 c01a.qxd 5/16/12 7:34 PM Page 137 167. 138 C HAP TE R 1 Equations and Inequalities 95. Speed of Sound. A man buys a house with an old well but does not know how deep the well is. To get an estimate he decides to drop a rock in the opening of the well and time how long it takes until he hears the splash. The total elapsed time T given by T t1 t2, is the sum of the time it takes for the rock to reach the water, t1, and the time it takes for the sound of the splash to travel to the top of the well, t2. The time (seconds) that it takes for the rock to reach the water is given by , where d is the depth of the well in feet. Since the speed of sound is 1100 ft/s, the time (seconds) it takes for the sound to reach the top of the well is If the splash is heard after 3 seconds, how deep is the well? 96. Speed of Sound. If the owner of the house in Exercise 91 forgot to account for the speed of sound, what would he have calculated the depth of the well to be? 97. Physics: Pendulum. The period (T) of a pendulum is related to the length (L) of the pendulum and acceleration due to gravity (g) by the formula . If gravity is 9.8 m/s2 and the period is 1 second, nd the approximate length of the pendulum. Round to the nearest centimeter. Note: 100 cm 1 m. 98. Physics: Pendulum. The period (T) of a pendulum is related to the length (L) of the pendulum and acceleration due to gravity (g) by the formula . If gravity is 32 ft/s2 and the period is 1 second, nd the approximate length of the pendulum. Round to the nearest inch. Note: 12 in. 1 ft. In Exercises 99 and 100, refer to the following: Einsteins special theory of relativity states that time is relative: Time speeds up or slows down, depending on how fast one object is moving with respect to another. For example, a space probe traveling at a velocity v near the speed of light c will have clocked a time t hours, but for a stationary observer on Earth that corresponds to a time t0. The formula governing this relativity is given by 99. Physics: Special Theory of Relativity. If the time elapsed on a space probe mission is 18 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light. 100. Physics: Special Theory of Relativity. If the time elapsed on a space probe mission is 5 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light. t = t0 A 1 - v2 c2 T = 2p A L g T = 2p A L g t2 = d 1100 . t1 = 2d 4 In Exercises 87 and 88 refer to the following: An analysis of sales indicates that demand for a product during a calendar year is modeled by where d is demand in millions of units and t is the month of the year where t 0 represents January. 87. Economics. During which month(s) is demand 3 million units? 88. Economics. During which month(s) is demand 4 million units? In Exercises 89 and 90 refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function where w is weight in kilograms and h is height in centimeters. 89. Health. The BSA of a 72 kilogram female is 1.8. Find the height of the female to the nearest centimeter. 90. Health. The BSA of a 177 centimeter tall male is 2.1. Find the weight of the male to the nearest kilogram. 91. Insurance: Health. Cost for health insurance with a private policy is given by where C is the cost per day and a is the insureds age in years. Health insurance for a 6-year-old, a 6, is $4 a day (or $1460 per year). At what age would someone be paying $9 a day (or $3285 per year)? 92. Insurance: Life. Cost for life insurance is given by where C is the cost per day and a is theC = 15a + 1, C = 110 + a, BSA = A wh 3600 d = 31t + 1 - 0.75t A P P L I C AT I O N S MOGN Daily Apr May 40 50 60 70 5/28/04 insureds age in years. Life insurance for a newborn, a 0, is $1 a day (or $365 per year). At what age would someone be paying $20 a day (or $7300 per year)? 93. Stock Value. The stock price of MGI Pharmaceutical (MOGN) from March 2004 to June 2004 can be approximately modeled by the equation where P is the price ofP = 52t2 + 1 + 50, the stock in dollars and t is the month with t 0 correspond- ing to March 2004. Assuming this trend continues, when would the stock be worth $85? 94. Grades. The average combined math and verbal SAT score of incoming freshmen at a university is given by the equation where t is in years and t 0 corresponds to 1990. What year will the incoming class have an average SAT score of 1230? S = 1000 + 1012t, c01a.qxd 5/16/12 7:34 PM Page 138 168. 1.4 Other Types of Equations 139 In Exercises 101104, explain the mistake that is made. 101. Solve the equation Solution: This is incorrect. What mistake was made? 102. Solve the equation Solution: This is incorrect. What mistake was made? x = 2x = -1, (x - 2)(x + 1) = 0 x2 - x - 2 = 0 x2 = x + 2 x = 1x + 2. t = 5 3t = 15 3t + 1 = 16 13t + 1 = -4. 115. Solve the equation Plot both sides of the equation in the same viewing screen, and and zoom in on the x-coordinate of the point of intersection. Does the graph agree with your solution? 116. Solve the equation Plot both sides of the equation in the same viewing screen, and and zoom in on the x-coordinate of the points of intersection. Does the graph agree with your solution? 117. Solve the equation Plot both sides of the equation in the same viewing screen, y1 4 and Does the graph agree or disagree with your solution? y2 = 1x + 3. -4 = 1x + 3. y2 = 1 + 13 - x,y1 = 21x + 1 21x + 1 = 1 + 13 - x. y2 = 4 - 1x + 2, y1 = 1x - 3 1x - 3 = 4 - 1x + 2. C AT C H T H E M I S TA K E C O N C E P T U A L C H A L L E N G E 118. Solve the equation x1/4 4x1/2 21. Plot both sides of the equation in the same viewing screen, y1 x1/4 and y2 4x1/2 21. Does the point(s) of intersection agree with your solution? 119. Solve the equation Plot both sides of the equation in the same viewing screen, and Does the point(s) of intersection agree with your solution? 120. Solve the equation Plot both sides of the equation in the same viewing screen, and Does the point(s) of intersection agree with your solution? 121. Solve the equation Plot both sides of the equation in the same viewing screen, and Does the point(s) of intersection agree with your solution? y2 = 3x-1 - 10. y1 = x-2 x-2 = 3x-1 - 10. y2 = 3x-2 - 10. y1 = x-1 x-1 = 3x-2 - 10. y2 = -4x1/4 + 21. y1 = x1/2 x1/2 = -4x1/4 + 21. 103. Solve the equation x2/3 x1/3 20 0. Solution: u x1/3 u2 u 20 0 (u 5)(u 4) 0 x 5, x 4 This is incorrect. What mistake was made? 104. Solve the equation x4 2x2 3. Solution: x4 2x2 3 0 u x2 u2 2u 3 0 (u 3)(u 1) 0 u 1, u 3 u x2 x2 1, x2 3 x ;1, x ;3 This is incorrect. What mistake was made? In Exercises 105108, determine whether each statement is true or false. 105. The equation (2x 1)6 4(2x 1)3 3 0 is quadratic in form. 106. The equation t25 2t5 1 0 is quadratic in form. 107. If two solutions are found and one does not check, then the other does not check. 108. Squaring both sides of leads to x 2 x x 5. 1x + 2 + 1x = 1x + 5 109. Solve . 110. Solve . 111. Solve the equation without squaring both sides. 3x2 + 2x = 23x2 + 2x 2x2 = -x 2x2 = x 112. Solve the equation 3x7/12 x5/6 2x1/3 0. 113. Solve the equation 114. Solve the equation .3 4 2x23 x1x = 2 1x + 6 + 111 + x = 513 + x. T E C H N O L O G Y c01a.qxd 5/16/12 7:34 PM Page 139 169. LINEAR INEQUALITIES SECTION 1.5 Graphing Inequalities and Interval Notation An example of a linear equation is 3x 2 7. On the other hand, 3x 2 7 is an example of a linear inequality. One difference between a linear equation and a linear inequality is that the equation has at most only one solution, or value of x, that makes the statement true, whereas the inequality can have a range or continuum of numbers that make the statement true. For example, the inequality x 4 denotes all real numbers x that are less than or equal to 4. Four inequality symbols are used. Symbol In Words Less than Greater than Less than or equal to Greater than or equal to We call and strict inequalities. For any two real numbers a and b, one of three things must be true: This property is called the trichotomy property of real numbers. If x is less than 5 (x 5) and x is greater than or equal to 2 (x 2), then we can represent this as a double (or combined) inequality, 2 x 5, which means that x is greater than or equal to 2 and less than 5. We will express solutions to inequalities in four ways: an inequality, a solution set, an interval, and a graph. The following are ways of expressing all real numbers greater than or equal to a and less than b. a 6 b or a = b or a 7 b CONCEPTUAL OBJECTIVES Apply intersection and union concepts. Compare and contrast equations and inequalities. Understand that linear inequalities may have one solution, no solution, or an interval solution. SKILLS OBJECTIVES Use interval notation. Solve linear inequalities. Solve application problems involving linear inequalities. a b or a b [ ) Inequality Solution Interval Notation Set Notation Graph/Number Line a x b {x | a x b} [a, b) In this example, a is referred to as the left endpoint and b is referred to as the right endpoint. If an inequality is a strict inequality ( or ), then the graph and interval notation use parentheses. If it includes an endpoint ( or ), then the graph and interval notation use brackets. Number lines are drawn with either closed/open circles or brackets/parentheses. In this text the brackets/parentheses notation will be used. Intervals are classied as follows: Open ( , ) Closed [ , ] Half open ( , ] or [ , ) 140 c01b.qxd 12/22/11 6:40 PM Page 140 170. Inequality Interval Graph a. x 3 (3, ) b. x 5 (, 5] c. 1 x 4 [1, 4) d. 0 x 4 [0, 4] 1.5 Linear Inequalities 141 LET X BE A REAL NUMBER. X IS... INEQUALITY SET NOTATION INTERVAL GRAPH greater than a and less than b a x b {x | a x b} (a, b) greater than or equal to a and less than b a x b {x | a x b} [a, b) greater than a and less than or equal to b a x b {x | a x b} (a, b] greater than or equal to a and less than or equal to b a x b {x | a x b} [a, b] less than a x a {x | x a} (, a) less than or equal to a x a {x | x a} (, a] greater than b x b {x | x b} (b, ) greater than or equal to b x b {x | x b} [b, ) all real numbers R R (, ) 1. Innity () is not a number. It is a symbol that means continuing indenitely to the right on the number line. Similarly, negative innity () means continuing indenitely to the left on the number line. Since both are unbounded, we use a parenthesis, never a bracket. 2. In interval notation, the lower number is always written to the left. Write the inequality in interval notation: 1 x 3. INCORRE CT (3, 1]CORRECT [1, 3) EXAMPLE 1 Expressing Inequalities Using Interval Notation and a Graph Express the following as an inequality, an interval, and a graph. a. x is greater than 3. b. x is less than or equal to 5. c. x is greater than or equal to 1 and less than 4. d. x is greater than or equal to 0 and less than or equal to 4. Solution: Since the solutions to inequalities are sets of real numbers, it is useful to discuss two operations on sets called intersection and union. 1 3 c01b.qxd 12/22/11 6:40 PM Page 141 171. 142 CHAPTER 1 Equations and Inequalities The union of sets A and B, denoted is the set formed by combining all the elements in A with all the elements in B. The intersection of sets A and B, denoted is the set formed by the elements that are in both A and B. The notation x | x is in is read all x such that x is in. The vertical line represents such that. {x | x is in A and B}A B = A B, {x | x is in A or B or both}AB = A B, Union and IntersectionDE F I N I T I O N As an example of intersection and union, consider the following sets of people: {Austin, Brittany, Jonathan} {Anthony, Brittany, Elise} Intersection: {Brittany} Union: {Anthony, Austin, Brittany, Elise, Jonathan}AB = A B = B =A = YOUR TURN If C [3, 3) and D (0, 5], nd and Express the intersection and union in interval notation, and graph. C D.CD Solving Linear Inequalities As mentioned at the beginning of this section, if we were to solve the equation 3x 2 7, we would add 2 to both sides, divide by 3, and nd that x 3 is the solution, the only value that makes the equation true. If we were to solve the linear inequality 3x 2 7, we would follow the same procedure: add 2 to both sides, divide by 3, and nd that x 3, which is an interval or range of numbers that make the inequality true. In solving linear inequalities we follow the same procedures that we used in solving linear equations with one general exception: if you multiply or divide an inequality by a Study Tip If you multiply or divide an inequality by a negative number, remember to change the direction of the inequality sign. Set Interval notation A [3, 2] B (1, 7) [3, 7) (1, 2]A B AB EXAMPLE 2 Determining Unions and Intersections: Intervals and Graphs If A [3, 2] and B (1, 7), determine and Write these sets in interval notation, and graph. Solution: A B.AB c01b.qxd 11/23/11 5:36 PM Page 142 172. 1.5 Linear Inequalities 143 Procedures That Do Not Change the Inequality Sign 1. Simplifying by eliminating parentheses 3(x 6) 6x x and collecting like terms. 3x 18 5x 2. Adding or subtracting the same 7x 8 29 quantity on both sides. 7x 21 3. Multiplying or dividing by the 5x 15 same positive real number. x 3 Procedures That Change (Reverse) the Inequality Sign 1. Interchanging the two sides x 4 is equivalent to 4 x of the inequality. 2. Multiplying or dividing by the 5x 15 is equivalent to x 3 same negative real number. INEQUALITY PROPERTIES EXAMPLE 3 Solving a Linear Inequality Solve and graph the inequality 5 3x 23. Solution: Write the original inequality. 5 3x 23 Subtract 5 from both sides. 3x 18 Divide both sides by 3 and reverse the inequality sign. Simplify. Solution set: {x | x 6} Interval notation: (6, ) Graph: x 7 -6 -3x -3 7 18 -3 YOUR TURN Solve the inequality 5 3 2x. Express the solution in set and interval notation, and graph. negative number, then you must change the direction of the inequality sign. For example, if 2x 10, then the solution set includes real numbers such as x 6 and x 7. Note that real numbers such as x 6 and x 7 are not included in the solution set. Therefore, when this inequality is divided by 2, the inequality sign must also be reversed: x 5. If a b, then ac bc if c 0 and ac bc if c 0. The most common mistake that occurs when solving an inequality is forgetting to change the direction of, or reverse, the inequality symbol when the inequality is multiplied or divided by a negative number. Technology Tip Use a graphing utility to display graphs of y1 5 3x and y2 23. The solutions are the x-values such that the graph of y1 5 3x is below that of y2 23. Answer: Solution set: {x | x 1} Interval notation: (,1] c01b.qxd 11/23/11 5:36 PM Page 143 173. 144 CHAPTER 1 Equations and Inequalities INCORRE CT Cross multiply. 3(4 3x) 2(5x) The error is in cross multiplying. CORRECT Eliminate the fractions by multiplying by the LCD, 6. Simplify. Eliminate the parentheses. Subtract 9x from both sides. x 12 10x 12 + 9x 10x 3(4 + 3x) 6a 5x 3 b 6a 4 + 3x 2 b A common mistake is using cross multiplication to solve inequalities. The reason cross multiplication should not be used is because the expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. C O M M O N M I S T A K E C A U T I O N Cross multiplication should not be used in solving inequalities. Although it is not possible to check inequalities since the solutions are often intervals, it is possible to conrm that some points that lie in your solution do satisfy the inequality. It is important to remember that cross multiplication cannot be used in solving inequalities. EXAMPLE 5 Solving a Double Linear Inequality Solve the inequality 2 3x 4 16. Solution: This double inequality can be written as two inequalities. 2 3x 4 16 Both inequalities must be satised. 2 3x 4 and 3x 4 16 Subtract 4 from both sides of each inequality. 6 3x and 3x 12 Divide each inequality by 3. 2 x and x 4 Combining these two inequalities gives us 2 x 4 in inequality notation; in interval notation we have or (2, 4]. Notice that the steps we took in solving these inequalities individually were identical. This leads us to a shortcut method in which we solve them together: Write the combined inequality. 2 3x 4 16 Subtract 4 from each part. 6 3x 12 Divide each part by 3. 2 x 4 Interval notation: (2, 4] For the remainder of this section we will use the shortcut method for solving inequalities. (2, )(, 4] Technology Tip Use a graphing utility to display graphs of y1 2, y2 3x 4, and y3 16. The solutions are the x-values such that the graph of y2 3x 4 is between the graphs of y1 2 and y3 16 and overlaps that of y3 16. EXAMPLE 4 Solving Linear Inequalities with Fractions Solve the inequality 5x 3 4 + 3x 2 . c01b.qxd 11/23/11 5:36 PM Page 144 174. 1.5 Linear Inequalities 145 EXAMPLE 6 Solving a Double Linear Inequality Solve the inequality Express the solution set in interval notation, and graph. Solution: Write the original double inequality. Multiply each part by 7. Add 2 to each part. Divide each part by 3 and reverse the inequality signs. 3 x 10 Write in standard form. 10 x 3 Interval notation: (10, 3] Graph: 9 -3x 6 30 7 -2 - 3x 6 28 1 -2 - 3x 7 6 4 1 -2 - 3x 7 6 4. Technology Tip Use a graphing utility to display graphs of y1 23, and y3 86. y2 = 32 + 9 5 x, The solutions are the x-values such that the graph of isy2 = 32 + 9 5 x between the graphs of y1 23 and y3 86 and overlaps both graphs. EXAMPLE 7 Solving a Double Linear Inequality Solve the inequality Express the solution in interval notation. Solution: Subtract x from all three parts. Add 4 to all three parts. Divide all three parts by 3. Express the solution in interval notation. [1, 4] 1 x 4 3 3x 12 -1 3x - 4 8 x - 1 4x - 4 x + 8. YOUR TURN Solve the inequality 2x 1 4x 2 2x 5. Express the solution in interval notation. Answer: Q-1 2, 3 2 R Applications Involving Linear Inequalities EXAMPLE 8 Temperature Ranges New York City on average has a yearly temperature range of 23 degrees Fahrenheit to 86 degrees Fahrenheit. What is the range in degrees Celsius given that the conversion relation is Solution: The temperature ranges from 23F to 86F. 23 F 86 Replace F using the Celsius conversion. Subtract 32 from all three parts. Multiply all three parts by 5 C 30 New York City has an average yearly temperature range of 5C to 30C . 5 9. -9 9 5 C 54 23 32 + 9 5 C 86 F = 32 + 9 5 C? c01b.qxd 11/23/11 5:36 PM Page 145 175. 146 CHAPTER 1 Equations and Inequalities Linear inequalities are solved using the same procedures as linear equations with one exception: When you multiply or divide by a negative number, you must reverse the inequality sign. Note: Cross multiplication cannot be used with inequalities. SUMMARY The solutions of linear inequalities are solution sets that can be expressed four ways: 1. Inequality notation a x b 2. Set notation {x | a x b} 3. Interval notation (a, b] 4. Graph (number line) SECTION 1.5 EXERCISES SECTION 1.5 In Exercises 116, rewrite in interval notation and graph. 1. x 3 2. x 2 3. x 5 4. x 7 5. 2 x 3 6. 4 x 1 7. 3 x 5 8. 0 x 6 9. 0 x 0 10. 7 x 7 11. x 6 and x 4 12. x 3 and x 2 13. x 6 and x 8 14. x 8 and x 2 15. x 4 and x 2 16. x 5 and x 6 In Exercises 1724, rewrite in set notation. 17. [0, 2) 18. (0, 3] 19. (7, 2) 20. [3, 2] 21. (, 6] 22. (5, ) 23. (, ) 24. [4, 4] EXAMPLE 9 Comparative Shopping Two car rental companies have advertised weekly specials on full-size cars. Hertz is advertising an $80 rental fee plus an additional $0.10 per mile. Thrifty is advertising $60 and $0.20 per mile. How many miles must you drive for the rental car from Hertz to be the better deal? Solution: Let x number of miles driven during the week. Write the cost for the Hertz rental. 80 0.1x Write the cost for the Thrifty rental. 60 0.2x Write the inequality if Hertz is less than Thrifty. 80 0.1x 60 0.2x Subtract 0.1x from both sides. 80 60 0.1x Subtract 60 from both sides. 20 0.1x Divide both sides by 0.1. 200 x You must drive more than 200 miles for Hertz to be the better deal. SKILLS c01b.qxd 11/23/11 5:36 PM Page 146 176. In Exercises 5158, write in interval notation. 51. 52. 53. 54. 55. 56. 57. 58. In Exercises 5990, solve and express the solution in interval notation. 59. x 3 7 60. x 4 9 61. 3x 2 4 62. 3x 7 8 63. 5p 10 64. 4u 12 65. 3 2x 7 66. 4 3x 17 67. 1.8x 2.5 3.4 68. 2.7x 1.3 6.8 69. 3(t 1) 2t 70. 2(y 5) 3(y 4) 71. 7 2(1 x) 5 3(x 2) 72. 4 3(2 x) 5 73. 74. 75. 76. 77. 78. 79. 2 x 3 5 80. 1 x 6 12 81. 8 4 2x 8 82. 0 2 x 5 83. 3 1 x 9 84. 3 2 5x 13 85. 86. 87. 88. 89. 0.7 0.4x 1.1 1.3 90. 7.1 4.7 1.2x 1.1-1 6 2 - z 4 1 5 1 2 1 + y 3 3 4 3 6 1 2A - 3 6 70 6 2 - 1 3y 6 4 s 2 - (s - 3) 3 7 s 4 - 1 12 2 3 y - 1 2 (5 - y) 6 5y 3 - (2 + y) 2p + 1 5 7 -3 t - 5 3 -4 y - 3 5 - 2 y 4 x + 2 3 - 2 x 2 1.5 Linear Inequalities 147 In Exercises 2532, write in inequality and interval notation. 25. 26. 27. 28. 29. 30. 31. 32. In Exercises 3350, graph the indicated set and write as a single interval, if possible. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. (-, -2)(-1, )(-6, -2)[1, 4) (-, )(-4, 7)(-, )(-3, 2](-2, 2)[-3, 1](-, -3][3, ) (-, -3][-3, )(-, 4)(4, )[-4, 5)[-2, 7)[-5, 2)[-1, 3] (-3, )[-5, )(-, 4)[1, )(-, -5)(-, 7](-, 1][-1, ) [-3, 1)[-6, 0)[-6, 4)[-2, 5)(2, 7)[-5, 3)(-5, 2](-1, 3) 0 5 1010 5 0 11 53 4 8 5 50 10 5 0 5 10 8 2 532 5 0 2 4 2 5 12 5 2 4 2 3 7 2 0 2 6 3 0 4 5 1 c01b.qxd 11/23/11 5:36 PM Page 147 177. 148 CHAPTER 1 Equations and Inequalities 99. Cost: Cell Phones. A cell phone company charges $50 for an 800-minute monthly plan, plus an additional $0.22 per minute for every minute over 800. If a customers bill ranged from a low of $67.16 to a high of $96.86 over a 6-month period, what were the most minutes used in a single month? What were the least? 100. Cost: Internet. An Internet provider charges $30 per month for 1000 minutes of DSL service plus $0.08 for each additional minute. In a one-year period the customers bill ranged from $36.40 to $47.20. What were the most and least minutes used? 101. Grades. In your general biology class, your rst three test scores are 67, 77, and 84. What is the lowest score you can get on the fourth test to earn at least a B for the course? Assume that each test is of equal weight and the minimum score required to earn a B is an 80. 102. Grades. In your Economics I class there are four tests and a nal exam, all of which count equally. Your four test grades are 96, 87, 79, and 89. What grade on your nal exam is needed to earn between 80 and 90 for the course? 103. Markups. Typical markup on new cars is 1530%. If the sticker price is $27,999, write an inequality that gives the range of the invoice price (what the dealer paid the manufacturer for the car). 104. Markups. Repeat Exercise 103 with a sticker price of $42,599. 105. Lasers. A circular laser beam with a radius rT is transmitted from one tower to another tower. If the received beam radius rR uctuates 10% from the transmitted beam radius due to atmospheric turbulence, write an inequality representing the received beam radius. 106. Electronics: Communications. Communication systems are often evaluated based on their signal-to-noise ratio (SNR), which is the ratio of the average power of received signal, S, to average power of noise, N, in the system. If the SNR is required to be at least 2 at all times, write an inequality representing the received signal power if the noise can uctuate 10%. 107. Real Estate. The Aguileras are listing their house with a real estate agent. They are trying to determine a listing price, L, for the house. Their realtor advises them that most buyers traditionally offer a buying price, B, that is 8595% of the listing price. Write an inequality that relates the buying price to the listing price. 108. Humidity. The National Oceanic and Atmospheric Administration (NOAA) has stations on buoys in the oceans to measure atmosphere and ocean characteristics such as temperature, humidity, and wind. The humidity sensors have an error of 5%. Write an inequality relating the measured humidity hm, and the true humidity ht. 91. Weight. A healthy weight range for a woman is given by the following formula: 110 pounds for the rst 5 feet (tall) 26 pounds per inch for every inch above 5 feet Write an inequality representing a healthy weight, w, for a 5 foot 9 inch woman. 92. Weight. NASA has more stringent weight allowances for its astronauts. Write an inequality representing allowable weight for a female 5 foot 9 inch mission specialist given 105 pounds for the rst 5 feet, and 15 pounds per inch for every additional inch. 93. Prot. A seamstress decides to open a dress shop. Her xed costs are $4000 per month, and it costs her $20 to make each dress. If the price of each dress is $100, how many dresses does she have to sell per month to make a prot? 94. Prot. Labrador retrievers that compete in eld trials typically cost $2000 at birth. Professional trainers charge $400 to $1000 per month to train the dogs. If the dog is a champion by age 2, it sells for $30,000. What is the range of prot for a champion at age 2? In Exercises 95 and 96 refer to the following: The annual revenue for a small company is modeled by where x is hundreds of units sold and R is revenue in thousands of dollars. 95. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $10 million in revenue. 96. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $7.5 million in revenue. In Exercises 97 and 98 refer to the following: The Target or Training Heart Rate (THR) is a range of heart rate (measured in beats per minute) that enables a persons heart and lungs to benet the most from an aerobic workout. THR can be modeled by the formula where HRmax is the maximum heart rate that is deemed safe for the individual, HRrest is the resting heart rate, and I is the intensity of the workout that is reported as a percentage. 97. Health. A female with a resting heart rate of 65 beats per minute has a maximum safe heart rate of 170 beats per minute. If her target heart rate is between 100 and 140 beats per minute, what percent intensities of workout can she consider? 98. Health. A male with a resting heart rate of 75 beats per minute has a maximum safe heart rate of 175 beats per minute. If his target heart rate is between 110 and 150 beats per minute, what percent intensities of workout can he consider? THR = (HRmax - HRrest) * I + HRrest R = 5000 + 1.75x A P P L I C A T I O N S c01b.qxd 12/22/11 6:40 PM Page 148 178. 109. Recreation: Golf. Two friends enjoy playing golf. Their favorite course charges $40 for greens fees (to play the course) and a $15 cart rental (per person), so it currently costs each of them $55 every time they play. The membership offered at that course is $160 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10 every time they play. What is the least number of times they should play a month in order for the membership to be the better deal? 110. Recreation: Golf. The same friends in Exercise 109 have a second favorite course. That golf course charges $30 for greens fees (to play the course) and a $10 cart rental (per person), so it currently costs each of them $40 every time they play. The membership offered at that course is $125 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10. What is the least number of times they should play a month in order for the membership to be the better deal? The following table is the 2007 Federal Tax Rate Schedule for people ling as single: TAX BRACKET # IF TAXABLE INCOME IS OVER BUT NOT OVER THE TAX IS: I $0 $7,825 10% of the amount over $0 II $7,825 $31,850 $782.50 plus 15% of the amount over 7,825 III $31,850 $77,100 $4,386.25 plus 25% of the amount over 31,850 IV $77,100 $160,850 $15,698.75 plus 28% of the amount over 77,100 V $160,850 $349,700 $39,148.75 plus 33% of the amount over 160,850 VI $349,700 No limit $101,469.25 plus 35% of the amount over 349,700 111. Federal Income Tax. What is the range of federal income taxes a person in tax bracket III will pay the IRS? 112. Federal Income Tax. What is the range of federal income taxes a person in tax bracket IV will pay the IRS? In Exercises 113116, explain the mistake that is made. 113. Rewrite in interval notation. This is incorrect. What mistake was made? 114. Graph the indicated set and write as a single interval if possible. This is incorrect. What mistake was made? C A T C H T H E M I S T A K E 115. Solve the inequality 2 3p 4 and express the solution in interval notation. Solution: This is incorrect. What mistake was made? 116. Solve the inequality 3 2x 7 and express the solution in interval notation. Solution: This is incorrect. What mistake was made? (-, -2] x -2 -2x 4 3 - 2x 7 (-, 2] p 2 -3p -6 2 - 3p -4 1.5 Linear Inequalities 149 2 0 2 4 6 2 0 2 4 6 c01b.qxd 11/23/11 5:36 PM Page 149 179. 150 CHAPTER 1 Equations and Inequalities In Exercises 123 and 124, select any of the statements a c that could be true. a. n 0 b. n 0 c. n 0 123. m n m n 124. m n m n In Exercises 117 and 118, determine whether each statement is true or false. 117. If x a, then a x. 118. If x a, then x a. In Exercises 119122, select any of the statements a d that could be true. a. m 0 and n 0 b. m 0 and n 0 c. m 0 and n 0 d. m 0 and n 0 119. mn 0 120. mn 0 121. 122. m n 6 0 m n 7 0 C O N C E P T U A L C H A L L E N G E 127. Solve the inequality ax b ax c, where 0 b c. 128. Solve the inequality ax b ax c, where 0 b c. 125. Solve the inequality x x mentally (without doing any algebraic manipulation). 126. Solve the inequality x x mentally (without doing any algebraic manipulation). 129. a. Solve the inequality 2.7x 3.1 9.4x 2.5. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do (a) and (b) agree? 130. a. Solve the inequality 0.5x 2.7 4.1x 3.6. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do (a) and (b) agree? 131. a. Solve the inequality x 3 2x 1 x 4. b. Graph all three expressions of the inequality in the same viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and below the graph of the right side. c. Do (a) and (b) agree? 132. a. Solve the inequality x 2 3x 4 2x 6. b. Graph all three expressions of the inequality in the same viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and on top of and below the graph of the right side. c. Do (a) and (b) agree? 133. a. Solve the inequality x 3 x 5. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do (a) and (b) agree? 134. a. Solve the inequality . b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do (a) and (b) agree? 1 2 x - 3 7 -2 3 x + 1 T E C H N O L O G Y c01b.qxd 1/20/12 4:55 PM Page 150 180. CONCEPTUAL OBJECTIVES Understand zeros and test intervals. Realize that a rational inequality has an implied domain restriction on the variable. POLYNOMIAL AND RATIONAL INEQUALITIES SKILLS OBJECTIVES Solve polynomial inequalities. Solve rational inequalities. SECTION 1.6 Polynomial Inequalities In this section we will focus primarily on quadratic inequalities, but the procedures outlined are also valid for higher degree polynomial inequalities. An example of a quadratic inequality is x2 x 2 0. This statement is true when the value of the polynomial on the left side is negative. For any value of x, a polynomial has either a positive, negative, or zero value. A polynomial must pass through zero before its value changes from positive to negative or from negative to positive. Zeros of a polynomial are the values of x that make the polynomial equal to zero. These zeros divide the real number line into test intervals where the value of the polynomial is either positive or negative. For example, if we set the above polynomial equal to zero and solve: we nd that x 2 and x 1 are the zeros. These zeros divide the real number line into three test intervals: (, 2), (2, 1), and (1, ). Since the polynomial is equal to zero at x 2 and x 1, the value of the polynomial in each of these three intervals is either positive or negative. We select one real number that lies in each of the three intervals and test to see whether the value of the polynomial at each point is either positive or negative. In this example, we select the real numbers: x 3, x 0, and x 2. At this point, there are two ways we can determine whether the value of the polynomial is positive or negative on the interval. One approach is to substitute each of the test points into the polynomial x2 x 2. The second approach is to simply determine the sign of the result as opposed to actually calculating the exact number. This alternate approach is often used when the expressions or test points get more complicated to evaluate. The polynomial is written as the product (x 2)(x 1); therefore, we simply look for the sign in each set of parentheses. (2 2)(2 1) (4)(1) S ()() ()x 2: (0 2)(0 1) (2)(1) S ()() ()x 0: (3 2)(3 1) (1)(4) S ()() ()x 3: (x + 2)(x - 1) x = 2 (2)2 + (2) - 2 = 4 + 2 - 2 = 4 Positive x = 0 (0)2 + (0) - 2 = 0 - 0 - 2 = -2 Negative x = -3 (-3)2 + (-3) - 2 = 9 - 3 - 2 = 4 Positive x = -2 or x = 1 (x + 2)(x - 1) = 0 x2 + x - 2 = 0 y x x = 1x = 2 x2 + x 2 < 0 x2 + x 2 > 0 x2 + x 2 > 0 3 0 2 151 c01b.qxd 11/23/11 5:36 PM Page 151 181. Step 1: Write the inequality in standard form. Step 2: Identify zeros. Step 3: Draw the number line with zeros labeled. Step 4: Determine the sign of the polynomial in each interval. Step 5: Identify which interval(s) make the inequality true. Step 6: Write the solution in interval notation. PROCEDURE FOR SOLVING POLYNOMIAL INEQUALITIES In this second approach we nd the same result: (, 2) and (1, ) correspond to a positive value of the polynomial, and (2, 1) corresponds to a negative value of the polynomial. In this example, the statement x2 x 2 0 is true when the value of the polynomial (in factored form), (x 2)(x 1), is negative. In the interval (2, 1), the value of the polynomial is negative. Thus, the solution to the inequality x2 x 2 0 is (2, 1). To check the solution, select any number in the interval and substitute it into the original inequality to make sure it makes the statement true. The value x 1 lies in the interval (2, 1). Upon substituting into the original inequality, we nd that x 1 satises the inequality (1)2 (1) 2 2 0. Study Tip If the original polynomial is 0, then the interval(s) that yield(s) negative products should be selected. If the original polynomial is 0, then the interval(s) that yield(s) positive products should be selected. Note: Be careful in Step 5. If the original polynomial is 0, then the interval(s) that correspond(s) to the value of the polynomial being negative should be selected. If the original polynomial is 0, then the interval(s) that correspond(s) to the value of the polynomial being positive should be selected. Technology Tip Use a graphing utility to display graphs of y1 x2 x and y2 12. The solutions are the x-values such that the graph of y1 lies above the graph of y2. EXAMPLE 1 Solving a Quadratic Inequality Solve the inequality x2 x 12. Solution: STEP 1 Write the inequality in standard form. x2 x 12 0 Factor the left side. (x 3)(x 4) 0 STEP 2 Identify the zeros. (x 3)(x 4) 0 x 3 or x 4 STEP 3 Draw the number line with the zeros labeled. STEP 4 Determine the sign of (x 3)(x 4) in each interval. STEP 5 Intervals in which the value of the polynomial is positive make this inequality true. (, 3) or (4, ) STEP 6 Write the solution in interval notation. (-, -3)(4, ) YOUR TURN Solve the inequality x2 5x 6 and express the solution in interval notation. Answer: [1, 6] 4 0 5 152 CHAPTER 1 Equations and Inequalities c01b.qxd 11/23/11 5:36 PM Page 152 182. 1.6 Polynomial and Rational Inequalities 153 The inequality in Example 1, x2 x 12, is a strict inequality, so we use parentheses when we express the solution in interval notation . It is important to note(-, -3)(4, ) that if we change the inequality sign from to , then the zeros x 3 and x 4 also make the inequality true. Therefore the solution to x2 x 12 is (-, -3][4, ). EXAMPLE 2 Solving a Quadratic Inequality Solve the inequality x2 4. INCORRE CT ERROR: Take the square root of both sides. x 2; CORRECT Step 1: Write the inequality in standard form. Factor. Step 2: Identify the zeros. Step 3: Draw the number line with the zeros labeled. Step 4: Determine the sign of , in each interval. Step 5: Intervals in which the value of the polynomial is negative make the inequality true. The endpoints, x 2 and x 2, satisfy the inequality, so they are included in the solution. Step 6: Write the solution in interval notation. [2, 2] When solving quadratic inequalities, you must rst write the inequality in standard form and then factor to identify zeros. (2, 2) 3 30 (x + 2)(x - 2) 2 2 x = 2 and x = -2 (x - 2)(x + 2) = 0 (x - 2)(x + 2) 0 x2 - 4 0 Do not take the square root of both sides. You must write the inequality in standard form and factor. C O M M O N M I S T A K E C A U T I O N The square root method cannot be used for quadratic inequalities. Not all inequalities have a solution. For example, x2 0 has no real solution. Any real number squared is always nonnegative, so there are no real values that when squared will yield a negative number. The zero is x 0, which divides the real number line into two intervals: (, 0) and (0, ). Both of these intervals, however, correspond to the value of x2 being positive, so there are no intervals that satisfy the inequality. We say that this inequality has no real solution. c01b.qxd 12/22/11 6:40 PM Page 153 183. 154 CHAPTER 1 Equations and Inequalities EXAMPLE 3 Solving a Quadratic Inequality Solve the inequality x2 2x 3. Solution: STEP 1 Write the inequality in standard form. x2 2x 3 0 STEP 2 Identify the zeros. x2 2x 3 0 Apply the quadratic formula. Simplify. Since there are no real zeros, the quadratic expression x2 2x 3 never equals zero; hence its value is either always positive or always negative. If we select any value for x, say, x 0, we nd that (0)2 2(0) 3 0. Therefore the quadratic expression is always positive, so the solution is the set of all real numbers, .(-, ) x = -2 ; 1-8 2 = -2 ; 2i12 2 = -1 ; i12 x = -2 ; 222 - 4(1)(3) 2(1) Study Tip When solving quadratic inequalities, you must rst write the inequality in standard form and then factor to identify zeros. INCORRE CT Write the original inequality. ERROR: Divide both sides by x. Dividing by x is the mistake. If x is negative, the inequality sign must be reversed. What if x is zero? x 7 -5 x2 7 -5x CORRECT Step 1: Write the inequality in standard form. Factor. Step 2: Identify the zeros. Step 3: Draw the number line with the zeros labeled. Step 4: Determine the sign of in each interval. Step 5: Intervals in which the value of the polynomial is positive satisfy the inequality. Step 6: Express the solution in interval notation. (-, -5)(0, ) (, 5) and (0, ) 5 0 ( ()( ) ( ) )( ) ( ) ( )( ) ( ) 6 1 1 x(x + 5) 5 0 x = 0, x = -5 x(x + 5) 7 0 x2 + 5x 7 0 A common mistake is to divide by x. Never divide by a variable, because the value of the variable might be zero. Always start by writing the inequality in standard form and then factor to determine the zeros. C O M M O N M I S T A K E C A U T I O N Do not divide inequalities by a variable. EXAMPLE 4 Solving a Quadratic Inequality Solve the inequality x2 5x. c01b.qxd 11/23/11 5:36 PM Page 154 184. 1.6 Polynomial and Rational Inequalities 155 EXAMPLE 5 Solving a Quadratic Inequality Solve the inequality x2 2x 1. Solution: Write the inequality in standard form. x2 2x 1 0 Identify the zeros. x2 2x 1 0 Apply the quadratic formula. Simplify. Draw the number line with the intervals labeled. Note: Test each interval. Intervals in which the value of the polynomial is negative make this inequality true. A -1 - 12, -1 + 12B (-, -1 - 12) x = -3: (-3)2 + 2(-3) - 1 = 2 7 0 (-1 - 12, -1 + 12) x = 0: (0)2 + 2(0) - 1 = -1 6 0 (-1 + 12, ) x = 1: (1)2 + 2(1) - 1 = 2 7 0 -1 + 12 L 0.41 -1 - 12 L -2.41 x = -2 ; 28 2 = -2 ; 222 2 = -1 ; 22 x = -2 ; 222 - 4(1)(-1) 2(1) YOUR TURN Solve the inequality x2 2x 1. YOUR TURN Solve the inequality x3 x2 6x 0. EXAMPLE 6 Solving a Polynomial Inequality Solve the inequality x3 3x2 10x. Solution: Write the inequality in standard form. x3 3x2 10x 0 Factor. x(x 5)(x 2) 0 Identify the zeros. x 0, x 5, x 2 Draw the number line with the zeros (intervals) labeled. Test each interval. Intervals in which the value of the polynomial is positive make this inequality true. [-2, 0][5, ) Answer: Q-, 1 - 12 D C1 + 12,R Answer: (- , -2 )(0, 3) Technology Tip Using a graphing utility, graph y1 x2 2x and y2 1. The solutions are the x-values such that the graph of y1 lies below the graph of y2. Note that: 2 c01b.qxd 11/23/11 5:36 PM Page 155 185. 156 CHAPTER 1 Equations and Inequalities Rational Inequalities Rational expressions have numerators and denominators. Recalling the properties of negative real numbers (Chapter 0), we see that the following possible combinations correspond to either positive or negative rational expressions. A rational expression can change signs if either the numerator or denominator changes signs. In order to go from positive to negative or vice versa, you must pass through zero. Therefore, to solve rational inequalities such as we use a similar procedure to the one used for solving polynomial inequalities, with one exception. You must eliminate values for x that make the denominator equal to zero. In this example, we must eliminate x 2 and x 2 because these values make the denominator equal to zero. Rational inequalities have implied domains. In this example, is a domain restriction and these values (x 2 and x 2) must be eliminated from a possible solution. We will proceed with a similar procedure involving zeros and test intervals that was outlined for polynomial inequalities. However, in rational inequalities once expressions are combined into a single fraction, any values that make either the numerator or the denominator equal to zero divide the number line into intervals. EXAMPLE 7 Solving a Rational Inequality Solve the inequality Solution: Factor the denominator. State the domain restrictions on the variable. Identify the zeros of numerator and denominator. x 2, x 2, x 3 Draw the number line and divide into intervals. Test the intervals. Intervals in which the value of the rational (2, 2) and (3, ) expression is positive satisfy this inequality. Since this inequality is greater than or equal to, we include x 3 in our solution because it satises the inequality. However, x 2 and x 2 are not included in the solution because they make the denominator equal to zero. The solution is .(-2, 2)[3, ) x Z 2, x Z -2 (x - 3) (x - 2)(x + 2) 0 x - 3 x2 - 4 0. x Z ;2 x - 3 x2 - 4 0 (+) (-) = (-) (-) (-) = (+) (-) (+) = (-) (+) (+) = (+) Study Tip Values that make the denominator equal to zero are always excluded. YOUR TURN Solve the inequality x + 2 x - 1 0. Technology Tip Use a graphing utility to display graph of .y1 = x - 3 x2 - 4 The solutions are the x-values such that the graph of y1 lies on top and above the x-axis, excluding x 2.; Answer: [2, 1) 3 0 2.5 4 c01b.qxd 12/22/11 6:40 PM Page 156 186. 1.6 Polynomial and Rational Inequalities 157 YOUR TURN Solve the inequality x + 4 x2 + 25 0. EXAMPLE 8 Solving a Rational Inequality Solve the inequality Solution: Identify the zero(s) of the numerator. x 5 Note that the denominator is never equal to zero when x is any real number. Draw the number line and divide into intervals. Test the intervals. The denominator is always positive. Intervals in which the value of the rational expression is negative satisfy the inequality. Notice that x 5 is not included in the solution because of the strict inequality. (-, 5) x - 5 x2 + 9 6 0. 5 5 (()) (()) (()) (()) (()) (()) Answer: [4, ) INCORRE CT ERROR: Do not cross multiply. x 3(x + 2) CORRECT Subtract 3 from both sides. Write as a single rational expression. Eliminate the parentheses. Simplify the numerator. Factor the numerator. -2(x + 3) x + 2 0 -2x - 6 x + 2 0 x - 3x - 6 x + 2 0 x - 3(x + 2) x + 2 0 x x + 2 - 3 0 Do not cross multiply. The LCD or expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. C O M M O N M I S T A K E EXAMPLE 9 Solving a Rational Inequality Solve the inequality . x x + 2 3 C A U T I O N Rational inequalities should not be solved using cross multiplication. c01b.qxd 11/23/11 5:36 PM Page 157 187. 158 CHAPTER 1 Equations and Inequalities Identify the zeros of the numerator and the denominator. Draw the number line and test the intervals. Intervals in which the value of the rational expression is negative satisfy the inequality, (, 3] and (2, ). Note that x 2 is not included in the solution because it makes the denominator zero, and x 3 is included because it satises the inequality. The solution is: (-, -3](-2, ) -2(x + 3) x + 2 0 x = -3 and x = -2 Applications EXAMPLE 10 Stock Prices From November 2003 until March 2004 Wal-Mart stock (WMT) was worth approximately P 2t2 12t 70, where P denotes the price of the stock in dollars and t corresponds to months. November 2003 is represented by t 1, December 2003 by t 2, January 2004 by t 3, and so on. During what months was the stock value at least $54? WMT Daily J J A S O N D 04 F M A M 50 55 60 65 5/27/04 Solution: Set the price greater than or equal to 54. 2t2 12t 70 54 1 t 5 Write in standard form. 2t2 12t 16 0 Divide by 2. (t2 6t 8) 0 Factor. (t 4)(t 2) 0 Identify the zeros. t 4 and t 2 Test the intervals. (t 4)(t 2) 0 Positive intervals satisfy the inequality. [1, 2] and [4, 5] The Wal-Mart stock price was at least $54 during November 2003, December 2003, February 2004, and March 2004. c01b.qxd 11/23/11 5:36 PM Page 158 188. 1.6 Polynomial and Rational Inequalities 159 Rational Inequality Write as a single fraction. Determine values that make the numerator or denominator equal to zero. Always exclude values that make the denominator 0. 3. Draw the number line labeling the intervals. 4. Test the intervals to determine whether they are positive or negative. 5. Select the intervals according to the sign of the inequality. 6. Write the solution in interval notation. SUMMARY SECTION 1.6 The following procedure can be used for solving polynomial and rational inequalities. 1. Write in standard formzero on one side. 2. Determine the zeros; if it is a rational function, note the domain restrictions. Polynomial Inequality Factor if possible. Otherwise, use the quadratic formula. In Exercises 128, solve the polynomial inequality and express the solution set in interval notation. 1. x2 3x 10 0 2. x2 2x 3 0 3. u2 5u 6 0 4. u2 6u 40 0 5. p2 4p 3 6. p2 2p 15 7. 2t2 3 t 8. 3t2 5t 2 9. 5v 1 6v2 10. 12t2 37t 10 11. 2s2 5s 3 12. 8s 12 s2 13. y2 2y 4 14. y2 3y 1 15. x2 4x 6 16. x2 2x 5 17. u2 3u 18. u2 4u 19. 2x x2 20. 3x x2 21. x2 9 22. x2 16 23. t2 81 24. t2 49 25. z2 16 26. z2 2 27. y2 4 28. y2 25 In Exercises 2958, solve the rational inequality and graph the solution on the real number line. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 2 2p - 3 - 1 p + 1 1 2p2 - p - 3 1 p - 2 - 1 p + 2 3 p2 - 4 1 p - 3 - 1 p + 3 2 1 p + 4 + 1 p - 4 7 p2 - 48 p2 - 16 2 x - 5 - 1 x - 1 0 3 x + 4 - 1 x - 2 0 1 t - 2 + 1 t + 2 0 2 t - 3 + 1 t + 3 0 v2 - 1 v + 1 0 v2 - 9 v - 3 0 - x2 + 2 x2 + 4 6 0 x2 + 10 x2 + 16 7 0 x2 5 + x2 0 x2 5 + x2 6 0 - 7p p2 - 100 p + 2 p + 10 3p - 2p2 4 - p2 6 3 + p 2 - p -2t - t2 4 - t t 3t2 t + 2 5t u2 - 3u 3 6 2u2 + u 3 6 1 1 - x x2 - 9 0 x - 3 x2 - 25 0 s + 5 4 - s2 0 s + 1 4 - s2 0 2t - 5 t - 6 6 0 t + 3 t - 4 0 y 2 - y 0 y y + 3 7 0 3 x 0- 3 x 0 SKILLS EXERCISES SECTION 1.6 c01b.qxd 11/23/11 5:36 PM Page 159 189. 160 CHAPTER 1 Equations and Inequalities 59. Prot. A Web-based embroidery company makes monogrammed napkins. The prot associated with producing x orders of napkins is governed by the equation Determine the range of orders the company should accept in order to make a prot. 60. Prot. Repeat Exercise 59 using P x2 130x 3600. 61. Car Value. The term upside down on car payments refers to owing more than a car is worth. Assume you buy a new car and nance 100% over 5 years. The difference between the value of the car and what is owed on the car is governed by the expression where t is age (in years) of the car. Determine the time period when the car is worth more than you owe . When do you owe more than its worth 62. Car Value. Repeat Exercise 61 using the expression . 63. Bullet Speed. A .22-caliber gun res a bullet at a speed of 1200 feet per second. If a .22-caliber gun is red straight upward into the sky, the height of the bullet in feet is given by the equation h 16t2 1200t, where t is the time in seconds with t 0 corresponding to the instant the gun is red. How long is the bullet in the air? 64. Bullet Speed. A .38-caliber gun res a bullet at a speed of 600 feet per second. If a .38-caliber gun is red straight upward into the sky, the height of the bullet in feet is given by the equation h 16t2 600t. How many seconds is the bullet in the air? 65. Geometry. A rectangular area is fenced in with 100 feet of fence. If the minimum area enclosed is to be 600 square feet, what is the range of feet allowed for the length of the rectangle? - 2 - t 4 - t a t t - 3 6 0b? a t t - 3 7 0b t t - 3 P = -x2 + 130x - 3000 A P P L I C A T I O N S 66. Stock Value. From June 2003 until April 2004, JetBlue airlines stock (JBLU) was approximately worth P 4t2 80t 360, where P denotes the price of the stock in dollars and t corresponds to months, with t 1 corresponding to January 2003. During what months was the stock value at least $36? In Exercises 67 and 68 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly prot. The function models the effect that a price increase of x dollars on a bottle of wine will have on the prot P measured in dollars. 67. Economics. What price increase will lead to a weekly prot of less than $460? 68. Economics. What price increases will lead to a weekly prot of more than $550? 69. Real Estate. A woman is selling a piece of land that she advertises as 400 acres ( 7 acres) for $1.36 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar. 70. Real Estate. A woman is selling a piece of land that she advertises as 1000 acres ( 10 acres) for $1 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar. ; ; P = -5(x + 3)(x - 24) JBLU Daily J J A S O N D 04 F M A M 20 30 40 50 5/27/04 In Exercises 7174, explain the mistake that is made. 71. Solve the inequality 3x x2 . Solution: Divide by x. 3 x Write the solution in interval notation. (3, ) This is incorrect. What mistake was made? 72. Solve the inequality u2 25. Solution: Take the square root of both sides. u 5 Write the solution in interval notation. (, 5) This is incorrect. What mistake was made? C A T C H T H E M I S T A K E c01b.qxd 11/23/11 5:36 PM Page 160 190. 1.6 Polynomial and Rational Inequalities 161 77. Assume the quadratic inequality ax2 bx c 0 is true. If b2 4ac 0, then describe the solution. 78. Assume the quadratic inequality ax2 bx c 0 is true. If b2 4ac 0, then describe the solution. In Exercises 75 and 76, determine whether each statement is true or false. Assume that a is a positive real number. C O N C E P T U A L 75. If x a2 , then the solution is (, a). 76. If x a2 , then the solution is [a, ). C H A L L E N G E In Exercises 7982, solve for x given that a and b are both positive real numbers. 79. x2 a2 80. 81. 82. a x2 6 -b x2 + a2 x2 + b2 0 x2 - b2 x + b 6 0 88. x2 3x 5 x2 2x 10 89. 90. 3p 4 - p 6 1 2p 5 - p 7 1 T E C H N O L O G Y In Exercises 8390, plot the left side and the right side of each inequality in the same screen and use the zoom feature to determine the range of values for which the inequality is true. 83. 1.4x2 7.2x 5.3 8.6x 3.7 84. 17x2 50x 19 9x2 2 85. 11x2 8x 16 86. 0.1x 7.3 0.3x2 4.1 87. x x2 3x 6 2x 73. Solve the inequality Solution: Factor the numerator and denominator. Cancel the (x 2) common factor. x 2 0 Solve. x 2 This is incorrect. What mistake was made? (x - 2)(x + 2) (x + 2) 6 0 x2 - 4 x + 2 7 0. 74. Solve the inequality Solution: Cross multiply. 3(x 4) 1(x) Eliminate the parentheses. 3x 12 x Combine like terms. 4x 12 Divide both sides by 4. x 3 This is incorrect. What mistake was made? x + 4 x 6 - 1 3 . c01b.qxd 12/22/11 6:40 PM Page 161 191. Equations Involving Absolute Value The absolute value of a real number can be interpreted algebraically and graphically. Algebraically, the absolute value of 5 is 5, or in mathematical notation, |5| 5; and the absolute value of 5 is 5 or |5| 5. Graphically, the absolute value of a real number is the distance on the real number line between the real number and the origin; thus the distance from 0 to either 5 or 5 is 5. CONCEPTUAL OBJECTIVE Understand absolute value in terms of distance on the number line. ABSOLUTE VALUE EQUATIONS AND INEQUALITIES SKILLS OBJECTIVES Solve absolute value equations. Solve absolute value inequalities. SECTION 1.7 The absolute value of a real number is never negative. When a 5, this denition says |5| (5) 5. The absolute value of a real number a, denoted by the symbol |a|, is dened by |a| = b a, if a 0 -a, if a 6 0 Absolute ValueD E F I N I T I O N 5 5 55 For all real numbers a and b, 1. |a| 0 2. |a| |a| 3. |ab| |a| |b| 4. ` a b ` = a b b Z 0 PROPERTIES OF ABSOLUTE VALUE Absolute value can be used to dene the distance between two points on the real number line. If a and b are real numbers, the distance between a and b is the absolute value of their difference given by |a b| or |b a|. DISTANCE BETWEEN TWO POINTS ON THE REAL NUMBER LINE 162 c01b.qxd 11/24/11 6:19 PM Page 162 192. 1.7 Absolute Value Equations and Inequalities 163 When absolute value is involved in algebraic equations, we interpret the denition of absolute value as follows. If |x| a, then x a or x a, where a 0. Absolute Value EquationDE F I N I T I O N Technology Tip Use a graphing utility to display graphs of y1 |x 3| and y2 8. The x-coordinates of the points of intersection are the solutions to |x 3| 8. EXAMPLE 1 Finding the Distance Between Two Points on the Number Line Find the distance between 4 and 3 on the real number line. Solution: The distance between 4 and 3 is given by the absolute value of the difference. Note that if we reverse the numbers the result is the same. We check this by counting the units between 4 and 3 on the number line. |3 - (-4)| = |7| = 7 |-4 - 3| = |-7| = 7 In words, if the absolute value of a number is a, then that number equals a or a. For example, the equation |x| 7 is true if x 7 or x 7. We say the equation |x| 7 has the solution set {7, 7}. Note: |x| 3 does not have a solution because there is no value of x such that its absolute value is 3. EXAMPLE 2 Solving an Absolute Value Equation Solve the equation |x 3| 8 algebraically and graphically. Solution: Using the absolute value equation denition, we see that if the absolute value of an expres- sion is 8, then that expression is either 8 or 8. Rewrite as two equations: or The solution set is {5, 11} . x = 11x = -5 x - 3 = 8x - 3 = -8 Graph: The absolute value equation |x 3| 8 is interpreted as what numbers are eight units away from 3 on the number line? We nd that eight units to the right of 3 is 11 and eight units to the left of 3 is 5. YOUR TURN Solve the equation |x 5| 7. Answer: x 2 or x 12. The solution set is {12, 2}. 4 3 7 5 3 11 88 c01b.qxd 12/22/11 6:40 PM Page 163 193. 164 CHAPTER 1 Equations and Inequalities EXAMPLE 5 Finding That an Absolute Value Equation Has No Solution Solve the equation |1 3x| 7. Solution: The absolute value of an expression is never negative. Therefore no values of x make this equation true. No solution EXAMPLE 3 Solving an Absolute Value Equation Solve the equation |1 3x| 7. Solution: If the absolute value of an expression is 7, then that expression is 7 or 7. Study Tip Rewrite an absolute value equation as two equations. YOUR TURN Solve the equation |1 2x| 5. EXAMPLE 4 Solving an Absolute Value Equation Solve the equation 2 3|x 1| 4|x 1| 7. Solution: Isolate the absolute value expressions to one side. Add 4|x 1| to both sides. 2 |x 1| 7 Subtract 2 from both sides. |x 1| 5 If the absolute value of an expression or is equal to 5, then the expression is equal to either 5 or 5. The solution set is .{-4, 6} x = 6x = -4 x - 1 = 5x - 1 = -5 YOUR TURN Solve the equation 3 2|x 4| 3|x 4| 11. Answer: x 4 or x 12. The solution set is {4, 12}. Answer: x 3 or x 2. The solution set is {3, 2}. x = -2x = 8 3 -3x = 6-3x = -8 1 - 3x = -7 or 1 - 3x = 7 The solution set is .E-2, 8 3 F c01b.qxd 11/23/11 5:36 PM Page 164 194. 1.7 Absolute Value Equations and Inequalities 165 Inequalities Involving Absolute Value To solve the inequality |x| 3, look for all real numbers that make this statement true. Some numbers that make it true are 2, , 1, 0, , 1, and 2. Some numbers that make it false are 7, 5, 3.5, 3, 3, and 4. If we interpret this inequality as distance, we ask what numbers are less than three units from the origin? We can represent the solution in the following ways. Inequality notation: 3 x 3 Interval notation: (3, 3) 1 5- 3 2 YOUR TURN Solve the equation |7 x2 | 2. Graph: Answer: . The solution set is E ; 25, ;3F. x = ; 25 or x = ;3 Similarly, to solve the inequality |x| 3, look for all real numbers that make the statement true. If we interpret this inequality as a distance, we ask what numbers are at least three units from the origin? We can represent the solution in the following three ways. Inequality notation: x 3 or x 3 Interval notation: Graph: This discussion leads us to the following equivalence relations. (-, -3][3, ) or x = ; 24 = ; 2x = ; 26 x2 = 4x2 = 6 -x2 = -4-x2 = -6 5 - x2 = 15 - x2 = -1 The solution set is .E ;2, ; 26F EXAMPLE 6 Solving a Quadratic Absolute Value Equation Solve the equation |5 x2 | 1. Solution: If the absolute value of an expression is 1, that expression is either 1 or 1, which leads to two equations. 1. |x| a is equivalent to a x a 2. |x| a is equivalent to a x a 3. |x| a is equivalent to x a or x a 4. |x| a is equivalent to x a or x a Note: a 0. PROPERTIES OF ABSOLUTE VALUE INEQUALITIES It is important to realize that in the above four properties the variable x can be any algebraic expression. 3 3 3 3 c01b.qxd 12/23/11 5:35 PM Page 165 195. 166 CHAPTER 1 Equations and Inequalities EXAMPLE 7 Solving an Inequality Involving an Absolute Value Solve the inequality |3x 2| 7. Solution: We apply property (2) and squeeze the absolute value expression between 7 and 7. 7 3x 2 7 Add 2 to all three parts. 5 3x 9 Divide all three parts by 3. The solution in interval notation is . Graph: C - 5 3, 3D - 5 3 x 3 Technology Tip Use a graphing utility to display graphs of y1 |3x 2| and y2 7. The values of x where the graph of y1 lies on top and below the graph of y2 are the solutions to this inequality. YOUR TURN Solve the inequality |2x 1| 11. It is often helpful to note that for absolute value inequalities, less than inequalities can be written as a single statement (see Example 7). greater than inequalities must be written as two statements (see Example 8). Answer: Inequality notation: 6 x 5. Interval notation: (6, 5). EXAMPLE 8 Solving an Inequality Involving an Absolute Value Solve the inequality |1 2x| 5. Solution: Apply property (3). 1 2x 5 or 1 2x 5 Subtract 1 from all expressions. 2x 6 2x 4 Divide by 2 and reverse the inequality sign. x 3 x 2 Express the solution in interval notation. Graph: (-, -2)(3, ) YOUR TURN Solve the inequality |5 2x| 1. Notice that if we change the problem in Example 8 to |1 2x| 5, the answer is all real numbers because the absolute value of any expression is greater than or equal to zero. Similarly, |1 2x| 5 would have no solution because the absolute value of an expression can never be negative. Study Tip Less than inequalities can be written as a single statement. Greater than inequalities must be written as two statements. Answer: Inequality notation: x 2 or x 3. Interval notation: .(-, 2][3, ) 30 3 5 2 3 c01b.qxd 11/23/11 5:36 PM Page 166 196. 1.7 Absolute Value Equations and Inequalities 167 EXAMPLE 9 Solving an Inequality Involving an Absolute Value Solve the inequality 2 |3x| 1. Solution: Subtract 2 from both sides. |3x| 1 Multiply by (1) and reverse the inequality sign. |3x| 1 Apply property (3). 3x 1 or 3x 1 Divide both inequalities by 3. Express in interval notation. a -, - 1 3 b a 1 3 , b x 7 1 3 x 6 - 1 3 Inequalities x 7 A is equivalent to x 6 -A or x 7 A x 6 A is equivalent to -A 6 x 6 A SUMMARY Absolute value equations and absolute value inequalities are solved by writing the equations or inequalities in terms of two equations or inequalities. Note: A 0. Equations x = A is equivalent to x = -A or x = A SECTION 1.7 SKILLS EXERCISES In Exercises 138, solve the equation. 1. |x| 3 2. |x| 2 3. |x| 4 4. |x| 2 5. |t 3| 2 6. |t 3| 2 7. |p 7| 3 8. |p 7| 3 9. |4 y| 1 10. |2 y| 11 11. |3x| 9 12. |5x| 50 13. |2x 7| 9 14. |2x 5| 7 15. |3t 9| 3 16. |4t 2| 2 17. |7 2x| 9 18. |6 3y| 12 19. |1 3y| 1 20. |5 x| 2 21. |4.7 2.1x| 3.3 22. |5.2x 3.7| 2.4 23. 24. 25. |x 5| 4 12 26. |x 3| 9 2 27. 3|x 2| 1 19 28. 2|1 x| 4 2 29. 5 7 |2 x| 30. 1 3 |x 3| 31. 2 |p 3| 15 5 32. 8 3 |p 4| 2 1 2x + 3 4 = 1 162 3x - 4 7 = 5 3 SECTION 1.7 Graph. c01b.qxd 12/23/11 6:35 PM Page 167 197. 168 CHAPTER 1 Equations and Inequalities In Exercises 81 and 82 refer to the following: A company is reviewing revenue for the prior sales year. The model for projected revenue and the model for actual revenue are Rprojected 200 5x Ractual 210 4.8x where x represents the number of units sold and R represents the revenue in thousands of dollars. Since the two revenue models are not identical, an error in projected revenue occurred. This error is represented by E |Rprojected Ractual| 81. Business. For what number of units sold was the error in projected revenue less than $5000? 82. Business. For what number of units sold was the error in projected revenue less than $3000? 77. Temperature. If the average temperature in Hawaii is 83 ( write an absolute value inequality representing the temperature in Hawaii. 78. Temperature. If the average temperature of a human is 97.8 (1.2), write an absolute value inequality describing normal human body temperature. 79. Sports. Two women tee off the green of a par-3 hole on a golf course. They are playing closest to the pin. If the rst woman tees off and lands exactly 4 feet from the hole, write an inequality that describes where the second woman must land in order to win the hole. What equation would suggest a tie? Let d the distance from where the second woman lands to the tee. 80. Electronics. A band-pass lter in electronics allows certain frequencies within a range (or band) to pass through to the receiver and eliminates all other frequencies. Write an absolute value inequality that allows any frequency f within 15 Hertz of the carrier frequency fc to pass. F ;15), F A P P L I C A T I O N S 33. 5|y 2| 10 4|y 2| 3 34. 3 |y 9| 11 3 |y 9| 35. |4 x2 | 1 36. |7 x2 | 3 37. |x2 1| 5 38. |x2 1| 5 In Exercises 3970, solve the inequality and express the solution in interval notation. 39. |x| 7 40. |y| 9 41. |y| 5 42. |x| 2 43. |x 3| 7 44. |x 2| 4 45. |x 4| 2 46. |x 1| 3 47. |4 x| 1 48. |1 y| 3 49. |2x| 3 50. |2x| 3 51. |2t 3| 5 52. |3t 5| 1 53. |7 2y| 3 54. |6 5y| 1 55. |4 3x| 0 56. |4 3x| 1 57. 2|4x| 9 3 58. 5|x 1| 2 7 59. 2|x 1| 3 7 60. 3|x 1| 5 4 61. 3 2|x 4| 5 62. 7 3|x 2| 14 63. 9 |2x| 3 64. 4 |x 1| 1 65. 66. 67. |2.6x 5.4| 1.8 68. |3.7 5.5x| 4.3 69. |x2 1| 8 70. |x2 4| 29 In Exercises 7176, write an inequality that ts the description. 71. Any real numbers less than seven units from 2. 72. Any real numbers more than three units from 2. 73. Any real numbers at least unit from . 74. Any real number no more than units from . 75. Any real numbers no more than two units from a. 76. Any real number at least a units from 3. 11 3 5 3 3 2 1 2 ` 2 - 3x 5 ` 2 5 1 - 2x 6 1 2 c01b.qxd 11/23/11 5:36 PM Page 168 198. 1.7 Absolute Value Equations and Inequalities 169 In Exercises 9196, assuming a and b are real positive numbers, solve the equation or inequality and express the solution in interval notation. 91. |x a| b 92. |a x| b 93. |x| a 94. |x| b 95. |x a| b 96. |x a| b In Exercises 8790, determine whether each statement is true or false. 87. |m| m |m| 88. |n2 | n2 89. |m n| |m| |n| is true only when m and n are both nonnegative. 90. For what values of x does the absolute value equation |x 7| x 7 hold? C O N C E P T U A L C H A L L E N G E 98. Solve the inequality |3x2 7x 2| 8.97. For what values of x does the absolute value equation |x 1| 4 |x 2| hold? 99. Graph y1 |x 7| and y2 x 7 in the same screen. Do the x-values where these two graphs coincide agree with your result in Exercise 90? 100. Graph y1 |x 1| and y2 |x 2| 4 in the same screen. Do the x-values where these two graphs coincide agree with your result in Exercise 97? 101. Graph y1 |3x2 7x 2| and y2 8 in the same screen. Do the x-values where y1 lies above y2 agree with your result in Exercise 98? 102. Solve the inequality |2.7x2 7.9x 5| |5.3x2 9.2| by graphing both sides of the inequality and identify which x-values make this statement true. 103. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. 104. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. ` x x + 1 ` 6 2 ` x x + 1 ` 6 1 T E C H N O L O G Y In Exercises 8386, explain the mistake that is made. 83. Solve the absolute value equation |x 3| 7. Solution: Eliminate the absolute value symbols. x 3 7 Add 3 to both sides. x 10 Check. |10 3| 7 This is incorrect. What mistake was made? 84. Solve the inequality |x 3| 7. Solution: Eliminate the absolute x 3 7 or x 3 7 value symbols. Add 3 to both sides. x 4 x 10 The solution is . This is incorrect. What mistake was made? (-, -4)(10, ) 85. Solve the inequality |5 2x| 1. Solution: Eliminate the absolute value symbols. 1 5 2x 1 Subtract 5. 6 2x 4 Divide by 2. 3 x 2 Write the solution in interval notation. This is incorrect. What mistake was made? 86. Solve the equation |5 2x| 1. Solution: or The solution is {2, 3}. This is incorrect. What mistake was made? x = 2x = 3 -2x = -4-2x = -6 5 - 2x = 15 - 2x = -1 (-, 2][3, ) C A T C H T H E M I S T A K E c01b.qxd 12/22/11 6:40 PM Page 169 199. CHAPTER 1 INQUIRY-BASED LEARNING PROJECT 170 Equivalent Equations and Extraneous Solutions A general strategy for solving all the various types of equations you encountered in this chapter can be summarized as follows: From a given equation, perform algebraic operations on both sides in order to generate equivalent equations. Remember, equivalent equations have the same solution set. 1. Consider rst a linear equation: 3x 1 5. a. Use a graphing utility to show y1 3x 1 and y2 5 and determine the point of intersection. Make a sketch and label it. b. How does the graph in part (a) relate to the solution set of the equation 3x 1 5? c. To solve the equation 3x 1 5 algebraically, the rst step is to add 1 to both sides of the equation, as follows: 3x 1 5 1 1 3x 6 Use a graphing utility to show y1 3x and y2 6, and determine the point of intersection. Make a sketch and label it. How does this graph relate to the equation 3x 6? d. The nal algebraic step to solve the equation is to divide both sides of the equation by 3. 3x 6 3 3 x 2 Use a graphing utility to show y1 x and y2 2, and determine the point of intersection. Make a sketch and label it. How does this graph relate to the equation x 2? e. The algebraic steps to solve the equation 3x 1 5 produce two equations: 3x 6 and x 2. How do the graphs you sketched above represent the fact that these equations are equivalent to the original? c01b.qxd 11/23/11 5:36 PM Page 170 200. 2. Next consider the equation x 2 a. Use a graphing utility to show y1 x 2 and y2 and determine any points of intersection. What do you learn about the solution set of the equation x 2 ? Make a sketch to explain. b. The algebraic steps to solve this equation are as follows: x 2 (x 2)2 ( )2 Square both sides. x2 4x 4 4 x Simplify. x2 3x 0 Write the quadratic equation in standard form. x(x 3) 0 Factor. x 0 or x 3 Use the zero product property. Use a graphing utility to show the rst step above: y1 (x 2)2 and y2 ( )2 . What do you learn about the solution set of (x 2)2 ( )2 ? c. Discuss whether x 2 and (x 2)2 ( )2 are equivalent equations. d. The algebraic process of squaring both sides introduced an extraneous solution. What do you think that means? e. Why is it important to always check the solutions you obtain when solving equations? 3. Consider x4 x2 0 a. A fellow student suggests dividing both sides of the equation by x2 . What will be the resulting equation? b. Is the equation you wrote in part (a) equivalent to the original equation? How can you use a graphing utility to illustrate this? c. Show the algebraic steps you should take to solve x4 x2 0. 14 - x14 - x 14 - x 14 - x 14 - x 14 - x 14 - x 14 - x 14 - x 171 c01b.qxd 11/23/11 5:36 PM Page 171 201. Year Global Temperatures 200019601840 1880 1920 TemperatureAnomaly(C) +0.6 +0.4 +0.2 0 0.2 0.4 0.6 Annual Average Five Year Average Year Global Warming Projections 21001900 1950 2000 2050 7 6 5 4 3 2 1 0 1 2 TemperatureAnomaly(C) CCSR/NIES CCCma CSIRO Hadley Centre GFDL MPIM NCAR PCM NCAR CSM The Earths near-surface global air temperature increased approximately 0.74 0.18C over 19052005.* *Climate Change 2007: The Physical Science Basis. Contribution of Working Group 1 to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change. Intergovernmental Panel on Climate Change (2007-02-05). Used in elds of study ranging from engineering to economics to sociology, a mathematical model is a tool that uses mathematical language to describe a system. There are many types of models, which help us not only better understand the world as it is, but by projecting different scenarios based on available data, allow us glimpses into possible futures. Current changes in the environment, which will affect all of our futures, have brought about a erce debate. Some scientists believe that human activities have played a large part in bringing about global warming, which impacts not only day-to-day temperatures, but species extinction, loss of glacial ice, and the quality of the air we breathe. Others feel that current changes in the climate are simply part of a natural cycle and are not a cause for concern. The Modeling Our World feature at the end of every chapter allows you to use modeling to explore the topic of global climate change and become an informed participant in this debate. 1. Write an absolute value equation that models the increase in the Earths near-surface air temperature from 1905 to 2005 in degrees Celsius (C). Let t represent the increase in temperature. 2. Use the temperature scale conversion F C 32 equation to write an absolute value equation that models the increase in the Earths near-surface air temperature from 1905 to 2005 in degrees Fahrenheit (F). Let t represent the increase in temperature. The following chart illustrates different global warming projections for the next 100 years. = 9 5 3. Let t represent the increase in temperature in degrees Celsius (let t 0 correspond to the year 2000) and write an approximate absolute value inequality such that the NCAR PCM projection is the lowest possible temperature anomaly and the CCSR/NIES projection is the highest possible temperature anomaly. 4. Repeat Problem 3 for temperature in degrees Fahrenheit. MODEL ING OUR WORLD 172 c01b.qxd 11/23/11 5:36 PM Page 172 202. 173 CHAPTER 1 Equations and Inequalities SECTION CONCEPT KEY IDEAS/FORMULAS 1.1 Linear equations ax b 0 Solving linear equations in one variable Isolate variable on one side and constants on the other side. Solving rational equations that are reducible Any values that make the denominator equal to 0 must be eliminated to linear equations as possible solutions. 1.2 Applications involving linear equations Solving application problems using Five-step procedure: mathematical models Step 1: Identify the question. Step 2: Make notes. Step 3: Set up an equation. Step 4: Solve the equation. Step 5: Check the solution. Geometry problems Formulas for rectangles, triangles, and circles Interest problems Simple interest: I Prt Mixture problems Whenever two distinct quantities are mixed, the result is a mixture. Distanceratetime problems 1.3 Quadratic equations ax2 bx c 0 a 0 Factoring If (x h)(x k) 0, then x h or x k. Square root method If x2 P, then . Completing the square Find half of b; square that quantity; add the result to both sides. Quadratic Formula 1.4 Other types of equations Radical equations Check solutions to avoid extraneous solutions. Equations quadratic in form: u-substitution Use a u-substitution to write the equation in quadratic form. Factorable equations Extract common factor or factor by grouping. 1.5 Linear inequalities Solutions are a range of real numbers. Graphing inequalities and interval notation a x b is equivalent to (a, b). x a is equivalent to (, a]. x a is equivalent to (a, ). Solving linear inequalities If an inequality is multiplied or divided by a negative number, the inequality sign must be reversed. 1.6 Polynomial and rational inequalities Polynomial inequalities Zeros are values that make the polynomial equal to 0. Rational inequalities The number line is divided into intervals. The endpoints of these intervals are values that make either the numerator or denominator equal to 0. Always exclude values that make the denominator 0. 1.7 Absolute value equations and inequalities |b a| is the distance between points a and b on the number line. Equations involving absolute value If |x| a, then x a or x a. Inequalities involving absolute value |x| a is equivalent to a x a. |x| a is equivalent to x a or x a. x = -b ; 2b2 - 4ac 2a x = ; 2P Z d = r # t CHAPTERREVIEW CHAPTER 1 REVIEW 173 c01b.qxd 11/23/11 5:36 PM Page 173 203. 1.1 Linear Equations Solve for the variable. 1. 7x 4 12 2. 13d 12 7d 6 3. 20p 14 6 5p 4. 4(x 7) 4 4 5. 3(x 7) 2 4(x 2) 6. 7c 3(c 5) 2(c 3) 14 7. 8. 9. 10. 11. 12. Specify any values that must be excluded from the solution set and then solve. 13. 14. 15. 16. 17. 18. 19. 7x (2 4x) 3[6 (4 2x 7)] 12 20. Solve for the specied variable. 21. Solve for x in terms of y: 3x 2[(y 4)3 7] y 2x 6(x 3) 22. If , nd in terms of x. 1.2 Applications Involving Linear Equations 23. Transportation. Maria is on her way from her home near Orlando to the Sundome in Tampa for a rock concert. She drives 16 miles to the Orlando park-n-ride, takes a bus of3 4 y + 2 1 - 2y y = x + 3 1 + 2x x 5 - x - 3 15 = -6 3 - (5/m) 2 + (5/m) = 1 3 2x - 6 x = 9 3 2x - 7 = -2 3x + 1 2 t + 4 - 7 t = 6 t(t + 4) 4 x + 1 - 8 x - 1 = 3 1 x - 4 = 3 x - 5 5b + b 6 = b 3 - 29 6 13x 7 - x = x 4 - 3 14 g 3 + g = 7 9 12 b - 3 = 6 b + 4 [6 - 4x + 2(x - 7)] - 52 = 3(2x - 4) + 6[3(2x - 3) + 6] 14 - [-3(y - 4) + 9] = [4(2y + 3) - 6] + 4 25. Numbers. Find a number such that 12 more than the number is the number. 26. Numbers. Find four consecutive odd integers such that the sum of the four numbers is equal to three more than three times the fourth integer. 27. Geometry. The length of a rectangle is one more than two times the width, and the perimeter is 20 inches. What are the dimensions of the rectangle? 28. Geometry. Find the perimeter of a triangle if one side is 10 inches, another side is of the perimeter, and the third side is of the perimeter. 29. Investments. You win $25,000 and you decide to invest the money in two different investments: one paying 20% and the other paying 8%. A year later you have $27,600 total. How much did you originally invest in each account? 30. Investments. A college student on summer vacation was able to make $5000 by working a full-time job every summer. He invested half the money in a mutual fund and half the money in a stock that yielded four times as much interest as the mutual fund. After a year he earned $250 in interest. What were the interest rates of the mutual fund and the stock? 31. Chemistry. For an experiment, a student requires 150 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 10% NaCl and 5% NaCl. How many milliliters of each available solution should be mixed to get 150 milliliters of 8% NaCl? 32. Chemistry. A mixture containing 8% salt is to be mixed with 4 ounces of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the rst solution must be used? 33. Grades. Going into the College Algebra nal exam, which will count as two tests, Danny has test scores of 95, 82, 90, and 77. If his nal exam is higher than his lowest test score, then it will count for the nal exam and replace the lowest test score. What score does Danny need on the nal in order to have an average score of at least 90? 34. Car Value. A car salesperson reduced the price of a model car by 20%. If the new price is $25,000, what was its original price? How much can be saved by purchasing the model? 1.3 Quadratic Equations Solve by factoring. 35. b2 4b 21 36. x(x 3) 54 37. x2 8x 38. 6y2 7y 5 0 1 6 1 3 1 3 1 4 CHAPTER 1 REVIEW EXERCISES REVIEWEXERCISES the way to a bus station in Tampa, and then takes a cab of the way to the Sundome. How far does Maria live from the Sundome? 24. Diet. A particular 2000 calorie per day diet suggests eating breakfast, lunch, dinner, and four snacks. Each snack is the calories of lunch. Lunch has 100 calories less than dinner. Dinner has 1.5 times as many calories as breakfast. How many calories are in each meal and snack? 1 4 1 12 174 c01b.qxd 11/23/11 5:36 PM Page 174 204. Review Exercises 175 Solve by the square root method. 39. q2 169 0 40. c2 36 0 41. (2x 4)2 64 42. (d 7)2 4 0 Solve by completing the square. 43. x2 4x 12 0 44. 2x2 5x 7 0 45. 46. 8m m2 15 Solve by the Quadratic Formula. 47. 3t2 4t 7 48. 4x2 5x 7 0 49. 50. x2 6x 6 Solve by any method. 51. 5q2 3q 3 0 52. (x 7)2 12 53. 2x2 3x 5 0 54. (g 2)(g 5) 7 55. 7x2 19x 6 56. 7 2b2 1 Solve for the indicated variable. 57. S r2 h for r 58. for r 59. h vt 16t2 for v 60. A 2r2 2rh for h 61. Geometry. Find the base and height of a triangle with an area of 2 square feet if its base is 3 feet longer than its height. 62. Falling Objects. A man is standing on top of a building 500 feet tall. If he drops a penny off the roof, the height of the penny is given by h 16t2 500, where t is in seconds. Determine how many seconds it takes until the penny hits the ground. 1.4 Other Types of Equations Solve the radical equation for the given variable. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. Solve the equation by introducing a substitution that transforms the equation to quadratic form. 75. 28 (3x 2)2 11(3x 2) 76. x4 6x2 9 0 315 + 21x - 4 + 1x = 5 -x = 13 - x 12x - 5 - 1x + 2 = 3 x - 2 = 249 - x2 4 + 1x - 3 = 1x - 5 1x + 3 = 2 - 13x + 2 12x - 7 = 1x + 3x - 4 = 2x2 + 5x + 6 x = 17x - 10(2x - 7)15 = 3 1x - 2 = -413 2x - 4 = 2 V = pr3 h 3 82 - 1 3 = 7 6 x2 2 = 4 + x 2 77. 78. 3(x 4)4 11(x 4)2 20 0 79. y2 5y1 4 0 80. p2 4p1 12 81. 3x1/3 2x2/3 5 82. 2x2/3 3x1/3 5 0 83. 84. 85. x4 5x2 36 86. 3 4x1/2 x1 0 Solve the equation by factoring. 87. x3 4x2 32x 0 88. 9t3 25t 0 89. p3 3p2 4p 12 0 90. 4x3 9x2 4x 9 0 91. p(2p 5)2 3(2p 5) 0 92. 2(t2 9 20(t2 9 0 93. y 81y1 0 94. 1.5 Linear Inequalities Rewrite using interval notation. 95. x 4 96. 1 x 7 97. 2 x 6 98. x 1 Rewrite using inequality notation. 99. (6, ) 100. (, 0] 101. [3, 7] 102. (5, 2] Express each interval using inequality and interval notation. 103. 104. Graph the indicated set and write as a single interval, if possible. 105. 106. 107. 108. (-, -2)[-2, 9)(3, 12][8, ) (-, -3)[-7, 2](4, 6][5, ) 9x3/2 - 37x1/2 + 4x-1/2 = 0 ) 2 ) 3 y-1/2 - 2y-1/4 + 1 = 0 x-2/3 + 3x-1/3 + 2 = 0 a x 1 - x b 2 = 15 - 2 a x 1 - x b REVIEWEXERCISESREVIEWEXERCISESREVIEWEXERCISES c01b.qxd 12/23/11 5:21 PM Page 175 205. 176 CHAPTER 1 Equations and Inequalities Solve and graph. 109. 2x 5 x 110. 6x 4 2 111. 4(x 1) 2x 7 112. 113. 6 2 x 11 114. 6 1 4(x 2) 16 115. 116. Applications 117. Grades. In your algebra class your rst four exam grades are 72, 65, 69, and 70. What is the lowest score you can get on the fth exam to earn a C for the course? Assume that each exam is equal in weight and a C is any score greater than or equal to 70. 118. Prot. A tailor decided to open a mens custom suit business. His xed costs are $8500 per month, and it costs him $50 for the materials to make each suit. If the price he charges per suit is $300, how many suits does he have to tailor per month to make a prot? 1.6 Polynomial and Rational Inequalities Solve the polynomial inequality and express the solution set using interval notation. 119. x2 36 120. 6x2 7x 20 121. 4x x2 122. x2 9x 14 123. x2 7x 124. x2 4 125. 4x2 12 13x 126. 3x x2 2 Solve the rational inequality and express the solution set using interval notation. 127. 128. 129. 130. 131. 132. 133. 134. 1.7 Absolute Value Equations and Inequalities Solve the equation. 135. |x 3| 4 136. |2 x| 5 137. |3x 4| 1.1 138. |x2 6| 3 x 6 5x + 6 x x2 + 9 x - 3 0 4 x - 1 2 x + 3 3 x - 2 - 1 x - 4 0 x2 - 49 x - 7 0 x2 - 3x 3 18 x - 1 x - 4 7 0 x x - 3 6 0 x 3 + x + 4 9 7 x 6 - 1 3 2 3 1 + x 6 3 4 x + 3 3 6 Solve the inequality and express the solution using interval notation. 139. |x| 4 140. |x 3| 6 141. |x 4| 7 142. |7 y| 4 143. |2x| 6 144. 145. |2 5x| 0 146. |1 2x| 4 Applications 147. Temperature. If the average temperature in Phoenix is 85F write an inequality representing the average temperature T in Phoenix. 148. Blood Alcohol Level. If a person registers a 0.08 blood alcohol level, he will be issued a DUI ticket in the state of Florida. If the test is accurate within 0.007,write a linear inequality representing an actual blood alcohol level that will not be issued a ticket. Technology Exercises Section 1.1 Graph the function represented by each side of the question in the same viewing rectangle, and solve for x. 149. 0.031x 0.017(4000 x) 103.14 150. Section 1.3 151. a. Solve the equation x2 4x b, b 5 by rst writing in standard form and then factoring. Now plot both sides of the equation in the same viewing screen (y1 x2 4x and y2 b). At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5, 0, 7, and 12. 152. a. Solve the equation x2 4x b, b 5 by rst writing in standard form and then factoring. Now plot both sides of the equation in the same viewing screen (y1 x2 4x and y2 b). At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5, 0, 7, and 12. Section 1.4 153. Solve the equation . Round your answer to two decimal places. Plot both sides of the equation in the same viewing screen, and Does the point(s) of intersection agree with your solution? 154. Solve the equation . Plot both sides of the equation in the same viewing screen, and Does the point(s) of intersection agree with your solution? y2 = x-14 + 6. y1 = 2x-12 2x-12 = x-14 + 6 y2 = -x12 + 6.y1 = 2x14 2x14 = -x12 + 6 1 0.16x - 0.2 x = 1 4 (;10) ` 4 + 2x 3 ` 1 7 REVIEWEXERCISES c01b.qxd 11/23/11 5:36 PM Page 176 206. Review Exercises 177 Section 1.5 155. a. Solve the inequality Express the solution set using interval notation. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do parts (a) and (b) agree? 156. a. Solve the inequality Express the solution set using interval notation. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do parts (a) and (b) agree? Section 1.6 Plot the left side and the right side of each inequality in the same screen, and use the zoom feature to determine the range of values for which the inequality is true. 157. 158. 12x2 - 7x - 10 6 2x2 + 2x - 1 0.2x2 - 2 7 0.05x + 3.25 - 1 2 x + 7 6 3 4 x - 5. -0.61x + 7.62 7 0.24x - 5.47. 159. 160. Section 1.7 161. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation and round to two decimal places. 162. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation and round to two decimal places. |0.8x2 - 5.4x| 7 4.5 |1.6x2 - 4.5| 6 3.2 7p 15 - 2p 6 1 3p 7 - 2p 7 1 REVIEWEXERCISESREVIEWEXERCISESREVIEWEXERCISES c01b.qxd 11/23/11 5:36 PM Page 177 207. AndyWashnik 25 15 20 29. Solve the equation Graph the function represented by each side in the same viewing rectangle and solve for x. 30. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation. 0.3 + |2.4x2 - 1.5| 6.3 1 0.75x - 0.45 x = 1 9 . Solve the equation. 1. 4p 7 6p 1 2. 2(z 1) 3 3z 3(z 1) 3. 3t t2 28 4. 8x2 13x 6 5. 6x2 13x 8 6. 7. 8. x4 5x2 36 0 9. 10. 2x2/3 3x1/3 2 0 11. 12. x(3x 5)3 2(3x 5)2 0 13. x7/3 8x4/3 12x1/3 0 Solve for the specied variable. 14. for C 15. P 2L 2W for L Solve the inequality and express the solution in interval notation. 16. 7 5x 18 17. 3x 19 5(x 3) 18. 1 3x 5 26 19. 20. 3x 2x2 21. 3p2 p 4 22. |5 2x| 1 23. 24. x + 4 x2 - 9 0 x - 3 2x + 1 0 2 5 6 x + 8 4 1 2 F = 9 5C + 32 13y - 2 = 3 - 13y + 1 12x + 1 + x = 7 5 y - 3 + 1 = 30 y2 - 9 3 x - 1 = 5 x + 2 is in the sand, 150 feet is in the water, and of the piling is in the air. What is the total height of the piling? 26. Real Estate. As a realtor you earn 7% of the sale price. The owners of a house you have listed at $150,000 will entertain offers within 10% of the list price. Write an inequality that models the commission you could make on this sale. 27. Costs: Cell Phones. A cell phone company charges $49 for a 600-minute monthly plan, plus an additional $0.17 per minute for every minute over 600. If a customers bill ranged from a low of $53.59 to a high of $69.74 over a 6-month period, write an inequality expressing the number of monthly minutes used over the 6-month period. 28. Television. Television and lm formats are classied as ratios of width to height. Traditional televisions have a 4:3 ratio (1.33:1), and movies are typically made in widescreen format with a 21:9 ratio (2.35:1). If you own a traditional 25-inch television (20 inch 15 inch screen) and you play a widescreen DVD on it, there will be black bars above and below the image. What are the dimensions of the movie and of the black bars? 3 5 CHAPTER 1 PRACTICE TEST PRACTICETEST 178 25. Puzzle. A piling supporting a bridge sits so that of the piling1 4 c01b.qxd 11/23/11 5:36 PM Page 178 208. Simplify. 1. 5 . (7 3 . 4 2) Simplify and express in terms of positive exponents. 2. 3. Perform the operations and simplify. 4. (x4 2x3 ) (x3 5x 6) (5x4 4x3 6x 8) 5. x2 (x 5)(x 3) Factor completely. 6. 3x3 3x2 60x 7. 2a3 2000 Perform the operations and simplify. 8. 9. Solve for x. 10. x3 x2 30x 0 11. 12. Perform the operation and express in standard form: . Solve for x. 13. 14. x - 6 6 - x = 3 2 6x 5 - 8x 3 = 4 - 7x 15 45 6 - 3i 2 7 x = 1 8 x + 9 6x x - 2 - 5x x + 2 3 - x x2 - 1 , 5x - 15 x + 1 (x2 y-2 )3 (x2 y)-3 (4x-3 b4 ) -3 15. Tim can paint the interior of a condo in 9 hours. If Chelsea is hired to help him, they can do a similar condo in 5 hours. Working alone, how long will it take Chelsea to paint a similar condo? 16. Solve using the square root method: y2 36 0. 17. Solve by completing the square: x2 12x 40 0. 18. Solve using the Quadratic Formula: x2 x 9 0. 19. Solve and check: . 20. Solve using substitution: 3x2 8x1 4 0. Solve and express the solution in interval notation. 21. 0 4 x 7 22. 4x2 9x 11 23. 24. 25. Solve for . 26. Solve the equation Plot both sides of the equation in the same viewing screen, and y2 27. Does the point(s) of intersection agree with your solution? 27. Solve the inequality by graphing both sides of the inequality, and identify which x-values make this statement true. ` 3x x - 2 ` 6 1 y1 = x6 + 37 8 x3 x6 + 37 8 x3 = 27. x: 1 5 x + 2 3 = 7 15 ` 4 - 5x 7 ` 3 14 x + 2 9 - x2 0 24 - x = x - 4 CHAPTER 1 CUMULATIVE TEST CUMULATIVETEST 179 c01b.qxd 11/23/11 5:36 PM Page 179 209. 2 Graphs G raphs are used in many ways. There is only one temperature that yields the same number in degrees Celsius and degrees Fahrenheit. Do you know what it is?* The penguins are a clue. HIV infection rates Stock prices uctuating throughout the day The conversion between degrees Fahrenheit and degrees Celsius is a linear relationship. Notice that 0C corresponds to 32F. Emperor penguins walking in a line, Weddell Sea, Antarctica Vetta/GettyImages *See Section 2.3, Exercise 107. 100 200 100 50 F C COKE 5Minute 10 11 12 1 2 3 54.50 54.75 55.00 55.25 4:53 PM Year 20001992 199819961994 Antenatalclinicprevalence(%) 10 5 15 20 25 South Africa Thailand c02a.qxd 11/22/11 8:49 PM Page 180 210. 181 I N TH I S C HAP TE R you will review the Cartesian plane. You will calculate the distance between two points and find the midpoint of a line segment joining two points. You will then apply point-plotting techniques to sketch graphs of equations. Special attention is given to two types of equations: lines and circles. GRAPHS 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 2.3 Lines 2.4 Circles 2.5* Linear Regression: Best Fit Point-Plotting Intercepts Symmetry Using Intercepts and Symmetry as Graphing Aids Cartesian Plane Distance Between Two Points Midpoint of a Line Segment Joining Two Points Graphing a Line Equations of Lines Parallel and Perpendicular Lines Standard Equation of a Circle Transforming Equations of Circles to the Standard Form by Completing the Square Scatterplots Identifying Patterns Linear Regression Calculate the distance between two points and the midpoint of a line segment joining two points. Sketch the graph of an equation using intercepts and symmetry as graphing aids. Find the equation of a line. Graph circles. Find the line of best fit for a given set of data.* L E A R N I N G O B J E C T I V E S *Optional Technology Required Section. c02a.qxd 11/22/11 8:49 PM Page 181 211. Cartesian Plane HIV infection rates, stock prices, and temperature conversions are all examples of relationships between two quantities that can be expressed in a two-dimensional graph. Because it is two dimensional, such a graph lies in a plane. Two perpendicular real number lines, known as the axes in the plane, intersect at a point we call the origin. Typically, the horizontal axis is called the x-axis, and the vertical axis is denoted as the y-axis. The axes divide the plane into four quadrants, numbered by Roman numerals and ordered counterclockwise. Points in the plane are represented by ordered pairs, denoted (x, y). The rst number of the ordered pair indicates the position in the horizontal direction and is often called the x-coordinate or abscissa. The second number indicates the position in the vertical direction and is often called the y-coordinate or ordinate. The origin is denoted (0, 0). Examples of other coordinates are given on the graph to the right. The point (2, 4) lies in quadrant I. To plot this point, start at the origin (0, 0) and move to the right two units and up four units. All points in quadrant I have positive coordinates, and all points in quadrant III have negative coordinates. Quadrant II has negative x-coordinates and positive y-coordinates; quadrant IV has positive x-coordinates and negative y-coordinates. This representation is called the rectangular coordinate system or Cartesian coordinate system, named after the French mathematician Ren Descartes. E X AM P LE 1 Plotting Points in a Cartesian Plane a. Plot and label the points (1, 4), (2, 2), (2, 3), (2, 3), (0, 5), and (3, 0) in the Cartesian plane. b. List the points and corresponding quadrant or axis in a table. Solution: a. b. x (2, 2) y (2, 3) (1, 4) (3, 0) (2, 3) (0, 5) C O N C E P TUAL O BJ E CTIVE Expand the concept of a one-dimensional number line to a two-dimensional plane. BAS I C TO O LS: CARTE S IAN P LAN E, D I STAN C E, AN D M I D P O I NT S K I LLS O BJ E CTIVE S Plot points in the Cartesian plane. Calculate the distance between two points. Find the midpoint of a line segment joining two points. S E CTI O N 2.1 x I x > 0, y > 0 II x < 0, y > 0 IV x > 0, y < 0 III x < 0, y < 0 y x-axis Origin III IVIII y-axis 182 x x-coordinate y-coordinate (4, 1) (3, 2) (0, 0) (0, 2) (3, 4) (5, 0) (2, 4) y POINT QUADRANT (2, 2) I (2, 3) II (1, 4) III (2, 3) IV (0, 5) y-axis (3, 0) x-axis c02a.qxd 11/22/11 8:49 PM Page 182 212. 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 183 Distance Between Two Points Suppose you want to nd the distance between any two points in the plane. In the previous graph, to nd the distance between the points (2, 3) and (2, 2), count the units between the two points. The distance is 5. What if the two points do not lie along a horizontal or vertical line? Example 2 uses the Pythagorean theorem to help nd the distance between any two points. E X AM P LE 2 Finding the Distance Between Two Points Find the distance between the points (2, 1) and (1, 3). Solution: STEP 1 Plot and label the two points in the Cartesian plane and draw a line segment indicating the distance d between the two points. STEP 2 Form a right triangle by connecting the points to a third point, (1, 1). STEP 3 Calculate the length of the horizontal segment. 3 1 (2) Calculate the length of the vertical segment. 4 3 (1) STEP 4 Use the Pythagorean theorem to d2 32 42 calculate the distance d. d2 25 d 5 WORDS MATH For any two points, (x1, y1) and (x2, y2): x (1, 3) d y (2, 1) x (1, 3) y (2, 1) (1, 1) d 4 3 x y (x1, y1) (x2, y2) |x2 x1| |y2 y1|d c02a.qxd 12/26/11 4:24 PM Page 183 213. 184 C HAP TE R 2 Graphs The distance along the horizontal segment is the absolute value of the difference between the x-values. x2 x1 The distance along the vertical segment is the absolute value of the difference between the y-values. y2 y1 Use the Pythagorean theorem to calculate the distance d. d2 x2 x1 2 y2 y1 2 a 2 a2 for all real numbers a. d2 (x2 x1)2 ( y2 y1)2 Use the square root property. Distance can be only positive. d = 2(x2 - x1)2 + (y2 - y1)2 d = ; 2(x2 - x1)2 + (y2 - y1)2 The distance d between two points P1 (x1, y1) and P2 (x2, y2) is given by The distance between two points is the square root of the sum of the square of the distance between the x-coordinates and the square of the distance between the y-coordinates. d = 2(x2 - x1)2 + (y2 - y1)2 Distance FormulaStudy Tip It does not matter which point is taken to be the rst point or the second point. E X AM P LE 3 Using the Distance Formula to Find the Distance Between Two Points Find the distance between (3, 7) and (5, 2). Solution: Write the distance formula. Substitute (x1, y1) (3, 7) and (x2, y2) (5, 2). Simplify. Solve for d. d = 2145 d = 282 + (-9)2 = 264 + 81 = 2145 d = 2[5 + 3]2 + [-2 - 7]2 d = 2[5 - (-3)]2 + [-2 - 7]2 d = 2[x2 - x1]2 + [y2 - y1]2 D E F I N I T I O N You will prove in the exercises that it does not matter which point you take to be the rst point when applying the distance formula. YO U R TU R N Find the distance between (4, 5) and (3, 2). Answer: d = 258 c02a.qxd 11/22/11 8:49 PM Page 184 214. 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 185 Midpoint of a Line Segment Joining Two Points The midpoint, (xm, ym), of a line segment joining two points (x1, y1) and (x2, y2) is dened as the point that lies on the segment which has the same distance d from both points. In other words, the midpoint of a segment lies halfway between the given endpoints. The coordinates of the midpoint are found by averaging the x-coordinates and averaging the y-coordinates. The midpoint, (xm, ym), of the line segment with endpoints (x1, y1) and (x2, y2) is given by The midpoint can be found by averaging the x-coordinates and averaging the y-coordinates. (xm, ym) = a x1 + x2 2 , y1 + y2 2 b Midpoint Formula E X AM P LE 4 Finding the Midpoint of a Line Segment Find the midpoint of the line segment joining the points (2, 6) and (4, 2). Solution: Write the midpoint formula. Substitute (x1, y1) (2, 6) and (x2, y2) (4, 2). Simplify. (xm, ym) (1, 2) One way to verify your answer is to plot the given points and the midpoint to make sure your answer looks reasonable. (xm, ym) = a 2 + (-4) 2 , 6 + (-2) 2 b (xm, ym) = a x1 + x2 2 , y1 + y2 2 b x (2, 6) y (4, 2) (1, 2) x P1 = (x1, y1) (xm, ym) P2 = (x2, y2) y d d D E F I N I T I O N YO U R TU R N Find the midpoint of the line segment joining the points (3, 4) and (5, 8). Answer: Midpoint (4, 2) Technology Tip Show a screen display of how to enter and . Scientic calculators: Or 2)26( 2)42( 2)2(-)6( 2)4(-)2( 6 + (-2) 2 2 + (-4) 2 ; ; c02a.qxd 11/22/11 8:49 PM Page 185 215. 186 C HAP TE R 2 Graphs Distance Between Two Points Midpoint of Line Segment Joining Two Points Midpoint = (xm, ym) = a x1 + x2 2 , y1 + y2 2 b d = 2(x2 - x1)2 + (y2 - y1)2 y-axis II (, ) (, ) (, )(, ) III IV I x-axis Origin In Exercises 16, give the coordinates for each point labeled. 1. Point A 2. Point B 3. Point C 4. Point D 5. Point E 6. Point F In Exercises 7 and 8, plot each point in the Cartesian plane and indicate in which quadrant or on which axis the point lies. 7. A: (2, 3) B: (1, 4) C: (3, 3) D: (5, 1) E: (0, 2) F: (4, 0) 8. A: (1, 2) B: (1, 3) C: (4, 1) D: (3, 2) E: (0, 5) F: (3, 0) 9. Plot the points (3, 1), (3, 4), (3, 2), (3, 0), (3, 4). Describe the line containing points of the form (3, y). 10. Plot the points (1, 2), (3, 2), (0, 2), (3, 2), (5, 2). Describe the line containing points of the form (x, 2). In Exercises 1132, calculate the distance between the given points, and nd the midpoint of the segment joining them. 11. (1, 3) and (5, 3) 12. (2, 4) and (2, 4) 13. (1, 4) and (3, 0) 14. (3, 1) and (1, 3) 15. (10, 8) and (7, 1) 16. (2, 12) and (7, 15) 17. (3, 1) and (7, 2) 18. (4, 5) and (9, 7) 19. (6, 4) and (2, 8) 20. (0, 7) and (4, 5) 21. and 22. and 23. and 24. and 25. (1.5, 3.2) and (2.1, 4.7) 26. (1.2, 2.5) and (3.7, 4.6) 27. (14.2, 15.1) and (16.3, 17.5) 28. (1.1, 2.2) and (3.3, 4.4) 29. 30. 31. 32. (2 15, 4) and (1, 2 13)(1, 13) and (- 12, -2) (3 15, -3 13) and (- 15, - 13)(13, 5 12) and (13, 12) A1 2, -7 3 BA7 5, 1 9 BA1 4, 1 3 BA- 2 3, - 1 5 B A9 5, -2 3 BA1 5, 7 3 BA7 2, 10 3 BA -1 2, 1 3 B S K I LL S E X E R C I S E S S E CTI O N 2.1 x F y E D C B A S E CTI O N 2.1 S U M MARY Cartesian Plane Plotting coordinates: (x, y) Quadrants: I, II, III, and IV Origin: (0, 0) c02a.qxd 12/23/11 9:03 PM Page 186 216. In Exercises 33 and 34, calculate (to two decimal places) the perimeter of the triangle with the following vertices: 33. Points A, B, and C 34. Points C, D, and E In Exercises 3538, determine whether the triangle with the given vertices is a right triangle, an isosceles triangle, neither, or both. (Recall that a right triangle satises the Pythagorean theorem and an isosceles triangle has at least two sides of equal length.) 35. (0, 3), (3, 3), and (3, 5) 36. (0, 2), (2, 2), and (2, 2) 37. (1, 1), (3, 1), and (2, 4) 38. (3, 3), (3, 3), and (3, 3) x C y D A B E 39. Cell Phones. A cellular phone company currently has three towers: one in Tampa, one in Orlando, and one in Gainesville to serve the central Florida region. If Orlando is 80 miles east of Tampa and Gainesville is 100 miles north of Tampa, what is the distance from Orlando to Gainesville? 40. Cell Phones. The same cellular phone company in Exercise 39 has decided to add additional towers at each halfway between cities. How many miles from Tampa is each halfway tower? 41. Travel. A retired couple who live in Columbia, South Carolina, decide to take their motor home and visit two children who live in Atlanta and in Savannah, Georgia. Savannah is 160 miles south of Columbia, and Atlanta is 215 miles west of Columbia. How far apart do the children live from each other? A P P L I C AT I O N S Albany Marietta Savannah Atlanta Columbia Macon Columbus Greenville Charleston 42. Sports. In the 1984 Orange Bowl, Doug Flutie, the 5 foot 9 inch quarterback for Boston College, shocked the world as he threw a hail Mary pass that was caught in the end zone with no time left on the clock, defeating the Miami Hurricanes 4745. Although the record books have it listed as a 48 yard pass, what was the actual distance the ball was thrown? The following illustration depicts the path of the ball. 43. NASCAR Revenue. Action Performance Inc., the leading seller of NASCAR merchandise, recorded $260 million in revenue in 2002 and $400 million in revenue in 2004. Calculate the midpoint to estimate the revenue Action Performance Inc. recorded in 2003. Assume the horizontal axis represents the year and the vertical axis represents the revenue in millions. RevenueinMillions (2004, 400) (2002, 260) Year 2001 2004 2007 2010 200 100 300 400 500 x y (5, 50) (10, 2) Midfield End Zone 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 187 c02a.qxd 11/22/11 8:49 PM Page 187 217. 188 C HAP TE R 2 Graphs 44. Ticket Price. In 1993, the average Miami Dolphins ticket price was $28, and in 2001 the average price was $56. Find the midpoint of the segment joining these two points to estimate the ticket price in 1997. AverageCostof MiamiDolphinsTicket (2001, 56) (1993, 28) Year 1995 2000 40 20 60 In Exercises 45 and 46, refer to the following: It is often useful to display data in visual form by plotting the data as a set of points. This provides a graphical display between the two variables. The following table contains data on the average monthly price of gasoline. U.S. All Grades Conventional Retail Gasoline Prices, 19942010 (Dollars per Gallon) YEAR JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC 1994 1.175 1.112 1995 1.107 1.099 1.099 1.143 1.213 1.226 1.189 1.161 1.148 1.122 1.098 1.105 1996 1.123 1.121 1.169 1.259 1.302 1.282 1.254 1.238 1.238 1.243 1.273 1.273 1997 1.270 1.263 1.237 1.228 1.229 1.227 1.206 1.250 1.254 1.222 1.198 1.159 1998 1.115 1.082 1.055 1.064 1.088 1.086 1.078 1.049 1.033 1.045 1.020 0.964 1999 0.957 0.940 1.000 1.137 1.143 1.134 1.177 1.237 1.279 1.271 1.280 1.302 2000 1.319 1.409 1.538 1.476 1.496 1.645 1.568 1.480 1.562 1.546 1.533 1.458 2001 1.467 1.471 1.423 1.557 1.689 1.586 1.381 1.422 1.539 1.312 1.177 1.111 2002 1.134 1.129 1.259 1.402 1.394 1.380 1.402 1.398 1.403 1.466 1.424 1.389 2003 1.464 1.622 1.675 1.557 1.477 1.489 1.519 1.625 1.654 1.551 1.512 1.488 2004 1.595 1.654 1.728 1.794 1.981 1.950 1.902 1.880 1.880 1.993 1.973 1.843 2005 1.852 1.927 2.102 2.251 2.155 2.162 2.287 2.489 2.907 2.736 2.265 2.216 2006 2.343 2.293 2.454 2.762 2.873 2.849 2.964 2.952 2.548 2.258 2.254 2.328 2007 2.237 2.276 2.546 2.831 3.157 3.067 2.989 2.821 2.858 2.838 3.110 3.032 2008 3.068 3.064 3.263 3.468 3.783 4.038 4.051 3.789 3.760 3.065 2.153 1.721 2009 1.821 1.942 1.987 2.071 2.289 2.645 2.530 2.613 2.530 2.549 2.665 2.620 2010 2.730 2.657 2.793 2.867 2.847 2.733 2.728 2.733 2.727 2.816 2.866 3.004 Source: http://www.eia.doe.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=EMM_EPM0U_PTE_NUS_DPG&f=M The following graph displays the data for the year 2000. 2 4 6 8 1210 $2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20 Month DollarsperGallon(2000) 45. Economics. Create a graph displaying the price of gasoline for the year 2008. 46. Economics. Create a graph displaying the price of gasoline for the year 2009. c02a.qxd 11/22/11 8:49 PM Page 188 218. 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 189 C HALLE N G E 57. Assume that two points (x1, y1) and (x2, y2) are connected by a segment. Prove that the distance from the midpoint of the segment to either of the two points is the same. 58. Prove that the diagonals of a parallelogram in the gure intersect at their midpoints. x y (0, 0) (b, 0) (a, c) (a + b, c) In Exercises 4750, explain the mistake that is made. 47. Calculate the distance between (2, 7) and (9, 10). Solution: Write the distance formula. Substitute (2, 7) and (9, 10). Simplify. This is incorrect. What mistake was made? 48. Calculate the distance between (2, 1) and (3, 7). Solution: Write the distance formula. Substitute (2, 1) and (3, 7). Simplify. This is incorrect. What mistake was made? d = 2(1)2 + (-8)2 = 265 d = 2(3 - 2)2 + (-7 - 1)2 d = 2(x2 - x1)2 + (y2 - y1)2 d = 2(5)2 + (1)2 = 226 d = 2(7 - 2)2 + (10 - 9)2 d = 2(x2 - x1)2 + (y2 - y1)2 49. Compute the midpoint of the segment with endpoints (3, 4) and (7, 9). Solution: Write the midpoint formula. (xm, ym) = a x1 + x2 2 , y1 + y2 2 b C AT C H T H E M I S TA K E Substitute (3, 4) and (7, 9). Simplify. This is incorrect. What mistake was made? 50. Compute the midpoint of the segment with endpoints (1, 2) and (3, 4). Solution: Write the midpoint formula. Substitute (1, 2) and (3, 4). Simplify. (xm, ym) (1, 1) This is incorrect. What mistake was made? (xm, ym) = a -1 - (-3) 2 , -2 - (-4) 2 b (xm, ym) = a x1 - x2 2 , y1 - y2 2 b (xm, ym) = a 1 2 , 16 2 b = a 1 2 , 4b (xm, ym) = a -3 + 4 2 , 7 + 9 2 b 51. The distance from the origin to the point (a, b) is 52. The midpoint of the line segment joining the origin and the point (a, a) is 53. The midpoint of any segment joining two points in quadrant I also lies in quadrant I. a a 2 , a 2 b. d = 2a2 + b2 . In Exercises 5154, determine whether each statement is true or false. C O N C E P T U A L 54. The midpoint of any segment joining a point in quadrant I to a point in quadrant III also lies in either quadrant I or III. 55. Calculate the length and the midpoint of the line segment joining the points (a, b) and (b, a). 56. Calculate the length and the midpoint of the line segment joining the points (a, b) and (a, b). c02a.qxd 11/22/11 8:49 PM Page 189 219. In Exercises 6164, calculate the distance between the two points. Use a graphing utility to graph the segment joining the two points and nd the midpoint of the segment. 61. (2.3, 4.1) and (3.7, 6.2) 62. (4.9, 3.2) and (5.2, 3.4) 63. (1.1, 2.2) and (3.3, 4.4) 64. (1.3, 7.2) and (2.3, 4.5) T E C H N O L O G Y C O N C E P TUAL O BJ E CTIVE Relate symmetry graphically and algebraically. G R AP H I N G E Q UATI O N S: P O I NT-P LOT TI N G, I NTE R C E P TS, AN D SYM M ETRY S K I LLS O BJ E CTIVE S Sketch graphs of equations by plotting points. Find intercepts for graphs of equations. Conduct a test for symmetry about the x-axis, y-axis, and origin. Use intercepts and symmetry as graphing aids. S E CTI O N 2.2 In this section, you will learn how to graph equations by plotting points. However, when we discuss graphing principles in Chapter 3, you will see that other techniques can be more efcient. Point-Plotting Most equations in two variables, such as y x2 , have an innite number of ordered pairs as solutions. For example, (0, 0) is a solution to y x2 because when x 0 and y 0, the equation is true. Two other solutions are (1, 1) and (1, 1). The graph of an equation in two variables, x and y, consists of all the points in the xy-plane whose coordinates (x, y) satisfy the equation. A procedure for plotting the graphs of equations is outlined below and is illustrated with the example y x2 . 190 C HAP TE R 2 Graphs 59. Assume that two points (a, b) and (c, d) are the endpoints of a line segment. Calculate the distance between the two points. Prove that it does not matter which point is labeled as the rst point in the distance formula. 60. Show that the points (1, 1), (0, 0), and (2, 2) are collinear (lie on the same line) by showing that the sum of the distance from (1, 1) to (0, 0) and the distance from (0, 0) to (2, 2) is equal to the distance from (1, 1) to (2, 2). c02a.qxd 11/22/11 8:49 PM Page 190 220. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 191 E X AM P LE 1 Graphing an Equation of a Line by Plotting Points Graph the equation y 2x 1. Solution: STEP 1 In a table, list several pairs of coordinates that make the equation true. STEP 2 Plot these points on a graph and connect the points, resulting in a line. WORDS MATH Step 1: In a table, list several pairs of coordinates that make the equation true. Step 2: Plot these points on a graph and connect the points with a smooth curve. Use arrows to indicate that the graph continues. In graphing an equation, rst select arbitrary values for x and then use the equation to nd the corresponding value of y, or vice versa. x y x2 (x, y) 0 0 (0, 0) 1 1 (1, 1) 1 1 (1, 1) 2 4 (2, 4) 2 4 (2, 4) x y (2, 4) (1, 1) (0, 0) (2, 4) (1, 1) x y 2x 1 (x, y) 0 1 (0, 1) 1 3 (1, 3) 1 1 (1, 1) 2 5 (2, 5) 2 3 (2, 3) x y (1, 3) (0, 1) (2, 5) (2, 3) (1, 1) YO U R TU R N The graph of the equation y x 1 is a line. Graph the line. Answer: x y (1, 2) (1, 0) c02a.qxd 11/22/11 8:49 PM Page 191 221. 192 C HAP TE R 2 Graphs E X AM P LE 2 Graphing an Equation by Plotting Points Graph the equation y x2 5. Solution: STEP 1 In a table, list several pairs of coordinates that make the equation true. STEP 2 Plot these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues. This graph is called a parabola and will be discussed in further detail in Chapter 8. x y x2 5 (x, y) 0 5 (0, 5) 1 4 (1, 4) 1 4 (1, 4) 2 1 (2, 1) 2 1 (2, 1) 3 4 (3, 4) 3 4 (3, 4) Technology Tip Graph of is shown in a by viewing rectangle.[-8, 8][-5, 5] y1 = x2 - 5 x y (3, 4) (3, 4) (2, 1) (1, 4) (0, 5) (2, 1) (1, 4) YO U R TU R N Graph the equation y x2 1. E X AM P LE 3 Graphing an Equation by Plotting Points Graph the equation y x3 . Solution: STEP 1 In a table, list several pairs of coordinates that satisfy the equation. STEP 2 Plot these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues in both the positive and negative directions. x y x3 (x, y) 0 0 (0, 0) 1 1 (1, 1) 1 1 (1, 1) 2 8 (2, 8) 2 8 (2, 8) x y (2, 8) (2, 8) (1, 1) (0, 0) (1, 1) 4 8 8 4 Answer: x y (1, 0) (0, 1) (2, 3) (2, 3) (1, 0) c02a.qxd 11/22/11 8:49 PM Page 192 222. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 193 One x-intercept No x-intercepts Two y-intercepts One y-intercept No x-intercepts Three x-intercepts No y-intercepts One y-intercept Note: The origin (0, 0) corresponds to both an x-intercept and a y-intercept. Intercepts When point-plotting graphs of equations, which points should be selected? Points where a graph crosses (or touches) either the x-axis or y-axis are called intercepts, and identifying these points helps dene the graph unmistakably. An x-intercept of a graph is a point where the graph intersects the x-axis. Specically, an x-intercept is the x-coordinate of such a point. For example, if a graph intersects the x-axis at the point (3, 0), then we say that 3 is the x-intercept. Since the value for y along the x-axis is zero, all points corresponding to x-intercepts have the form (a, 0). A y-intercept of a graph is a point where the graph intersects the y-axis. Specically, a y-intercept is the y-coordinate of such a point. For example, if a graph intersects the y-axis at the point (0, 2), then we say that 2 is the y-intercept. Since the value for x along the y-axis is zero, all points corresponding to y-intercepts have the form (0, b). It is important to note that graphs of equations do not have to have intercepts, and if they do have intercepts, they can have one or more of each type. x y x y x y x y The graph given to the right has two y-intercepts and one x-intercept. The x-intercept is 1, which corresponds to the point (1, 0). The y-intercepts are 1 and 1, which correspond to the points (0, 1) and (0, 1), respectively. Algebraically, how do we nd intercepts from an equation? The graph in the margin corresponds to the equation x y2 1. The x-intercepts are located on the x-axis, which corresponds to y 0. If we let y 0 in the equation x y2 1 and solve for x, the result is x 1. This corresponds to the x-intercept we identied above. Similarly, the y-intercepts are located on the y-axis, which corresponds to x 0. If we let x 0 in the equation x y2 1 and solve for y, the result is y ;1. These correspond to the y-intercepts we identied above. Study Tip Identifying the intercepts helps dene the graph unmistakably. x y (1, 0) (0, 1) (3, 2) (3, 2)(0, 1) c02a.qxd 12/23/11 9:05 PM Page 193 223. 194 C HAP TE R 2 Graphs E X AM P LE 4 Finding Intercepts from an Equation Given the equation y x2 1, nd the indicated intercepts of its graph, if any. a. x-intercept(s) b. y-intercept(s) Solution (a): Let y 0. 0 x2 1 Solve for x. x2 1 no real solution There are no x-intercepts. Solution (b): Let x 0. y 02 1 Solve for y. y 1 The y-intercept is located at the point (0, 1) . YO U R TU R N For the equation y x2 4, a. nd the x-intercept(s), if any. b. nd the y-intercept(s), if any. Symmetry The word symmetry conveys balance. Suppose you have two pictures to hang on a wall. If you space them equally apart on the wall, then you prefer a symmetric dcor. This is an example of symmetry about a line. The word (water) written below is identical if you rotate the word 180 degrees (or turn the page upside down). This is an example of symmetry about a point. Symmetric graphs have the characteristic that their mirror image can be obtained about a reference, typically a line or a point. Answer: a. x-intercepts: 2 and 2 b. y-intercept: 4 x y (3, 4) (3, 4) (2, 1) (1, 4) (0, 5) (2, 1) y = x2 5 (1, 4) x y (2, 8) (2, 8) (1, 1) (0, 0) (1, 1) 4 8 8 4 y = x3 In Example 2, the points (2, 1) and (2, 1) both lie on the graph of y x2 5, as do the points (1, 4) and (1, 4). Notice that the graph on the right side of the y-axis is a mirror image of the part of the graph to the left of the y-axis. This graph illustrates symmetry with respect to the y-axis (the line x 0). In the graph of the equation x y2 1 in the margin, the points (0, 1) and (0, 1) both lie on the graph, as do the points (3, 2) and (3, 2). Notice that the part of the graph above the x-axis is a mirror image of the part of the graph below the x-axis. This graph illustrates symmetry with respect to the x-axis (the line y 0). In Example 3, the points (1, 1) and (1, 1) both lie on the graph. Notice that rotating this graph 180 degrees (or turning your page upside down) results in an identical graph. This is an example of symmetry with respect to the origin (0, 0). Symmetry aids in graphing by giving information for free. For example, if a graph is symmetric about the y-axis, then once the graph to the right of the y-axis is found, the left side of the graph is the mirror image of that. If a graph is symmetric about the origin, then once the graph is known in quadrant I, the graph in quadrant III is found by rotating the known graph 180 degrees. x y (1, 0) (0, 1) (3, 2) (3, 2)(0, 1) x = y2 1 c02a.qxd 11/22/11 8:49 PM Page 194 224. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 195 It would be benecial to know whether a graph of an equation is symmetric about a line or point before the graph of the equation is sketched. Although a graph can be symmetric about any line or point, we will discuss only symmetry about the x-axis, y-axis, and origin. These types of symmetry and the algebraic procedures for testing for symmetry are outlined below. Types and Tests for Symmetry IF THE POINT (a, b) TYPE OF IS ON THE GRAPH, ALGEBRAIC TEST FOR SYMMETRY GRAPH THEN THE POINT . . . SYMMETRY Symmetric (a, b) is on the graph. Replacing y with y with respect leaves the equation to the x-axis unchanged. Symmetric (a, b) is on the graph. Replacing x with x with respect leaves the equation to the y-axis unchanged. Symmetric (a, b) is on the graph. Replacing x with x with respect and y with y leaves to the origin the equation unchanged. a b b x y (a, b) (a, b) x a a b y (a, b)(a, b) a a b b x y (0, 0) (a, b) (a, b) Study Tip Symmetry gives us information about the graph for free. Technology Tip To enter the graph of y2 x3 , solve for y rst. The graphs of are shown.and y2 = - 2x3 y1 = 2x3 E X AM P LE 5 Testing for Symmetry with Respect to the Axes Test the equation y2 x3 for symmetry with respect to the axes. Solution: Test for symmetry with respect to the x-axis. Replace y with y. (y)2 x3 Simplify. y2 x3 The resulting equation is the same as the original equation, y2 x3 . Therefore y2 x3 is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis. Replace x with x. y2 (x)3 Simplify. y2 x3 The resulting equation, y2 x3 , is not the same as the original equation, y2 x3 . Therefore y2 x3 is not symmetric with respect to the y-axis. c02a.qxd 11/22/11 8:49 PM Page 195 225. 196 C HAP TE R 2 Graphs Technology Tip Graph of y1 x2 1 is shown. When testing for symmetry about the x-axis, y-axis, and origin, there are ve possibilities: No symmetry Symmetry with respect to the x-axis Symmetry with respect to the y-axis Symmetry with respect to the origin Symmetry with respect to the x-axis, y-axis, and origin Graph of y1 x3 1 is shown. YO U R TU R N Determine the symmetry (if any) for x y2 1. Answer: The graph of the equation is symmetric with respect to the x-axis. E X AM P LE 6 Testing for Symmetry Determine what type of symmetry (if any) the graphs of the equations exhibit. a. y x2 1 b. y x3 1 Solution (a): Replace x with x. y (x)2 1 Simplify. y x2 1 The resulting equation is equivalent to the original equation, so the graph of the equation y x2 1 is symmetric with respect to the y-axis. Replace y with y. (y) x2 1 Simplify. y x2 1 The resulting equation y x2 1 is not equivalent to the original equation y x2 1, so the graph of the equation y x2 1 is not symmetric with respect to the x-axis. Replace x with x and y with y. (y) (x)2 1 Simplify. y x2 1 y x2 1 The resulting equation y x2 1 is not equivalent to the original equation y x2 1, so the graph of the equation y x2 1 is not symmetric with respect to the origin. The graph of the equation y x2 1 is symmetric with respect to the y-axis. Solution (b): Replace x with x. y (x)3 1 Simplify. y x3 1 The resulting equation y x3 1 is not equivalent to the original equation y x3 1. Therefore, the graph of the equation y x3 1 is not symmetric with respect to the y-axis. Replace y with y. (y) x3 1 Simplify. y x3 1 The resulting equation y x3 1 is not equivalent to the original equation y x3 1. Therefore, the graph of the equation y x3 1 is not symmetric with respect to the x-axis. Replace x with x and y with y. (y) (x)3 1 Simplify. y x3 1 y x3 1 The resulting equation y x3 1 is not equivalent to the original equation y x3 1. Therefore, the graph of the equation y x3 1 is not symmetric with respect to the origin. The graph of the equation y x3 1 exhibits no symmetry. c02a.qxd 11/22/11 8:49 PM Page 196 226. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 197 Using Intercepts and Symmetry as Graphing Aids How can we use intercepts and symmetry to assist us in graphing? Intercepts are a good starting pointthough not the only one. For symmetry, look back at Example 2, y x2 5. We selected seven x-coordinates and solved the equation to find the corresponding y-coordinates. If we had known that this graph was symmetric with respect to the y-axis, then we would have had to find the solutions to only the positive x-coordinates, since we get the negative x-coordinates for free. For example, we found the point (1, 4) to be a solution to the equation. The rules of symmetry tell us that (1, 4) is also on the graph. E X AM P LE 7 Using Intercepts and Symmetry as Graphing Aids For the equation x2 y2 25, use intercepts and symmetry to help you graph the equation using the point-plotting technique. Solution: STEP 1 Find the intercepts. For the x-intercepts, let y 0. x2 02 25 Solve for x. x ;5 The two x-intercepts correspond to the points (5, 0) and (5, 0). For the y-intercepts, let x 0. 02 y2 25 Solve for y. y ;5 The two y-intercepts correspond to the points (0, 5) and (0, 5). STEP 2 Identify the points on the graph corresponding to the intercepts. STEP 3 Test for symmetry with respect to the y-axis, x-axis, and origin. Test for symmetry with respect to the y-axis. Replace x with x. (x)2 y2 25 Simplify. x2 y2 25 The resulting equation is equivalent to the original, so the graph of x2 y2 25 is symmetric with respect to the y-axis. y x 55 5 5 c02a.qxd 11/22/11 8:49 PM Page 197 227. 198 C HAP TE R 2 Graphs Technology Tip To enter the graph of x2 y2 25, solve for y rst. The graphs of and are shown.y2 = - 225 - x2 y1 = 225 - x2 Test for symmetry with respect to the x-axis. Replace y with y. x2 (y)2 25 Simplify. x2 y2 25 The resulting equation is equivalent to the original, so the graph of x2 y2 25 is symmetric with respect to the x-axis. Test for symmetry with respect to the origin. Replace x with x and y with y. (x)2 (y)2 25 Simplify. x2 y2 25 The resulting equation is equivalent to the original, so the graph of x2 y2 25 is symmetric with respect to the origin. Since the graph is symmetric with respect to the y-axis, x-axis, and origin, we need to determine solutions to the equation on only the positive x- and y-axes and in quadrant I because of the following symmetries: Symmetry with respect to the y-axis gives the solutions in quadrant II. Symmetry with respect to the origin gives the solutions in quadrant III. Symmetry with respect to the x-axis yields solutions in quadrant IV. Solutions to x2 y2 25. Quadrant I: (3, 4), (4, 3) Additional points due to symmetry: Quadrant II: (3, 4), (4, 3) Quadrant III: (3, 4), (4, 3), Quadrant IV: (3, 4), (4, 3) Connecting the points with a smooth curve yields a circle. We will discuss circles in more detail in Section 2.4. S U M MARY S E CTI O N 2.2 POINT WHERE THE INTERCEPTS GRAPH INTERSECTS THE ... HOW TO FIND INTERCEPTS POINT ON GRAPH x-intercept x-axis Let y 0 and solve for x. (a, 0) y-intercept y-axis Let x 0 and solve for y. (0, b) Symmetry about the x-axis, y-axis, and origin is dened both algebraically and graphically. Intercepts and symmetry provide much of the information useful for sketching graphs of equations. Sketching graphs of equations can be accomplished using a point-plotting technique. Intercepts are dened as points where a graph intersects an axis or the origin. Intercepts x (3, 4) (3, 4) (4, 3) (4, 3) (5, 0) (5, 0) (0, 5) (0, 5) (4, 3)(4, 3) (3, 4)(3, 4) y c02a.qxd 11/22/11 8:49 PM Page 198 228. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 199 a b b x y (a, b) (a, b) x a a b y (a, b)(a, b) a a b b x y (0, 0) (a, b) (a, b) Types and Tests for Symmetry IF THE POINT (a, b) TYPE OF IS ON THE GRAPH, ALGEBRAIC TEST FOR SYMMETRY GRAPH THEN THE POINT . . . SYMMETRY Symmetric (a, b) is on the graph. Replacing y with y with respect leaves the equation to the x-axis unchanged. Symmetric (a, b) is on the graph. Replacing x with x with respect leaves the equation to the y-axis unchanged. Symmetric (a, b) is on the graph. Replacing x with x with respect and y with y leaves to the origin the equation unchanged. In Exercises 18, determine whether each point lies on the graph of the equation. 1. y 3x 5 a. (1, 2) b. (2, 11) 2. y 2x 7 a. (1, 9) b. (2, 4) 3. a. (5, 2) b. (5, 6) 4. a. (8, 5) b. (4, 4) 5. y x2 2x 1 a. (1, 4) b. (0, 1) 6. y x3 1 a. (1, 0) b. (2, 9) 7. a. (7, 3) b. (6, 4) 8. y 2 3 x a. (9, 4) b. (2, 7) In Exercises 914, complete the table and use the table to sketch a graph of the equation. 9. 10. y = 2x + 2 y = -3 4x + 1y = 2 5x - 4 S K I LL S E X E R C I S E S S E C TI O N 2.2 x y 2 x (x, y) 2 0 1 x y 3x 1 (x, y) 1 0 2 c02a.qxd 11/22/11 8:49 PM Page 199 229. 11. 12. 13. 14. In Exercises 1522, graph the equation by plotting points. 15. y 3x 2 16. y 4 x 17. y x2 x 2 18. y x2 2x 1 19. x y2 1 20. x y 1 2 21. 22. y 0.5 x 1 In Exercises 2332, nd the x-intercept(s) and y-intercepts(s) (if any) of the graphs of the given equations. 23. 2x y 6 24. 4x 2y 10 25. y x2 9 26. y 4x2 1 27. 28. 29. 30. 31. 4x2 y2 16 32. x2 y2 9 In Exercises 3338, match the graph with the corresponding symmetry. a. No symmetry b. Symmetry with respect to the x-axis c. Symmetry with respect to the y-axis d. Symmetry with respect to the origin e. Symmetry with respect to the x-axis, y-axis, and origin 33. 34. 35. y = x2 - x - 12 x y = 1 x2 + 4 y = 2 3 x - 8y = 2x - 4 y = 1 2x - 3 2 x y (x, y) 1 2 5 10 2x 1 x y12xx2 (x, y) 3 2 1 0 1 x y x2 x (x, y) 1 0 1 2 1 2 x (x, y) 2 1 2 7 y 2x 2 x y x (1, 1) (1, 1) y x y 200 C HAP TE R 2 Graphs c02a.qxd 11/22/11 8:49 PM Page 200 230. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 201 36. 37. 38. In Exercises 3944, a point that lies on a graph is given along with that graphs symmetry. State the other known points that must also lie on the graph. Point on The Graph Is Symmetric Point on The Graph Is Symmetric a Graph about the a Graph about the 39. (1, 3) x-axis 40. (2, 4) y-axis 41. (7, 10) origin 42. (1, 1) origin 43. (3, 2) x-axis, y-axis, and origin 44. (1, 7) x-axis, y-axis, and origin In Exercises 4558, test algebraically to determine whether the equations graph is symmetric with respect to the x-axis, y-axis, or origin. 45. x y2 4 46. x 2y2 3 47. y x3 x 48. y x5 1 49. x y 50. x y 2 51. x2 y2 100 52. x2 2y2 30 53. y x2/3 54. x y2/3 55. x2 y3 1 56. 57. 58. xy 1 In Exercises 5972, plot the graph of the given equation. 59. y x 60. 61. y x2 1 62. y 9 4x2 63. 64. x y2 1 65. 66. xy 1 67. y x 68. x y 69. x2 y2 16 70. 71. x2 y2 16 72. x2 - y2 25 = 1 x2 4 + y2 9 = 1 y = 1 x y = x3 2 y = - 1 2 x + 3 y = 2 x y = 21 + x2 75. Electronic Signals: Radio Waves. The received power of an electromagnetic signal is a fraction of the power transmitted. The relationship is given by where R is the distance that the signal has traveled in meters. Plot the percentage of transmitted power that is received for R 100 m, 1 km, and 10,000 km. Preceived = Ptransmitted . 1 R2 73. Sprinkler. A sprinkler will water a grassy area in the shape of x2 y2 9. Apply symmetry to draw the watered area, assuming the sprinkler is located at the origin. 74. Sprinkler. A sprinkler will water a grassy area in the shape of Apply symmetry to draw the watered area, assuming the sprinkler is located at the origin. x2 + y2 9 = 1. A P P L I C AT I O N S x y (2, 2) (3, 3) (3, 3) (2, 2) (0, 0) x y x y c02a.qxd 11/22/11 8:49 PM Page 201 231. 202 C HAP TE R 2 Graphs In Exercises 8184, explain the mistake that is made. C AT C H T H E M I S TA K E 81. Graph the equation y x2 1. Solution: 82. Test y x2 for symmetry with respect to the y-axis. Solution: Replace x with x. y (x)2 Simplify. y x2 The resulting equation is not equivalent to the original equation; y x2 is not symmetric with respect to the y-axis. This is incorrect. What mistake was made? 83. Test x y for symmetry with respect to the y-axis. Solution: Replace y with y. x y Simplify. x y The resulting equation is equivalent to the original equation; x y is symmetric with respect to the y-axis. This is incorrect. What mistake was made? x y x2 1 (x, y) 0 1 (0, 1) 1 2 (1, 2) x y (1, 2)(0, 1) 76. Electronic Signals: Laser Beams. The wavelength and the frequency f of a signal are related by the equation where c is the speed of light in a vacuum, c 3.0 108 meters per second. For the values, 0.001, 1, and 100 mm, plot the points corresponding to frequency, f. What do you notice about the relationship between frequency and wavelength? Note that the frequency will have units Hz 1/seconds. 77. Prot. The prot associated with making a particular product is given by the equation where y represents the prot in millions of dollars and x represents the number of thousands of units sold. (x 1 corresponds to 1000 units and y 1 corresponds to $1M.) Graph this equation and determine how many units must be sold to break even (prot 0). Determine the range of units sold that correspond to making a prot. 78. Prot. The prot associated with making a particular product is given by the equation where y represents the prot in millions of dollars and x represents the number of thousands of units sold. (x 1 corresponds to 1000 units and y 1 corresponds to $1M.) Graph this equation and determine how many units must be sold to break even (prot 0). Determine the range of units sold that correspond to making a prot. y = -x2 + 4x - 3 y = -x2 + 6x - 8 f = c l 79. Economics. The demand for an electronic device is modeled by where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b. Plot the demand equation. 80. Economics. The demand for a new electronic game is modeled by where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b. Plot the demand equation. p = 39.95 - 20.01x - 0.4 p = 2.95 - 20.01x - 0.01 This is incorrect. What mistake was made? c02a.qxd 11/22/11 8:49 PM Page 202 232. 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry 203 85. If the point (a, b) lies on a graph that is symmetric about the x-axis, then the point (a, b) also must lie on the graph. 86. If the point (a, b) lies on a graph that is symmetric about the y-axis, then the point (a, b) also must lie on the graph. C O N C E P T U A L C H A L L E N G E 89. Determine whether the graph of has any symmetry, where a, b, and c are real numbers. y = ax2 + b cx3 91. y 16.7x4 3.3x2 7.1 94. 3.2x2 5.1y2 1.3 92. y 0.4x5 8.2x3 1.3x 95. 1.2x2 4.7y2 19.4 93. 2.3x2 5.5 y 96. 2.1y2 0.8 x 1 T E C H N O L O G Y 87. If the point (a, b) lies on a graph that is symmetric about the x-axis, y-axis, and origin, then the points (a, b), (a, b), and (a, b) must also lie on the graph. 88. Two points are all that is needed to plot the graph of an equation. 84. Use symmetry to help you graph x2 y 1. Solution: Replace x with x. (x)2 y 1 Simplify. x2 y 1 x2 y 1 is symmetric with respect to the x-axis. Determine points that lie on the graph in quadrant I. x y 5 5 5 5 y x2 y 1 (x, y) 1 0 (0, 1) 2 1 (1, 2) 5 2 (2, 5) Symmetry with respect to the x-axis implies that (0, 1), (1, 2), and (2, 5) are also points that lie on the graph. This is incorrect. What mistake was made? In Exercises 9196, graph the equation using a graphing utility and state whether there is any symmetry. 90. Find the intercepts of y (x a)2 b2 , where a and b are real numbers. In Exercises 8588, determine whether each statement is true or false. c02a.qxd 11/22/11 8:49 PM Page 203 233. x y (x, y) 2 2 (2, 2) 1 4 (1, 4) x y (1, 4) 2x y = 2 (2, 2) If A, B, and C are constants and x and y are variables, then the equation is in general form, and its graph is a straight line. Note: A or B (but not both) can be zero. Ax + By = C EQUATION OF A STRAIGHT LINE: GENERAL* FORM *Some books refer to this as standard form. The equation 2x y 2 is a rst-degree equation, so its graph is a straight line. To graph this line, list two solutions in a table, plot those points, and use a straight edge to draw the line. Graphing a Line What is the shortest path between two points? The answer is a straight line. In this section, we will discuss characteristics of lines such as slope and intercepts. We will also discuss types of lines such as horizontal, vertical, falling, and rising, and recognize relationships between lines such as perpendicular and parallel. At the end of this section, you should be able to nd the equation of a line when given two specic pieces of information about the line. First-degree equations, such as and have graphs that are straight lines. The rst two equations given represent inclined or slant lines, whereas y 2 represents a horizontal line and x 3 represents a vertical line. One way of writing an equation of a straight line is called general form. x = -3,y = 2,3x + y = 6,y = -2x + 4, C O N C E P TUAL O BJ E CTIVE S Classify lines as rising, falling, horizontal, and vertical. Understand slope as a rate of change. Associate two lines having the same slope with the graph of parallel lines. Associate two lines having negative reciprocal slopes with the graph of perpendicular lines. LI N E S S K I LLS O BJ E CTIVE S Determine x- and y-intercepts of a line. Calculate the slope of a line. Find the equation of a line using slopeintercept form. Find the equation of a line using pointslope form. Find the equation of a line that is parallel or perpendicular to a given line. S E CTI O N 2.3 204 c02a.qxd 11/22/11 8:49 PM Page 204 234. 2.3 Lines 205 Coordinates Axis Crossed Algebraic Method x-intercept: (a, 0) x-axis Set y 0 and solve for x. y-intercept: (0, b) y-axis Set x 0 and solve for y. DETERMINING INTERCEPTS x y (a, 0) (0, b) TYPE OF LINE EQUATION x-INTERCEPT y-INTERCEPT GRAPH Horizontal y b None b Vertical x a a None Note: The special cases of (y-axis) and (x-axis) have innitely many y-intercepts and x-intercepts, respectively. y = 0x = 0 Horizontal lines and vertical lines, however, each have only one intercept. x y (0, b) (a, 0) x y Intercepts The point where a line crosses, or intersects, the x-axis is called the x-intercept. The point where a line crosses, or intersects, the y-axis is called the y-intercept. By inspecting the graph of the previous line, we see that the x-intercept is (1, 0) and the y-intercept is (0, 2). Intercepts were discussed in Section 2.2. There, we found that the graphs of some equations could have no intercepts or one or multiple intercepts. Slant lines, however, have exactly one x-intercept and exactly one y-intercept. Study Tip x-intercept: Set y 0. y-intercept: Set x 0. x y (1, 4) (1, 0) (0, 2) (2, 2) xintercept yintercept x y (1, 4) (1, 0) (0, 2) (2, 2) xintercept yintercept c02a.qxd 11/22/11 8:49 PM Page 205 235. 206 C HAP TE R 2 Graphs x y x = 2 (2, 0) YO U R TU R N Determine the x- and y-intercepts (if they exist) for the lines given by the following equations. a. 3x y 2 b. y 5 Slope If the graph of 2x y 2 represented an incline that you were about to walk on, would you classify that incline as steep? In the language of mathematics, we use the word slope as a measure of steepness. Slope is the ratio of the change in y over the change in x. An easy way to remember this is rise over run. A nonvertical line passing through two points (x1, y1) and (x2, y2) has slope m, given by the formula or Note: Always start with the same point for both the x-coordinates and the y-coordinates. m = rise run = vertical change horizontal change m = y2 - y1 x2 - x1 , where x1 Z x2 SLOPE OF A LINE x rise run y (x2, y2) (x1, y1) Study Tip To get the correct sign (;) for the slope, remember to start with the same point for both x and y. Solution (b): x 2 This vertical line consists of all points (2, y). The graph shows that the x-intercept is 2. We also nd that the line never crosses the y-axis, so the y-intercept does not exist. Answer: a. x-intercept: ; y-intercept: (0, 2) b. x-intercept does not exist; y-intercept: (0, 5) A2 3, 0B E X AM P LE 1 Determining x- and y-Intercepts Determine the x- and y-intercepts (if they exist) for the lines given by the following equations. a. 2x 4y 10 b. x 2 Solution (a): 2x 4y 10 To nd the x-intercept, set y 0. 2x 4(0) 10 Solve for x. 2x 10 x 5 The x-intercept corresponds to the point (5, 0) . To nd the y-intercept, set x 0. 2(0) 4y 10 Solve for y. 4y 10 The y-intercept corresponds to the point .A0, 5 2 B y = 5 2 c02a.qxd 11/22/11 8:49 PM Page 206 236. 2.3 Lines 207 Notice that if we had chosen the two intercepts (x1, y1) (0, 2) and (x2, y2) (1, 0) instead, we still would have found the slope to be m 2. I N C O R R E CT The ERROR is interchanging the coordinates of the rst and second points. The calculated slope is INCORRECT by a negative sign. m = 2 -2 = -1 m = 4 - 2 1 - 3 C O R R E CT Label the points. Write the slope formula. Substitute the coordinates. Simplify. m = 2 2 = 1 m = 4 - 2 3 - 1 m = y2 - y1 x2 - x1 (x2, y2) = (3, 4)(x1, y1) = (1, 2) C O M M O N M I S TA K E The most common mistake in calculating slope is writing the coordinates in the wrong order, which results in the slope being opposite in sign. Find the slope of the line containing the two points (1, 2) and (3, 4). C A U T I O N Interchanging the coordinates can result in a sign error in the slope. Lets nd the slope of our graph 2x y 2. Well let (x1, y1) (2, 2) and (x2, y2) (1, 4) in the slope formula: m = y2 - y1 x2 - x1 = [4 - (-2)] [1 - (-2)] = 6 3 = 2 x y (1, 4) 2x y = 2 (2, 2) When interpreting slope, always read the graph from left to right. Since we have determined the slope to be 2, or we can interpret this as rising two units and running (to the right) one unit. If we start at the point (2, 2) and move two units up and one unit to the right, we end up at the x-intercept, (1, 0). Again, moving two units up and one unit to the right puts us at the y-intercept, (0, 2). Another rise of two and run of one takes us to the point (1, 4). See the gure on the right. Lines fall into one of four categories: rising, falling, horizontal, or vertical. Line Slope Rising Positive (m 0) Falling Negative (m 0) Horizontal Zero (m 0), hence y b Vertical Undened, hence x a < > 2 1, x y a b x y (1, 4) (0, 2) (2, 2) (1, 0)Rise = 2 Run = 1 The slope of a horizontal line is 0 because the y-coordinates of any two points are the same. The change in y in the slope formulas numerator is 0, hence m 0. The slope of a vertical line is undened because the x-coordinates of any two points are the same. The change in x in the slope formulas denominator is zero; hence m is undened. c02a.qxd 11/22/11 8:49 PM Page 207 237. 208 C HAP TE R 2 Graphs YO U R TU R N For each pair of points classify the line that passes through them as rising, falling, vertical, or horizontal, and determine its slope. Do not graph. a. (2, 0) and (1, 5) b. (2, 3) and (2, 5) c. (3, 1) and (3, 4) d. (1, 2) and (3, 2) Answer: a. m 5, falling b. m 2, rising c. slope is undened, vertical d. m 0, horizontal E X AM P LE 2 Graph, Classify the Line, and Determine the Slope Sketch a line through each pair of points, classify the line as rising, falling, vertical, or horizontal, and determine its slope. a. (1, 3) and (1, 1) b. (3, 3) and (3, 1) c. (1, 2) and (3, 2) d. (1, 4) and (1, 3) Solution (a): (1, 3) and (1, 1) This line is rising, so its slope is positive. Solution (b): (3, 3) and (3, 1) This line is falling, so its slope is negative. m = 3 - 1 -3 - 3 = - 2 6 = - 1 3 m = 1 - (-3) 1 - (-1) = 4 2 = 2 1 = 2 x y (3, 1) (3, 3) x y (1, 1) (1, 3) x y (3, 2)(1, 2) x y (1, 4) (1, 3) Solution (c): (1, 2) and (3, 2) This is a horizontal line, so its slope is zero. Solution (d): (1, 4) and (1, 3) This is a vertical line, so its slope is undened. which is undened.m = 3 - (-4) 1 - 1 = 7 0 , m = -2 - (-2) 3 - (-1) = 0 4 = 0 c02a.qxd 11/22/11 8:49 PM Page 208 238. 2.3 Lines 209 Equations of Lines SlopeIntercept Form As mentioned earlier, the general form for an equation of a line is Ax By C. A more standard way to write an equation of a line is in slopeintercept form, because it identies the slope and the y-intercept. The slopeintercept form for the equation of a nonvertical line is Its graph has slope m and y-intercept b. y = mx + b EQUATION OF A STRAIGHT LINE: SLOPEINTERCEPT FORM For example, 2x y 3 is in general form. To write this equation in slopeintercept form, we isolate the y variable: The slope of this line is 2 and the y-intercept is 3. y = 2x + 3 x y (0, 5) (3, 3) Technology Tip To graph the equation 2x 3y 15, solve for y rst. The graph of is shown.y1 = 2 3x - 5 YO U R TU R N Write 3x 2y 12 in slopeintercept form and graph it. x y (0, 6) (2, 3) Answer: y = 3 2 x - 6 E X AM P LE 3 Using SlopeIntercept Form to Graph an Equation of a Line Write 2x 3y 15 in slopeintercept form and graph it. Solution: STEP 1 Write in slopeintercept form. Subtract 2x from both sides. 3y 2x 15 Divide both sides by 3. STEP 2 Graph. Identify the slope and y-intercept. Slope: y-intercept: b 5 Plot the point corresponding to the y-intercept (0, 5). From the point (0, 5), rise two units and run (to the right) three units, which corresponds to the point (3, 3). Draw the line passing through the two points. m = 2 3 y = 2 3 x - 5 c02a.qxd 1/2/12 2:27 PM Page 209 239. YO U R TU R N Find the equation of the line that has slope and y-intercept (0, 2).- 3 2 Answer: y = - 3 2x + 2 210 C HAP TE R 2 Graphs Instead of starting with equations of lines and characterizing them, let us now start with particular features of a line and derive its governing equation. Suppose that you are given the y-intercept and the slope of a line. Using the slopeintercept form of an equation of a line, y mx b, you could nd its equation. E X AM P LE 4 Using SlopeIntercept Form to Find the Equation of a Line Find the equation of a line that has slope and y-intercept (0, 1). Solution: Write the slopeintercept form of an equation of a line. y mx b Label the slope. Label the y-intercept. b 1 The equation of the line in slopeintercept form is .y = 2 3x + 1 m = 2 3 2 3 PointSlope Form Now, suppose that the two pieces of information you are given about an equation are its slope and one point that lies on its graph. You still have enough information to write an equation of the line. Recall the formula for slope: We are given the slope m, and we know a particular point that lies on the line (x1, y1). We refer to all other points that lie on the line as (x, y). Substituting these values into the slope formula gives us Cross multiplying yields This is called the pointslope form of an equation of a line. y - y1 = m(x - x1) m = y - y1 x - x1 m = y2 - y1 x2 - x1 , where x2 Z x1 The pointslope form for the equation of a line is Its graph passes through the point (x1, y1), and its slope is m. Note: This formula does not hold for vertical lines since their slope is undened. y - y1 = m(x - x1) EQUATION OF A STRAIGHT LINE: POINTSLOPE FORM c02a.qxd 11/22/11 8:49 PM Page 210 240. 2.3 Lines 211 E X AM P LE 5 Using PointSlope Form to Find the Equation of a Line Find the equation of the line that has slope and passes through the point (1, 2). Solution: Write the pointslope form of an equation of a line. y y1 m(x x1) Substitute the values and (x1, y1) (1, 2). Distribute. Isolate y. We can also express the equation in general form . YO U R TU R N Derive the equation of the line that has slope and passes through the point . Give the answer in general form.A1, - 1 2 B 1 4 x + 2y = 3 y = - 1 2 x + 3 2 y - 2 = - 1 2 x - 1 2 y - 2 = - 1 2 (x - (-1))m = - 1 2 - 1 2 Suppose the slope of a line is not given at all. Instead, two points that lie on the line are given. If we know two points that lie on the line, then we can calculate the slope. Then, using the slope and either of the two points, the equation of the line can be derived. Answer: y = 1 4x - 3 4 Answer: or 7x 3y 2 y = -7 3x + 2 3 E X AM P LE 6 Finding the Equation of a Line Given Two Points Find the equation of the line that passes through the points (2, 1) and (3, 2). Solution: Write the equation of a line. y mx b Calculate the slope. Substitute (x1, y1) (2, 1) and (x2, y2) (3, 2). Substitute for the slope. Let (x, y) (3, 2). (Either point satises the equation.) Solve for b. Write the equation in slopeintercept form. Write the equation in general form. YO U R TU R N Find the equation of the line that passes through the points (1, 3) and (2, 4). -3x + 5y = 1 y = 3 5 x + 1 5 b = 1 5 2 = 3 5 (3) + b y = 3 5 x + b 3 5 m = 2 - (-1) 3 - (-2) = 3 5 m = y2 - y1 x2 - x1 Study Tip When two points that lie on a line are given, rst calculate the slope of the line, then use either point and the slopeintercept form (shown in Example 6) or the pointslope form: -3x + 5y = 1 5y - 10 = 3x - 9 5y - 10 = 3(x - 3) y - 2 = 3 5(x - 3) y - y1 = m(x - x1) m = 3 5, (3, 2) c02b.qxd 12/23/11 5:53 PM Page 211 241. 212 C HAP TE R 2 Graphs Parallel and Perpendicular Lines Two distinct nonintersecting lines in a plane are parallel. How can we tell whether the two lines in the graph on the left are parallel? Parallel lines must have the same steepness. In other words, parallel lines must have the same slope. The two lines shown on the left are parallel because they have the same slope, 2. Two distinct lines in a plane are parallel if and only if their slopes are equal. Parallel Lines In other words, if two lines in a plane are parallel, then their slopes are equal, and if the slopes of two lines in a plane are equal, then the lines are parallel. WORDS MATH Lines L1 and L2 are parallel. L1 L2 Two parallel lines have the same slope. m1 m2 x y y = 2x 1y = 2x + 3 D E F I N I T I O N Answer: a. y 7 b. x 5 E X AM P LE 7 Finding the Equation of a Horizontal or Vertical Line Find the equation of each of the following lines given their slope and a point that the line passes through: a. Slope: undened; b. Slope: Solution (a): A vertical line has undened slope. The x-coordinate of the point the line passes through is . Graph of the line and the point indicated. Solution (b): A horizontal line has slope . The y-coordinate of the point the line passes through is 5. Graph of the line with the point indicated. YO U R TU R N Find the equation of each of the following lines given their slope and a point that the line passes through: a. Slope: ; (3, 7) b. Slope: undened; (5, -2) m = 0 (-4, 5)y = 5 y = 5(-4, 5) y = bm = 0 (-3, 2)x = -3 x = -3-3(-3, 2) x = a m = 0; (-4, 5) (-3, 2) x y (4, 5) y = 5 x y (3, 2) x = 3 c02b.qxd 11/22/11 9:15 PM Page 212 242. 2.3 Lines 213 Two perpendicular lines form a right angle at their point of intersection. Notice the slopes of the two perpendicular lines in the gure to the right. They are and 2, negative reciprocals of each other. It turns out that almost all perpendicular lines share this property. Horizontal (m 0) and vertical (m undened) lines do not share this property. -1 2 E X AM P LE 9 Finding an Equation of a Parallel Line Find the equation of the line that passes through the point (1, 1) and is parallel to the line y 3x 1. Solution: Write the slopeintercept equation of a line. y mx b Parallel lines have equal slope. m 3 Substitute the slope into the equation of the line. y 3x b Since the line passes through (1, 1), this point must satisfy the equation. 1 3(1) b Solve for b. b 2 The equation of the line is y 3x 2 . YO U R TU R N Find the equation of the line parallel to y 2x 1 that passes through the point (1, 3). In other words, if two lines in a plane are perpendicular, their slopes are negative reciprocals, provided their slopes are dened. Similarly, if the slopes of two lines in a plane are negative reciprocals, then the lines are perpendicular. Except for the special case of a vertical and a horizontal line, two lines in a plane are perpendicular if and only if their slopes are negative reciprocals of each other. Perpendicular Lines x y y = 2xy = x 1 2 D E F I N I T I O N Answer: y 2x 5 E X AM P LE 8 Determining Whether Two Lines Are Parallel Determine whether the lines x 3y 3 and are parallel. Solution: Write the rst line in slopeintercept form. x 3y 3 Add x to both sides. 3y x 3 Divide by 3. Compare the two lines. and Both lines have the same slope, . These are distinct lines because they have different y-intercepts. Thus, the two lines are parallel . 1 3 y = 1 3 x - 6y = 1 3 x - 1 y = 1 3 x - 1 y = 1 3x - 6 WORDS MATH Lines L1 and L2 are perpendicular. Two perpendicular lines have negative reciprocal slopes. m1 = - 1 m2 m1 Z 0, m2 Z 0 L1 L2 Study Tip If a line has slope equal to 3, then a line perpendicular to it has slope - 1 3. c02b.qxd 11/22/11 9:15 PM Page 213 243. 214 C HAP TE R 2 Graphs E X AM P LE 11 Slope as a Rate of Change The average age that a person rst marries has been increasing over the last several decades. In 1970 the average age was 20, in 1990 it was 25 years old, and in 2010 it is expected that the average age will be 30 years old at the time of a persons rst marriage. Find the slope of the line passing through these points. Describe what that slope represents. Solution: If we let x represent the year and y represent the age, then two points* that lie on the line are (1970, 20) and (2010, 30). Write the slope formula. Substitute the points into the slope formula. The slope is and can be interpreted as the rate of change of the average age when a person is rst married. Every 4 years the average age at the rst marriage is 1 year older. 1 4 m = 30 - 20 2010 - 1970 = 10 40 = 1 4 m = change in y change in x Year 20101990197019501930 Age 15 10 5 20 25 30 35 Applications Involving Linear Equations Slope is the ratio of the change in y over the change in x. In applications, slope can often be interpreted as the rate of change, as illustrated in the next example. E X AM P LE 10 Finding an Equation of a Line That Is Perpendicular to Another Line Find the equation of the line that passes through the point (3, 0) and is perpendicular to the line y 3x 1. Solution: Identify the slope of the given line y 3x 1. The slope of a line perpendicular to the given line is the negative reciprocal of the slope of the given line. Write the equation of the line we are looking for in slopeintercept form. Substitute into . Since the desired line passes through (3, 0), this point must satisfy the equation. 0 1 b Solve for b. b 1 The equation of the line is . YO U R TU R N Find the equation of the line that passes through the point (1, 5) and is perpendicular to the line .y = - 1 2x + 4 y = - 1 3x + 1 0 = - 1 3 (3) + b y = - 1 3 x + by = m2x + bm2 = - 1 3 y = m2x + b m2 = - 1 m1 = - 1 3 m1 = 3 Answer: y 2x 7 *In Example 11 we chose to use the points (1970, 20) and (2010, 30). We could have also used the point (1990, 25) with either of the other points. c02b.qxd 11/22/11 9:15 PM Page 214 244. 2.3 Lines 215 Either point must satisfy 100 30(2) b the equation. Use (2, 100). 100 60 b b 40 The service charge y is given by y 30x 40. Substitute x 5 into this equation for a 5-hour job. y 30(5) 40 190 The 5-hour job will cost $190 . STEP 5 Check the solution. The service charge y 30x 40 can be interpreted as a fixed cost of $40 for coming to your home and a $30 per hour fee for the job. A 5-hour job would cost $190. Additionally, a 5-hour job should cost less than the sum of a 2-hour job and a 3-hour job ($100 $130 $230), since the $40 fee is charged only once. YO U R T U R N You decide to hire a tutor that some of your friends recommended. The tutor comes to your home, so she charges a at fee per session and then an hourly rate. One friend prefers 2-hour sessions, and the charge is $60 per session. Another friend has 5-hour sessions that cost $105 per session. How much should you be charged for a 3-hour session? Answer: $75 E X AM P LE 12 Service Charges Suppose that your two neighbors both use the same electrician. One neighbor had a 2-hour job that cost her $100, and another neighbor had a 3-hour job that cost him $130. Assuming that a linear equation governs the service charge of this electrician, what will your cost be for a 5-hour job? Solution: STEP 1 Identify the question. Determine the linear equation for this electricians service charge and calculate the charge for a 5-hour job. STEP 2 Make notes. A 2-hour job costs $100 and a 3-hour job costs $130. STEP 3 Set up an equation. Let x equal the number of hours and y equal the service charge in dollars. Linear equation: y mx b Two points that must satisfy this equation are (2, 100) and (3, 130). STEP 4 Solve the equation. Calculate the rate of change (slope). Substitute the slope into the linear y 30x b equation. m = 130 - 100 3 - 2 = 30 1 = 30 c02b.qxd 11/22/11 9:15 PM Page 215 245. 216 C HAP TE R 2 Graphs For each graph in Exercises 1116, identify (by inspection) the x- and y-intercepts and slope if they exist, and classify the line as rising, falling, horizontal, or vertical. 11. 12. 13. 14. 15. 16. x y (4, 3) (4, 4) x y (3, 1)(1, 1) x y (2, 2) (3, 3) x y (2, 1) (2, 3) x y (4, 3) (0, 3) x y (1, 3) (2, 3) SMH Horizontal lines: m 0 Vertical lines: m is undened We found equations of lines, given either two points or the slope and a point. We found the pointslope form, y y1 m(x x1), useful when the slope and a point are given. We also discussed both parallel (nonintersecting) and perpendicular (forming a right angle) lines. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, provided their slopes are dened. S U M MARY In this section we discussed graphs and equations of lines. Lines are often expressed in two forms: General Form: Ax By C SlopeIntercept Form: y mx b All lines (except horizontal and vertical) have exactly one x-intercept and exactly one y-intercept. The slope of a line is a measure of steepness. Slope of a line passing through (x1, y1) and (x2, y2): m = y2 - y1 x2 - x1 = rise run S E CTI O N 2.3 In Exercises 110, nd the slope of the line that passes through the given points. 1. (1, 3) and (2, 6) 2. (2, 1) and (4, 9) 3. (2, 5) and (2, 3) 4. (1, 4) and (4, 6) 5. (7, 9) and (3, 10) 6. (11, 3) and (2, 6) 7. (0.2, 1.7) and (3.1, 5.2) 8. (2.4, 1.7) and (5.6, 2.3) 9. 10. A1 2, 3 5 B and A-3 4, 7 5 BA2 3, - 1 4 B and A5 6, - 3 4 B S K I LL S E X E R C I S E S S E CTI O N 2.3 c02b.qxd 11/22/11 9:15 PM Page 216 246. 2.3 Lines 217 In Exercises 1730, nd the x- and y-intercepts if they exist and graph the corresponding line. 17. y 2x 3 18. y 3x 2 19. 20. 21. 2x 3y 4 22. x y 1 23. 24. 25. x 1 26. y 3 27. y 1.5 28. x 7.5 29. 30. In Exercises 3142, write the equation in slopeintercept form. Identify the slope and the y-intercept. 31. 2x 5y 10 32. 3x 4y 12 33. x 3y 6 34. x 2y 8 35. 4x y 3 36. x y 5 37. 12 6x 3y 38. 4 2x 8y 39. 0.2x 0.3y 0.6 40. 0.4x 0.1y 0.3 41. 42. In Exercises 4350, write the equation of the line, given the slope and intercept. 43. Slope: m 2 44. Slope: m 2 45. Slope: 46. Slope: y-intercept: (0, 3) y-intercept: (0, 1) y-intercept: (0, 0) y-intercept: (0, 3) 47. Slope: m 0 48. Slope: m 0 49. Slope: undened 50. Slope: undened y-intercept: (0, 2) y-intercept: (0, 1.5) x-intercept: x-intercept: (3.5, 0) In Exercises 5160, write an equation of the line in slopeintercept form, if possible, given the slope and a point that lies on the line. 51. Slope: m 5 52. Slope: m 2 53. Slope: m 3 54. Slope: m 1 (1, 3) (1, 1) (2, 2) (3, 4) 55. Slope: 56. Slope: 57. Slope: m 0 58. Slope: m 0 (1, 1) (5, 3) (2, 4) (3, 3) 59. Slope: undened 60. Slope: undened (1, 4) (4, 1) In Exercises 6180, write the equation of the line that passes through the given points. Express the equation in slopeintercept form or in the form x a or y b. 61. (2, 1) and (3, 2) 62. (4, 3) and (5, 1) 63. (3, 1) and (2, 6) 64. (5, 8) and (7, 2) 65. (20, 37) and (10, 42) 66. (8, 12) and (20, 12) 67. (1, 4) and (2, 5) 68. (2, 3) and (2, 3) 69. and 70. and 71. (3, 5) and (3, 7) 72. (5, 2) and (5, 4) 73. (3, 7) and (9, 7) 74. (2, 1) and (3, 1) 75. (0, 6) and (5, 0) 76. (0, 3) and (0, 2) 77. (6, 8) and (6, 2) 78. (9, 0) and (9, 2) 79. and 80. and In Exercises 8186, write the equation corresponding to each line. Express the equation in slopeintercept form. 81. 82. 83. x y x y x y A1 3, 1 2 BA1 3, 2 5 BA2 5, 1 2 BA2 5, - 3 4 B A7 3, 1 2 BA- 2 3, - 1 2 BA3 2, 9 4 BA1 2, 3 4 B m = - 1 7m = 3 4 A3 2, 0B m = 1 2m = - 1 3 1 4 x + 2 5 y = 21 2 x + 2 3 y = 4 y = 5 3x = - 7 2 1 3 x - 1 4 y = 1 12 1 2 x + 1 2 y = -1 y = 1 3 x - 1y = - 1 2 x + 2 c02b.qxd 11/22/11 9:15 PM Page 217 247. 218 C HAP TE R 2 Graphs 105. Business. The operating costs for a local business are a xed amount of $1300 plus $3.50 per unit sold, while revenue is $7.25 per unit sold. How many units does the business have to sell in order to break even? 106. Business. The operating costs for a local business are a xed amount of $12,000 plus $13.50 per unit sold, while revenue is $27.25 per unit sold. How many units does the business have to sell in order to break even? 107. Weather: Temperature. The National Oceanic and Atmospheric Administration (NOAA) has an online conversion chart that relates degrees Fahrenheit, F, to degrees Celsius, C. 77F is equivalent to 25C, and 68F is equivalent to 20C. Assuming the relationship is linear, write the equation relating degrees Celsius C to degrees Fahrenheit F. What temperature is the same in both degrees Celsius and degrees Fahrenheit? 108. Weather: Temperature. According to NOAA, a standard day is 15C at sea level, and every 500 feet elevation above sea level corresponds to a 1C temperature drop. Assuming the relationship between temperature and elevation is linear, write an equation that models this relationship. What is the expected temperature at 2500 feet on a standard day? 101. Budget: Home Improvement. The cost of having your bathroom remodeled is the combination of material costs and labor costs. The materials (tile, grout, toilet, xtures, etc.) cost is $1200, and the labor cost is $25 per hour. Write an equation that models the total cost C of having your bathroom remodeled as a function of hours h. How much will the job cost if the worker estimates 32 hours? 102. Budget: Rental Car. The cost of a one-day car rental is the sum of the rental fee, $50, plus $0.39 per mile. Write an equation that models the total cost associated with the car rental. 103. Budget: Monthly Driving Costs. The monthly costs associated with driving a new Honda Accord are the monthly loan payment plus $25 every time you ll up with gasoline. If you ll up 5 times in a month, your total monthly cost is $500. How much is your loan payment? 104. Budget: Monthly Driving Costs. The monthly costs associated with driving a Ford Explorer are the monthly loan payment plus the cost of lling up your tank with gasoline. If you ll up 3 times in a month, your total monthly cost is $520. If you ll up 5 times in a month, your total monthly cost is $600. How much is your monthly loan, and how much does it cost every time you ll up with gasoline? A P P L I C AT I O N S 84. 85. 86. In Exercises 87100, nd the equation of the line that passes through the given point and also satises the additional piece of information. Express your answer in slopeintercept form, if possible. 87. (3, 1); parallel to the line y 2x 1 88. (1, 3); parallel to the line y x 2 89. (0, 0); perpendicular to the line 2x 3y 12 90. (0, 6); perpendicular to the line x y 7 91. (3, 5); parallel to the x-axis 92. (3, 5); parallel to the y-axis 93. (1, 2); perpendicular to the y-axis 94. (1, 2); perpendicular to the x-axis 95. (2, 7); parallel to the line 96. (1, 4); perpendicular to the line 97. ; perpendicular to the line 98. ; perpendicular to the line 99. ; parallel to the line 100. ; parallel to the line 10x + 45y = -9A-1 4, -13 9 B-15x + 35y = 7A7 2, 4B 6x + 14y = 7A6 5, 3B8x + 10y = -45A-2 3, 2 3 B -2 3x + 3 2y = -21 2x - 1 3y = 5 x y x y x y c02b.qxd 11/22/11 9:15 PM Page 218 248. 2.3 Lines 219 109. Life Sciences: Height. The average height of a man has increased over the last century. What is the rate of change in inches per year of the average height of men? 110. Life Sciences: Height. The average height of a woman has increased over the last century. What is the rate of change in inches per year of the average height of women? 111. Life Sciences: Weight. The average weight of a baby born in 1900 was 6 pounds 4 ounces. In 2000, the average weight of a newborn was 6 pounds 10 ounces. What is the rate of change of birth weight in ounces per year? What do we expect babies to weigh at birth in 2040? 112. Sports. The fastest a man could run a mile in 1906 was 4 minutes and 30 seconds. In 1957, Don Bowden became the rst American to break the 4-minute mile. Calculate the rate of change in mile speed per year. 113. Monthly Phone Costs. Mikes home phone plan charges a at monthly fee plus a charge of $0.05 per minute for long-distance calls. The total monthly charge is represented by y 0.05x 35, x 0, where y is the total monthly charge and x is the number of long-distance minutes used. Interpret the meaning of the y-intercept. 114. Cost: Automobile. The value of a Daewoo car is given by y 11,100 1850x, x 0, where y is the value of the car and x is the age of the car in years. Find the x-intercept and y-intercept and interpret the meaning of each. 115. Weather: Rainfall. The average rainfall in Norfolk, Virginia, for July was 5.2 inches in 2003. The average July rainfall for Norfolk was 3.8 inches in 2007. What is the rate of change of rainfall in inches per year? If this trend continues, what is the expected average rainfall in 2010? 116. Weather: Temperature. The average temperature for Boston in January 2005 was 43F. In 2007 the average January temperature was 44.5F. What is the rate of change of the temperature per year? If this trend continues, what is the expected average temperature in January 2010? Year 20001900 1950 Height(inches) 63 62 61 60 64 65 66 67 Year 20001900 1950 Height(inches) 68 67 66 65 69 70 71 72 117. Environment. In 2000, Americans used approximately 380 billion plastic bags. In 2005, approximately 392 billion were used. What is the rate of change of plastic bags used per year? How many plastic bags will be expected to be used in 2010? 118. Finance: Debt. According to the Federal Reserve, Americans individually owed $744 in revolving credit in 2004. In 2006, they owed approximately $788. What is the rate of change of the amount of revolving credit owed per year? How much should Americans be expected to owe in 2008? 119. Business. A website that supplies Asian specialty foods to restaurants advertises a 64 ounce bottle of Hoisin Sauce for $16.00. Shipping cost for one bottle is $15.93. The shipping cost for two bottles is $19.18. The cost for ve bottles, including shipping, is $111.83. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a. Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost (including shipping) for one, two, or ve bottles purchased. b. Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of Hoisin Sauce. Explain what this amount means in terms of the sauce purchase. c. Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of Hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. d. Calculate the slope between the origin and the ordered pair that represents the purchase of ve bottles of Hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. 120. Business. A website that supplies Asian specialty foods to restaurants advertises an 8 ounce bottle of Plum Sauce for $4.00, but shipping for one bottle is $14.27. The shipping cost for two bottles is $14.77. The cost for ve bottles, including shipping, is $35.93. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a. Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost, including shipping for one, two, or ve bottles purchased. b. Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of Plum Sauce. Explain what this amount means in terms of the sauce purchase. c. Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of Plum Sauce (including shipping). Explain what this amount means in terms of the sauce purchase. d. Calculate the slope between the origin and the ordered pair that represents the purchase of ve bottles of Plum Sauce (including shipping). Explain what this amount means in terms of the sauce purchase. c02b.qxd 11/22/11 9:15 PM Page 219 249. 125. A line can have at most one x-intercept. 126. A line must have at least one y-intercept. 127. If the slopes of two lines are and 5, then the lines are parallel. -1 5 C O N C E P T U A L 128. If the slopes of two lines are 1 and 1, then the lines are perpendicular. 129. If a line has slope equal to zero, describe a line that is perpendicular to it. 130. If a line has no slope, describe a line that is parallel to it. In Exercises 125128, determine whether each statement is true or false. 134. Find an equation of a line that passes through the point (A, B 1) and is perpendicular to the line Ax By C. Assume that A and B are both nonzero. 135. Show that two lines with equal slopes and different y-intercepts have no point in common. Hint: Let y1 mx b1 and y2 mx b2 with . What equation must be true for there to be a point of intersection? Show that this leads to a contradiction. 136. Let y1 m1x b1 and y2 m2 x b2 be two nonparallel lines . What is the x-coordinate of the point where they intersect? (m1 Z m2 ) b1 Z b2 131. Find an equation of a line that passes through the point (B, A 1) and is parallel to the line Ax By C. Assume that B is not equal to zero. 132. Find an equation of a line that passes through the point (B, A 1) and is parallel to the line Ax By C. Assume that B is not equal to zero. 133. Find an equation of a line that passes through the point (A, B 1) and is perpendicular to the line Ax By C. Assume that A and B are both nonzero. C HALLE N G E 220 C HAP TE R 2 Graphs In Exercises 121124, explain the mistake that is made. C AT C H T H E M I S TA K E 121. Find the x- and y-intercepts of the line with equation 2x 3y 6. Solution: x-intercept: set x 0 and solve for y. 3y 6 y 2 The x-intercept is (0, 2). y-intercept: set y 0 and solve for x. 2x 6 x 3 The y-intercept is (3, 0). This is incorrect. What mistake was made? 122. Find the slope of the line that passes through the points (2, 3) and (4, 1). Solution: Write the slope formula. Substitute (2, 3) and (4, 1). This is incorrect. What mistake was made? m = 1 - 3 -2 - 4 = -2 -6 = 1 3 m = y2 - y1 x2 - x1 123. Find the slope of the line that passes through the points (3, 4) and (3, 7). Solution: Write the slope formula. Substitute (3, 4) and (3, 7). This is incorrect. What mistake was made? 124. Given the slope, classify the line as rising, falling, horizontal, or vertical. a. m 0 b. m undened c. m 2 d. m 1 Solution: a. vertical line b. horizontal line c. rising d. falling These are incorrect. What mistakes were made? m = -3 - (-3) 4 - 7 = 0 m = y2 - y1 x2 - x1 c02b.qxd 12/23/11 5:53 PM Page 220 250. 137. y1 17x 22 y2 = - 1 17x - 13 139. y1 0.25x 3.3 y2 4x 2 140. y2 2x 3 y1 = 1 2x + 5 141. y1 0.16x 2.7 y2 6.25x 1.4 142. y1 3.75x 8.2 y2 = 4 15x + 5 6 138. y1 0.35x 2.7 y2 0.35x 1.2 T E C H N O L O G Y For Exercises 137142, determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to conrm your answer. 2.4 Circles 221 C O N C E P TUAL O BJ E CTIVE Understand algebraic and graphical representations of circles. C I R C LE S S K I LLS O BJ E CTIVE S Identify the center and radius of a circle from the standard equation. Graph a circle. Transform equations of circles to the standard form by completing the square. S E C TI O N 2.4 Standard Equation of a Circle Most people understand the shape of a circle. The goal in this section is to develop the equation of a circle. A circle is the set of all points in a plane that are a xed distance from a point, the center. The center, C, is typically denoted by (h, k), and the xed distance, or radius, is denoted by r. Circle y x(h, k) (x, y) r D E F I N I T I O N What is the equation of a circle? Well use the distance formula from Section 2.1. Distance formula: d = 2(x2 - x1)2 + (y2 - y1)2 c02b.qxd 11/22/11 9:15 PM Page 221 251. For the special case of a circle with center at the origin (0, 0), the equation simplies to .x2 + y2 = r2 The standard form of the equation of a circle with radius r and center (h, k) is (x - h)2 + (y - k)2 = r2 EQUATION OF A CIRCLE A circle with radius 1 and center (0, 0) is called the unit circle: x2 + y2 = 1 UNIT CIRCLE The unit circle plays an important role in the study of trigonometry. Note that if x2 y2 0, the radius is 0, so the circle is just a point. 222 C HAP TE R 2 Graphs y x (2, 3) (0, 1) (4, 1)(2, 1) (2, 1) E X AM P LE 1 Finding the Center and Radius of a Circle Identify the center and radius of the given circle and graph. Solution: Rewrite this equation in standard form. [x 2]2 [y (1)]2 22 Identify h, k, and r by comparing this equation with the standard form of a circle: (x h)2 (y k)2 r2 . h 2, k 1, and r 2 Center (2, 1) and r 2 To draw the circle, label the center (2, 1). Label four additional points two units (the radius) away from the center: . Note that the easiest four points to get are those obtained by going out from the center both horizontally and vertically. Connect those four points with a smooth curve. YO U R TU R N Identify the center and radius of the given circle and graph. (x + 1)2 + (y + 2)2 = 9 (4, -1), (0, -1), (2, 1), and (2, -3) (x - 2)2 + (y + 1)2 = 4 The distance between the center (h, k) and any point (x, y) on the circle is the radius r. Substitute these values d r, (x1, y1) (h, k), and (x2, y2) (x, y) into the distance formula and square both sides: . All circles can be written in standard form, which makes it easy to identify the center and radius. (x - h)2 + (y - k)2 = r2 r = 2(x - h)2 + (y - k)2 Technology Tip To enter the graph of (x 2)2 (y 1)2 4, solve for y rst. The graphs of and are shown.y2 = - 24 - (x - 2)2 - 1 y1 = 24 - (x - 2)2 - 1 x y (1, 5) (4, 2) (2, 2) (1, 2) (1, 1) Answer: Center: (1, 2) Radius: 3 c02b.qxd 11/22/11 9:15 PM Page 222 252. 2.4 Circles 223 E X AM P LE 2 Graphing a Circle: Fractions and Radicals Identify the center and radius of the given circle and sketch its graph. Solution: If r2 20, then . Write the equation in standard form. Identify the center and radius. Center and To graph the circle, well use decimal approximations of the fractions and radicals: (0.5, 0.3) for the center and 4.5 for the radius. Four points on the circle that are 4.5 units from the center are (4, 0.3), (5, 0.3), (0.5, 4.2), and (0.5, 4.8). Connect them with a smooth curve. r = 225a 1 2 , - 1 3 b = A215B 2 ax - 1 2 b 2 + cy - a- 1 3 b d 2 r = 220 = 225 ax - 1 2 b 2 + ay + 1 3 b 2 = 20 x y (0.5, 4.8) (4.0, 0.3) (5.0, 0.3) (0.5, 4.2) (0.5, 0.3) Technology Tip To enter the graph of , solve for y rst. The graphs of and are shown.y2 = -220 - Ax - 1 2B2 - 1 3 y1 = 220 - Ax - 1 2 B 2 - 1 3 Ax - 1 2 B 2 + Ay + 1 3 B 2 = 20 x y (2, 2) 5 (7, 3) (3, 3) (2, 3) (2, 8) E X AM P LE 3 Determining the Equation of a Circle Given the Center and Radius Find the equation of a circle with radius 5 and center (2, 3). Graph the circle. Solution: Substitute (h, k) (2, 3) and r 5 into the standard equation of a circle. [x (2)]2 (y 3)2 52 Simplify. (x 2)2 (y 3)2 25 To graph the circle, plot the center (2, 3) and four points 5 units away from the center: . Connect them with a smooth curve. YO U R TU R N Find the equation of a circle with radius 3 and center (0, 1) and graph. (-7, 3), (3, 3), (-2, -2), and (-2, 8) x y (0, 2) 3(3, 1) (3, 1)(0, 1) (0, 4) Answer: x2 (y 1)2 9 c02b.qxd 11/22/11 9:15 PM Page 223 253. 224 C HAP TE R 2 Graphs Answer: x2 y2 10x 6y 66 0 Transforming Equations of Circles to the Standard Form by Completing the Square If the equation of a circle is given in general form, it must be rewritten in standard form in order to identify its center and radius. To transform equations of circles from general to standard form, complete the square (Section 1.3) on both the x- and y-variables. Lets change the look of the equation given in Example 1. In Example 1 the equation of the circle was given as: (x 2)2 (y 1)2 4 Eliminate the parentheses. x2 4x 4 y2 2y 1 4 Group like terms and subtract 4 from both sides. x2 y2 4x 2y 1 0 We have written the general form of the equation of the circle in Example 1. The general form of the equation of a circle is x2 + y2 + ax + by + c = 0 Suppose you are given a point that lies on a circle and the center of the circle. Can you nd the equation of the circle? E X AM P LE 4 Finding the Equation of a Circle Given Its Center and One Point The point (10, 4) lies on a circle centered at (7, 8). Find the equation of the circle in general form. Solution: This circle is centered at (7, 8), so its standard equation is (x 7)2 (y 8)2 r2 . All that remains is to nd the radius. Approach 1: Since the point (10, 4) lies on the circle, it must satisfy the equation of the circle. Substitute (x, y) (10, 4). Simplify. 32 42 r2 The distance from (10, 4) to (7, 8) is ve units. r 5 Approach 2: Find the distance between (10, 4) and (7, 8). Substitute r 5 into the standard equation. (x 7)2 (y 8)2 52 Eliminate the parentheses and simplify. x2 14x 49 y2 16y 64 25 Write in general form. x2 y2 14x 16y 88 0 YO U R TU R N The point (1, 11) lies on a circle centered at (5, 3). Find the equation of the circle in general form. = 5 = 225 = 232 + 42 r = d = 2(10 - 7)2 + (-4 - (-8))2 (10 - 7)2 + (-4 + 8)2 = r2 x2 2 4 10 12 14 4 6 8 10 12 14 y (7, 13) (2, 8) (10, 4) (12, 8)(7, 8) (7, 3) c02b.qxd 11/22/11 9:15 PM Page 224 254. 2.4 Circles 225 E X AM P LE 5 Finding the Center and Radius of a Circle by Completing the Square Find the center and radius of the circle with the equation: Solution: Our goal is to transform this equation into standard form Group x and y terms, respectively, on the left side of the equation; move constants to the right side. (x2 8x) (y2 20y) 107 Complete the square on both the x and y expressions. (x2 8x ) (y2 20y ) 107 Add and to both sides. Factor the perfect squares on the left side and simplify the right side. (x 4)2 (y 10)2 9 Write in standard form. (x 4)2 [y (10)]2 32 The center is (4, 10) and the radius is 3. YO U R TU R N Find the center and radius of the circle with the equation: x2 + y2 + 4x - 6y - 12 = 0 = -107 + 16 + 100Ax2 - 8x + 42 B + A y2 + 20y + 102 B A20 2 B 2 = 100A- 8 2 B 2 16 nn (x - h)2 + (y - k)2 = r2 x2 - 8x + y2 + 20y + 107 = 0 Answer: Center: (2, 3) Radius: 5 Technology Tip To graph x2 8x y2 20y 107 0 without transforming it into a standard form, solve for y using the Quadratic Formula: Next, set the window to [5, 30] by [30, 5] and use ZSquare under y = -20 ; 2202 - 4(1)Ax2 - 8x + 107B 2 ZOOM to adjust the window variable to make the circle look circular. The graphs of and are shown. y2 = -20 - 2202 - 4Ax2 - 8x + 107B 2 y1 = -20 + 2202 - 4Ax2 - 8x +107B 2 I N C O R R E CT ERROR: Dont forget to add 16 and 64 to the right. = -44 Ax2 + 16x + 64B + A y2 + 8y + 16B C O R R E CT Center: (8, 4) Radius: 6 (x + 8)2 + (y + 4)2 = 36 = -44 + 64 + 16 Ax2 + 16x + 64B + A y2 + 8y + 16B = -44 Ax2 + 16x + nB + A y2 + 8y + nB Ax2 + 16xB + A y2 + 8yB = -44 x2 + y2 + 16x + 8y + 44 = 0 C O M M O N M I S TA K E A common mistake is forgetting to add both constants to the right side of the equation. Identify the center and radius of the circle with the equation: x2 + y2 + 16x + 8y + 44 = 0 C A U T I O N Dont forget to add both constants to each side of the equation when completing the square for x and y. 16 100 { { c02b.qxd 11/22/11 9:15 PM Page 225 255. 226 C HAP TE R 2 Graphs SMH General form: x2 y2 ax by c 0. Complete the square to transform the equation to standard form. S U M MARY The equation of a circle is given by Standard form: (x h)2 (y k)2 r2 . Center: (h, k) Radius: r S E CTI O N 2.4 In Exercises 120, write the equation of the circle in standard form. S K I LL S E X E R C I S E S 1. Center (1, 2) 2. Center (3, 4) 3. Center (3, 4) 4. Center (1, 2) r 3 r 5 r 10 r 4 5. Center (5, 7) 6. Center (2, 8) 7. Center (11, 12) 8. Center (6, 7) r 9 r 6 r 13 r 8 9. Center (0, 0) 10. Center (0, 0) 11. Center (0, 2) 12. Center (3, 0) r 2 r 3 13. Center (0, 0) 14. Center (1, 2) 15. Center (5, 3) 16. Center (4, 1) 17. 18. 19. Center (1.3, 2.7) 20. Center (3.1, 4.2) r 3.2 r 5.5 In Exercises 2132, nd the center and radius of the circle with the given equations. 21. (x 1)2 (y 3)2 25 22. (x 1)2 (y 3)2 11 23. (x 2)2 (y 5)2 49 24. (x 3)2 (y 7)2 81 25. (x 4)2 (y 9)2 20 26. (x 1)2 (y 2)2 8 27. 28. 29. (x 1.5)2 (y 2.7)2 1.69 30. (x 3.1)2 (y 7.4)2 56.25 31. x2 y2 50 0 32. x2 y2 8 0 In Exercises 3350, state the center and radius of each circle. 33. x2 y2 4x 6y 3 0 34. x2 y2 2x 10y 17 0 35. x2 y2 6x 8y 75 0 36. x2 y2 2x 4y 9 0 37. x2 y2 10x 14y 7 0 38. x2 y2 4x 16y 32 0 39. x2 y2 2y 15 0 40. x2 y2 2x 8 0 41. x2 y2 2x 6y 1 0 42. x2 y2 8x 6y 21 0 43. x2 y2 10x 6y 22 0 44. x2 y2 8x 2y 28 0 Ax - 1 2 B 2 + A y - 1 3 B 2 = 9 25Ax - 2 5 B 2 + A y - 1 7 B 2 = 4 9 r = 2 5r = 1 4 Center A -1 3, -2 7 BCenter A2 3, -3 5 B r = 325r = 223r = 27r = 22 r = 2r = 3 S E CTI O N 2.4 c02b.qxd 11/22/11 9:15 PM Page 226 256. 2.4 Circles 227 45. x2 y2 6x 4y 1 0 46. x2 y2 2x 10y 2 0 47. 48. 49. x2 y2 2.6x 5.4y 1.26 0 50. x2 y2 6.2x 8.4y 3 0 In Exercises 5156, nd the equation of each circle. 51. Centered at (1, 2) and passing through the point (1, 0). 52. Centered at (4, 9) and passing through the point (2, 5). 53. Centered at (2, 3) and passing through the point (3, 7). 54. Centered at (1, 1) and passing through the point (8, 5). 55. Centered at (2, 5) and passing through the point (1, 9). 56. Centered at (3, 4) and passing through the point (1, 8). x2 + y2 - x 2 - 3y 2 + 3 8 = 0x2 + y2 - x + y + 1 4 = 0 57. Cell Phones. If a cellular phone tower has a reception radius of 100 miles and you live 95 miles north and 33 miles east of the tower, can you use your cell phone while at home? 58. Cell Phones. Repeat Exercise 57, assuming you live 45 miles south and 87 miles west of the tower. 59. Construction/Home Improvement. A couple and their dog moved into a new house that does not have a fenced-in backyard. The backyard is square with dimensions 100 feet by 100 feet. If they put a stake in the center of the backyard with a long leash, write the equation of the circle that will map out the dogs outer perimeter. 60. Construction/Home Improvement. Repeat Exercise 59 except that the couple put in a pool and a garden and want to restrict the dog to quadrant I. What coordinates represent the center of the circle? What is the radius? y r (h, k) x 5050 50 50 100 ft 100 ft 61. Design. A university designs its campus with a master plan of two concentric circles. All of the academic buildings are within the inner circle (so that students can get between classes in less than 10 minutes), and the outer circle contains all the dormitories, the Greek park, cafeterias, the gymnasium, and intramural elds. Assuming the center of campus is the origin, write an equation for the inner circle if the diameter is 3000 feet. 62. Design. Repeat Exercise 61 for the outer circle with a diameter of 6000 feet. 63. Cell Phones. A cellular phone tower has a reception radius of 200 miles. Assuming the tower is located at the origin, write the equation of the circle that represents the reception area. 64. Environment. In a state park, a re has spread in the form of a circle. If the radius is 2 miles, write an equation for the circle. x y A P P L I C AT I O N S c02b.qxd 11/22/11 9:15 PM Page 227 257. 228 C HAP TE R 2 Graphs In Exercises 6770, explain the mistake that is made. 67. Identify the center and radius of the circle with equation (x 4)2 (y 3)2 25. Solution: The center is (4, 3) and the radius is 5. This is incorrect. What mistake was made? 68. Identify the center and radius of the circle with equation (x 2)2 (y 3)2 2. Solution: The center is (2, 3) and the radius is 2. This is incorrect. What mistake was made? 69. Graph the solution to the equation (x 1)2 (y 2)2 16. Solution: The center is (1, 2) and the radius is 4. This is incorrect. What mistake was made? x y (1, 6) 4 (3, 2) (5, 2) (1, 2) (1, 2) C AT C H T H E M I S TA K E 70. Find the center and radius of the circle with the equation x2 y2 6x 4y 3 0. Solution: Group like terms. (x2 6x) (y2 4y) 3 Complete the (x2 6x 9) (y2 4y 4) 12 square. The center is (3, 2) and the radius is . This is incorrect. What mistake was made? 213 (x - 3)2 + (y + 2)2 = (213) 2 For Exercises 65 and 66, refer to the following: A cell phone provider is expanding its coverage and needs to place four cell phone towers to provide complete coverage of a 100-square-mile area formed by a 10 mile by 10 mile square. This area can be represented by a region on the Cartesian coordinate system; see gure. The placement of the four towers is very important in that the cell phone provider needs to provide coverage of the entire 100-square-mile area. The cell phone towers being installed can process signals from cell phones within a 3.5-mile radius. x y 10 10 Cell Phone Coverage Area 65. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points (2.5, 2.5), (2.5, 7.5), (7.5, 2.5), and (7.5, 7.5) on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage? 66. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points (3, 3), (3, 7), (7, 3), and (7, 7) on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage? c02b.qxd 11/22/11 9:15 PM Page 228 258. 2.4 Circles 229 C O N C E P T U A L 71. The equation whose graph is depicted has innitely many solutions. x y (0, 5) 5 (5, 0) (5, 0) (0, 5) In Exercises 7174, determine whether each statement is true or false. 72. The equation (x 7)2 (y 15)2 64 has no solution. 73. The equation (x 2)2 (y 5)2 20 has no solution. 74. The equation (x 1)2 (y 3)2 0 has only one solution. 75. Describe the graph (if it exists) of: x2 y2 10x 6y 34 0 76. Describe the graph (if it exists) of: x2 y2 4x 6y 49 0 77. Find the equation of a circle that has a diameter with endpoints (5, 2) and (1, 6). 78. Find the equation of a circle that has a diameter with endpoints (3, 0) and (1, 4). 79. For the equation x2 y2 ax by c 0, specify conditions on a, b, and c so that the graph is a single point. 80. For the equation x2 y2 ax by c 0, specify conditions on a, b, and c so that there is no corresponding graph. 81. Determine the center and radius of the circle given by the equation x2 y2 2ax 100 a2 . 82. Determine the center and radius of the circle given by the equation x2 y2 2by 49 b2 . C HALLE N G E T E C H N O L O G Y In Exercises 8386, use a graphing utility to graph each equation. Does this agree with the answer you gave in the Conceptual section? 83. (x 2)2 (y 5)2 20 (See Exercise 73 for comparison.) 84. (x 1)2 (y 3)2 0 (See Exercise 74 for comparison.) 85. x2 y2 10x 6y 34 0 (See Exercise 75 for comparison.) 86. x2 y2 4x 6y 49 0 (See Exercise 76 for comparison.) In Exercises 8788, (a) with the equation of the circle in standard form, state the center and radius, and graph; (b) use the quadratic formula to solve for y; and (c) use a graphing utility to graph each equation found in (b). Does the graph in (a) agree with the graphs in (c)? 87. x2 y2 11x 3y 7.19 0 88. x2 y2 1.2x 3.2y 2.11 0 c02b.qxd 12/23/11 5:53 PM Page 229 259. E X AM P LE 1 Drawing a Scatterplot of Olympic Decathlon Data The 2004 Mens Olympic Decathlon consisted of the following 10 events: 100 meter, long jump, shot put, high jump, 400 meter, 110 meter hurdles, discus, pole vault, javelin throw, and 1500 meter. Actual scores are converted to a point system where points are assigned to each of these events based on performance. Events are equally weighted when converting to points. These points are then summed to obtain total scores, and, in turn, medals are assigned based on these total scores. It would be interesting to know if certain events are more predictive of the total scores than are others. If someone does exceedingly well in the javelin throw, for example, is that person more likely to do well across all events and therefore obtain a large total points score? Data from the Mens 2004 Olympic Decathlon are presented on the next page and were retrieved from the following Web source: http://rss.acs.unt.edu/Rdoc/library/FactoMineR/ html/decathlon.html. Lets consider the paired data set {(x, y)} where x score on the 400 m and y total score. Scatterplots An important aspect of applied research across disciplines is to discover and understand relationships between variables, and often how to use such a relationship to predict values of one variable in terms of another. You have likely encountered such issues while watching TV, reading a magazine or newspaper, or simply talking with friends. Some questions include Is age predictive of texting speed? Is the level of pollution in a country related to the prevalence of asthma in that country? Do the ratings of car reliability necessarily increase with the price of the car? In this section we focus on situations involving relationships between two variables x and y, so that the experimental data gathered consists of ordered pairs (x1, y1), . . . , (xn, yn). A rst step in understanding a data set of the form {(x1, y1), . . . , (xn, yn)} is to create a pictorial representation of it. Identifying the rst coordinates of these ordered pairs as values of an independent variable (or predictor variable) x and the second coordinates as the values of a dependent variable (or response variable) y, we simply plot them all on a single xy-plane. The resulting picture is called a scatterplot. C O N C E P TUAL O BJ E CTIV E S Recognize positive or negative association. Recognize linear or nonlinear association. Understand what best t means. LI N EAR R E G R E S S I O N: B E ST F IT S K I LLS O BJ E CTIVE S Draw a scatterplot. Use linear regression to determine the line of best t associated with some data. Use the line of best t to predict values of one variable from the values of another. S E CTI O N 2.5* 230 *Optional Technology Required Section. c02b.qxd 12/23/11 5:53 PM Page 230 260. One scatterplot using the 400 m and total points information from this data set is shown in the following graph. Several natural questions arise: How are different pairs of these data related? Are there any discernible patterns present, and if so, how strong are they? Is there a single curve that can be used to describe the general trend present in the data? We shall answer these questions one by one in this section. 2.5* Linear Regression: Best Fit 231 LONG HIGH X110M POLE TOTAL OLYMPIANS X100M JUMP SHOT-PUT JUMP X400M HURDLE DISCUS VAULT JAVELIN X1500M RANK SCORE Sebrle 10.85 7.84 16.36 2.12 48.36 14.05 48.72 5.00 70.52 280.01 1 8893 Clay 10.44 7.96 15.23 2.06 49.19 14.13 50.11 4.90 69.71 282.00 2 8820 Karpov 10.50 7.81 15.93 2.09 46.81 13.97 51.65 4.60 55.54 278.11 3 8725 Macey 10.89 7.47 15.73 2.15 48.97 14.56 48.34 4.40 58.46 265.42 4 8414 Warners 10.62 7.74 14.48 1.97 47.97 14.01 43.73 4.90 55.39 278.05 5 8343 Zsivoczky 10.91 7.14 15.31 2.12 49.40 14.95 45.62 4.70 63.45 269.54 6 8287 Hernu 10.97 7.19 14.65 2.03 48.73 14.25 44.72 4.80 57.76 264.35 7 8237 Nool 10.80 7.53 14.26 1.88 48.81 14.80 42.05 5.40 61.33 276.33 8 8235 Bernard 10.69 7.48 14.80 2.12 49.13 14.17 44.75 4.40 55.27 276.31 9 8225 Schwarzl 10.98 7.49 14.01 1.94 49.76 14.25 42.43 5.10 56.32 273.56 10 8102 Pogorelov 10.95 7.31 15.10 2.06 50.79 14.21 44.60 5.00 53.45 287.63 11 8084 Schoenbeck 10.90 7.30 14.77 1.88 50.30 14.34 44.41 5.00 60.89 278.82 12 8077 Barras 11.14 6.99 14.91 1.94 49.41 14.37 44.83 4.60 64.55 267.09 13 8067 Smith 10.85 6.81 15.24 1.91 49.27 14.01 49.02 4.20 61.52 272.74 14 8023 Averyanov 10.55 7.34 14.44 1.94 49.72 14.39 39.88 4.80 54.51 271.02 15 8021 Ojaniemi 10.68 7.50 14.97 1.94 49.12 15.01 40.35 4.60 59.26 275.71 16 8006 Smirnov 10.89 7.07 13.88 1.94 49.11 14.77 42.47 4.70 60.88 263.31 17 7993 Qi 11.06 7.34 13.55 1.97 49.65 14.78 45.13 4.50 60.79 272.63 18 7934 Drews 10.87 7.38 13.07 1.88 48.51 14.01 40.11 5.00 51.53 274.21 19 7926 Parkhomenko 11.14 6.61 15.69 2.03 51.04 14.88 41.90 4.80 65.82 277.94 20 7918 Terek 10.92 6.94 15.15 1.94 49.56 15.12 45.62 5.30 50.62 290.36 21 7893 Gomez 11.08 7.26 14.57 1.85 48.61 14.41 40.95 4.40 60.71 269.70 22 7865 Turi 11.08 6.91 13.62 2.03 51.67 14.26 39.83 4.80 59.34 290.01 23 7708 Lorenzo 11.10 7.03 13.22 1.85 49.34 15.38 40.22 4.50 58.36 263.08 24 7592 Karlivans 11.33 7.26 13.30 1.97 50.54 14.98 43.34 4.50 52.92 278.67 25 7583 Korkizoglou 10.86 7.07 14.81 1.94 51.16 14.96 46.07 4.70 53.05 317.00 26 7573 Uldal 11.23 6.99 13.53 1.85 50.95 15.09 43.01 4.50 60.00 281.70 27 7495 Casarsa 11.36 6.68 14.92 1.94 53.20 15.39 48.66 4.40 58.62 296.12 28 7404 400 m Mens 2004 Olympic Decathlon 400 m and Total Points 47 49 51 53 55 TotalPoints 9000 8750 8500 8250 8000 7750 7500 7250 7000 x y Creating a scatterplot by hand can be tedious, especially for large data sets. You can also very easily lose precision and detail. Using technology to create a scatterplot is very appropriate and quite easy. Below are the procedures for how you would create the scatterplot shown in Example 1 using the TI-83 (or TI-84) and Excel 2007. c02b.qxd 6/6/12 7:26 PM Page 231 261. 232 C HAP TE R 2 Graphs Creating a Scatterplot Using the TI-83 (or TI-84) INSTRUCTION SCREENSHOT Entering Step 1 Press STAT, followed by 1:Edit. . . . the Data Clear any data already present in columns L1 and L2 so that the screen looks like the one to the right. Step 2 Input the values of the x-variable (rst entries in the ordered pairs) in column L1, pressing ENTER after each entry. Then, right arrow over to column L2 and input the values of the y-variable. The screen (starting from the beginning of the data set) should look like the one to the right when you are done. Plotting Step 3 Press Y= and then select Plot1 the Data in the top row of the screen. If either Plot2 or Plot3 is darkened, move the cursor onto it and press ENTER to undarken it. The screen should look like the one to the right when you are done. Step 4 Press 2nd, followed by Y= (for StatPlot). Select 1: and modify the entries to make the screen look like the one to the right. Step 5 Press 2nd, followed by Y= and make certain that both Plot2 and Plot3 are OFF. The screen should look like the one to the right. Step 6 Make certain the ranges for the x and y values are appropriate for the given data set. Here, we use the window shown to the right. Step 7 Press GRAPH and you should get the scatterplot shown to the right. c02b.qxd 11/22/11 9:15 PM Page 232 262. Creating a Scatterplot Using Excel 2007 INSTRUCTION SCREENSHOT Entering Step 1 Open a new Excel spreadsheet. the Data Input the values of the x-variable (rst entries in the ordered pairs) in column A, starting with cell 1A. Then, input the values of the y-variable in column B, starting with cell 1B. The screen should look like the one to the right when you are done. Step 4 Select the leftmost choice in the top row. Press ENTER. The scatterplot shown to the right should appear on the screen. Plotting Step 2 Highlight the data. The screen the Data should look like the one to the right when you are done. Step 3 Go to the Insert tab and select the icon labeled Scatter. A window list of ve possible choices pops up. 2.5* Linear Regression: Best Fit 233 Step 2 Step 3 c02b.qxd 6/6/12 7:26 PM Page 233 263. 234 C HAP TE R 2 Graphs Identifying Patterns While scatterplots are comprised merely of clusters of ordered pairs, patterns of various types can emerge that can provide insight into how the variables x and y are related. Direction of Association This characteristic is analogous to the concept of slope of a line. If the cluster of points tends to rise from left to right, we say that x and y are positively associated, whereas if the cluster of points falls from left to right, we say that x and y are negatively associated. Certainly, the more closely packed together the points are to an identiable curve, the easier it is to make such a determination. Some examples of scatterplots of varying degrees of positive and negative association are shown in the following table. SCATTERPLOT DIRECTION OF ASSOCIATION VERBAL DESCRIPTION Positive Association A shape that increases from left to right is very discernible in each case. The association is a bit loose, but you can still tell the data points tend to rise from left to right. x y y x y x INSTRUCTION SCREENSHOT Step 5 You can alter the format of the scatterplot with various bells and whistles by right-clicking anywhere near the data points and then selecting Format Plot Area at the bottom of the pop-up window. YO U R TU R N Using the data in Example 1, identify x score on the pole vault and z total score. Use technology to create a scatterplot for the data set consisting of the ordered pairs (x, z). Answer: Mens 2004 Olympic Decathlon Pole Vault and Total Points 4.50 4.75 5.00 5.25 7000 7250 7750 7500 8000 8250 8500 8750 9000 Pole Vault TotalPoints c02b.qxd 12/23/11 5:53 PM Page 234 264. 2.5* Linear Regression: Best Fit 235 Linearity Depending on the phenomena being studied and the actual sample being used, the data points comprising a scatterplot can conform very closely to an actual curve. If the curve is a line, we say that the relationship between x and y is linear; otherwise, we say the relationship is nonlinear. Some illustrative examples follow. SCATTERPLOT DIRECTION OF ASSOCIATION VERBAL DESCRIPTION No Association Haphazard scattering of points suggests neither positive nor negative association. Negative Association The association is a bit loose, but you can still tell the data points tend to fall from left to right. Negative Association A shape that decreases from left to right is very discernible in each case. y x y x SCATTERPLOT LINEARITY VERBAL DESCRIPTION Linear Perfect linear relationship; positive association Linear Perfect linear relationship; negative association y x y x y x y x c02b.qxd 6/6/12 7:26 PM Page 235 265. 236 C HAP TE R 2 Graphs y x y x y x y x SCATTERPLOT LINEARITY VERBAL DESCRIPTION Linear Fairly tight linear relationship; positive association Linear Fairly tight linear relationship; negative association Linear Rather loose, but still somewhat discernible, linear relationship; positive association Linear Rather loose, but still somewhat discernible, linear relationship; negative association Nonlinear Perfect nonlinear relationship Nonlinear Fairly tight nonlinear relationship No specically identiable No discernible linear relationship relationship or degree of association y x y x y x c02b.qxd 12/23/11 5:53 PM Page 236 266. 2.5* Linear Regression: Best Fit 237 EXAMPLE 2 Describing Patterns in a Data Set Describe the patterns present in the paired data set (x, y) considered in Example 1, where x points on the 400 m and y total score. Is this intuitive? Solution: We can surmise that the variable score on the 400 m has a relatively strong linear, negative association with the variable total score. This means that as the 400 m score decreases, the total score tends to increase. A negative relationship makes sense here in that 400 m scores reect how much time it took to complete this race. So, lower scores (less time) reect better performance and therefore more total points. YO U R TU R N Using the data in Example 1, identify x score on the pole vault and z total score. Comment on the degree of association and linearity of the scatterplot consisting of the ordered pairs (x, z). Is this intuitive? Strength of Linear Relationship The variability in the data can render it difcult to determine if there is a linear relationship between two variables.As such, it is useful to have a way of measuring how tightly a paired data set conforms to a linear shape. This measure is called the correlation coefcient, r, and is dened by the following formula: Answer: The variable score the pole vault has a rather weak linear, positive association with the variable total score. This means that as the pole vault score increases, the total score tends to increase. In this case, a positive relationship makes sense in that pole vault scores reect the height achieved. So, higher scores (greater height) reect better performance and therefore more total points. For a paired data set {(x1, y1), . . . , (xn, yn)}, the correlation coefcient, r, is dened by The symbol is a shorthand way of writing . So, for instance, This is tedious to calculate by hand but is easily computed using technology. Below are the procedures for how you would compute the correlation coefcient for the data set introduced in Example 1 using the TI-83+ (or TI-84) and Excel 2007. ax2 = x2 1 + + x2 n. z1 + + znaz r = na xy - A axB A ayB A nax2 - A axB2 # A nay2 - A ayB2 D E F I N I T I O N Computing a Correlation Coefcient Using the TI-83 (or TI-84) INSTRUCTION SCREENSHOT Step 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. c02b.qxd 6/6/12 7:26 PM Page 237 267. 238 C HAP TE R 2 Graphs INSTRUCTION SCREENSHOT Step 2 Set up what will display! In order for the desired output to display once we execute the commands to follow, we must tell the calculator to do so. As such, do the following: i. Press 2nd, followed by 0 to get CATALOG. ii. Scroll down until you get to DiagnosticOn. Press ENTER. Then, this command will appear on the home screen. Press ENTER again. The resulting screen should look like the one to the right. Step 3 Press STAT, followed by CALC, and then by 4:LinReg(axb). The resulting screen should look like the one to the right. Press ENTER. Step 4 Press ENTER again. After a brief moment, your screen should look like the one to the right. The value we want is in the bottom row of the screen, about r 0.7045. Note: The other information provided will be pertinent once we dene the best t line in the next subsection. Computing a Correlation Coefcient Using Excel 2007 INSTRUCTION SCREENSHOT Step 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. Step 2 Select the Formulas tab at the top of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right. Step 1 Step 2 c02b.qxd 12/26/11 4:36 PM Page 238 268. 2.5* Linear Regression: Best Fit 239 Step 3 From here, select Statistical, and then from this list, scroll down and choose CORREL. A pop-up window should appear, as shown to the right. Step 4 Enter A1:A28 in Array 1 and B1:B28 in Array 2, as shown to the right. Press OK. You will notice that the correlation coefcient appears directly beneath Array 2. In this case, r is about 0.7045. The square of the correlation coefcient is interpreted as a signed percentage of the variability among the y-values that is actually explained by the linear relationship, where the sign corresponds to the direction of the association. For instance, an r-value of 1 means that 100% of the variability among the y-values is explained by a line with positive slope; in such a case, all of the points in the data set actually lie on a single line. An r-value of 1 means the same thing, but the line has a negative slope. As the r-values get closer to zero, the more dispersed the points become from a line describ- ing the pattern, so that an r-value very close to zero suggests no linear relationship whatsoever is discernible. The following sample of scatterplots with the associated correlation coefcients should provide you with a feel for the strength of linearity suggested by various values of r. c02b.qxd 6/6/12 7:26 PM Page 239 269. 240 C HAP TE R 2 Graphs SCATTERPLOT CORRELATION COEFFICIENT r STRENGTH OF LINEARITY r 1.0 Perfect positive linear relationship r 1.0 Perfect negative linear relationship r 0.80 Reasonably strong, though not perfect, positive linear relationship r 0.45 Pretty weak, barely discernible, negative linear relationship r 0.10 Essentially no discernible linear relationship whatsoever y x y x y x y x y x E X AM P LE 3 Calculating the Correlation Coefcient Associated with a Data Set Use technology to calculate the correlation coefcient r for the paired data set (x, y) considered in Example 1, where x points on the 400 m and y total score. Interpret the strength of the linear relationship. Solution: We see that using either form of technology yields r 0.7045. This suggests that the data follow a relatively strong negative (i.e., negative slope) linear pattern. YO U R TU R N Using the data in Example 1, identify x score on the pole vault and z total score, and calculate the correlation coefcient for the data set consisting of the ordered pairs (x, z) using technology. Interpret the strength of the linear relationship. Answer: The correlation coefcient is approximately r 0.28. This suggests that while the data follow a positive (i.e., positive slope) pattern, the degree to which an actual line describes the trend in the data is rather weak. c02b.qxd 11/22/11 9:15 PM Page 240 270. 2.5* Linear Regression: Best Fit 241 Linear Regression Determining the Best Fit Line Assuming that a data set follows a reasonably strong linear pattern, it is natural to ask which single straight line best describes this pattern. Having such a line would enable us to not only describe the relationship between the two variables x and y precisely, but it would also enable us to predict values of y from values of x not present among the points of the data set. Consider the paired data set (x, y) from Example 1, where x points on the 400 m and y total score. You learned in Section 2.3 that between any two points there is a unique line whose equation can be determined. Three such lines passing through various pairs of points in the data set are illustrated below. The unavoidable shortcoming of all of these lines, however, is that not all of the data points lie on a single one of them. Each has a negative slope, which is characteristic of the data set, and eachofthelinesisclosetosomeofthedatapoints,butnotclosetoothers.Infact,wecoulddraw innitely many such lines and make a similar assessment. But which one best ts the data? The answer to this question depends on how you dene best. Reasonably, for the line that best ts the data, the error incurred in using it to describe all of the points in the data set should be as small as possible. The conventional approach is to dene this error by summing the n distances di between the y-coordinates of the data points and the corresponding y-value on the line y Mx B (that is, the y-value of the point on the line corresponding to the same x-value). These distances are, in effect, the error in making the approximation. This is illustrated below: Using the distance formula, we nd that di = 2(xi - xi)2 + ( yi - (Mxi + B))2 = yi - (Mxi + B). x y x1 (x1 , y1 ) (x3 , y3 ) (x6 , y6 ) (x8 , y8 ) (x4 , y4 ) (x7 , y7 ) (x5 , y5 ) (x2 , y2 ) (x1 , Mx1 +B) (x2 , Mx2 +B) (x3 , Mx3 +B) y = Mx + B d1 d3 d4 d5 d6 d8 d7 x2 x3 x4 x5 x6 x7 x8 400 m Mens 2004 Olympic Decathlon 400 m and Total Points 47 49 51 53 5548 50 52 54 TotalPoints 9000 8750 8500 8250 8000 7750 7500 7250 7000 x y line 1 line 3 line 2 c02b.qxd 6/6/12 7:26 PM Page 241 271. Note that di 0 precisely when the data point (xi, yi) lies directly on the line y Mx B, and that the closer di is to 0, the closer the point (xi, yi) is to the line y Mx B. As such, the goal is to determine the values of the slope M and y-intercept B for which the sum d1 dn is as small as possible. Then, the resulting straight line y Mx B best ts the data set {(x1, y1), (xn, yn)}. This is ne, in theory, but it turns out to be inconvenient to work with a sum of absolute value expressions. It is actually much more convenient to work with the squared distances d2 i . The values of M and B that minimize d1 ... dn are precisely the same as those that minimize d2 1 ... d2 n. Using calculus, it can be shown that the formulas for M and B are as follows: , The resulting line y Mx B is called the best t least-squares regression line for the data set {(x1, y1), (xn, yn)}. Again, it is tedious to compute these by hand, but their values are actually produced easily using technology. B = ay n - M ax n M = naxy - A axB A ayB na x2 - A axB2 E X AM P LE 4 Finding the Line of Best Fit by Linear Regression Find the line of best t (best t least-squares regression line) for the paired data set (x, y) from Example 1, where x points on the 400 m and y total score, using (a) the TI-83 (or TI-84) and (b) Excel 2007. Solution (a): Determining the Best Fit Least-Squares Regression Line Using the TI-83 (or TI-84). INSTRUCTION SCREENSHOT STEP 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. STEP 2 Press STAT, followed by CALC, and then by 4:LinReg(axb). The resulting screen should look like the one to the right. 242 C HAP TE R 2 Graphs c02b.qxd 11/22/11 9:15 PM Page 242 272. INSTRUCTION SCREENSHOT STEP 3 For this example, the data is stored in lists L1 and L2. And, since we will want to graph our best t line on the scatterplot, it will need to be stored as a function of x, say as Y1. In order to do this, proceed as follows: Directly next to LinReg(axb) on the home screen, we need to type the following: L1, L2, Y1. Use the following key strokes: 2nd, 1, , 2nd, 2, , VARS, Y-VARS, 1:FUNCTION, Y1 The resulting screen should look like the one to the right. STEP 4 Press ENTER. The equation of the best t least-squares line with the slope (labeled as a) and the y-intercept (labeled as b) appears on the screen as shown to the right. So, the equation of the best t least-squares regression line is approximately y 206.9x 18317.03. STEP 5 In order to obtain a graph of the scatterplot with the best t line from Step 4 superimposed on it, press ZOOM, then 9:ZoomStat. The resulting screen should look like the one to the right. Solution (b): Determining the Best Fit Least-Squares Regression Line Using Excel 2007. INSTRUCTION SCREENSHOT STEP 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. 2.5* Linear Regression: Best Fit 243 Step 1 c02b.qxd 6/6/12 7:26 PM Page 243 273. INSTRUCTION SCREENSHOT STEP 2 Select the Formulas tab at the top of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right. STEP 3 From here, select Statistical, and then from this list, scroll down and choose LINEST. A pop-up window should appear, as shown to the right. STEP 4 Enter B1:B28 in Known_ys and A1:A28 in Known_xs, as shown to the right. You will notice that a set of two values occurs directly beneath the entry boxesthe output is about {206.92, 18,317.03}. The rst value is the slope M, and the second value is the y-intercept B of the best t line. So, the equation of the best t least-squares regression line is approximately y 206.92x 18,317.03. 244 C HAP TE R 2 Graphs c02b.qxd 11/22/11 9:15 PM Page 244 274. 2.5* Linear Regression: Best Fit 245 INSTRUCTION SCREENSHOT STEP 5 In order to obtain a graph of the scatterplot with the best t line from Step 4 superimposed on it, construct the scatterplot as before, right-click on the scatterplot near the data points, choose Add Trendline, and press ENTER. STEP 6 The best t line will appear on the scatterplot, along with a pop-up window allowing you to change the format of the line/ curve that is displayed. YO U R TU R N Using the data in Example 1, identify x score on the pole vault and z total score, and determine the best t least-squares regression line for the data set consisting of the ordered pairs (x, z). Superimpose the graph of this line on the scatterplot. Answer: Here, the best t line displays a positive relationship, and its equation is z 364.97x 6324.46. Mens 2004 Olympic Decathlon: Pole Vault and Total Points with Best Fit Line R2 Linear = 0.08 4.50 4.75 5.00 5.25 7000 7250 7750 7500 8000 8225 8500 8750 9000 Pole Vault TotalPoints c02b.qxd 6/6/12 7:26 PM Page 245 275. 246 C HAP TE R 2 Graphs Using the Best Fit Line for Prediction It is important to realize that for any scatterplot, no matter how haphazardly dispersed the data, a best t least-squares regression line can be created. This is true even when the relationship between x and y is nonlinear. However, the utility of such a line in these instances is very limited. In fact, a best t line should only be created when the linear relationship is reasonably strong, which means the correlation coefcient is reasonably far away from 0. This criterion can be made more precise using statistical methods, but for our present purposes, we shall make the blanket assumption that it makes sense to form the best t line in all of the scenarios we present. Once we have the best t line in hand, we can use it to predict y-values for values of x that do not correspond to any of the points in the data set. For instance, consider the following: E X AM P LE 5 Making Predictions Using the Line of Best Fit Consider the data set from Example 1. a. Use the best t line to predict the total score given that an Olympian scored 50.05 points in the 400 m. Is it reasonable to use the best t line to make such a prediction? b. Use the best t line to predict the total score given that an Olympian scored 40.05 points in the 400 m. Is it reasonable to use the best t line to make such a prediction? Solution: a. Using the line y 206.92x 18,317.03, we see that y is approximately 7961 when x 50.05. This means that if an Olympian were to score 50.05 on the 400 m, then his predicted total score would be approximately 7961. Using the best t line to predict the total score in this case is reasonable because the value 50.05 is well within the range of x-values already present in the data set. b. Using the line y 206.92x 18,317.03, we see that y is approximately 10,030 when x 40.05. This means that if an Olympian were to score 40.05 on the 400 m, then his predicted total score would be approximately 10,030. This prediction is questionable because the x-value at which you are using the best t line to predict y is sufciently far away from the rest of the data points that were used to construct the line. As such, there is no reason to believe that the line is valid for such x-values. YO U R TU R N Using the data in Example 1, identify x score on the pole vault and z total score, and use the best t line to predict the total score given that an Olympian scored 4.65 points on the pole vault and then, given that an Olympian scored 5.9 points on the pole vault. Comment on the validity of these predictions. Answer: Using the line z 364.97x 6324.46, we see that z is approximately 8022 when x 4.65. This means that if an Olympian were to score 4.65 on the pole vault, then his predicted total score would be approximately 8022. Using the best t line to predict the total score in this case is reasonable because the value 4.65 is well within the range of x-values already present in the data set. Next, using the same line, we see that z is approximately 8478 when x 5.9. This prediction is less reliable than the former one because 5.9 is outside of the range of x-values corresponding to the rest of the data points used to construct the line. But it isnt too far outside this range, so there is a degree of validity to the prediction. It is important to realize that the correlation coefcient is NOT equal, or even related to, the actual slope of the best t line. In fact, two distinct perfectly linear, positively associated scatterplots will both have r 1, even though the actual lines that t the data might have slope M 15 and M 0.04. c02b.qxd 11/22/11 9:15 PM Page 246 276. 2.5* Linear Regression: Best Fit 247 E X AM P LE 6 The Power of TV Advertisement Video Board Tests, Inc., an advertising testing agency, collected data based on 4000 adult participants of a survey. The participants (who were regular product users) were asked to recall a commercial that they had viewed for a given product category in the previous week. The goal was to examine the relationship between retained impressions of commercials and the corresponding TV advertising budget for a given product. The data were published in the Wall Street Journal in March 1984. The following is an adaptation of the original data set (TV Ad Yields was obtained from the following Web source: http://lib.stat.cmu.edu/DASL/Datales/tvadsdat.html), but contains three modications made for illustrative purposes. Specically, ATT/BELL, FORD, and MCDONALDS have been replaced with DIALTONE USA, CARZ, and HAPPY BURGERS, respectively. These changes have been highlighted in green. TV ADVERTISING BUDGET, MILLIONS RETAINED COMPANY 1983 ($ MILLIONS) IMPRESSIONS PER WEEK MILLER_LITE 50.1 32.1 PEPSI 74.1 99.6 STROHS 19.3 11.7 FEDERAL_EXPRESS 22.9 21.9 BURGER_KING 82.4 60.8 COCA-COLA 40.1 78.6 HAPPY_BURGERS 165.0 10.0 MCI 26.9 50.7 DIET_COLA 20.4 21.4 CARZ 165.0 50.0 LEVIS 27.0 40.8 BUD_LITE 45.6 10.4 DIALTONE_USA 70.0 88.9 CALVIN_KLEIN 5.0 12.0 WENDYS 49.7 29.2 POLAROID 26.9 38.0 SHASTA 5.7 10.0 MEOW_MIX 7.6 12.3 OSCAR_MEYER 9.2 23.4 CREST 32.4 71.1 KIBBLES_N_BITS 6.1 4.4 c02b.qxd 6/6/12 7:26 PM Page 247 277. 248 C HAP TE R 2 Graphs SMH measured using the correlation coefcient r. If r is sufciently far from 0, a best t least-squares regression line can be formed to precisely describe the linear relationship and used for reasonable prediction purposes. S U M MARY Two variables x and y can be related in different ways. A paired data set {(x1, y1), (xn, yn)} obtained experimentally can be illustrated using a scatterplot. Patterns concerning the direction of association and linearity can be used to describe the relationship between x and y, and the strength of the linear relationship can be S E CTI O N 2.5* First, a scatterplot for this data set (formed using PASW Statistics 18) is shown below. The approximate regression line is y 0.18x 29.04 and r 0.28. It appears that there might be a relationship between the budget and the retained impressions. Both CARZ and HAPPY_BURGERS are pretty far away from the bulk of the data and might be skewing an otherwise tighter relationship between x and y; such points are potential outliers. What would happen to the regression line if we removed each of these points, one at a time? Would the new line be dramatically different from the original one, or might there be very little change? Lets start by removing CARZ. The resulting regression line is y 0.21x 27.96. We observe only a small change in both the slope and intercept. Next, lets put CARZ back into the data set and remove HAPPY_BURGERS instead. In this case, the best t regression line is y 0.40x 22.61. This time, we have a slightly larger change in the y-intercept, but more importantly, the slope is more than double the slope of the regression line from the original data set. The resulting best t regression line is displayed to the right. Clearly, HAPPY_BURGERS was a very inuential data point since removing it dramatically changed the slope between the budget and retained impressions, thereby considerably changing the mathematical description of the relationship between these two variables. But why did this happen? There was a large distance between it and both the x- and y-directions from the bulk of the data set. When a potential outlier is distant in only one of the two directions (x or y), as was the case with CARZ, it is far less likely to be inuential. $50 100 150 200 250 100 90 80 70 60 50 40 30 20 10 TV Advertising Budget, 1983 (millions of dollars) TV Advertising Budget and Retained Impressions MillionsofRetained ImpressionsperWeek R 2 Linear = 0.08 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 PEPSI 2 DIALTONE_USA 3 COCA-COLA 4 CREST 5 BURGER_KING 6 MCI 7 LEVIS 8 POLAROID 9 MILLER_LITE 10 WENDY'S 11 OSCAR_MEYER 12 DIET_COLA 13 FEDERAL_EXPRESS 14 CALVIN_KLEIN 15 MEOW_MIX 16 STROHS 17 SHASTA 18 KIBBLES_'N_BITS 19 BUD_LITE 20 CARZ 21 HAPPY_BURGERS $50 100 150 200 250 100 90 80 70 60 50 40 30 20 10 TV Advertising Budget, 1983 (millions of dollars) MillionsofRetained ImpressionsperWeek R 2 Linear = 0.278 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 PEPSI 2 DIALTONE_USA 3 COCA-COLA 4 CREST 5 BURGER_KING 6 MCI 7 LEVIS 8 POLAROID 9 MILLER_LITE 10 WENDY'S 11 OSCAR_MEYER 12 DIET_COLA 13 FEDERAL_EXPRESS 14 CALVIN_KLEIN 15 MEOW_MIX 16 STROHS 17 SHASTA 18 KIBBLES_'N_BITS 19 BUD_LITE 20 CARZ TV Advertising Budget and Retained Impressions c02b.qxd 6/6/12 7:26 PM Page 248 278. 2.5* Linear Regression: Best Fit 249 In Exercises 58, match the following scatterplots with the following correlation coefcients. a. r 0.90 c. r 0.68 b. r 0.80 d. r 0.20 S K I L L S E X E R C I S E S S E CTI O N 2.5* 1. 2. In Exercises 14, for each of the following scatterplots, identify the pattern as a. having a positive association, negative association, or no identiable association. b. being linear or nonlinear. y x y x y x y x y x y x y x y x For each of the following data sets, a. create a scatterplot. b. guess the value of the correlation coefcient r. c. use technology to determine the equation of the best t line and to calculate r. d. give a verbal description of the relationship between x and y. 9. 10.x y 3 14 1 8 0 5 1 2 3 4 5 10 x y 8 16 6 12 4 8 2 4 1 2 3 6 11. 12. x y 10 1 6 0 0 2 8 10 14 11 20 16 13. 14. x y 1 17 1/2 11 1/4 5 1/10 0 0 1 1/10 1 1/5 8 1 12 x y 3 6 2 3 1 1 0 1 1 5 2 1 4 1 x y 6 1 3 4 1 3 1 0 2 6 5 4 8 1 3. 4. 5. 6. 7. 8. c02b.qxd 6/6/12 7:26 PM Page 249 279. 250 C HAP TE R 2 Graphs For Exercises 1922, a. use technology to create a scatterplot, to determine the best t line, and to compute r for the entire data set. b. repeat (a), but with the data set obtained by removing the starred (***) data points. c. compare the r-values from (a) and (b), as well as the slopes of the best t lines. Comment on any differences, whether they are substantive, and why this seems reasonable. 18. x y 15 4 15 12 15 16 13 3 10 4 10 8 10 12 5 4 2 3 2 2 2 3 2 6 4 1 4 0 4 4 7 2 7 3 10 4 10 2 10 3 19. 20. x y 3 14 1 8 0 5 1 2 3 4 *** 5 10 x y 10 1 6 0 0 2 8 10 14 11 *** 20 16 21. 22. x y 3 14 1 8 0 5 1 2 *** 3 4 5 15 *** 6 16 x y 0 0 1 2 2 4 3 6 4 8 6 12 *** 7 25 In Exercises 1518, for each of the data sets, a. use technology to create a scatterplot, to determine the best t line, and to compute r. b. indicate whether or not the best t line can be used for predictive purposes for the following x-values. For those for which it can be used, give the predicted value of y: i. x 0 iii. x 12 ii. x 6 iv. x 15 c. Using the best t line, at what x-value would you expect y to be equal to 2? 15. 16. 17. x y 20 15 18 10 14 3 14 8 13 3 8 0 8 3 5 6 1 11 1 15 x y 5 8 3 0 2 0 2 1.5 5 4 7 2 10 8 x y 5 0 5 3 6 3 7 6 7 9 8 9 8 15 9 9 9 15 10 15 10 18 c02b.qxd 11/22/11 9:16 PM Page 250 280. 2.5* Linear Regression: Best Fit 251 27. The following screenshot was taken when using the TI-83 to determine the equation of the best t line for paired data (x, y): Using the regression line, we observe that there is a strong positive linear association between x and y, and that for every unit increase in x, the y-value increases by about 1.257 units. 28. The following scatterplot was produced using the TI-83 for paired data (x, y). The equation of the best t line was reported to be y 3.207x 0.971 with r2 0.9827. Thus, the correlation coefcient is given by r 0.9913, which indicates a strong linear association between x and y. 23. Consider the data set from Exercise 17. a. Reverse the roles of x and y so that now y is the explanatory variable and x is the response variable. Create a scatterplot for the ordered pairs of the form (y, x) using this data set. b. Compute r. How does it compare to the r-value from Exercise 17? Why does this make sense? c. The best t line for the scatterplot in (a) will be of the form x my b. Determine this line. d. Using the line from (c), nd the predicted x-value for the following y-values, if appropriate. If it is not appropriate, tell why. i. y 23 ii. y 2 iii. y 16 24. Consider the data set from Exercise 16. Redo the parts in Exercise 23. 25. Consider the following data set. C O N C E P T U A L x y 3 0 3 1 3 1 3 2 3 4 3 15 3 6 3 8 3 10 x y 5 2 4 2 1 2 0 2 1 2 3 2 8 2 17 2 Guess the values of r and the best t line. Then, check your answers using technology. What happens? Can you reason why this is the case? 26. Consider the following data set. Guess the values of r and the best t line. Then, check your answers using technology. What happens? Can you reason why this is the case? C AT C H T H E M I S TA K E c02b.qxd 6/6/12 7:26 PM Page 251 281. 252 C HAP TE R 2 Graphs For Exercises 31 and 32, refer to the following scenario: Texting Speed. 31. What is the relationship between the variables left thumb length and total both scores? a. Create a scatterplot to show the relationship between left thumb length and total both scores. b. What is the correlation coefcient between left thumb length and total both scores? c. Describe the strength of the relationship between left thumb length and total both scores. d. What is the equation of the best t line that describes the relationship between left thumb length and total both scores? e. Could you use the best t line to produce accurate predictions of total both scores using left thumb length? 32. Repeat Exercise 31 for right thumb length and total both scores. According to the CTIAThe Wireless Association, as of December 2010, 187.7 billion messages were sent per month or 2.1 trillion messages in that year.1 According to a 2010 Pew Internet survey, 72% of all teensor 88% of teen cell phone usersare text-messagers. Teens make and receive far fewer phone calls than text messages on their cell phones. A number of competitions regarding texting speed have taken place worldwide. According to the Guinness World Records, The fastest completion of a prescribed 160-character text message is 34.65 seconds and was achieved by Frode Ness (Norway) at the Norwegian SMS championships held at the Oslo City shopping centre in Oslo, stlandet, Norway, on 13 November 2010.2 The data set regarding texting speed on the next page was provided by AP Central. (http://apcentral.collegeboard.com/apc/public/ courses/teachers_corner/195435.html) In the data given, the A total score is the amount of time (in seconds) it took to text the following message, Statistics students are above average. The B total score is the amount of time (in seconds) to type, Meet me at my car after school today. The Total both scores is the sum of the A total score and B total score. What inuences texting speed in this group? Lets consider thumb length. 1 http://www.cita.org/advocacy/research/index.cfm/aid/10323 2 http://www.guinnessworldrecords.com/Search/Details/Fastest-text- message/57979.htm For Exercises 29 and 30, refer to the data set in Example 1. A P P L I C AT I O N S 29. a. Examine the relationship between each of the decathlon events and the total points by computing the correlation coefcient in each case. b. Using the information from part (a), which event has the strongest relationship to the total points? c. What is the equation of the best t line that describes the relationship between the event from part (b) and the total points? d. Using the best t line, if you had a score of 40 for this event, what would the predicted total points score be? 30. a. Using the information from part (a), which event has the second strongest relationship to the total points? b. What is the equation of the best t line that describes the relationship between the event in part (b) and the total points? c. Is it reasonable to expect the best t line from part (c) to produce accurate predictions of total points using this event? d. Using the best t line, if you had a score of 40 for this event, what would the total points score be? c02b.qxd 11/22/11 9:16 PM Page 252 282. 2.5* Linear Regression: Best Fit 253 TEXTING LEFT THUMB RIGHT THUMB A TOTAL B TOTAL TOTAL BOTH AVG DIFF GENDER STYLE LENGTH LENGTH SCORE SCORE SCORES A MINUS B THUMB THUMB Male Char 6.5 6.5 35 25 60 10 6.5 0 Female Char 5 5 61 57 118 4 5 0 Female Word 6 6 24 20 44 4 6 0 Male Word 7 7 43 60 103 17 7 0 Female Word 6 6 14 19 33 5 6 0 Male Word 7 6 15 18 33 3 6.5 1 Female Word 6 6 13 14 27 1 6 0 Female Word 6 6 22 10 22 12 6 0 Male Word 6.5 6 13 15 28 2 6.25 0.5 Female Word 5.5 5.5 16 16 32 0 5.5 0 Male Char 6 5 85 78 163 7 5.5 1 Male Char 6 6 126 120 246 6 6 0 Male Word 7.5 6.5 67 69 136 2 7 1 Female Char 5.5 5.5 11 7 18 4 5.5 0 Female Word 5.5 5.7 14 17 31 3 5.6 0.2 Female Word 6 6 17 14 31 3 6 0 Female Word 5 5 20 15 35 5 5 0 Male Word 6.5 6.5 15 13 28 2 6.5 0 Male Word 7 7 30 31 61 1 7 0 Male Word 6 6.1 120 117 237 3 6.05 0.1 Male Word 6 6 74 25 99 49 6 0 Male Word 6.3 6 23 21 44 2 6.15 0.3 Male Char 6 5.9 45 50 95 5 5.95 0.1 Male Word 6 6.1 86 100 186 14 6.05 0.1 Male Char 6 6 25 23 48 2 6 0 Male Char 6.3 6.3 81 57 138 24 6.3 0 Female Char 5.5 5.5 88 66 154 22 5.5 0 Female Word 7 6.8 10 9 19 1 6.9 0.2 Male Char 6.5 7 21 18 39 3 6.75 0.5 Female Char 5.4 5.2 72 48 121 24 5.3 0.2 Female Char 8 8 36 23 59 13 8 0 Male Char 7 6.5 46 45 91 1 6.75 0.5 Female Char 6 6.8 48 39 87 9 6.4 0.8 Female Char 7.1 7.1 84 57 141 27 7.1 0 Female Word 5.9 5.5 25 23 48 2 5.7 0.4 Female Char 7.6 7.2 32 45 77 13 7.4 0.4 Male Word 6.9 7 23 28 51 5 6.95 0.1 Female Char 7.7 7.5 18 15 33 3 7.6 0.2 Male Char 8.6 (cast) 22 20 42 2 N/A N/A Male Char 7.3 7.1 54 50 104 4 7.2 0.2 c02b.qxd 6/6/12 7:26 PM Page 253 283. 254 C HAP TE R 2 Graphs 33. What is the relationship between the variables % residents immunized and % residents with inuenza? a. Create a scatterplot to illustrate the relationship between % residents immunized and % residents with inuenza. b. What is the correlation coefcient between % residents immunized and % residents with inuenza? c. Describe the strength of the relationship between % residents immunized and % residents with inuenza. d. What is the equation of the best t line that describes the relationship between % residents immunized and % residents with inuenza? e. Could you use the best t line to produce accurate predictions of % residents with inuenza using % residents immunized? 34. What is the impact of the outlier(s) on this data set? a. Identify the outlier in this data set. What is the nursing home number for this outlier? b. Remove the outlier and re-create the scatterplot to show the relationship between % residents immunized and % residents with inuenza. c. What is the revised correlation coefcient between % residents immunized and % residents with inuenza? d. By removing the outlier is the strength of the relationship between % residents immunized and % residents with inuenza increased or decreased? e. What is the revised equation of the best t line that describes the relationship between % residents immunized and % residents with inuenza? For Exercises 3538, refer to the following data set: Amusement Park Rides. According to the International Association of Amusement Parks and Attractions (IAAPA), There are more than 400 amusement parks and traditional attractions in the United States alone. In 2008, amusement parks in the United States entertained 300 million visitors who safely enjoyed more than 1.7 billion rides.4 Despite the popularity of amusement parks, the wait times, especially for the most popular rides, are not so highly regarded. There are different approaches and tactics that people take to get the most rides in their visit to the park. Now, there are even apps for the iPhone and Android to track waiting times at various amusement parks. One might ask, Are the wait times worth it? Are the rides with the longest wait times, the most enjoyable? Consider the following ctional data. For Exercises 33 and 34, refer to the following data set: Herd Immunity. According to the U.S. Department of Health and Human Services, herd immunity is dened as a concept of protecting a community against certain diseases by having a high percentage of the communitys population immunized. Even if a few members of the community are unable to be immunized, the entire community will be indirectly protected because the disease has little opportunity for an outbreak. However, with a low percentage of population immunity, the disease would have great opportunity for an outbreak.3 Suppose a study is conducted in the year 2016 looking at the outbreak of Haemophilus inuenzae type b in the winter of 2015 across 22 nursing homes. We might look at the percentage of residents in each of the nursing homes that were immunized and the percentage of residents who were infected with this type of inuenza. The ctional data set is as follows. % RESIDENTS % RESIDENTS NURSING HOME IMMUNIZED WITH INFLUENZA 1 70 11 2 68 9 3 80 8 4 10 34 5 12 30 6 18 31 7 27 22 8 64 18 9 73 6 10 9 31 11 35 19 12 56 16 13 57 22 14 83 10 15 74 13 16 64 15 17 16 28 18 23 25 19 29 24 20 33 20 21 82 28 22 67 9 4 http://www.iaapa.org/pressroom/Amusement_Park_Industry_Statistics.asp3 http://www.hhs.gov/nvpo/glossary1.htm c02b.qxd 12/23/11 5:53 PM Page 254 284. 2.5* Linear Regression: Best Fit 255 AVG AVG WAIT ENJOYMENT PARK RIDE ID RIDE NAME TIME RATING PARK LOCATION 1 Xoom 45 58 1 Florida 2 Accentuator 35 40 1 Florida 3 Wobbler 15 15 1 Florida 4 Arctic_Attack 75 75 1 Florida 5 Gusher 60 70 1 Florida 6 Alley_Cats 5 60 1 Florida 7 Moon_Swing 10 15 1 Florida 8 Speedster 70 50 1 Florida 9 Hailstorm 80 90 1 Florida 10 DragonFire 70 88 1 Florida 1 Xoom 50 10 2 California 2 Accentuator 35 40 2 California 3 Wobbler 20 75 2 California 4 Arctic_Attack 70 60 2 California 5 Gusher 70 80 2 California 6 Alley_Cats 10 18 2 California 7 Moon_Swing 15 80 2 California 8 Speedster 80 35 2 California 9 Hailstorm 95 40 2 California 10 DragonFire 55 60 2 California The data shows 10 popular rides in two sister parks located in Florida and California. For each ride in each park, average wait times (in minutes) in the summer of 2010 and the average rating of ride enjoyment (on a scale of 1100) are provided. 35. What is the relationship between the variables average wait times and average rating of enjoyment? a. Create a scatterplot to show the relationship between average wait times and average rating of enjoyment. b. What is the correlation coefcient between average wait times and average rating of enjoyment? c. Describe the strength of the relationship between average wait times and average rating of enjoyment. d. What is the equation of the best t line that describes the relationship between average wait times and average rating of enjoyment? e. Could you use the best t line to produce accurate predictions of average wait times using average rating of enjoyment? 36. Examine the relationship between average wait times and average rating of enjoyment for Park 1 in Florida by repeating Exercise 35 for only Park 1. 37. Examine the relationship between average wait times and average rating of enjoyment for Park 2 in California repeating Exercise 35 for only Park 2. 38. Compare the relationship between average wait times and average rating of enjoyment for Park 1 in Florida versus Park 2 in California. C H A L L E N G E For Exercises 3942, refer to the following: Exploring other types of best-t curves When describing the patterns that emerge in paired data sets, there are many more possibilities other than best t lines. Indeed, once you have drawn a scatterplot and are ready to identify the curve that best ts the data, there is a substantive collection of other curves that might more accurately describe the data. The following are listed among those in STATS/CALC on the TI-83, along with some comments: NAME OF REGRESSION CURVE FORM OF THE CURVE COMMENTS 5: QuadReg y ax2 bx c The data set must have at least 3 points to be able to select this option. 6: CubicReg y ax3 bx2 cx d The data set must have at least 4 points to be able to select this option. 7: QuartReg y ax4 bx3 cx2 dx e The data set must have at least 5 points to be able to select this option. 9: LnReg y a b ln x The data set must have at least 2 points to be able to select this option, and x cannot take on negative values. 0: ExpReg y a* bx The data set must have at least 2 points to be able to select this option, and y cannot take on the value of 0. A: PwrReg y a* xb The data set must have at least 2 points to be able to select this option. c02b.qxd 6/6/12 7:26 PM Page 255 285. 256 C HAP TE R 2 Graphs For each of the following data sets, a. Create a scatterplot. b. Use LinReg(axb) to determine the best t line and r. Does the line seem to accurately describe the pattern in the data? c. For each of the different choices listed in the above chart, nd the equation of the best t curve and its associated r2 value. Of all of the curves, which seems to provide the best t? Note: The r2 -value reported in each case is NOT the linear correlation coefcient reported when running LinReg(axB). Rather, the value will typically change depending on the curve. The reason why is that each time, the r2 -value is measuring how accurate the t is between the data and that type of curve. A value of r2 close to 1 still corresponds to a good t with whichever curve you are tting to the data. 39. 40.x y 1 16.2 2 21 3 23.7 4 24.8 5 23.9 6 20.7 7 15.8 8 9.1 9 0.3 x y 0.5 1.20 1.0 0.760 1.5 0.412 2.1 0.196 2.9 0.131 3.3 0.071 41. 42.x y 1 0.2 1.5 0.93 2 1.46 3 2.25 10 4.51 15 5.50 x y 1 32.3 2 8.12 3 16.89 5 45.2 6 0.89 8 62.1 c02b.qxd 11/22/11 9:16 PM Page 256 286. C HAP TE R 2 I N Q U I RY- BAS E D LE AR N I N G P R OJ E CT 257 The following table shows U.S. population estimates for the years 1991 to 2002, along with the number of tons of municipal waste generated (in 100 million tons) and the percentage that was recycled in the United States during those years. U.S. POPULATION ESTIMATE MUNICIPAL WASTE GENERATED PERCENTAGE YEAR (100 MILLION PEOPLE) (100 MILLION TONS) RECYCLED 1991 2.46 2.69 8% 1992 2.49 2.93 11.5% 1993 2.52 2.80 14% 1994 2.55 2.91 17% 1995 2.58 3.06 19% 1996 2.60 3.22 23% 1997 2.63 3.26 27% 1998 2.65 3.27 28% 1999 2.68 3.40 30% 2000 2.73 3.74 31.5% 2001 2.80 3.82 33% 2002 2.86 4.09 32% Adapted from: A Yearly Snapshot of U.S. (Municipal) Waste and Recycling (Data Source: BIOCYCLE/ Table & Conversion: ZWA http://www.zerowasteamerica.org/statistics.htm) 1. For parts (a)(f), consider the columns for U.S. Population Estimate, x, and Municipal Waste Generated, y. a. Write the equation (in slopeintercept form) of the line that passes through the points (2.60, 3.22) and (2.80, 3.82). b. How can you interpret the slope of the line in part (a)? c. Now choose two other data points and write the equation of the line that passes through your chosen points. How can you interpret the slope of this line? d. How does the slope of this line compare to the line in part (a)? e. Sketch a graph of the two lines. What do you notice about their y-intercepts? f. Finally, plot all the other ordered points (Population, Waste Generated). The graph of all the data points is called a scatterplot. Since the data fall in approximately a straight line, each of the lines you graphed above is an approximation of the data. Section 2.5* presented methods for nding the line that best ts a set of data points, called the least-squares regression line. 2. For parts (a)(c) below, consider the columns U.S. Population Estimate, x, and Percentage Recycled, y. a. Graph the scatterplot for this data. b. Sketch the graph of a line that contains two data points. Choose a line you think ts the data well. c. Write the equation of the line. d. How can you interpret the slope of this line? c02b.qxd 6/6/12 7:26 PM Page 257 287. 258 The Intergovernmental Panel on Climate Change (IPCC) claims that carbon dioxide (CO2) production from industrial activity (such as fossil fuel burning and other human activities) has increased the CO2 concentrations in the atmosphere. Because it is a greenhouse gas, elevated CO2 levels will increase global mean (average) temperature. In this section, we will examine the increasing rate of carbon emissions on Earth. In 1955, there were (globally) 2 billion tons of carbon emitted per year. In 2005, the carbon emissions more than tripled to reach approximately 7 billion tons of carbon emitted per year. Currently, we are on the path to doubling our current carbon emissions in the next 50 years. Two Princeton professors* (Stephen Pacala and Rob Socolow) introduced the Climate Carbon Wedge concept. A wedge is a strategy to reduce carbon emissions over a 50-year time period by 1.0 GtC/yr (gigatons of carbon per year). 1. Draw the Cartesian plane. Label the vertical axis C, where C represents the number of gigatons (billions of tons) of carbon emitted, and label the horizontal axis t, where t is the number of years. Let t 0 correspond to 2005. 2. Find the equations of the at path and the seven lines corresponding to the seven wedges. a. Flat path (no increase) over 50 years (2005 to 2055) b. Increase of 1 GtC over 50 years (2005 to 2055) c. Increase of 2 GtC over 50 years (2005 to 2055) d. Increase of 3 GtC over 50 years (2005 to 2055) e. Increase of 4 GtC over 50 years (2005 to 2055) f. Increase of 5 GtC over 50 years (2005 to 2055) g. Increase of 6 GtC over 50 years (2005 to 2055) h. Increase of 7 GtC over 50 years (2005 to 2055) [projected path] Total = 25 Gigatons Carbon 50 years 1 GtC/yr Year 1975 1995 2015 2035 2055 TonsofCarbonEmitted/Year (inbillions) 20 18 16 14 12 10 8 6 4 2 The Stabilization Triangle Flat Path Interim Goal Stabilization TriangleHistorical Emissions Currently Projected Path = Ramp M O D E L I N G O U R W O R LD c02b.qxd 11/22/11 9:16 PM Page 258 288. 259 3. For each of the seven wedges and the at path, determine how many total gigatons of carbon will be reduced over a 50-year period. In other words, how many gigatons of carbon would the world have to reduce in each of the eight cases? a. Flat path b. Increase of 1 GtC over 50 years c. Increase of 2 GtC over 50 years d. Increase of 3 GtC over 50 years e. Increase of 4 GtC over 50 years f. Increase of 5 GtC over 50 years g. Increase of 6 GtC over 50 years h. Increase of 7 GtC over 50 years (projected path) 4. Research the climate carbon wedge concept and discuss the types of changes (transportation efciency, transportation conservation, building efciency, efciency in electricity production, alternate energies, etc.) the world would have to make that would correspond to each of the seven wedges and the at path. a. Flat path b. Wedge 1 c. Wedge 2 d. Wedge 3 e. Wedge 4 f. Wedge 5 g. Wedge 6 *S. Pacala and R. Socolow, Stabilization Wedges: Solving the Climate Problem for the Next 50 Years with Current Technologies, Science, Vol. 305 (2004). 259 c02b.qxd 11/22/11 9:16 PM Page 259 289. SECTION CONCEPT KEY IDEAS/FORMULAS 2.1 Basic tools: Cartesian plane, Two points in the xy-plane: (x1, y1) and (x2, y2) distance, and midpoint Cartesian plane x-axis, y-axis, origin, and quadrants Distance between two points Midpoint of a line segment joining two points 2.2 Graphing equations: Point- plotting, intercepts, and symmetry Point-plotting List a table with several coordinates that are solutions to the equation; plot and connect. Intercepts x-intercept: let y 0 y-intercept: let x 0 Symmetry The graph of an equation can be symmetric about the x-axis, y-axis, or origin. Using intercepts and symmetry If (a, b) is on the graph of the equation, then (a, b) is on the graph if as graphing aids symmetric about the y-axis, (a, b) is on the graph if symmetric about the x-axis, and (a, b) is on the graph if symmetric about the origin. 2.3 Lines General form: Ax By C Graphing a line Vertical: x a Slant: Ax By C Horizontal: y b where and Intercepts x-intercept (a, 0) y-intercept (0, b) Slope , where Equations of lines Slopeintercept form: y mx b m is the slope and b is the y-intercept. Pointslope form: y y1 m(x x1) Parallel and perpendicular lines L1 || L2 if and only if m1 m2 (slopes are equal). if and only if (slopes are negative reciprocals). 2.4 Circles Standard equation of a circle (x h)2 (y k)2 r2 C: (h, k) Transforming equations of circles to General form: x2 y2 ax by c 0 the standard form by completing the square 2.5* Linear regression: Best t Fitting data with a line Scatterplots Creating a scatterplot: Using Microsoft Excel Using a graphing calculator Identifying patterns Association Positive Negative Linear regression Determine the best t line Making predictions m1 = - 1 m2 e m1 Z 0 m2 Z 0 L1 L2 rise run x1 Z x2m = y2 - y1 x2 - x1 B Z 0A Z 0 (xm, ym) = a x1 + x2 2 , y1 + y2 2 b d = 2(x2 - x1)2 + (y2 - y1)2 260 C HAP TE R 2 R EVI EW Linearity Linear Nonlinear Correlation coefcient, r CHAPTERREVIEW c02b.qxd 6/6/12 7:26 PM Page 260 290. C HAP TE R 2 R EVI EW E X E R C I S E S 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint Plot each point and indicate which quadrant the point lies in. 1. (4, 2) 2. (4, 7) 3. (1, 6) 4. (2, 1) Calculate the distance between the two points. 5. (2, 0) and (4, 3) 6. (1, 4) and (4, 4) 7. (4, 6) and (2, 7) 8. and Calculate the midpoint of the segment joining the two points. 9. (2, 4) and (3, 8) 10. (2, 6) and (5, 7) 11. (2.3, 3.4) and (5.4, 7.2) 12. (a, 2) and (a, 4) Applications 13. Sports. A quarterback drops back to pass. At the point (5, 20) he throws the ball to his wide receiver located at (10, 30). Find the distance the ball has traveled. Assume the width of the football eld is [15, 15] and the length is [50, 50]. Units of measure are yards. 14. Sports. Suppose that in the above exercise a defender was midway between the quarterback and the receiver. At what point was the defender located when the ball was thrown over his head? 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry Find the x-intercept(s) and y-intercept(s) if any. 15. x2 4y2 4 16. y x2 x 2 17. 18. y = x2 - x - 12 x - 12 y = 2x2 - 9 A1 3, -7 3 BA1 4, 1 12 B Use algebraic tests to determine symmetry with respect to the x-axis, y-axis, or origin. 19. x2 y3 4 20. y x2 2 21. xy 4 22. y2 5 x Use symmetry as a graphing aid and point-plot the given equations. 23. y x2 3 24. y x 4 25. 26. x y2 2 27. 28. x2 y2 36 Applications 29. Sports. A track around a high school football eld is in the shape of the graph 8x2 y2 8. Graph using symmetry and by plotting points. 30. Transportation. A bypass around a town follows the graph y x3 2, where the origin is the center of town. Graph the equation. 2.3 Lines Express the equation for each line in slopeintercept form. Identify the slope and y-intercept of each line. 31. 6x 2y 12 32. 3x 4y 9 33. 34. Find the x- and y-intercepts and the slope of each line if they exist and graph. 35. y 4x 5 36. 37. x y 4 38. x 4 39. y 2 40. -1 2x - 1 2 y = 3 y = - 3 4x - 3 -2 3x - 1 4 y = 1 8 -1 2x - 1 3 y = 1 6 y = x29 - x2 y = 3 2x REVIEWEXERCISES 261 c02b.qxd 12/23/11 5:53 PM Page 261 291. 58. Budget: Car Repair. The cost of having the air conditioner in your car repaired is the combination of material costs and labor costs. The materials (tubing, coolant, etc.) are $250, and the labor costs $38 per hour. Write an equation that models the total cost C of having your air conditioner repaired as a function of hours t. Graph this equation with t as the horizontal axis and C representing the vertical axis. How much will the job cost if the mechanic works 1.5 hours? 2.4 Circles Write the equation of the circle in standard form. 59. center (2, 3) r 6 60. center (6, 8) 61. 62. center (1.2, 2.4) r 3.6 Find the center and the radius of the circle given by the equation. 63. (x 2)2 (y 3)2 81 64. (x 4)2 (y 2)2 32 65. 66. x2 y2 4x 2y 0 67. x2 y2 2y 4x 11 0 68. 3x2 3y2 6x 7 0 69. 9x2 9y2 6x 12y 76 0 70. x2 y2 3.2x 6.6y 2.4 0 71. Find the equation of a circle centered at (2, 7) and passing through (3, 6). 72. Find the equation of a circle that has the diameter with endpoints (2, 1) and (5, 5). Technology Exercises Section 2.1 Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 73. (10, 5), (20, 45), (10, 10) 74. (4.2, 8.4), (4.2, 2.1), (6.3, 10.5) Ax + 3 4 B 2 + Ay - 1 2 B 2 = 16 36 r = 2 5 center A3 4, 5 2 B r = 326 PRETEST POSTTEST 1020 1324 950 1240 Write the equation of the line, given the slope and the intercepts. 41. Slope: m 4 y-intercept: (0, 3) 42. Slope: m 0 y-intercept: (0, 4) 43. Slope: m is undened x-intercept: (3, 0) 44. Slope: y-intercept: Write an equation of the line, given the slope and a point that lies on the line. 45. m 2 (3, 4) 46. (2, 16) 47. m 0 (4, 6) 48. m is undened (2, 5) Write the equation of the line that passes through the given points. Express the equation in slopeintercept form or in the form of x a or y b. 49. (4, 2) and (2, 3) 50. (1, 4) and (2, 5) 51. 52. (3, 2) and (9, 2) Find the equation of the line that passes through the given point and also satises the additional piece of information. 53. (2, 1) parallel to the line 2x 3y 6 54. (5, 6) perpendicular to the line 5x 3y 0 55. perpendicular to the line 56. (a 2, b 1) parallel to the line Ax By C Applications 57. Grades. For a GRE prep class, a student must take a pretest and then a posttest after the completion of the course. Two students results are shown below. Give a linear equation to represent the given data. 2 3x - 1 2y = 12A- 3 4, 5 2 B A-3 4, 1 2 B and A-7 4, 5 2 B m = 3 4 A0, 3 4 B m = -2 3 REVIEWEXERCISESREVIEWEXERCISES 262 C HAP TE R 2 Graphs c02b.qxd 11/22/11 9:16 PM Page 262 292. Review Exercises 263 Section 2.2 Graph the equation using a graphing utility and state whether there is any symmetry. 75. 76. Section 2.3 Determine whether the lines are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a graphing utility to conrm your answer. 77. 78. y2 = 5 6 - 9 20x y1 = -0.45x - 2.1 y2 = -8 7x - 9 14 y1 = 0.875x + 1.5 0.8x2 - 1.5y2 = 4.8 y2 = x2 - 4 Section 2.4 79. Use the Quadratic Formula to solve for y, and use a graphing utility to graph each equation. Do the graphs agree with the graph in Exercise 69? 80. Use the Quadratic Formula to solve for y, and use a graphing utility to graph each equation. Do the graphs agree with the graph in Exercise 70? x2 + y2 + 3.2x - 6.6y - 2.4 = 0 9x2 + 9y2 - 6x + 12y - 76 = 0 REVIEWEXERCISES c02b.qxd 11/22/11 9:16 PM Page 263 293. 264 C HAP TE R 2 Graphs 1.1 Linear Equations 264 1. Find the distance between the points (7, 3) and (2, 2). 2. Find the midpoint between (3, 5) and (5, 1). 3. Determine the length and the midpoint of a segment that joins the points (2, 4) and (3, 6). 4. Research Triangle. The Research Triangle in North Carolina was established as a collaborative research center among Duke University (Durham), North Carolina State University (Raleigh), and the University of North Carolina (Chapel Hill). Durham is 10 miles north and 8 miles east of Chapel Hill, and Raleigh is 28 miles east and 15 miles south of Chapel Hill. What is the perimeter of the research triangle? Round your answer to the nearest mile. 5. Determine the two values for y so that the point (3, y) is 5 units away from the point (6, 5). 6. If the point (3, 4) is on a graph that is symmetric with respect to the y-axis, what point must also be on the graph? 7. Determine whether the graph of the equation x y2 5 has any symmetry (x-axis, y-axis, and origin). 8. Find the x-intercept(s) and the y-intercept(s), if any: 4x2 9y2 36. Graph the following equations. 9. 2x2 y2 8 10. 11. Find the x-intercept and the y-intercept of the line x 3y 6. y = 4 x2 + 1 Raleigh Durham Chapel Hill Gorman Cary New Hope Creedmoor Hillsborough 12. Express the line in slopeintercept form: 4x 6y 12. 13. Express the line in slopeintercept form: . Find the equation of the line that is characterized by the given information. Graph the line. 14. Slope 4; y-intercept (0, 3) 15. Passes through the points (3, 2) and (4, 9) 16. Parallel to the line y 4x 3 and passes through the point (1, 7) 17. Perpendicular to the line 2x 4y 5 and passes through the point (1, 1) 18. x-intercept (3, 0); y-intercept (0, 6) For Exercises 19 and 20, write the equation of the line that corresponds to the graph. 19. 20. 21. Write the equation of a circle that has center (6, 7) and radius r 8. 22. Determine the center and radius of the circle x2 y2 10x 6y 22 0. 23. Find the equation of the circle that is centered at (4, 9) and passes through the point (2, 5). 24. Solar System. Earth is approximately 93 million miles from the Sun. Approximating Earths orbit around the Sun as circular, write an equation governing Earths path around the Sun. Locate the Sun at the origin. 25. Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 26. Graph the given equation using a graphing utility and state whether there is any symmetry. 0.25y2 + 0.04x2 = 1 (-8.4, 16.8), (0, 37.8), (12.6, 8.4) x y (2, 1) (2, 3) x y (1, 4) (2, 2) 2 3 x - 1 4 y = 2 C HAP TE R 2 P R ACTI C E TE ST PRACTICETEST 264 c02b.qxd 11/22/11 9:16 PM Page 264 294. 1.1 Linear Equations 265265 C HAP TE R 2 Graphs 1. Simplify . 7 - 2 7 + 3 16. Solve using substitution: . Solve and express the solution in interval notation: 17. 18. x2 x 20 19. 2 x 4 20. Solve for x: 5 4x 23. 21. Use algebraic tests to determine whether the graph of the equation is symmetric with respect to the x-axis, y-axis, or origin. 22. Write an equation of a line in slopeintercept form with slope that passes through the point (5, 1). 23. Write an equation of a line that is perpendicular to the x-axis and passes through the point (5, 3). 24. Write an equation of a line in slopeintercept form that passes through the two points and . 25. Find the center and radius of the circle: (x 5)2 (y 3)2 30. 26. Calculate the distance between the two points A- 111, 5B A-6 7, -2 3 BA1 7, 5 3 B m = 4 5 y = 4x 6 6 1 4x + 6 6 9 1 (x + 2)2 - 5 x + 2 + 4 = 0 2. Simplify and express in terms of positive exponents: . 3. Perform the operation and simplify: (x 4)2 (x 4)2 . 4. Factor completely 8x3 27y3 . 5. Perform the operations and simplify: . 6. Solve for x: x3 5x2 4x 20 0. 7. Perform the operations and write in standard form: . Solve for x. 8. 15 [5 3x 4(2x 6)] 4(6x 7) [5(3x 7) 6x 10] 9. 10. Ashley inherited $17,000. She invested some money in a CD that earns 5% and the rest in a stock that earns 8%. How much was invested in each account, if the interest for the rst year is $1075? 11. Solve by factoring: 5x2 45. 12. Solve by completing the square: 3x2 6x 7. 13. Use the discriminant to determine the number and type of roots: 5x2 2x 7 0. 14. Solve for r: p2 q2 r2 . 15. Solve and check: .2x2 + 3x - 10 = x - 2 5 4x + 1 = 3 4x - 1 2-36(5 - 2i) 1/x - 1/5 1/x + 1/5 A 5x3/4 B 4 25x-1/4 and , and nd the midpoint of the segment joining the two points. Round your answers to one decimal place. 27. Determine whether the lines andy1 = 0.32x + 1.5 A2, 17B are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a graphing utility to conrm your answer. y2 = - 5 16x + 1 4 C HAP TE R S 12 C U M U LATIVE TE ST CUMULATIVETEST 265 c02b.qxd 11/22/11 9:16 PM Page 265 295. Functions and Their Graphs On a sales rack of clothes at a department store, you see a shirt you like. The original price of the shirt was $100, but it has been discounted 30%. As a preferred shopper, you get an automatic additional 20% off the sale price at the register. How much will you pay for the shirt? Nave shoppers might be lured into thinking this shirt will cost $50 because they add the 20% and 30% to get 50% off, but they will end up paying more than that. Experienced shoppers know that they first take 30% off of $100, which results in a price of $70, and then they take an additional 20% off of the sale price, $70, which results in a final discounted price of $56. Experienced shoppers have already learned composition of functions. A composition of functions can be thought of as a function of a function. One function takes an input (original price, $100) and maps it to an output (sale price, $70), and then another function takes that output as its input (sale price, $70) and maps that to an output (checkout price, $56). 3 JohnGiustina/Superstock c03a.qxd 11/14/12 9:47 PM Page 266 296. I N TH I S C HAP TE R you will find that functions are part of our everyday thinking: converting from degrees Celsius to degrees Fahrenheit, DNA testing in forensic science, determining stock values, and the sale price of a shirt. We will develop a more complete, thorough understanding of functions. First, we will establish what a relation is, and then we will determine whether a relation is a function. We will discuss common functions, domain and range of functions, and graphs of functions. We will determine whether a function is increasing or decreasing on an interval and calculate the average rate of change of a function. We will perform operations on functions and composition of functions. We will discuss one-to-one functions and inverse functions. Finally, we will model applications with functions using variation. 267 Recognizing and Classifying Functions Increasing and Decreasing Functions Average Rate of Change Piecewise- Defined Functions Relations and Functions Functions Defined by Equations Function Notation Domain of a Function Horizontal and Vertical Shifts Reflection about the Axes Stretching and Compressing Adding, Subtracting, Multiplying, and Dividing Functions Composition of Functions Determine Whether a Function Is One-to-One Inverse Functions Graphical Interpretation of Inverse Functions Finding the Inverse Function Direct Variation Inverse Variation Joint Variation and Combined Variation FUNCTIONS AND THEIR GRAPHS L E A R N I N G O B J E C T I V E S Find the domain and range of a function. Sketch the graphs of common functions. Sketch graphs of general functions employing translations of common functions. Perform composition of functions. Find the inverse of a function. Model applications with functions using variation. 3.1 Functions 3.2 Graphs of Functions; Piecewise- Defined Functions; Increasing and Decreasing Functions; Average Rate of Change 3.3 Graphing Techniques: Transformations 3.4 Operations on Functions and Composition of Functions 3.5 One-to-One Functions and Inverse Functions 3.6 Modeling Functions Using Variation c03a.qxd 1/18/12 1:11 PM Page 267 297. Relations and Functions What do the following pairs have in common? Every person has a blood type. Temperature is some specic value at a particular time of day. Every working household phone in the United States has a 10-digit phone number. First-class postage rates correspond to the weight of a letter. Certain times of the day are start times of sporting events at a university. They all describe a particular correspondence between two groups. A relation is a correspondence between two sets. The rst set is called the domain, and the corresponding second set is called the range. Members of these sets are called elements. A relation is a set of ordered pairs. The domain is the set of all the rst components of the ordered pairs, and the range is the set of all the second components of the ordered pairs. C O N C E P TUAL O BJ E CTIVE S Think of function notation as a placeholder or mapping. Understand that all functions are relations but not all relations are functions. F U N CTI O N S S K I LLS O BJ E CTIVE S Determine whether a relation is a function. Determine whether an equation represents a function. Use function notation. Find the value of a function. Determine the domain and range of a function. S E CTI O N 3.1 WORDS MATH The domain is the set of all the {Michael, Tania, Dylan, Trevor, Megan} rst components. The range is the set of all the {A, AB, O} second components. A relation in which each element in the domain corresponds to exactly one element in the range is a function. PERSON BLOOD TYPE ORDERED PAIR Michael A (Michael, A) Tania A (Tania, A) Dylan AB (Dylan, AB) Trevor O (Trevor, O) Megan O (Megan, O) A relation is a correspondence between two sets where each element in the rst set, called the domain, corresponds to at least one element in the second set, called the range. D E F I N I T I O N Relation 268 c03a.qxd 11/24/11 4:52 PM Page 268 298. 3.1 Functions 269 Note that the denition of a function is more restrictive than the denition of a relation. For a relation, each input corresponds to at least one output, whereas, for a function, each input corresponds to exactly one output. The blood-type example given is both a relation and a function. Also note that the range (set of values to which the elements of the domain correspond) is a subset of the set of all blood types. However, although all functions are relations, not all relations are functions. For example, at a university, four primary sports typically overlap in the late fall: football, volleyball, soccer, and basketball. On a given Saturday, the following table indicates the start times for the competitions. WORDS MATH The 1:00 start time corresponds (1:00 P.M., Football) to exactly one event, Football. The 2:00 start time corresponds (2:00 P.M., Volleyball) to exactly one event, Volleyball. The 7:00 start time corresponds (7:00 P.M., Soccer) to two events, Soccer and Basketball. (7:00 P.M., Basketball) Because an element in the domain, 7:00 P.M., corresponds to more than one element in the range, Soccer and Basketball, this is not a function. It is, however, a relation. Domain Range Not a Function Football Volleyball Soccer Basketball START TIME ATHLETIC EVENT 7:00 P.M. 1:00 P.M. 2:00 P.M. Domain PEOPLE Michael Megan Dylan Trevor Tania Range BLOOD TYPE Function A O AB B A function is a correspondence between two sets where each element in the rst set, called the domain, corresponds to exactly one element in the second set, called the range. D E F I N I T I O N Function TIME OF DAY COMPETITION 1:00 P.M. Football 2:00 P.M. Volleyball 7:00 P.M. Soccer 7:00 P.M. Basketball Study Tip All functions are relations but not all relations are functions. E X AM P LE 1 Determining Whether a Relation Is a Function Determine whether the following relations are functions. a. {(3, 4), (2, 4), (3, 5), (6, 4)} b. {(3, 4), (2, 4), (3, 5), (2, 2)} c. Domain Set of all items for sale in a grocery store; Range Price Solution: a. No x-value is repeated. Therefore, each x-value corresponds to exactly one y-value. This relation is a function. b. The value x 2 corresponds to both y 2 and y 4. This relation is not a function. c. Each item in the grocery store corresponds to exactly one price. This relation is a function. YO U R TU R N Determine whether the following relations are functions. a. {(1, 2), (3, 2), (5, 6), (7, 6)} b. {(1, 2), (1, 3), (5, 6), (7, 8)} c. {(11:00 A.M., 83F), (2:00 P.M., 89F), (6:00 P.M., 85F)} Answer: a. function b. not a function c. function c03a.qxd 11/24/11 4:52 PM Page 269 299. 270 C HAP TE R 3 Functions and Their Graphs C A U T I O N Not all equations are functions. All of the examples we have discussed thus far are discrete sets in that they represent a countable set of distinct pairs of (x, y). A function can also be dened algebraically by an equation. Functions Defined by Equations Lets start with the equation y x2 3x, where x can be any real number. This equation assigns to each x-value exactly one corresponding y-value. x y x2 3x y 1 y (1)2 3(1) 2 5 y (5)2 3(5) 10 1.2 y (1.2)2 3(1.2) 2.16 22 9y = A-2 3 B 2 - 3A-2 3 B-2 3 Since the variable y depends on what value of x is selected, we denote y as the dependent variable. The variable x can be any number in the domain; therefore, we denote x as the independent variable. Although functions are dened by equations, it is important to recognize that not all equations are functions. The requirement for an equation to dene a function is that each element in the domain corresponds to exactly one element in the range. Throughout the ensuing discussion, we assume x to be the independent variable and y to be the dependent variable. Equations that represent functions of x: y x2 y x y x3 Equations that do not represent functions of x: x y2 x2 y2 1 x y In the equations that represent functions of x, every x-value corresponds to exactly one y-value. Some ordered pairs that correspond to these functions are y x2 : (1, 1) (0, 0) (1, 1) y x : (1, 1) (0, 0) (1, 1) y x3 : (1, 1) (0, 0) (1, 1) The fact that x 1 and x 1 both correspond to y 1 in the rst two examples does not violate the denition of a function. In the equations that do not represent functions of x, some x-values correspond to more than one y-value. Some ordered pairs that correspond to these equations are Study Tip We say that x y2 is not a function of x. However, if we reverse the independent and dependent variables, then x y2 is a function of y. RELATION SOLVE RELATION FOR Y POINTS THAT LIE ON THE GRAPH x y2 (1, 1) (0, 0) (1, 1) x 1 maps to both y 1 and y 1 x2 y2 1 (0, 1) (0, 1) (1, 0) (1, 0) x 0 maps to both y 1 and y 1 x y (1, 1) (0, 0) (1, 1) x 1 maps to both y 1 and y 1y = ;x y = ; 21 - x2 y = ; 1x c03a.qxd 8/22/12 8:39 AM Page 270 300. Lets look at the graphs of the three functions of x: x y y = x2 y = |x| y = x3 x y x y x y x y x y x = y2 x2 + y2 = 1 x = |y| Lets take any value for x, say x a. The graph of x a corresponds to a vertical line. A function of x maps each x-value to exactly one y-value; therefore, there should be at most one point of intersection with any vertical line. We see in the three graphs of the functions above that if a vertical line is drawn at any value of x on any of the three graphs, the vertical line only intersects the graph in one place. Look at the graphs of the three equations that do not represent functions of x. 3.1 Functions 271 A vertical line can be drawn on any of the three graphs such that the vertical line will intersect each of these graphs at two points. Thus, there are two y-values that correspond to some x-value in the domain, which is why these equations do not dene y as a function of x. Study Tip If any x-value corresponds to more than one y-value, then y is not a function of x. Given the graph of an equation, if any vertical line that can be drawn intersects the graph at no more than one point, the equation denes y as a function of x. This test is called the vertical line test. D E F I N I T I O N Vertical Line Test c03a.qxd 11/24/11 4:52 PM Page 271 301. To recap, a function can be expressed one of four ways: verbally, numerically, algebraically, and graphically. This is sometimes called the Rule of 4. Expressing a Function VERBALLY NUMERICALLY ALGEBRAICALLY GRAPHICALLY Every real number has a corresponding {(3, 3), (1, 1), (0, 0), (1, 1), (5, 5)} y x absolute value. 272 C HAP TE R 3 Functions and Their Graphs E X AM P LE 2 Using the Vertical Line Test Use the vertical line test to determine whether the graphs of equations dene functions of x. a. b. x y x y Solution: Apply the vertical line test. a. b. a. Because the vertical line intersects the graph of the equation at two points, this equation does not represent a function . b. Because any vertical line will intersect the graph of this equation at no more than one point, this equation represents a function . YO U R TU R N Determine whether the equation (x 3)2 (y 2)2 16 is a function of x. x y x y Answer: The graph of the equation is a circle, which does not pass the vertical line test. Therefore, the equation does not dene a function. x y c03a.qxd 11/24/11 4:52 PM Page 272 302. x (x) 2x 5 (x) 0 (0) 2(0) 5 (0) 5 1 (1) 2(1) 5 (1) 7 2 (2) 2(2) 5 (2) 9 INPUT FUNCTION OUTPUT EQUATION x (x) (x) 2x 5 Independent Mapping Dependent Mathematical variable variable rule Function Notation We know that the equation y 2x 5 denes y as a function of x because its graph is a nonvertical line and thus passes the vertical line test. We can select x-values (input) and determine unique corresponding y-values (output). The output is found by taking 2 times the input and then adding 5. If we give the function a name, say, , then we can use function notation: The symbol (x) is read evaluated at x or of x and represents the y-value that corresponds to a particular x-value. In other words, y (x). f(x) = 2x + 5 It is important to note that is the function name, whereas (x) is the value of the function. In other words, the function maps some value x in the domain to some value in the range. f(x) 3.1 Functions 273 Domain Range Function x 2x 5 f(0) 5 f(1) 7 f(2) 9 x 1 x 0 x 2 f f f The independent variable is also referred to as the argument of a function. To evaluate functions, it is often useful to think of the independent variable or argument as a placeholder. For example, (x) x2 3x can be thought of as f( ) ( )2 3( ) In other words, of the argument is equal to the argument squared minus 3 times the argument. Any expression can be substituted for the argument: It is important to note: (x) does not mean f times x. The most common function names are and F since the word function begins with an f. Other common function names are g and G, but any letter can be used. The letter most commonly used for the independent variable is x. The letter t is also common because in real-world applications it represents time, but any letter can be used. Although we can think of y and (x) as interchangeable, the function notation is useful when we want to consider two or more functions of the same independent variable. f(-x) = (-x)2 - 3(-x) f(x + 1) = (x + 1)2 - 3(x + 1) f(1) = (1)2 - 3(1) Study Tip It is important to note that (x) does not mean times x. c03a.qxd 11/24/11 4:53 PM Page 273 303. YO U R TU R N For the following graph of a function, nd: a. (1) b. (0) c. 3(2) d. the value of x that corresponds to (x) 0 x (0, 1) (1, 2) (1, 0) (2, 7) (2, 9) y 5 5 10 10 274 C HAP TE R 3 Functions and Their Graphs E X AM P LE 3 Evaluating Functions by Substitution Given the function (x) 2x3 3x2 6, nd (1). Solution: Consider the independent variable x to be a placeholder. ( ) 2( )3 3( )2 6 To nd (1), substitute x 1 into the function. (1) 2(1)3 3(1)2 6 Evaluate the right side. (1) 2 3 6 Simplify. (1) 1 E X AM P LE 4 Finding Function Values from the Graph of a Function The graph of is given on the right. a. Find (0). b. Find (1). c. Find (2). d. Find 4(3). e. Find x such that (x) 10. f. Find x such that (x) 2. Solution (a): The value x 0 corresponds to the value y 5. (0) 5 Solution (b): The value x 1 corresponds to the value y 2. (1) 2 Solution (c): The value x 2 corresponds to the value y 1. (2) 1 Solution (d): The value x 3 corresponds to the value y 2. 4(3) 4 . 2 8 Solution (e): The value y 10 corresponds to the value x 5 . Solution (f): The value y 2 corresponds to the values x 1 and x 3 . Answer: a. (1) 2 b. (0) 1 c. 3(2) 21 d. x 1 x (1, 2) (2, 1) (3, 2) (4, 5)(0, 5) (5, 10) y 5 10 c03a.qxd 11/24/11 4:53 PM Page 274 304. I N C O R R E CT The ERROR is in interpreting the notation as a sum. Z x2 - 3x - 2 f(x + 1) Z f(x) + f(1) C O R R E CT Write the original function. Replace the argument x with a placeholder. f( ) ( )2 3( ) Substitute x 1 for the argument. Eliminate the parentheses. Combine like terms. f(x + 1) = x2 - x - 2 f(x + 1) = x2 + 2x + 1 - 3x - 3 f(x + 1) = (x + 1)2 - 3(x + 1) f(x) = x2 - 3x A common misunderstanding is to interpret the notation (x 1) as a sum: f(x + 1) Z f(x) + f(1). C O M M O N M I S TA K E C A U T I O N f(x + 1) Z f(x) + f(1) YO U R TU R N For the given function g(x) x2 2x 3, evaluate g(x 1). E X AM P LE 6 Evaluating Functions: Sums For the given function H(x) x2 2x, evaluate: a. H(x 1) b. H(x) H(1) Solution (a): Write the function H in placeholder notation. H( ) ( )2 2( ) Substitute x 1 for the argument of H. H(x 1) (x 1) 2 2(x 1) Eliminate the parentheses on the right side. H(x 1) x2 2x 1 + 2x 2 Combine like terms on the right side. Solution (b): Write H(x). H(x) x2 2x Evaluate H at x 1. H(1) (1)2 2(1) 3 Evaluate the sum H(x) H(1). H(x) H(1) x2 2x 3 Note: Comparing the results of part (a) and part (b), we see that H(x 1) H(x) H(1). H(x) + H(1) = x2 + 2x + 3 H(x + 1) = x2 + 4x + 3 3.1 Functions 275 E X AM P LE 5 Evaluating Functions with Variable Arguments (Inputs) For the given function (x) x2 3x, evaluate (x 1) and simplify if possible. Answer: g(x - 1) = x2 - 4x + 6 Technology Tip Use a graphing utility to display graphs of y1 H(x 1) (x 1)2 2(x 1) and y2 H(x) H(1) x2 2x 3. The graphs are not the same. c03a.qxd 11/24/11 4:53 PM Page 275 305. C A U T I O N f a a b b Z f (a) f (b) 276 C HAP TE R 3 Functions and Their Graphs E X AM P LE 7 Evaluating Functions: Negatives For the given function G(t) t2 t, evaluate: a. G(t) b. G(t) Solution (a): Write the function G in placeholder notation. G( ) ( )2 ( ) Substitute t for the argument of G. G(t) (t)2 (t) Eliminate the parentheses on the right side. Solution (b): Write G(t). G(t) t2 t Multiply by 1. G(t) (t2 t) Eliminate the parentheses on the right side. Note: Comparing the results of part (a) and part (b), we see that E X AM P LE 8 Evaluating Functions: Quotients For the given function F(x) 3x 5, evaluate: a. b. Solution (a): Write F in placeholder notation. F( ) 3( ) 5 Replace the argument with . Simplify the right side. Solution (b): Evaluate F(1). F(1) 3(1) 5 8 Evaluate F(2). F(2) 3(2) 5 11 Divide F(1) by F(2). Note: Comparing the results of part (a) and part (b), we see that YO U R TU R N Given the function G(t) 3t 4, evaluate: a. G(t 2) b. G(t) G(2) c. d. Examples 6, 7, and 8 illustrate the following: f(a + b) Z f(a) + f (b) f(-t) Z -f(t) f a a b b Z f(a) f(b) Ga 1 3 b G(1) G(3) F a 1 2 b F(1) F(2) . F(1) F(2) = 8 11 Fa 1 2 b = 13 2 Fa 1 2 b = 3a 1 2 b + 51 2 F(1) F(2) Fa 1 2 b G(t) G(t). -G(t) = -t2 + t G(-t) = t2 + t Technology Tip Use a graphing utility to display graphs of y1 G(x) (x)2 (x) and y2 G(x) (x2 x). The graphs are not the same. Answer: a. G(t 2) 3t 10 b. G(t) G(2) 3t 6 c. d. Ga 1 3 b = -3 - 1 5 G(1) G(3) c03a.qxd 11/24/11 7:34 PM Page 276 306. Now that we have shown that , we turn our attention to one of the fundamental expressions in calculus: the difference quotient. Example 9 illustrates the difference quotient, which will be discussed in detail in Section 3.2. For now, we will concentrate on the algebra involved when nding the difference quotient. In Section 3.2, the application of the difference quotient will be the emphasis. E X AM P LE 9 Evaluating the Difference Quotient For the function nd Solution: Use placeholder notation for the function (x) x2 x. ( ) ( )2 ( ) Calculate (x h). (x h) (x h)2 (x h) Write the difference quotient. Let (x h) (x h)2 (x h) and (x) x2 x. Eliminate the parentheses inside the rst set of brackets. Eliminate the brackets in the numerator. Combine like terms. Factor the numerator. Divide out the common factor, h. 2x h 1 YO U R TU R N Evaluate the difference quotient for (x) x2 1. Domain of a Function Sometimes the domain of a function is stated explicitly. For example, f (x) x x 0 Here, the explicit domain is the set of all negative real numbers, . Every negative real number in the domain is mapped to a positive real number in the range through the absolute value function. (-, 0) h Z 0 = h(2x + h - 1) h = 2xh + h2 - h h = x2 + 2xh + h2 - x - h - x2 + x h = [x2 + 2xh + h2 - x - h] - [x2 - x] h h Z 0= C(x + h)2 - (x + h)D - Cx2 - xD h f(x + h) - f(x) h f(x + h) - f(x) h f(x + h) - f(x) h , h Z 0.f(x) = x2 - x, f(x + h) - f(x) h h Z 0 f(x + h) Z f(x) + f(h) 3.1 Functions 277 f(x+h) f(x) u r domain r Domain (, 0) f(x) |x| Range (0, ) 1 7 4 1 7 4 Answer: 2x + h c03a.qxd 11/24/11 4:53 PM Page 277 307. 278 C HAP TE R 3 Functions and Their Graphs The excluded x-values are 5 and 5. If the expression that denes the function is given but the domain is not stated explicitly, then the domain is implied. The implicit domain is the largest set of real numbers for which the function is dened and the output value (x) is a real number. For example, does not have the domain explicitly stated. There is, however, an implicit domain. Note that if the argument is negative, that is, if x 0, then the result is an imaginary number. In order for the output of the function, (x), to be a real number, we must restrict the domain to nonnegative numbers, that is, if x 0. f(x) = 1x In general, we ask the question, what can x be? The implicit domain of a function excludes values that cause a function to be undened or have outputs that are not real numbers. FUNCTION IMPLICIT DOMAIN [0, )f(x) = 1x EXPRESSION THAT DEFINES THE FUNCTION EXCLUDED X-VALUES EXAMPLE IMPLICIT DOMAIN Polynomial None (x) x3 4x2 All real numbers Rational x-values that make the denominator x Z ; 3 or (-, -3)(-3, 3)(3, ) g(x) = 2 x2 - 9 E X AM P LE 10 Determining the Domain of a Function State the domain of the given functions. a. b. c. Solution (a): Write the original equation. Determine any restrictions on the values of x. Solve the restriction equation. State the domain restrictions. Write the domain in interval notation. (-, -5)(-5, 5)(5, ) x Z ;5 x2 Z 25 or x Z ; 125 = ;5 x2 - 25 Z 0 F(x) = 3 x2 - 25 G(x) = 3 1x - 1H(x) = 4 19 - 2xF(x) = 3 x2 - 25 equal to 0 Radical x-values that result in a square (even) root of a negative number x 5 or [5, )h(x) = 1x - 5 Technology Tip To visualize the domain of each function, ask the question: What are the excluded x-values in the graph? a. Graph of F(x) is shown.= 3 x2 - 25 c03a.qxd 11/24/11 4:53 PM Page 278 308. 3.1 Functions 279 Solution (b): Write the original equation. Determine any restrictions on the values of x. 9 2x 0 Solve the restriction inequality. 9 2x State the domain restrictions. Write the domain in interval notation. Solution (c): Write the original equation. Determine any restrictions on the values of x. no restrictions State the domain. R Write the domain in interval notation. YO U R TU R N State the domain of the given functions. a. b. g(x) = 1 x2 - 4 f(x) = 1x - 3 (-, ) G(x) = 3 1x - 1 a-, 9 2 d x 9 2 H(x) = 4 19 - 2x Answer: a. x 3 or b. (-2, 2)(2, )(-, -2) x Z ;2 or [3, ) Applications Functions that are used in applications often have restrictions on the domains due to physical constraints. For example, the volume of a cube is given by the function V(x) x3 , where x is the length of a side. The function (x) x3 has no restrictions on x, and therefore the domain is the set of all real numbers. However, the volume of any cube has the restriction that the length of a side can never be negative or zero. E X AM P LE 11 Price of Gasoline Following the capture of Saddam Hussein in Iraq in 2003, gas prices in the United States escalated and then nally returned to their precapture prices. Over a 6-month period, the average price of a gallon of 87 octane gasoline was given by the function C(x) 0.05x2 0.3x 1.7, where C is the cost function and x represents the number of months after the capture. a. Determine the domain of the cost function. b. What was the average price of gas per gallon 3 months after the capture? Solution (a): Since the cost function C(x) 0.05x2 0.3x 1.7 modeled the price of gas only for 6 months after the capture, the domain is 0 x 6 or . Solution (b): Write the cost function. C(x) 0.05x2 0.3x 1.7 0 x 6 Find the value of the function when x 3. C(3) 0.05(3)2 0.3(3) 1.7 Simplify. The average price per gallon 3 months after the capture was .$2.15 C(3) = 2.15 [0, 6] c03a.qxd 11/24/11 6:02 PM Page 279 309. E X AM P LE 12 The Dimensions of a Pool Express the volume of a 30 ft 10 ft rectangular swimming pool as a function of its depth. Solution: The volume of any rectangular box is V lwh, where V is the volume, l is the length, w is the width, and h is the height. In this example, the length is 30 ft, the width is 10 ft, and the height represents the depth d of the pool. Write the volume as a function of depth d. Simplify. Determine any restrictions on the domain. d 0 V(d) = 300d V(d) = (30)(10)d 280 C HAP TE R 3 Functions and Their Graphs S U M MARY S E CTI O N 3.1 Relations and Functions (Let x represent the independent variable and y the dependent variable.) All functions are relations, but not all relations are functions. Functions can be represented by equations. In the following table, each column illustrates an alternative notation. TYPE MAPPING/CORRESPONDENCE EQUATION GRAPH Relation Every x-value in the domain maps to at least one y-value in the range. x y2 Function Every x-value in the domain maps to exactly one y-value in the range. y x2 Passes vertical line test x y x y INPUT CORRESPONDENCE OUTPUT EQUATION x Function y y 2x 5 Independent Mapping Dependent Mathematical variable variable rule Argument (x) (x) 2x 5 The domain is the set of all inputs (x-values), and the range is the set of all corresponding outputs (y-values). Placeholder notation is useful when evaluating functions. f( ) 3( )2 2( ) Explicit domain is stated, whereas implicit domain is found by excluding x-values that: make the function undened (denominator 0). result in a nonreal output (even roots of negative real numbers). f (x) = 3x2 + 2x c03a.qxd 11/24/11 4:53 PM Page 280 310. In Exercises 124, determine whether each relation is a function. Assume that the coordinate pair (x, y) represents the independent variable x and the dependent variable y. 1. 2. 3. 4. 5. 6. 7. {(0, 3), (0, 3), (3, 0), (3, 0)} 8. {(2, 2), (2, 2), (5, 5), (5, 5)} 9. {(0, 0), (9, 3), (4, 2), (4, 2), (9, 3)} 10. {(0, 0), (1, 1), (2, 8), (1, 1), (2, 8)} 11. {(0, 1), (1, 0), (2, 1), (2, 1), (5, 4), (3, 4)} 12. {(0, 1), (1, 1), (2, 1), (3, 1)} 13. x2 y2 9 14. x y 15. x y2 16. y x3 17. y x 1 18. y 3 19. 20. 21. 22. 23. 24. x y x y x y x y x y x y (0, 5) (0, 5) (5, 0) (5, 0) Domain Range MATH SAT SCORE 500 650 Carrie Michael Jennifer Sean PERSON Domain Range COURSE GRADE A B Carrie Michael Jennifer Sean PERSON Domain Range Jordan Pat Chris Alex Morgan DATE THIS WEEKENDPERSON Domain Range START TIME 1:00 P.M. 4:00 P.M. 7:00 P.M. Bucs/Panthers Bears/Lions Falcons/Saints Rams/Seahawks Packers/Vikings NFL GAME Domain Range Mary Jason Chester PERSON (202) 5551212 (307) 1234567 (878) 7996504 10-DIGIT PHONE # Domain Range 78F 68F October January April MONTH AVERAGE TEMPERATURE 3.1 Functions 281 E X E R C I S E S S E CTI O N 3.1 S K I LL S c03a.qxd 12/27/11 12:25 PM Page 281 311. 282 C HAP TE R 3 Functions and Their Graphs In Exercises 2532, use the given graphs to evaluate the functions. 25. y (x) 26. y g(x) 27. y p(x) 28. y r(x) 29. y C(x) 30. y q(x) 31. y S(x) 32. y T(x) 33. Find x if f(x) 3 in Exercise 25. 34. Find x if g(x) 2 in Exercise 26. 35. Find x if p(x) 5 in Exercise 27. 36. Find x if C(x) 7 in Exercise 29. 37. Find x if C(x) 5 in Exercise 29. 38. Find x if q(x) 2 in Exercise 30. 39. Find x if S(x) 1 in Exercise 31. 40. Find x if T(x) 4 in Exercise 32. In Exercises 4156, evaluate the given quantities applying the following four functions. (x) 2x 3 F(t) 4 t2 g(t) 5 t G(x) x2 2x 7 41. (2) 42. G(3) 43. g(1) 44. F(1) 45. (2) g(1) 46. G(3) F(1) 47. 3f(2) 2g(1) 48. 2F(1) 2G(3) 49. 50. 51. 52. 53. (x 1) (x 1) 54. F(t 1) F(t 1) 55. g(x a) (x a) 56. G(x b) F(b) In Exercises 5764, evaluate the difference quotients using the same , F, G, and g given for Exercises 4156. 57. 58. 59. 60. 61. 62. 63. 64. G(-3 + h) - G(-3) h g(1 + h) - g(1) h F(-1 + h) - F(-1) h f(-2 + h) - f(-2) h G(x + h) - G(x) h g(t + h) - g(t) h F(t + h) - F(t) h f(x + h) - f(x) h G(0) - G(-3) F(-1) f(0) - f(-2) g(1) G(-3) F(-1) f(-2) g(1) a. (2) b. (0) c. (2) x (4, 5) (4, 5) y 5 5 10 x (2, 4) (6, 0) y 4 6 6 4 a. C(2) b. C(0) c. C(2) a. q(4) b. q(0) c. q(2) x (6, 1) (2, 5) y 4 6 8 2 x (4, 3) (5, 1) (3, 4) y 5 5 5 5 a. S(3) b. S(0) c. S(2) a. T(5) b. T(2) c. T(4) x(0, 1) (1, 3) (2, 5) (1, 1) (2, 3) y 5 5 5 5 x (0, 5) (3, 2) (5, 0) (3, 1) y 5 5 5 5 x (1, 5) (0, 2) (1, 3) (3, 5) (5, 7) y 7 3 3 7 a. g(3) b. g(0) c. g(5) a. p(1) b. p(0) c. p(1) x (3, 5) (7, 3) (4, 0) (1, 4) (6, 6) y 10 10 10 10 a. r(4) b. r(1) c. r(3) c03a.qxd 12/27/11 12:25 PM Page 282 312. 3.1 Functions 283 In Exercises 6596, nd the domain of the given function. Express the domain in interval notation. 65. (x) 2x 5 66. f(x) 2x 5 67. g(t) t2 3t 68. h(x) 3x4 1 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. (x) (x2 16) 1/2 92. g(x) (2x 5)1/3 93. r(x) x2 (3 2x)1/2 94. p(x) (x 1)2 (x2 9) 3/5 95. 96. 97. Let g(x) x2 2x 5 and nd the values of x that correspond to g(x) 3. 98. Let and nd the value of x that corresponds to . 99. Let (x) 2x(x 5)3 12(x 5)2 and nd the values of x that correspond to (x) 0. 100. Let (x) 3x(x 3)2 6(x 3)3 and nd the values of x that correspond to (x) 0. g(x) = 2 3g(x) = 5 6 x - 3 4 g(x) = 2 3x2 - 1 6x - 3 4f(x) = 2 5x - 2 4 f(t) = t - 3 4 2t2 + 9 H(t) = t 2t2 - t - 6 p(x) = x2 225 - x2 R(x) = x + 1 4 13 - 2x Q(x) = x 3 2x2 - 9 P(x) = 1 5 1x + 4 g(x) = 5 17 - 5xf(x) = 3 11 - 2xG(x) = 2 15 - x F(x) = 1 1x - 3 F(x) = 2x2 - 25G(t) = 2t2 - 4g(x) = 15 - 2xf(x) = 12x + 5 k(t) = 1t - 7q(x) = 17 - xG(t) = 2 t2 + 4 F(x) = 1 x2 + 1 R(x) = 1 x2 - 1 T(x) = 2 x2 - 4 Q(t) = 2 - t2 t + 3 P(x) = x + 5 x - 5 104. Falling Objects: Firecrackers. A recracker is launched straight up, and its height is a function of time, h(t) 16t2 128t, where h is the height in feet and t is the time in seconds with t 0 corresponding to the instant it launches. What is the height 4 seconds after launch? What is the domain of this function? 105. Collectibles. The price of a signed Alex Rodriguez baseball card is a function of how many are for sale. When Rodriguez was traded from the Texas Rangers to the New York Yankees in 2004, the going rate for a signed baseball card on eBay was , where x represents the number of signed cards for sale. What was the value of the card when there were 10 signed cards for sale? What was the value of the card when there were 100 signed cards for sale? 1400,000 - 100xP(x) = 10 + 101. Budget: Event Planning. The cost associated with a catered wedding reception is $45 per person for a reception for more than 75 people. Write the cost of the reception in terms of the number of guests and state any domain restrictions. 102. Budget: Long-Distance Calling. The cost of a local home phone plan is $35 for basic service and $.10 per minute for any domestic long-distance calls. Write the cost of monthly phone service in terms of the number of monthly long-distance minutes and state any domain restrictions. 103. Temperature. The average temperature in Tampa, Florida, in the springtime is given by the function T(x)0.7x2 16.8x 10.8, where T is the temperature in degrees Fahrenheit and x is the time of day in military time and is restricted to 6 x 18 (sunrise to sunset). What is the temperature at 6 A.M.? What is the temperature at noon? A P P L I C AT I O N S c03a.qxd 11/24/11 4:53 PM Page 283 313. 284 C HAP TE R 3 Functions and Their Graphs 106. Collectibles. In Exercise 105, what was the lowest price on eBay, and how many cards were available then? What was the highest price on eBay, and how many cards were available then? 107. Volume. An open box is constructed from a square 10-inch piece of cardboard by cutting squares of length x inches out of each corner and folding the sides up. Express the volume of the box as a function of x, and state the domain. 108. Volume. A cylindrical water basin will be built to harvest rainwater. The basin is limited in that the largest radius it can have is 10 feet. Write a function representing the volume of water V as a function of height h. How many additional gallons of water will be collected if you increase the height by 2 feet? Hint: 1 cubic foot 7.48 gallons. For Exercises 109110, refer to the following: The weekly exchange rate of the U.S. dollar to the Japanese yen is shown in the graph as varying over an 8-week period. Assume the exchange rate E(t) is a function of time (week); let E(1) be the exchange rate during Week 1. 109. Economics. Approximate the exchange rates of the U.S. dollar to the nearest yen during Weeks 4, 7, and 8. 110. Economics. Find the increase or decrease in the number of Japanese yen to the U.S. dollar exchange rate, to the nearest yen, from (a) Week 2 to Week 3 and (b) Week 6 to Week 7. For Exercises 111112, refer to the following: An epidemiological study of the spread of malaria in a rural area nds that the total number P of people who contracted malaria t days into an outbreak is modeled by the function 111. Medicine/Health. How many people have contracted malaria 14 days into the outbreak? 112. Medicine/Health. How many people have contracted malaria 6 days into the outbreak? P(t) = - 1 4 t2 + 7t + 180 1 t 14 Week 2 4 6 8 101 3 5 7 9 JapaneseYentoOneU.S.Dollar 90 89 88 87 86 85 84 83 82 t E 113. Environment: Tossing the Envelopes. The average American adult receives 24 pieces of mail per week, usually of some combination of ads and envelopes with windows. Suppose each of these adults throws away a dozen envelopes per week. a. The width of the window of an envelope is 3.375 inches less than its length x. Create the function A(x) that represents the area of the window in square inches. Simplify, if possible. b. Evaluate A(4.5) and explain what this value represents. c. Assume the dimensions of the envelope are 8 inches by 4 inches. Evaluate A(8.5). Is this possible for this particular envelope? Explain. 114. Environment: Tossing the Envelopes. Each month, Jack receives his bank statement in a 9.5 inch by 6 inch envelope. Each month, he throws away the envelope after removing the statement. a. The width of the window of the envelope is 2.875 inches less than its length x. Create the function A(x) that represents the area of the window in square inches. Simplify, if possible. b. Evaluate A(5.25) and explain what this value represents. c. Evaluate A(10). Is this possible for this particular envelope? Explain. Refer to the table below for Exercises 115 and 116. It illustrates the average federal funds rate for the month of January (2000 to 2008). 115. Finance. Is the relation whose domain is the year and whose range is the average federal funds rate for the month of January a function? Explain. 116. Finance. Write ve ordered pairs whose domain is the set of even years from 2000 to 2008 and whose range is the set of corresponding average federal funds rate for the month of January. YEAR FED. RATE 2000 5.45 2001 5.98 2002 1.73 2003 1.24 2004 1.00 2005 2.25 2006 4.50 2007 5.25 2008 3.50 c03a.qxd 11/25/11 4:46 PM Page 284 314. 3.1 Functions 285 122. Given the function H(x) 3x 2, evaluate the quantity H(3) H(1). Solution: H(3) H(1) H(3) H(1) 7 1 8 This is incorrect. What mistake was made? 123. Given the function (x) x2 x, evaluate the quantity (x 1). Solution: (x 1) (x) (1) x2 x 0 (x 1) x2 x This is incorrect. What mistake was made? 124. Determine the domain of the function and express it in interval notation. Solution: What can t be? Any nonnegative real number. 3 t 0 3 t or t 3 Domain: (, 3) This is incorrect. What mistake was made? g(t) = 13 - t In Exercises 121126, explain the mistake that is made. 121. Determine whether the relationship is a function. Solution: Apply the horizontal line test. Because the horizontal line intersects the graph in two places, this is not a function. This is incorrect. What mistake was made? x y x y 118. Health Care Costs. Using the table found in Exercise 117, let the years correspond to the domain and the total costs correspond to the range. Is this relation a function? Explain. For Exercises 119 and 120, use the following information: Let the functions f, F, g, G, and H represent the number of tons of carbon emitted per year as a function of year corresponding to cement production, natural gas, coal, petroleum, and the total amount, respectively. Let t represent the year, with t 0 corresponding to 1900. 119. Environment: Global Climate Change. Estimate (to the nearest thousand) the value of a. F(50) b. g(50) c. H(50) 120. Environment: Global Climate Change. Explain what the sum represents.F(100) + g(100) + G(100)Write the ve ordered pairs resulting from the table. For Exercises 117 and 118, use the following gure: TOTAL HEALTH CARE COST YEAR FOR FAMILY PLANS 1989 1993 1997 2001 2005 Year Source: Kaiser Family Foundation Health Research and Education Trust. Note: The following years were interpolated: 19891992; 19941995; 19971998. 19951990 2000 2005 Employer-ProvidedHealthInsurance PremiumsforFamilyPlans (19882005,adjustedforinflation) 4,000 2,000 6,000 8,000 10,000 $12,000 Employee Share Employer Share Year 1800 1850 1900 1950 2000 MetricTonsofCarbon/Year (inmillions) 2,000 1,000 3,000 4,000 7,000 6,000 5,000 Total Global Fossil Carbon Emissions Petroleum Coal Natural Gas Cement Production Source: http:/www.naftc.wvu.edu 117. Health Care Costs. Fill in the following table. Round dollars to the nearest $1000. C AT C H T H E M I S TA K E c03a.qxd 11/24/11 4:53 PM Page 285 315. In Exercises 127130, determine whether each statement is true or false. 130. If (a) (a), then may or may not represent a function. 131. If (x) Ax2 3x and (1) 1, nd A. 132. If and g(3) is undened, nd b.g(x) = 1 b - x 127. If a vertical line does not intersect the graph of an equation, then that equation does not represent a function. 128. If a horizontal line intersects a graph of an equation more than once, the equation does not represent a function. 129. If (a) (a), then does not represent a function. 125. Given the function G(x) x2 , evaluate Solution: This is incorrect. What mistake was made? = G(h) h = h2 h = h G(-1 + h) - G(-1) h = G(-1) + G(h) - G(-1) h G(-1 + h) - G(-1) h . 126. Given the functions (x) x A 1 and (1) 1, nd A. Solution: Since (1) 1, the point (1, 1) must satisfy the function. 1 1 A 1 Add 1 to both sides of the equation. 1 A 0 The absolute value of zero is zero, so there is no need for the absolute value signs: This is incorrect. What mistake was made? -1 - A = 0 1 A = -1. 133. If is undened, and F(1) 4, nd C and D. 134. Construct a function that is undened at x 5 and whose graph passes through the point (1, 1). F(x) = C - x D - x , F(-2) In Exercises 135 and 136, nd the domain of each function, where a is any positive real number. 135. 136. f(x) = -52x2 - a2 f(x) = -100 x2 - a2 137. Using a graphing utility, graph the temperature function in Exercise 103. What time of day is it the warmest? What is the temperature? Looking at this function, explain why this model for Tampa, Florida, is valid only from sunrise to sunset (6 to 18). 138. Using a graphing utility, graph the height of the recracker in Exercise 104. How long after liftoff is the recracker airborne? What is the maximum height that the recracker attains? Explain why this height model is valid only for the rst 8 seconds. 139. Using a graphing utility, graph the price function in Exercise 105. What are the lowest and highest prices of the cards? Does this agree with what you found in Exercise 106? 140. The makers of malted milk balls are considering increasing the size of the spherical treats. The thin chocolate coating on a malted milk ball can be approximated by the surface area, S(r) 4r2 . If the radius is increased 3 mm, what is the resulting increase in required chocolate for the thin outer coating? 141. Let . Graph and in the same viewing window. Describe how the graph of can be obtained from the graph of . 142. Let Graph and in the same viewing window. Describe how the graph of can be obtained from the graph of .y1 y2 y2 = f(x + 2)y1 = f(x)f(x) = 4 - x2 . y1 y2 y2 = f(x - 2)y1 = f(x)f(x) = x2 + 1 C O N C E P T U A L C HALLE N G E T E C H N O L O G Y 286 C HAP TE R 3 Functions and Their Graphs c03a.qxd 11/24/11 4:53 PM Page 286 316. Recognizing and Classifying Functions Common Functions Point-plotting techniques were introduced in Section 2.2, and we noted there that we would explore some more efcient ways of graphing functions in Chapter 3. The nine main functions you will read about in this section will constitute a library of functions that you should commit to memory. We will draw on this library of functions in the next section when graphing transformations are discussed. Several of these functions have been shown previously in this chapter, but now we will classify them specically by name and identify properties that each function exhibits. In Section 2.3, we discussed equations and graphs of lines. All lines (with the exception of vertical lines) pass the vertical line test, and hence are classied as functions. Instead of the traditional notation of a line, y mx b, we use function notation and classify a function whose graph is a line as a linear function. m and b are real numbers.f(x) = mx + b LINEAR FUNCTION C O N C E P TUAL O BJ E CTIVE S Identify common functions. Develop and graph piecewise-dened functions. Identify and graph points of discontinuity. State the domain and range. Understand that even functions have graphs that are symmetric about the y-axis. Understand that odd functions have graphs that are symmetric about the origin. S K I LLS O BJ E CTIVE S Classify functions as even, odd, or neither. Determine whether functions are increasing, decreasing, or constant. Calculate the average rate of change of a function. Evaluate the difference quotient for a function. Graph piecewise-dened functions. G R AP H S O F F U N CTI O N S; P I E C EWI S E- D E F I N E D F U N CTI O N S; I N C R EAS I N G AN D D E C R EAS I N G F U N CTI O N S; AVE R AG E R ATE O F C HAN G E LINEAR FUNCTION: f(x) mx b SLOPE: m y-INTERCEPT: b (x) 2x 7 m 2 b 7 (x) x 3 m 1 b 3 (x) x m 1 b 0 (x) 5 m 0 b 5 The domain of a linear function (x) mx b is the set of all real numbers R. The graph of this function has slope m and y-intercept b. S E CTI O N 3.2 287 c03a.qxd 11/24/11 4:53 PM Page 287 317. 288 C HAP TE R 3 Functions and Their Graphs The graph of the square function is called a parabola and will be discussed in further detail in Chapters 4 and 8. The domain of the square function is the set of all real numbers R. Because squaring a real number always yields a positive number or zero, the range of the square function is the set of all nonnegative numbers. Note that the intercept is the origin and the square function is symmetric about the y-axis. This graph is contained in quadrants I and II. The graph of a constant function (x) b is a horizontal line. The y-intercept corresponds to the point (0, b). The domain of a constant function is the set of all real numbers R. The range, however, is a single value b. In other words, all x-values correspond to a single y-value. Points that lie on the graph of a constant function (x) b are (5, b) (1, b) (0, b) (2, b) (4, b) . . . (x, b) x y (0, b) (4, b)(5, b) Another specific example of a linear function is the function having a slope of one (m 1) and a y-intercept of zero (b 0). This special case is called the identity function. (x) x IDENTITY FUNCTION x y (0, 0) (2, 2) (3, 3) (2, 2) (3, 3) Identity Function Domain: (, ) Range: (, ) Domain: (, ) Range: [b, b] or {b} The graph of the identity function has the following properties: It passes through the origin, and every point that lies on the line has equal x- and y-coordinates. Both the domain and the range of the identity function are the set of all real numbers R. A function that squares the input is called the square function. f(x) = x2 SQUARE FUNCTION x y (1, 1)(1, 1) (2, 4)(2, 4) Square Function Domain: (, ) Range: [0, ) One special case of the linear function is the constant function (m 0). b is any real number.f(x) = b CONSTANT FUNCTION c03a.qxd 11/24/11 4:53 PM Page 288 318. Some points that are on the graph of the absolute value function are (1, 1), (0, 0), and (1, 1). The domain of the absolute value function is the set of all real numbers R, yet the range is the set of nonnegative real numbers. The graph of this function is symmetric with respect to the y-axis and is contained in quadrants I and II. x y (2, 2)(2, 2) Absolute Value Function Domain: (, ) Range: [0, ) f(x) = x ABSOLUTE VALUE FUNCTION The domain of the cube function is the set of all real numbers R. Because cubing a negative number yields a negative number, cubing a positive number yields a positive number, and cubing 0 yields 0, the range of the cube function is also the set of all real numbers R. Note that the only intercept is the origin and the cube function is symmetric about the origin. This graph extends only into quadrants I and III. The next two functions are counterparts of the previous two functions: square root and cube root. When a function takes the square root of the input or the cube root of the input, the function is called the square root function or the cube root function, respectively. In Section 3.1, we found the domain to be [0, ). The output of the function will be all real numbers greater than or equal to zero. Therefore, the range of the square root function is [0, ). The graph of this function will be contained in quadrant I. x y (8, 2) (8, 2) 10 5 Cube Root Function Domain: (, ) Range: (, ) x y 10 5 (4, 2) (9, 3) Square Root Function Domain: [0, ) Range: [0, ) In Section 3.1, we stated the domain of the cube root function to be (, ). We see by the graph that the range is also (, ). This graph is contained in quadrants I and III and passes through the origin. This function is symmetric about the origin. In Section 1.7, you read about absolute value equations and inequalities. Now we shift our focus to the graph of the absolute value function. f(x) = 1 3 x or f (x) = x1/3 CUBE ROOT FUNCTION f(x) = 1x or f(x) = x1/2 SQUARE ROOT FUNCTION x y (2, 8) (2, 8) Cube Function Domain: (, ) Range: (, ) 10 10 5 5 A function that cubes the input is called the cube function. f(x) = x3 CUBE FUNCTION 3.2 Graphs of Functions 289 c03a.qxd 11/24/11 4:53 PM Page 289 319. 290 C HAP TE R 3 Functions and Their Graphs x (1, 1) (1, 1) y f(x) = 1 x Reciprocal Function Domain: (, 0) (0, ) Range: (, 0) (0, ) The only restriction on the domain of the reciprocal function is that . Therefore, we say the domain is the set of all real numbers excluding zero. The graph of the reciprocal function illustrates that its range is also the set of all real numbers except zero. Note that the reciprocal function is symmetric with respect to the origin and is contained in quadrants I and III. Even and Odd Functions Of the nine functions discussed above, several have similar properties of symmetry. The constant function, square function, and absolute value function are all symmetric with respect to the y-axis. The identity function, cube function, cube root function, and reciprocal function are all symmetric with respect to the origin. The term even is used to describe functions that are symmetric with respect to the y-axis, or vertical axis, and the term odd is used to describe functions that are symmetric with respect to the origin. Recall from Section 2.2 that symmetry can be determined both graphically and algebraically. The box below summarizes the graphic and algebraic characteristics of even and odd functions. x Z 0 A function whose output is the reciprocal of its input is called the reciprocal function. The algebraic method for determining symmetry with respect to the y-axis, or vertical axis, is to substitute x for x. If the result is an equivalent equation, the function is symmetric with respect to the y-axis. Some examples of even functions are (x) b, (x) x2 , (x) x4 ; and (x) x . In any of these equations, if x is substituted for x, the result is the same; that is, (x) (x). Also note that, with the exception of the absolute value function, these examples are all even-degree polynomial equations. All constant functions are degree zero and are even functions. The algebraic method for determining symmetry with respect to the origin is to substitute x for x. If the result is the negative of the original function, that is, if (x) (x), then the function is symmetric with respect to the origin and, hence, classied as an odd function. Examples of odd functions are (x) x, (x) x3 , (x) x5 , and (x) x1/3 . In any of these functions, if x is substituted for x, the result is the negative of the original function. Note that with the exception of the cube root function, these equations are odd-degree polynomials. f(x) = 1 x x Z 0 RECIPROCAL FUNCTION Function Symmetric with Respect to On Replacing x with x Even y-axis or vertical axis (x) (x) Odd origin (x) (x) EVEN AND ODD FUNCTIONS c03a.qxd 11/24/11 4:53 PM Page 290 320. E X AM P LE 1 Determining Whether a Function Is Even, Odd, or Neither Determine whether the functions are even, odd, or neither. a. (x) x2 3 b. g(x) x5 x3 c. h(x) x2 x Solution (a): Original function. (x) x2 3 Replace x with x. (x) (x)2 3 Simplify. (x) x2 3 (x) Because (x) (x), we say that (x) is an even function . Solution (b): Original function. g(x) x5 x3 Replace x with x. g(x) (x)5 (x)3 Simplify. g(x) x5 x3 (x5 x3 ) g(x) Because g(x) g(x), we say that g(x) is an odd function . Solution (c): Original function. h(x) x2 x Replace x with x. h(x) (x)2 (x) Simplify. h(x) x2 x h(x) is neither h(x) nor h(x); therefore the function h(x) is neither even nor odd . In parts (a), (b), and (c), we classied these functions as either even, odd, or neither, using the algebraic test. Look back at them now and reect on whether these classications agree with your intuition. In part (a), we combined two functions: the square function and the constant function. Both of these functions are even, and adding even functions yields another even function. In part (b), we combined two odd functions: the fth-power function and the cube function. Both of these functions are odd, and adding two odd functions yields another odd function. In part (c), we combined two functions: the square function and the identity function. The square function is even, and the identity function is odd. In this part, combining an even function with an odd function yields a function that is neither even nor odd and, hence, has no symmetry with respect to the vertical axis or the origin. YO U R TU R N Classify the functions as even, odd, or neither. a. (x) x 4 b. (x) x3 1 Technology Tip a. Graph .y1 = f(x) = x2 - 3 Answer: a. even b. neither Even; symmetric with respect to the y-axis. b. Graph y1 = g(x) = x5 + x3 . Odd; symmetric with respect to origin. c. Graph .y1 = h(x) = x2 - x No symmetry with respect to y-axis or origin. Be careful, though, because functions that are combinations of even- and odd-degree polynomials can turn out to be neither even nor odd, as we will see in Example 1. 3.2 Graphs of Functions 291 c03a.qxd 11/24/11 4:53 PM Page 291 321. 292 C HAP TE R 3 Functions and Their Graphs Increasing and Decreasing Functions Look at the gure in the margin to the left. Graphs are read from left to right. If we start at the left side of the graph and trace the red curve with our pen, we see that the function values (values in the vertical direction) are decreasing until arriving at the point (2, 2). Then, the function values increase until arriving at the point (1, 1). The values then remain constant (y 1) between the points (1, 1) and (0, 1). Proceeding beyond the point (0, 1), the function values decrease again until the point (2, 2). Beyond the point (2, 2), the function values increase again until the point (6, 4). Finally, the function val- ues decrease and continue to do so. When specifying a function as increasing, decreasing, or constant, the intervals are classied according to the x-coordinate. For instance, in this graph, we say the function is increasing when x is between x 2 and x 1 and again when x is between x 2 and x 6. The graph is classied as decreasing when x is less than 2 and again when x is between 0 and 2 and again when x is greater than 6. The graph is classied as constant when x is between 1 and 0. In interval notation, this is summarized as Decreasing Increasing Constant (1, 0) An algebraic test for determining whether a function is increasing, decreasing, or constant is to compare the value (x) of the function for particular points in the intervals. (-2, -1)(2, 6)(-, -2)(0, 2)(6, ) Study Tip Graphs are read from left to right. Intervals correspond to the x-coordinates. x y (6, 4) (0, 1)(1, 1) (2, 2) (2, 2) 1. A function is increasing on an open interval I if for any x1 and x2 in I, where x1 x2, then (x1) (x2). 2. A function is decreasing on an open interval I if for any x1 and x2 in I, where x1 x2, then (x1) (x2). 3. A function f is constant on an open interval I if for any x1 and x2 in I, then (x1) (x2). INCREASING, DECREASING, AND CONSTANT FUNCTIONS In addition to classifying a function as increasing, decreasing, or constant, we can also determine the domain and range of a function by inspecting its graph from left to right: The domain is the set of all x-values (from left to right) where the function is dened. The range is the set of all y-values (from bottom to top) that the graph of the function corresponds to. A solid dot on the left or right end of a graph indicates that the graph terminates there and the point is included in the graph. An open dot indicates that the graph terminates there and the point is not included in the graph. Unless a dot is present, it is assumed that a graph continues indenitely in the same direction. (An arrow is used in some books to indicate direction.) c03b.qxd 11/24/11 4:55 PM Page 292 322. 3.2 Graphs of Functions 293 Average Rate of Change How do we know how much a function is increasing or decreasing? For example, is the price of a stock slightly increasing or is it doubling every week? One way we determine how much a function is increasing or decreasing is by calculating its average rate of change. Let (x1, y1) and (x2, y2) be two points that lie on the graph of a function . Draw the line that passes through these two points (x1, y1) and (x2, y2). This line is called a secant line. Note that the slope of the secant line is given by , and recall that the slope of a line is the rate of change of that line. The slope of the secant line is used to represent the average rate of change of the function. m = y2 - y1 x2 - x1 E X AM P LE 2 Finding Intervals When a Function Is Increasing or Decreasing Given the graph of a function: a. State the domain and range of the function. b. Find the intervals when the function is increasing, decreasing, or constant. Solution (a): Domain: [5, ) Range: [0, ) Solution (b): Reading the graph from left to right, we see that the graph decreases from the point (5, 7) to the point (2, 4). is constant from the point (2, 4) to the point (0, 4). decreases from the point (0, 4) to the point (2, 0). increases from the point (2, 0) on. The intervals of increasing and decreasing correspond to the x-coordinates. We say that this function is increasing on the interval (2, ). decreasing on the interval constant on the interval (2, 0). Note: The intervals of increasing or decreasing are dened on open intervals. This should not be confused with the domain. For example, the point x 5 is included in the domain of the function but not in the interval where the function is classied as decreasing. (-5, -2)(0, 2). x (2, 0) (0, 4) (2, 4) (5, 7) y 5 5 10 Decreasing Decreasing Constant Increasing x (2, 0) (0, 4) (2, 4) (5, 7) y 5 5 10 Decreasing Decreasing Constant Increasing x x1 x2 y secant (x1, y1) (x2, y2) f x (2, 0) (0, 4) (2, 4) (5, 7) y 5 5 10 c03b.qxd 11/24/11 4:55 PM Page 293 323. 294 C HAP TE R 3 Functions and Their Graphs E X AM P LE 3 Average Rate of Change Find the average rate of change of (x) x4 from: a. x 1 to x 0 b. x 0 to x 1 c. x 1 to x 2 Solution (a): Write the average rate of change formula. Let x1 1 and x2 0. Substitute (1) (1)4 1 and (0) 04 0. Simplify. Solution (b): Write the average rate of change formula. Let x1 0 and x2 1. Substitute (0) 04 0 and (1) (1)4 1. Simplify. Solution (c): Write the average rate of change formula. Let x1 1 and x2 2. Substitute (1) 14 1 and (2) (2)4 16. Simplify. 15 = 16 - 1 2 - 1 = f(2) - f(1) 2 - 1 f(x2) - f(x1) x2 - x1 1 = 1 - 0 1 - 0 = f(1) - f(0) 1 - 0 f(x2) - f(x1) x2 - x1 -1 = 0 - 1 0 - (-1) = f(0) - f(-1) 0 - (-1) f(x2) - f(x1) x2 - x1 Let (x1, (x1)) and (x2, (x2)) be two distinct points, , on the graph of the function . The average rate of change of between x1 and x2 is given by Average rate of change f(x2) - f(x1) x2 - x1 (x1 Z x2) AVERAGE RATE OF CHANGE x x1 x2 x2 x1 f(x2) f (x1) y secant (x1, f(x1)) (x2, f(x2)) c03b.qxd 11/24/11 4:55 PM Page 294 324. 3.2 Graphs of Functions 295 WORDS Let the difference between x1 and x2 be h. Solve for x2. Substitute x2 x1 h into the denominator and x2 x1 h into the numerator of the average rate of change. Let x1 x. MATH x2 x1 h x2 x1 h = f(x + h) - f(x) h = f(x1 + h) - f(x1) h Average rate of change = f(x2) - f(x1) x2 - x1 Graphical Interpretation: Slope of the Secant Line a. Average rate of change of f from x 1 to x 0: Decreasing at a rate of 1 b. Average rate of change of f from x 0 to x 1: Increasing at a rate of 1 c. Average rate of change of f from x 1 to x 2: Increasing at a rate of 15 YO U R TU R N Find the average rate of change of (x) x2 from: a. x 2 to x 0 b. x 0 to x 2 x y 22 20 (0, 0) (1, 1) (2, 16) x y 22 3 2 (0, 0) (1, 1) x y 22 3 2 (0, 0) (1, 1) Answer: a. 2 b. 2 The average rate of change can also be written in terms of the difference quotient. c03b.qxd 11/24/11 4:55 PM Page 295 325. 296 C HAP TE R 3 Functions and Their Graphs When written in this form, the average rate of change is called the difference quotient. The expression , where , is called the difference quotient.h Z 0 f(x + h) - f(x) h D E F I N I T I O N Difference Quotient The difference quotient is more meaningful when h is small. In calculus the difference quotient is used to dene a derivative. E X AM P LE 4 Calculating the Difference Quotient Calculate the difference quotient for the function (x) 2x2 1. Solution: Find (x h). = 2x2 + 4xh + 2h2 + 1 = 2(x2 + 2xh + h2 ) + 1 f(x + h) = 2(x + h)2 + 1 x x x + h y f(x) f(x + h) h Study Tip Use brackets or parentheses around (x) to avoid forgetting to distribute the negative sign: f(x + h) - [ f(x)] h Find the difference quotient. Simplify. Factor the numerator. Cancel (divide out) the common h. YO U R TU R N Calculate the difference quotient for the function (x) x2 2. h Z 0 f(x + h) - f(x) h = 4x + 2h f(x + h) - f(x) h = h(4x + 2h) h f(x + h) - f(x) h = 4xh + 2h2 h f(x + h) - f(x) h = 2x2 + 4xh + 2h2 + 1 - 2x2 - 1 h 4 f(x + h) - f(x) h = 2x2 + 4xh + 2h2 + 1 - (2x2 + 1) h f(x+h) f(x) Answer: f (x + h) - f (x) h = -2x - h Piecewise-Defined Functions Most of the functions that we have seen in this text are functions dened by polynomials. Sometimes the need arises to dene functions in terms of pieces. For example, most plumbers charge a at fee for a house call and then an additional hourly rate for the job. For instance, if a particular plumber charges $100 to drive out to your house and work for 1 hour and then an additional $25 an hour for every additional hour he or she works on your job, we would dene this function in pieces. If we let h be the number of hours worked, then the charge is dened as Plumbing charge = b 100 h 1 100 + 25(h - 1) h 7 1 PlumberCharge Time (hours) 1 32 4 5 76 9 108 100 50 150 200 $250 y c03b.qxd 12/27/11 12:39 PM Page 296 326. x f(x) f(x) = |x| Technology Tip Plot a piecewise-dened function using the TEST menu operations to dene the inequalities in the function. Press: Y ( X, T, , n x2 ) ( X, T, , n 2nd MATH Test 5 () 1 ) ( 1 ) ( X, T, , n 2nd MATH Test 4 () 1 ) ( X, T, , n 2nd MATH Test 6 1 ) ( X, T, , n ) ( X, T, , n 2nd MATH Test 3 1 ) x f(x) y = xy = x If we were to graph this function, we would see that there is 1 hour that is constant and after that the function continually increases. Another piecewise-dened function is the absolute value function. The absolute value function can be thought of as two pieces: the line y x (when x is negative) and the line y x (when x is nonnegative). We start by graphing these two lines on the same graph. The absolute value function behaves like the line y x when x is negative (erase the blue graph in quadrant IV) and like the line y x when x is positive (erase the red graph in quadrant III). Absolute value function The next example is a piecewise-dened function given in terms of functions in our library of functions. Because the function is dened in terms of pieces of other functions, we draw the graph of each individual function, and then for each function, darken the piece corresponding to its part of the domain. This is like the procedure above for the absolute value function. E X AM P LE 5 Graphing Piecewise-Dened Functions Graph the piecewise-dened function, and state the domain, range, and intervals when the function is increasing, decreasing, or constant. Solution: Graph each of the functions on the same plane. Square function: Constant function: Identity function: The points to focus on in particular are the x-values where the pieces change overthat is, x 1 and x 1. Lets now investigate each piece. When x 1, this function is dened by the square function, (x) x2 , so darken that particular function to the left of x 1. When 1 x 1, the function is dened by the constant function, (x) 1, so darken that particular function between the x values of 1 and 1. When x 1, the function is dened by the identity function, (x) x, so darken that function to the right of x 1. Erase everything that is not darkened, and the resulting graph of the piecewise-dened function is given on the right. f(x) x f(x) 1 f(x) x2 G(x) = c x2 x 6 -1 1 -1 x 1 x x 7 1 f(x) = x = b -x x < < f(x) = c 1 - x x 6 0 x 0 x 6 2 -1 x 7 2 (1, ) (-1, 1) (-, -1) [1, ) x y Technology Tip To plot a piecewise-dened function using a graphing utility, use the TEST menu operations to dene the inequalities in the function. Press: Y ( 1 X, T, , n ) ( X, T, , n 2nd MATH 5 0 ) ( X, T, , n ) ( X, T, , n 2nd MATH 4 0 ) ( X, T, , n 2nd MATH 5 2 ) ( () 1 ) ( X, T, , n 2nd MATH 3 2 ) To avoid connecting graphs of the pieces, press MODE and Dot . Set the viewing rectangle as [3, 4] by [2, 5]; then press GRAPH . Be sure to include the open circle and closed circle at the appropriate endpoints of each piece in the function. The table of values supports the graph, except at x 2. The function is not dened at x 2. c03b.qxd 6/12/12 2:07 PM Page 298 328. 3.2 Graphs of Functions 299 Piecewise-dened functions whose pieces are constants are called step functions. The reason for this name is that the graph of a step function looks like steps of a staircase. A common step function used in engineering is the Heaviside step function (also called the unit step function): This function is used in signal processing to represent a signal that turns on at some time and stays on indenitely. A common step function used in business applications is the greatest integer function. H(t) = b 0 t 6 0 1 t 0 At what intervals is the function increasing, decreasing, or constant? Remember that the intervals correspond to the x-values. Decreasing: Increasing: (0, 2) Constant: The function is dened for all values of x except x 2. The output of this function (vertical direction) takes on the y-values y 0 and the additional single value y 1. or We mentioned earlier that a discontinuous function has a graph that exhibits holes or jumps. In this example, the point x 0 corresponds to a jump, because you would have to pick up your pencil to continue drawing the graph. The point x 2 corresponds to both a hole and a jump. The hole indicates that the function is not dened at that point, and there is still a jump because the identity function and the constant function do not meet at the same y-value at x 2. {-1}[0, )Range: [-1, -1][0, ) Domain: (-, 2)(2, ) (2, )(, 0) YO U R TU R N Graph the piecewise-dened function, and state the intervals where the function is increasing, decreasing, or constant, along with the domain and range. f(x) = c -x x -1 2 -1 6 x 6 1 x x 7 1 t H(t) 1 x y Answer: Increasing: (1, ) Decreasing: (, 1) Constant: (1, 1) Domain: Range: [1, ) (-, 1) (1, ) x 1.0 1.3 1.5 1.7 1.9 2.0 f(x) = [[x]] 1 1 1 1 1 2 x 5 5 5 5 f(x) = x f(x) = [[x]] = greatest integer less than or equal to x. GREATEST INTEGER FUNCTION c03b.qxd 11/24/11 4:55 PM Page 299 329. 300 C HAP TE R 3 Functions and Their Graphs S U M MARY S E CTI O N 3.2 NAME FUNCTION DOMAIN RANGE GRAPH EVEN/ODD Linear (x) mx b, (, ) (, ) Neither (unless ) Constant (x) c (, ) [c, c] or {c} Even Identity (x) x (, ) (, ) Odd Square (x) x2 (, ) [0, ) Even Cube (x) x3 (, ) (, ) Odd Square Root [0, ) [0, ) Neither Cube Root (, ) (, ) Odd Absolute Value (x) x (, ) [0, ) Even Reciprocal Odd(-, 0)(0, )(-, 0)(0, )f(x) = 1 x f(x) = 3 1x f(x) = 1x y = x m Z 0 Domain and Range of a Function Implied domain: Exclude any values that lead to the function being undened (dividing by zero) or imaginary outputs (square root of a negative real number). Inspect the graph to determine the set of all inputs (domain) and the set of all outputs (range). x y x y x y x y x y x y x y x y x y c03b.qxd 11/24/11 4:55 PM Page 300 330. 3.2 Graphs of Functions 301 In Exercises 124, determine whether the function is even, odd, or neither. 1. G(x) x 4 2. h(x) 3 x 3. (x) 3x2 1 E X E R C I S E S 4. F(x) x4 2x2 5. g(t) 5t3 3t 6. (x) 3x5 4x3 7. h(x) x2 2x 8. G(x) 2x4 3x3 9. h(x) x1/3 x 10. g(x) x1 x 11. (x) x 5 12. (x) x x2 13. (x) x 14. (x) x3 15. G(t) t 3 16. g(t) t 2 17. G(t) = 1t - 3 18. f(x) = 12 - x 19. g(x) = 2x2 + x 20. f (x) = 2x2 + 2 21. h(x) = 1 x + 3 22. h(x) = 1 x - 2x 23. x y 5 5 10 24. x y 5 5 9 S E C TI O N 3.2 Finding Intervals Where a Function Is Increasing, Decreasing, or Constant Increasing: Graph of function rises from left to right. Decreasing: Graph of function falls from left to right. Constant: Graph of function does not change height from left to right. Average Rate of Change f(x2) - f(x1) x2 - x1 x1 Z x2 Difference Quotient Piecewise-Dened Functions Continuous: You can draw the graph of a function without picking up the pencil. Discontinuous: Graph has holes and/or jumps. f(x + h) - f(x) h h Z 0 In Exercises 2536, state the (a) domain, (b) range, and (c) x-interval(s) where the function is increasing, decreasing, or constant. Find the values of (d) f(0), (e) f(2), and (f) f(2). 25. 26. 27. 28. x f(x) y (3, 3) (2, 2) (0, 1) (1, 1) (2, 1) S K I LL S x f(x) y (3, 3) (2, 1) (1, 1) 5 5 x (0, 4) (7, 1) (4, 2) (2, 5) y 8 2 5 5 x (3, 4) (3, 4) (6, 0) (6, 0) (0, 0) y 10 10 5 5 c03b.qxd 11/24/11 6:05 PM Page 301 331. 302 C HAP TE R 3 Functions and Their Graphs In Exercises 3744, nd the difference quotient for each function. 37. (x) x2 x 38. (x) x2 2x 39. (x) 3x x2 40. (x) 5x x2 41. (x) x2 3x 2 42. (x) x2 2x 5 43. (x) 3x2 5x 4 44. (x) 4x2 2x 3 In Exercises 4552, nd the average rate of change of the function from x 1 to x 3. 45. (x) x3 46. 47. (x) x 48. (x) 2x 49. (x) 1 2x 50. (x) 9 x2 51. (x) 5 2x 52. In Exercises 5378, graph the piecewise-dened functions. State the domain and range in interval notation. Determine the intervals where the function is increasing, decreasing, or constant. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. G(x) = c -1 x 6 -1 x -1 6 x 6 3 3 x 7 3 G(x) = c -1 x 6 -1 x -1 x 3 3 x 7 3 f(x) = d 3 - 1 2 x x 6 -2 4 + 3 2 x x 7 -2 f(x) = b 5 - 2x x 6 2 3x - 2 x 7 2 f(x) = b 2 + x x -1 x2 x 7 -1 f(x) = b - x + 2 x 6 1 x2 x 1 f(x) = b -x x 0 x2 x 7 0 f(x) = b x x 6 0 x2 x 0 f(x) = b x2 x 6 2 4 x 2 f(x) = b 1 x 6 -1 x2 x -1 f (x) = b -x x 6 -1 -1 x -1 f(x) = b x x 6 2 2 x 2 f(x) = 2x2 - 1 f(x) = 1 x f(x h) f(x) h 29. 30. 31. 32. x (4, 2) (3, 2) y 5 5 5 5 x y 5 5 5 5 x (2, 0) (0, 4) (2, 0) y 5 5 5 5 x (3, 0)(3, 0) y 5 5 5 5 33. 34. 35. 36. x (2, 3) (2, 3) y 10 10 10 10 x (4, 2) (4, 0) (0, 4) y 5 5 5 5 x(5, 0) (0, 5) (0, 7) (5, 0) (2, 3) y 5 5 2 8 x (4, 3) (5, 3) (8, 0) (0, 4) y 10 10 10 10 c03b.qxd 11/24/11 4:55 PM Page 302 332. 3.2 Graphs of Functions 303 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. f(x) = c x2 x -1 x3 -1 6 x 6 1 x x 1 f(x) = c x x -1 x3 -1 6 x 6 1 x2 x 7 1 f(x) = c x x 6 -1 1 -1 6 x 6 1 x x 7 1 f(x) = c x + 3 x -2 x -2 6 x 6 2 x2 x 2 G(x) = c - 3 1x x 6 -1 x -1 x 6 1 1x x 7 1 G(x) = c - 3 1x x -1 x -1 6 x 6 1 - 1x x 7 1 G(x) = c 0 x = 0 - 1 x x Z 0 G(x) = c 0 x = 0 1 x x Z 0 G(x) = b 1 x 6 1 3 1x x 7 1 G(x) = b 0 x 6 0 1x x 0 f(x) = c -x - 1 x -2 x + 1 -2 6 x 6 1 -x + 1 x 7 1 f(x) = c -x - 1 x 6 -2 x + 1 -2 6 x 6 1 -x + 1 x 1 G(t) = c 1 t 6 1 t2 1 6 t 6 2 4 t 7 2 G(t) = c 1 t 6 1 t2 1 t 2 4 t 7 2 81. Budget: Costs. The Kappa Kappa Gamma sorority decides to order custom-made T-shirts for its Kappa Krush mixer with the Sigma Alpha Epsilon fraternity. If the sorority orders 50 or fewer T-shirts, the cost is $10 per shirt. If it orders more than 50 but less than or equal to 100, the cost is $9 per shirt. If it orders more than 100, the cost is $8 per shirt. Find the cost function C(x) as a function of the number of T-shirts x ordered. 82. Budget: Costs. The marching band at a university is ordering some additional uniforms to replace existing uniforms that are worn out. If the band orders 50 or fewer, the cost is $176.12 per uniform. If it orders more than 50 but less than or equal to 100, the cost is $159.73 per uniform. Find the cost function C(x) as a function of the number of new uniforms x ordered. 83. Budget: Costs. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to gure out how much money to raise. The entry fee is $250 per boat for the rst 10 boats and $175 for each additional boat. Find the cost function C(x) as a function of the number of boats x the club enters. For Exercises 79 and 80, refer to the following: A manufacturer determines that his prot and cost functions over one year are represented by the following graphs. 79. Business. Find the intervals on which prot is increasing, decreasing, and constant. 80. Business. Find the intervals on which cost is increasing, decreasing, and constant. A P P L I C AT I O N S 2 4 6 8 1210 $5 4 3 2 1 t (time in months) C (cost in millions of dollars) 2 4 6 8 1210 $20 18 16 14 12 10 8 6 4 2 t (time in months) P (profit in millions of dollars) c03b.qxd 11/24/11 4:55 PM Page 303 333. 304 C HAP TE R 3 Functions and Their Graphs 92. Postage Rates. The following table corresponds to rst-class postage rates for the U.S. Postal Service. Write a piecewise-dened function in terms of the greatest integer function that models this cost of mailing parcels rst class. WEIGHT LESS FIRST-CLASS RATE THAN (OUNCES) (FLAT ENVELOPES) 1 $0.80 2 $0.97 3 $1.14 4 $1.31 5 $1.48 6 $1.65 7 $1.82 8 $1.99 9 $2.16 10 $2.33 11 $2.50 12 $2.67 13 $2.84 WEIGHT LESS FIRST-CLASS RATE THAN (OUNCES) (PARCELS) 1 $1.13 2 $1.30 3 $1.47 4 $1.64 5 $1.81 6 $1.98 7 $2.15 8 $2.32 9 $2.49 10 $2.66 11 $2.83 12 $3.00 13 $3.17 84. Phone Cost: Long-Distance Calling. A phone company charges $.39 per minute for the rst 10 minutes of an international long-distance phone call and $.12 per minute every minute after that. Find the cost function C(x) as a function of the length of the phone call x in minutes. 85. Event Planning. A young couple are planning their wedding reception at a yacht club. The yacht club charges a at rate of $1000 to reserve the dining room for a private party. The cost of food is $35 per person for the rst 100 people and $25 per person for every additional person beyond the rst 100. Write the cost function C(x) as a function of the number of people x attending the reception. 86. Home Improvement. An irrigation company gives you an estimate for an eight-zone sprinkler system. The parts are $1400, and the labor is $25 per hour. Write a function C(x) that determines the cost of a new sprinkler system if you choose this irrigation company. 87. Sales. A famous author negotiates with her publisher the monies she will receive for her next suspense novel. She will receive $50,000 up front and a 15% royalty rate on the rst 100,000 books sold, and 20% on any books sold beyond that. If the book sells for $20 and royalties are based on the selling price, write a royalties function R(x) as a function of total number x of books sold. 88. Sales. Rework Exercise 87 if the author receives $35,000 up front, 15% for the rst 100,000 books sold, and 25% on any books sold beyond that. 89. Prot. Some artists are trying to decide whether they will make a prot if they set up a Web-based business to market and sell stained glass that they make. The costs associated with this business are $100 per month for the website and $700 per month for the studio they rent. The materials cost $35 for each work in stained glass, and the artists charge $100 for each unit they sell. Write the monthly prot as a function of the number of stained-glass units they sell. 90. Prot. Philip decides to host a shrimp boil at his house as a fundraiser for his daughters AAU basketball team. He orders gulf shrimp to be own in from New Orleans. The shrimp costs $5 per pound. The shipping costs $30. If he charges $10 per person, write a function F(x) that represents either his loss or prot as a function of the number of people x that attend. Assume that each person will eat 1 pound of shrimp. 91. Postage Rates. The following table corresponds to rst-class postage rates for the U.S. Postal Service. Write a piecewise-dened function in terms of the greatest integer function that models this cost of mailing at envelopes rst class. c03b.qxd 11/24/11 4:55 PM Page 304 334. 3.2 Graphs of Functions 305 A square wave is a waveform used in electronic circuit testing and signal processing. A square wave alternates regularly and instantaneously between two levels. sciencephotos/Alamy 93. Electronics: Signals. Write a step function f(t) that represents the following square wave. t 5 5 5 f(t) 94. Electronics: Signals. Write a step function f(x) that represents the following square wave, where x represents frequency in Hz. 1000 1 1 MILLIONS OF TONS YEAR OF CARBON 1900 500 1925 1000 1950 1500 1975 5000 2000 7000 For Exercises 95 and 96, refer to the following table: Global Carbon Emissions from Fossil Fuel Burning 95. Climate Change: Global Warming. What is the average rate of change in global carbon emissions from fossil fuel burning from a. 1900 to 1950? b. 1950 to 2000? 96. Climate Change: Global Warming. What is the average rate of change in global carbon emissions from fossil fuel burning from a. 1950 to 1975? b. 1975 to 2000? For Exercises 97 and 98, use the following information: The height (in feet) of a falling object with an initial velocity of 48 feet per second launched straight upward from the ground is given by h(t) 16t2 48t, where t is time (in seconds). 97. Falling Objects. What is the average rate of change of the height as a function of time from t 1 to t 2? 98. Falling Objects. What is the average rate of change of the height as a function of time from t 1 to t 3? For Exercises 99 and 100, refer to the following: An analysis of sales indicates that demand for a product during a calendar year (no leap year) is modeled by where d is demand in thousands of units and t is the day of the year and represents January 1. 99. Economics. Find the average rate of change of the demand of the product over the rst quarter. 100. Economics. Find the average rate of change of the demand of the product over the fourth quarter. t = 1 d(t) = 32t2 + 1 - 2.75t c03b.qxd 11/24/11 4:56 PM Page 305 335. 306 C HAP TE R 3 Functions and Their Graphs 102. Graph the piecewise-dened function. State the domain and range. Solution: Draw the graphs of (x) x and (x) x. f(x) = b -x x 1 x x 7 1 Darken the function (x) x when x 1 and the function (x) x when x 1. The resulting graph is as shown. Domain: (, ) or R Range: (1, ) This is incorrect. What mistake was made? 103. The cost of airport Internet access is $15 for the rst 30 minutes and $1 per minute for each additional minute. Write a function describing the cost of the service as a function of minutes used online. Solution: This is incorrect. What mistake was made? 104. Most money market accounts pay a higher interest with a higher principal. If the credit union is offering 2% on accounts with less than or equal to $10,000 and 4% on the additional money over $10,000, write the interest function I(x) that represents the interest earned on an account as a function of dollars in the account. Solution: This is incorrect. What mistake was made? I(x) = b 0.02x x 10,000 0.02(10,000) + 0.04x x 7 10,000 C(x) = b 15 x 30 15 + x x 7 30 x y x y In Exercises 101104, explain the mistake that is made. 101. Graph the piecewise-dened function. State the domain and range. Solution: Draw the graphs of (x) x and (x) x. f (x) = b -x x 6 0 x x 7 0 Darken the function (x) x when x 0 and the function (x) x when x 0. This gives us the familiar absolute value graph. Domain: (, ) or R Range: [0, ) This is incorrect. What mistake was made? x y x y C AT C H T H E M I S TA K E c03b.qxd 11/24/11 4:56 PM Page 306 336. 112. In trigonometry you will learn about the cosine function, cos x. Plot the function (x) cos x, using a graphing utility. It should look like the graph on the right. Is the cosine function even, odd, or neither? 111. In trigonometry you will learn about the sine function, sin x. Plot the function (x) sin x, using a graphing utility. It should look like the graph on the right. Is the sine function even, odd, or neither? x y 1010 1 1 x y 10 10 1 1 3.2 Graphs of Functions 307 In Exercises 109 and 110, for a and b real numbers, can the function given ever be a continuous function? If so, specify the value for a and b that would make it so. 109. f(x) = b ax x 2 bx2 x 7 2 110. f(x) = d - 1 x x 6 a 1 x x a In Exercises 105108, determine whether each statement is true or false. 105. The identity function is a special case of the linear function. 106. The constant function is a special case of the linear function. 107. If an odd function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing. 108. If an even function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing. 113. In trigonometry you will learn about the tangent function, tan x. Plot the function (x) tan x, using a graphing utility. If you restrict the values of x so that , the graph should resemble the graph below. Is the tangent function even, odd, or neither? 114. Plot the function . What function is this? 115. Graph the function using a graphing utility. State the domain and range. 116. Graph the function using a graphing utility. State the domain and range. f(x) = C C1 3xD D f(x) = [[3x]] f(x) = sin x cos x x y 1.5 1.5 15 15 - p 2 6 x 6 p 2 C O N C E P T U A L C HALLE N G E T E C H N O L O G Y c03b.qxd 11/24/11 4:56 PM Page 307 337. C O N C E P TUAL O BJ E CTIV E S Identify the common functions by their graphs. Apply multiple transformations of common functions to obtain graphs of functions. Understand that domain and range also are transformed. G R AP H I N G TE C H N I Q U E S: TR AN S F O R MATI O N S S K I LLS O BJ E CTIVE S Sketch the graph of a function using horizontal and vertical shifting of common functions. Sketch the graph of a function by reecting a common function about the x-axis or y-axis. Sketch the graph of a function by stretching or compressing a common function. Sketch the graph of a function using a sequence of transformations. S E CTI O N 3.3 Horizontal and Vertical Shifts The focus of the previous section was to learn the graphs that correspond to particular functions such as identity, square, cube, square root, cube root, absolute value, and reciprocal. Therefore, at this point, you should be able to recognize and generate the graphs of x (x) 2 2 1 1 0 0 1 1 2 2 x h(x) 2 3 1 2 0 1 1 0 2 1 x g(x) 2 4 1 3 0 2 1 3 2 4 x y f (x) g (x) h(x) Instead of point-plotting the function g(x) x 2, we could have started with the function f(x) x and shifted the entire graph up 2 units. Similarly, we could have generated the graph of the function h(x) x 1 by shifting the function f(x) x to the right 1 unit. In both cases, the base or starting function is f(x) x . Why did we go up for g(x) and to the right for h(x)? Note that we could rewrite the functions g(x) and h(x) in terms of (x): h(x) x 1 = f(x - 1) g(x) x 2 = f(x) + 2 308 , and . In this section, we willy = 1 x y = x, y = x2 , y = x3 , y = 1x, y = 1 3 x, y = x discuss how to sketch the graphs of functions that are very simple modications of these functions. For instance, a common function may be shifted (horizontally or vertically), reected, or stretched (or compressed). Collectively, these techniques are called transformations. Lets take the absolute value function as an example. The graph of (x) x was given in the last section. Now look at two examples that are much like this function: g(x) x 2 and h(x) x 1 . Graphing these functions by point-plotting yields c03b.qxd 11/24/11 6:05 PM Page 308 338. x y f (x) h(x) f(x) = |x| h(x) = |x 1| x y f (x) g(x) f(x) = |x| g(x) = |x|+ 2 3.3 Graphing Techniques: Transformations 309 In the case of g(x), the shift (2) occurs outside the functionthat is, outside the parentheses showing the argument. Therefore, the output for g(x) is two more than the typical output for (x). Because the output corresponds to the vertical axis, this results in a shift upward of two units. In general, shifts that occur outside the function correspond to a vertical shift corresponding to the sign of the shift. For instance, had the function been G(x) x 2, this graph would have started with the graph of the function (x) and shifted down two units. In the case of h(x), the shift occurs inside the functionthat is, inside the parentheses showing the argument. Note that the point (0, 0) that lies on the graph of (x) was shifted to the point (1, 0) on the graph of the function h(x). The y-value remained the same, but the x-value shifted to the right one unit. Similarly, the points (1, 1) and (1, 1) were shifted to the points (0, 1) and (2, 1), respectively. In general, shifts that occur inside the function correspond to a horizontal shift opposite the sign. In this case, the graph of the function h(x) x 1 shifted the graph of the function f(x) to the right one unit. If, instead, we had the function H(x) x 1 , this graph would have started with the graph of the function (x) and shifted to the left one unit. Study Tip Shifts outside the function are vertical shifts with the sign. Up () Down () Assuming that c is a positive constant, To Graph Shift the Graph of (x) (x) c c units upward (x) c c units downward Adding or subtracting a constant outside the function corresponds to a vertical shift that goes with the sign. VERTICAL SHIFTS Study Tip Shifts inside the function are horizontal shifts opposite the sign. Left () Right () Assuming that c is a positive constant, To Graph Shift the Graph of (x) (x c) c units to the left (x c) c units to the right Adding or subtracting a constant inside the function corresponds to a horizontal shift that goes opposite the sign. HORIZONTAL SHIFTS c03b.qxd 11/24/11 4:56 PM Page 309 339. 310 C HAP TE R 3 Functions and Their Graphs a. g(x) x2 1 can be rewritten as g(x) (x) 1. 1. The shift (one unit) occurs outside of the function. Therefore, we expect a vertical shift that goes with the sign. 2. Since the sign is negative, this corresponds to a downward shift. 3. Shifting the graph of the function (x) x2 down one unit yields the graph of g(x) x2 1. b. H(x) (x 1)2 can be rewritten as H(x) (x 1). 1. The shift (one unit) occurs inside of the function. Therefore, we expect a horizontal shift that goes opposite the sign. 2. Since the sign is positive, this corresponds to a shift to the left. 3. Shifting the graph of the function (x) x2 to the left one unit yields the graph of H(x) (x 1)2 . x y 2 2 4 1 f (x) (0, 0) (1, 1)g(x) (1, 0) (0, 1) x y 2 2 4 1 f(x) H(x) YO U R TU R N Sketch the graphs of the given functions using horizontal and vertical shifts. a. g(x) x2 1 b. H(x) (x 1)2 Answer: a. b. x y 5 5 10 x y 5 5 10 It is important to note that the domain and range of the resulting function can be thought of as also being shifted. Shifts in the domain correspond to horizontal shifts, and shifts in the range correspond to vertical shifts. b. Graphs of and are shown.y2 = H(x) = (x + 1)2 y1 = x2 E X AM P LE 1 Horizontal and Vertical Shifts Sketch the graphs of the given functions using horizontal and vertical shifts: a. g(x) x2 1 b. H(x) (x 1)2 Solution: In both cases, the function to start with is (x) x2 . x y 2 2 4 f (x) (0, 0) (1, 1) (2, 4)Technology Tip a. Graphs of and are shown.y2 = g(x) = x2 - 1 y1 = x2 c03b.qxd 11/24/11 4:56 PM Page 310 340. 3.3 Graphing Techniques: Transformations 311 Answer: a. x y 2 8 5 5 G(x) = 1x - 2 Domain: [2, ) Range: [0, ) E X AM P LE 2 Horizontal and Vertical Shifts and Changes in the Domain and Range Graph the functions using translations and state the domain and range of each function. a. b. Solution: In both cases the function to start with is . Domain: [0, ) Range: [0, ) a. can be rewritten as g(x) (x 1). g(x) 1x 1 f (x) 1x G(x) = 1x - 2g(x) = 1x + 1 YO U R TU R N Sketch the graph of the functions using shifts and state the domain and range. a. b. h(x) = x + 1G(x) = 1x - 2 Domain: (, ) Range: [1, ) b. h(x) x 1 x y 5 5 5 5 x y 10 5 (1, 1) (4, 2) (9, 3) f(x) = x (0, 0) x y 9 5 (1, 0) (0, 1) (3, 2) (8, 3) 1. The shift (one unit) is inside the function, which corresponds to a horizontal shift opposite the sign. 2. Shifting the graph of to the left one unit yields the graph of . Notice that the point (0, 0), which lies on the graph of (x), gets shifted to the point (1, 0) on the graph of g(x). Although the original function had an implicit restriction on the domain: [0, ), the function has the implicit restriction that x 1. We see that the output or range of g(x) is the same as the output of the original function f(x). Domain: [1, ) Range: [0, ) b. can be rewritten as . 1. The shift (two units) is outside the function, which corresponds to a vertical shift with the sign. 2. The graph of is found by shifting down two units. Note that the point (0, 0), which lies on the graph of (x), gets shifted to the point (0, 2) on the graph of G(x). The original function has an implicit restriction on the domain: [0, ). The function also has the implicit restriction that x 0. The output or range of G(x) is always two units less than the output of the original function (x). Domain: [0, ) Range: [2, ) G(x) 1x 2 f (x) 1x f (x) 1x G(x) 1x 2 G(x) = f(x) - 2 G(x) 1x 2 g(x) 1x 1 f (x) 1x g(x) 1x 1 f (x) 1x x y 10 3 2 (0, 2) (4, 0)(0, 0) (4, 2) (9, 3) c03b.qxd 11/24/11 4:56 PM Page 311 341. x (x) 2 4 1 1 0 0 1 1 2 4 x g(x) 2 4 1 1 0 0 1 1 2 4 x y 5 5 5 5 312 C HAP TE R 3 Functions and Their Graphs x y 5 5 8 (1, 2) Answer: (x) x 2 1 (x) x Domain: (, ) Range: [1, ) x y 5 5 10 YO U R TU R N Sketch the graph of the function (x) x 2 1. State the domain and range of f. All of the previous transformation examples involve starting with a common function and shifting the function in either the horizontal or the vertical direction (or a combination of both). Now, lets investigate reections of functions about the x-axis or y-axis. Reflection about the Axes To sketch the graphs of (x) x2 and g(x) x2 start by rst listing points that are on each of the graphs and then connecting the points with smooth curves. The previous examples have involved graphing functions by shifting a known function either in the horizontal or vertical direction. Let us now look at combinations of horizontal and vertical shifts. E X AM P LE 3 Combining Horizontal and Vertical Shifts Sketch the graph of the function F(x) (x 1)2 2. State the domain and range of F. Solution: The base function is y x2 . 1. The shift (one unit) is inside the function, so it represents a horizontal shift opposite the sign. 2. The 2 shift is outside the function, which represents a vertical shift with the sign. 3. Therefore, we shift the graph of y x2 to the left one unit and down two units. For instance, the point (0, 0) on the graph of y x2 shifts to the point (1, 2) on the graph of F(x) (x 1)2 2. Domain: (, ) Range: [2, ) Technology Tip a. Graphs of , , and are shown. y3 = F(x) = (x + 1)2 - 2 y2 = (x + 1)2 y1 = x2 c03b.qxd 11/24/11 4:56 PM Page 312 342. x f(x) 0 0 1 1 4 2 9 3 x g(x) 9 3 4 2 1 1 0 0 x y 55 5 3.3 Graphing Techniques: Transformations 313 Note that if the graph of is reected about the y-axis, the result is the graph of . Also note that the function g(x) can be written as g(x) (x). In general, reection about the y-axis is produced by replacing x with x in the function. Notice that the domain of is [0, ), whereas the domain of g is (, 0]. g(x) 1x f (x) 1x The graph of (x) is obtained by reecting the graph of (x) about the x-axis. The graph of (x) is obtained by reecting the graph of (x) about the y-axis. REFLECTION ABOUT THE AXES E X AM P LE 4 Sketching the Graph of a Function Using Both Shifts and Reections Sketch the graph of the function . Solution: Start with the square root function. Shift the graph of (x) to the left one unit to arrive at the graph of (x 1). Reect the graph of (x 1) about the x-axis to arrive at the graph of (x 1). f(x 1) 1x 1 f(x 1) 1x 1 f (x) 1x G(x) = - 1x + 1 x y 9 5 5 Note that if the graph of (x) x2 is reected about the x-axis, the result is the graph of g(x) x2 . Also note that the function g(x) can be written as the negative of the function (x); that is, g(x) (x). In general, reection about the x-axis is produced by multiplying a function by 1. Lets now investigate reection about the y-axis. To sketch the graphs of and start by listing points that are on each of the graphs and then connecting the points with smooth curves. g(x) 1x f (x) 1x c03b.qxd 11/24/11 4:56 PM Page 313 343. 3. Vertical shifts: (x) c; 2. Reection: (x) and/or (x) 314 C HAP TE R 3 Functions and Their Graphs x y 55 5 Look back at the order in which transformations were performed in Example 5: horizontal shift, reection, and then vertical shift. Let us consider an alternate order of transformations. WORDS MATH Start with the square root function. Shift the graph of g(x) up one unit to arrive at the graph of g(x) 1. Reect the graph of g(x) 1 about the y-axis to arrive at the graph of g(x) 1. Replace x with x 2, which corresponds to a shift of the graph of g(x) 1 to the right two units to arrive at the graph of g[(x 2)] 1. In the last step we replaced x with x 2, which required us to think ahead knowing the desired result was 2 x inside the radical. To avoid any possible confusion, follow this order of transformations: 1. Horizontal shifts: (x c); g(x 2) 1 22 x 1 g(x) 1 1x 1 g(x) 1 1x 1 g(x) 1x x y 55 5 YO U R TU R N Use shifts and reections to sketch the graph of the function . State the domain and range of (x).f(x) = - 2x - 1 + 2 Answer: Domain: [1, ) Range: (, 2] x y 5 5 10 E X AM P LE 5 Sketching the Graph of a Function Using Both Shifts and Reections Sketch the graph of the function . Solution: Start with the square root function. Shift the graph of g(x) to the left two units to arrive at the graph of g(x 2). Reect the graph of g(x 2) about the y-axis to arrive at the graph of g(x 2). Shift the graph g(x 2) up one unit to arrive at the graph of g(x 2) 1. g(x 2) 1 12 x 1 g(x 2) 1x 2 g(x 2) 1x 2 g(x) 1x f(x) = 12 - x + 1 Technology Tip a. Graphs of , , , and are shown. 12 - x + 1y4 = f(x) = y3 = 1-x + 2 y2 = 1x + 2y1 = 1x c03b.qxd 12/27/11 12:39 PM Page 314 344. 3.3 Graphing Techniques: Transformations 315 Stretching and Compressing Horizontal shifts, vertical shifts, and reections change only the position of the graph in the Cartesian plane, leaving the basic shape of the graph unchanged. These transformations (shifts and reections) are called rigid transformations because they alter only the position. Nonrigid transformations, on the other hand, distort the shape of the original graph. We now consider stretching and compressing of graphs in both the vertical and the horizontal direction. A vertical stretch or compression of a graph occurs when the function is multiplied by a positive constant. For example, the graphs of the functions (x) x2 , g(x) 2(x) 2x2 , and are illustrated below. Depending on if the constant is larger than 1 or smaller than 1 will determine whether it corresponds to a stretch (expansion) or compression (contraction) in the vertical direction. h(x) 1 2 f(x) 1 2 x2 x y 5 5 20 x g(x) 2 8 1 2 0 0 1 2 2 8 x (x) 2 4 1 1 0 0 1 1 2 4 x h(x) 2 2 1 0 0 1 2 2 1 2 1 2 Note that when the function (x) x2 is multiplied by 2, so that g(x) 2(x) 2x2 , the result is a graph stretched in the vertical direction. When the function (x) x2 is multiplied by , so that , the result is a graph that is compressed in the vertical direction. h(x) 1 2 f(x) 1 2 x21 2 The graph of cf(x) is found by: Vertically stretching the graph of (x) if c 1 Vertically compressing the graph of (x) if 0 c 1 Note: c is any positive real number. VERTICAL STRETCHING AND VERTICAL COMPRESSING OF GRAPHS E X AM P LE 6 Vertically Stretching and Compressing Graphs Graph the function Solution: 1. Start with the cube function. (x) x3 2. Vertical compression is expected because is less than 1.1 4 h(x) 1 4 x3 h(x) = 1 4 x3 . c03b.qxd 11/24/11 4:56 PM Page 315 345. 316 C HAP TE R 3 Functions and Their Graphs Solution (a): Since the function is multiplied (on the outside) by 2, the result is that each y-value of (x) is multiplied by 2, which corresponds to vertical stretching. x y 22 10 10 (2, 2) (2, 2) Conversely, if the argument x of a function is multiplied by a positive real number c, then the result is a horizontal stretch of the graph of if 0 c 1. If c 1, then the result is a horizontal compression of the graph of . E X AM P LE 7 Vertically Stretching and Horizontally Compressing Graphs Given the graph of (x), graph: a. 2(x) b. (2x) The graph of (cx) is found by: Horizontally stretching the graph of (x) if 0 c 1 Horizontally compressing the graph of (x) if c 1 Note: c is any positive real number. HORIZONTAL STRETCHING AND HORIZONTAL COMPRESSING OF GRAPHS x y 5 2 1 2 1 (, 0) 2 3 2( , 1) 2( , 1) x y 5 2 1 2 1 2 3 2( , 2) 2( , 2) 3. Determine a few points that lie on the graph of h. (0, 0) (2, 2) (2, 2) c03b.qxd 11/24/11 4:56 PM Page 316 346. 3.3 Graphing Techniques: Transformations 317 E X AM P LE 8 Sketching the Graph of a Function Using Multiple Transformations Sketch the graph of the function H(x) 2(x 3)2 . Solution: Start with the square function. (x) x2 Shift the graph of f(x) to the right three units to arrive at the graph of (x 3). (x 3) (x 3)2 Vertically stretch the graph of (x 3) by a factor of 2 to arrive at the graph of 2(x 3). 2(x 3) 2(x 3)2 Reect the graph 2f (x 3) about the x-axis to arrive at the graph of 2(x 3). 2(x 3) 2(x 3)2 YO U R TU R N Graph the function g(x) 4x3 . Answer: Stretching of the graph (x) x3 . x y g(x) f(x)2 2 40 40 Technology Tip Graphs of and are shown.-2(x - 3)2 y4 = H(x) = y3 = 2(x - 3)2 , y1 = x2 , y2 = (x - 3)2 , x y 5 2 1 2 1 2 3 4( , 1) 4( , 1) 2( , 0) (, 0) x y 4 6 5 5 Solution (b): Since the argument of the function is multiplied (on the inside) by 2, the result is that each x-value of (x) is divided by 2, which corresponds to horizontal compression. In Example 8 we followed the same inside out approach with the functions to determine the order for the transformations: horizontal shift, vertical stretch, and reection. c03b.qxd 12/27/11 12:39 PM Page 317 347. In Exercises 112, match the function to the graph. E X E R C I S E S S E CTI O N 3.3 a. x y b. x y c. x y 318 C HAP TE R 3 Functions and Their Graphs S U M MARY S E CTI O N 3.3 TRANSFORMATION TO GRAPH THE FUNCTION DRAW THE GRAPH OF f AND THEN DESCRIPTION Horizontal shifts (x c) Shift the graph of to the left c units. Replace x by x c. (x c) Shift the graph of to the right c units. Replace x by x c. Vertical shifts (x) c Shift the graph of up c units. Add c to (x). (x) c Shift the graph of down c units. Subtract c from (x). Reection about the x-axis (x) Reect the graph of about the x-axis. Multiply (x) by 1. Reection about the y-axis (x) Reect the graph of about the y-axis. Replace x by x. Vertical stretch c(x), where c 1 Vertically stretch the graph of . Multiply (x) by c. Vertical compression c(x), where 0 c 1 Vertically compress the graph of . Multiply (x) by c. Horizontal stretch (cx), where 0 c 1 Horizontally stretch the graph of . Replace x by cx. Horizontal compression (cx), where c 1 Horizontally compress the graph of . Replace x by cx. 1. (x) x2 1 2. (x) (x 1)2 3. (x) (1 x)2 4. (x) x2 1 5. (x) (x 1)2 6. (x) (1 x)2 1 7. f(x) = 1x - 1 + 1 8. f(x) = - 1x - 1 9. f(x) = 11 - x - 1 10. f(x) = 1-x + 1 11. f(x) = - 1-x + 1 12. f(x) = - 11 - x - 1 d. x y S K I LL S c03b.qxd 11/24/11 4:56 PM Page 318 348. 3.3 Graphing Techniques: Transformations 319 13. Shifted up three units 14. Shifted to the left four units 15. Reected about the y-axis 16. Reected about the x-axis 17. Vertically stretched by a factor of 3 18. Vertically compressed by a factor of 3 19. Shifted down four units 20. Shifted to the right three units 21. Shifted up three units and to the left one unit 22. Reected about the x-axis 23. Reected about the y-axis 24. Reected about both the x-axis and the y-axis In Exercises 1318, write the function whose graph is the graph of y x , but is transformed accordingly. In Exercises 1924, write the function whose graph is the graph of y x3 , but is transformed accordingly. e. x y f. x y g. x y h. x y i. x y j. x y k. x y l. x y 25. x y 26. x y 27. x y In Exercises 2548, use the given graph to sketch the graph of the indicated functions. a. y f (x 2) b. y f (x) 2 a. y f(x 2) b. y f(x) 2 a. y f(x) 3 b. y f(x 3) 28. x y a. y f(x) 3 b. y f(x 3) c03b.qxd 11/24/11 4:56 PM Page 319 349. 320 C HAP TE R 3 Functions and Their Graphs 29. x y 30. x y a. y f(x) b. y f (x) a. y 2f(x) b. y f(2x) a. y 2f(x) b. y f(2x) a. y f(x) b. y f(x) 31. x y 32. x y 33. y f(x 2) 3 34. y f(x 1) 2 35. y f(x 1) 2 36. y 2f(x) 1 37. 38. 39. y g(2x) 40. y = gA1 2xB y = 1 4g(-x) y = -1 2g(x) x f(x) y x g(x) y 41. 42. 43. 44. y = -F(x - 2) - 1 y = -F(1 - x) y = 1 2 F(-x) y = 1 2F(x - 1) + 2 45. 46. 47. 48. y = -G(x - 2) - 1 y = -2G(x - 1) + 3 y = 2G(-x) + 1 y = 2G(x + 1) - 4 In Exercises 4974, graph the function using transformations. 49. y x2 2 50. y x2 3 51. y (x 1)2 52. y (x 2)2 53. y (x 3)2 2 54. y (x 2)2 1 55. y (1 x)2 56. y (x 2)2 57. y x 58. y x 59. y x 2 1 60. y 1 x 2 61. y 2x2 1 62. y 2 x 1 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. y = -1 5 1xy = 51-x y = 2 - 1 1 - x y = 2 - 1 x + 2 y = 1 3 - x y = 1 x + 3 + 2 y = 3 1x + 2 - 1y = 3 1x - 1 + 2y = 12 - x + 3y = - 12 + x - 1 y = 12 - xy = - 1x - 2 x F(x) y x G(x) y c03b.qxd 11/24/11 4:56 PM Page 320 350. 3.3 Graphing Techniques: Transformations 321 In Exercises 7580, transform the function into the form f(x) c(x h)2 k, where c, k, and h are constants, by completing the square. Use graph-shifting techniques to graph the function. 75. y x2 6x 11 76. f(x) x2 2x 2 77. f(x) x2 2x 78. f(x) x2 6x 7 79. f(x) 2x2 8x 3 80. f(x) 3x2 6x 5 For Exercises 85 and 86, refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function where w is weight in kilograms and h is height in centimeters. Since BSA depends on weight and height, it is often thought of as a function of both weight and height. However, for an individual adult height is generally considered constant; thus BSA can be thought of as a function of weight alone. 85. Health/Medicine. (a) If an adult female is 162 centimeters tall, nd her BSA as a function of weight. (b) If she loses 3 kilograms, nd a function that represents her new BSA. 86. Health/Medicine. (a) If an adult male is 180 centimeters tall, nd his BSA as a function of weight. (b) If he gains 5 kilograms, nd a function that represents his new BSA. BSA = A wh 3600 81. Salary. A manager hires an employee at a rate of $10 per hour. Write the function that describes the current salary of the employee as a function of the number of hours worked per week, x. After a year, the manager decides to award the employee a raise equivalent to paying him for an additional 5 hours per week. Write a function that describes the salary of the employee after the raise. 82. Prot. The prot associated with St. Augustine sod in Florida is typically P(x) x2 14,000x 48,700,000, where x is the number of pallets sold per year in a normal year. In rainy years Sod King gives away 10 free pallets per year. Write the function that describes the prot of x pallets of sod in rainy years. 83. Taxes. Every year in the United States each working American typically pays in taxes a percentage of his or her earnings (minus the standard deduction). Karens 2011 taxes were calculated based on the formula T(x) 0.22(x 6500). That year the standard deduction was $6500 and her tax bracket paid 22% in taxes. Write the function that will determine her 2012 taxes, assuming she receives the raise that places her in the 33% bracket. 84. Medication. The amount of medication that an infant requires is typically a function of the babys weight. The number of milliliters of an antiseizure medication A is given by , where x is the weight of the infant in ounces. In emergencies there is often not enough time to weigh the infant, so nurses have to estimate the babys weight. What is the function that represents the actual amount of medication the infant is given if his weight is overestimated by 3 ounces? A(x) = 1x + 2 A P P L I C AT I O N S 87. Describe a procedure for graphing the function Solution: a. Start with the function b. Shift the function to the left three units. c. Shift the function up two units. This is incorrect. What mistake was made? f(x) = 1x. f(x) = 1x - 3 + 2. 88. Describe a procedure for graphing the function Solution: a. Start with the function b. Shift the function to the left two units. c. Reect the function about the y-axis. d. Shift the function down three units. This is incorrect. What mistake was made? f(x) = 1x. f(x) = - 1x + 2 - 3. In Exercises 8790, explain the mistake that is made. C AT C H T H E M I S TA K E c03b.qxd 11/24/11 6:05 PM Page 321 351. 322 C HAP TE R 3 Functions and Their Graphs In Exercises 9194, determine whether each statement is true or false. 91. The graph of y x is the same as the graph of y x . 92. The graph of is the same as the graph of .y = 1xy = 1-x 93. If the graph of an odd function is reected about the x-axis and then the y-axis, the result is the graph of the original odd function. 94. If the graph of is reected about the x-axis, it produces the same graph as if it had been reected about the y-axis. y = 1 x 97. Use a graphing utility to graph: a. y x2 2 and y x2 2 b. y x3 1 and y x3 1 What is the relationship between (x) and (x) ? 98. Use a graphing utility to graph: a. y x2 2 and y x 2 2 b. y x3 1 and y x 3 1 What is the relationship between (x) and ( x )? 99. Use a graphing utility to graph: a. and b. and What is the relationship between (x) and (ax), assuming that a is positive? y = 110xy = 1x y = 10.1xy = 1x 100. Use a graphing utility to graph: a. and b. and What is the relationship between (x) and a(x), assuming that a is positive? 101. Use a graphing utility to graph Use transforms to describe the relationship between f(x) and 102. Use a graphing utility to graph Use transforms to describe the relationship between g(x) and y = [[x]]. y = g(x) = 0.5[[x]] + 1. y = [[x]]. y = f(x) = [[0.5x]] + 1. y = 101xy = 1x y = 0.11xy = 1x 95. The point (a, b) lies on the graph of the function y f(x). What point is guaranteed to lie on the graph of (x 3) 2? 96. The point (a, b) lies on the graph of the function y (x). What point is guaranteed to lie on the graph of (x) 1? 89. Describe a procedure for graphing the function f(x) 3 x 1. Solution: a. Start with the function f(x) x . b. Reect the function about the y-axis. c. Shift the function to the left three units. d. Shift the function up one unit. This is incorrect. What mistake was made? 90. Describe a procedure for graphing the function f(x) 2x2 1. Solution: a. Start with the function f(x) x2 . b. Reect the function about the y-axis. c. Shift the function up one unit. d. Expand in the vertical direction by a factor of 2. This is incorrect. What mistake was made? C O N C E P T U A L C HALLE N G E T E C H N O L O G Y c03b.qxd 11/24/11 4:56 PM Page 322 352. Two different functions can be combined using mathematical operations such as addition, subtraction, multiplication, and division. Also, there is an operation on functions called composition, which can be thought of as a function of a function. When we combine functions, we do so algebraically. Special attention must be paid to the domain and range of the combined functions. Adding, Subtracting, Multiplying, and Dividing Functions Consider the two functions f(x) x2 2x 3 and g(x) x 1. The domain of both of these functions is the set of all real numbers. Therefore, we can add, subtract, or multiply these functions for any real number x. Addition: (x) g(x) x2 2x 3 x 1 x2 3x 2 The result is in fact a new function, which we denote: ( f g) (x) x2 3x 2 This is the sum function. Subtraction: (x) g(x) x2 2x 3 (x 1) x2 x 4 The result is in fact a new function, which we denote: ( f g) (x) x2 x 4 This is the difference function. Multiplication: (x) g(x) (x2 2x 3) (x 1) x3 3x2 x 3 The result is in fact a new function, which we denote: ( f g) (x) x3 3x2 x 3 This is the product function. Although both f and g are dened for all real numbers x, we must restrict x so that to form the quotient . The result is in fact a new function, which we denote: This is called the quotient function. Two functions can be added, subtracted, and multiplied. The resulting function domain is therefore the intersection of the domains of the two functions. However, for division, any value of x (input) that makes the denominator equal to zero must be eliminated from the domain. x Z -1a f gb(x) = x2 + 2x - 3 x + 1 , Division: f(x) g(x) = x2 + 2x - 3 x + 1 , x Z -1 f g x Z -1 C O N C E P TUAL O BJ E CTIVE S Understand domain restrictions when dividing functions. Realize that the domain of a composition of functions excludes values that are not in the domain of the inside function. O P E R ATI O N S O N F U N CTI O N S AN D C O M P O S ITI O N O F F U N CTI O N S S E CTI O N 3.4 S K I LLS O BJ E CTIVE S Add, subtract, multiply, and divide functions. Evaluate composite functions. Determine the domain of functions resulting from operations on and composition of functions. 323 c03c.qxd 11/24/11 4:58 PM Page 323 353. Answer: Domain: [-3, 1] 21 - x( f + g)(x) = 2x + 3 + 324 C HAP TE R 3 Functions and Their Graphs The previous examples involved polynomials. The domain of any polynomial is the set of all real numbers. Adding, subtracting, and multiplying polynomials result in other polynomials, which have domains of all real numbers. Lets now investigate operations applied to functions that have a restricted domain. The domain of the sum function, difference function, or product function is the intersection of the individual domains of the two functions. The quotient function has a similar domain in that it is the intersection of the two domains. However, any values that make the denominator zero must also be eliminated. Function Notation Domain Sum ( f g)(x) (x) g(x) {domain of } {domain of g} Difference ( f g)(x) (x) g(x) {domain of } {domain of g} Product ( f g)(x) (x) g(x) {domain of } {domain of g} Quotient {domain of } {domain of g} We can think of this in the following way: Any number that is in the domain of both the functions is in the domain of the combined function. The exception to this is the quotient function, which also eliminates values that make the denominator equal to zero. E X AM P LE 1 Operations on Functions: Determining Domains of New Functions For the functions and , determine the sum function, difference function, product function, and quotient function. State the domain of these four new functions. Solution: Sum function: Difference function: Product function: Quotient function: The domain of the square root function is determined by setting the argument under the radical greater than or equal to zero. Domain of (x): Domain of g(x): The domain of the sum, difference, and product functions is The quotient function has the additional constraint that the denominator cannot be zero. This implies that , so the domain of the quotient function is [1, 4).x Z 4 (- , 4] = [1, 4][1, ) (-, 4] [1, ) f (x) g(x) = 2x - 1 24 - x = A x - 1 4 - x = 2(x - 1)(4 - x) = 2-x2 + 5x - 4 f (x) # g(x) = 1x - 1 # 14 - x f (x) - g(x) = 2x - 1 - 24 - x f (x) + g(x) = 2x - 1 + 24 - x g(x) = 24 - xf (x) = 2x - 1 {g(x) Z 0}a f g b(x) = f (x) g(x) YO U R TU R N Given the function and , ndg(x) = 11 - xf (x) = 1x + 3 ( f g)(x) and state its domain. c03c.qxd 11/24/11 6:08 PM Page 324 354. 3.4 Operations on Functions and Composition of Functions 325 E X AM P LE 2 Quotient Function and Domain Restrictions Given the functions and G(x) x 3 , nd the quotient function, , and state its domain. Solution: The quotient function is written as Domain of Domain of The real numbers that are in both the domain for F(x) and the domain for G(x) are represented by the intersection . Also, the denominator of the quotient[0, ) (-, ) = [0, ) G(x): (-, )F(x): [0, ) a F G b(x) = F(x) G(x) = 1x x - 3 a F G b(x) F(x) = 1x Composition of Functions Recall that a function maps every element in the domain to exactly one corresponding element in the range as shown in the gure on the right. Suppose there is a sales rack of clothes in a department store. Let x correspond to the original price of each item on the rack. These clothes have recently been marked down 20%. Therefore, the function g(x) 0.80x represents the current sale price of each item. You have been invited to a special sale that lets you take 10% off the current sale price and an additional $5 off every item at checkout. The function f(g(x)) 0.90g(x) 5 determines the checkout price. Note that the output of the function g is the input of the function f as shown in the gure below. This is an example of a composition of functions, when the output of one function is the input of another function. It is commonly referred to as a function of a function. An algebraic example of this is the function . Suppose we let g(x) x2 2 and . Recall that the independent variable in function notation is a placeholder. Since , then . Substituting the expression for g(x), we nd . The function is said to be a composite function, y (g(x)). y = 2x2 - 2f (g(x)) = 2x2 - 2 f (g(x)) = 2(g(x))f (n) = 2(n) f(x) = 2x y = 2x2 - 2 x g(x) = 0.80x g(x) Domain of g Domain of f Range of f Range of g f(x) = 0.90 g(x) 5 f(g(x)) Sale price 20% off original price Additional 10% off sale price and $5 off at checkout Original price function is equal to zero when x 3, so we must eliminate this value from the domain. Domain of a F G b(x): [0, 3)(3, ) Answer: Domain: (0, ) = x - 3 1x a G F b(x) = G(x) F(x) YO U R TU R N For the functions given in Example 2, determine the quotient function and state its domain.a G F b(x), Technology Tip The graphs of , and are shown.y3 = F(x) G(x) = 1x x - 3 y2 = G(x) = x - 3, y1 = F(x) = 1x x f f(x) Domain Range c03c.qxd 11/24/11 4:58 PM Page 325 355. 326 C HAP TE R 3 Functions and Their Graphs E X AM P LE 3 Finding a Composite Function Given the functions (x) x2 1 and g(x) x 3, nd Solution: Write f (x) using placeholder notation. Express the composite function (g(x)) (g(x))2 1 Substitute g(x) x 3 into f. (g(x)) (x 3)2 1 Eliminate the parentheses on the right side. (g(x)) x2 6x 10 ( f g)(x) = f(g(x)) = x2 - 6x + 10 f g. f (n) = (n)2 + 1 ( f g)(x). Study Tip Order is important: (g f )(x) = g(f(x)) ( f g)(x) = f(g(x)) Study Tip The domain of is always a subset of the domain of g, and the range of is always a subset of the range of f. f g f g YO U R TU R N Given the functions in Example 3, nd .(g f )(x) Note that the domain of g(x) is the set of all real numbers, and the domain of (x) is the set of all nonnegative numbers. The domain of a composite function is the set of all x such that g(x) is in the domain of f. For instance, in the composite function y (g(x)), we know that the allowable inputs into f are all numbers greater than or equal to zero. Therefore, we restrict the outputs of g(x) 0 and nd the corresponding x-values. Those x-values are the only allowable inputs and constitute the domain of the composite function y (g(x)). The symbol that represents composition of functions is a small open circle; thus and is read aloud as f of g. It is important not to confuse this with( f g)(x) = f (g(x)) It is important to realize that there are two lters that allow certain values of x into the domain. The rst lter is g(x). If x is not in the domain of g(x), it cannot be in the domain of . Of those values for x that are in the domain of g(x), only some pass through, because we restrict the output of g(x) to values that are allowable as input into f. This adds an additional lter. The domain of is always a subset of the domain of g, and the range of is always a subset of the range of f. f g f g (f g)(x) = f(g(x)) C A U T I O N f g Z f # g x g(x) f(g(x)) ( f g)(x) = f(g(x)) Answer: g f = g( f(x)) = x2 - 2 NOTATION WORDS DEFINITION DOMAIN f composed (g(x)) The set of all real numbers x in the with g domain of g such that g(x) is also in the domain of f. g composed g( f (x)) The set of all real numbers x in the with f domain of f such that (x) is also in the domain of g. g f f g Given two functions f and g, there are two composite functions that can be formed. COMPOSITION OF FUNCTIONS the multiplication sign: ( f g)(x) (x)g(x). c03c.qxd 11/24/11 4:58 PM Page 326 356. 3.4 Operations on Functions and Composition of Functions 327 E X AM P LE 4 Determining the Domain of a Composite Function Given the functions and , determine and state its domain. Solution: Write f(x) using placeholder notation. Express the composite function Substitute into f. Multiply the right side by . What is the domain of ( f g)(x) f(g(x))? By inspecting the nal result of f(g(x)), we see that the denominator is zero when x 1. Therefore, x Z 1. Are there any other values for x that are not allowed? The function g(x) has the domain x Z 0; therefore we must also exclude zero. The domain of ( f g)(x) f(g(x)) excludes x 0 and x 1 or, in interval notation, (- , 0) (0, 1) (1, ) ( f g) = f(g(x)) = x 1 - x f(g(x)) = 1 1 x - 1 # x x = x 1 - x x x f(g(x)) = 1 1 x - 1 g(x) = 1 x f(g(x)) = 1 g(x) - 1 f g. f(n) = 1 (n) - 1 f g,g(x) = 1 x f(x) = 1 x - 1 YO U R TU R N For the functions f and g given in Example 4, determine the composite function g f and state its domain. C A U T I O N The domain of the composite func- tion cannot always be determined by examining the nal form of .f g E X AM P LE 5 Determining the Domain of a Composite Function (Without Finding the Composite Function) Let and . Find the domain of (g(x)). Do not nd the composite function. Solution: Find the domain of g. Find the range of g. In (g(x)), the output of g becomes the input for f. Since the domain of f is the set of all real numbers except 2, we eliminate any values of x in the domain of g that correspond to g(x) 2. Let g(x) 2. Square both sides. x 3 4 Solve for x. x 1 Eliminate x 1 from the domain of g, . State the domain of (g(x)). [-3, 1) (1, ) [-3, ) 1x + 3 = 2 [0, ) [-3, ) g(x) = 1x + 3f(x) = 1 x - 2 Answer: Domain of is , or in interval notation, .(-, 1) (1, ) x Z 1g f g( f (x)) = x - 1. Technology Tip The graphs of and are shown.= 1 1/x - 1 = x 1 - x y3 = ( f g)(x)y2 = g(x) = 1 x , y1 = f(x) = 1 x - 1 , The domain of the composite function cannot always be determined by examining the nal form of f g. c03c.qxd 11/24/11 4:58 PM Page 327 357. 328 C HAP TE R 3 Functions and Their Graphs YO U R TU R N Given the functions f(x) x3 3 and g(x) 1 x3 , evaluate (g(1)) and g((1)). E X AM P LE 6 Evaluating a Composite Function Given the functions (x) x2 7 and g(x) 5 x2 , evaluate: a. (g(1)) b. (g(2)) c. g((3)) d. g((4)) Solution: One way of evaluating these composite functions is to calculate the two individual composites in terms of x: (g(x)) and g((x)). Once those functions are known, the values can be substituted for x and evaluated. Another way of proceeding is as follows: a. Write the desired quantity. (g(1)) Find the value of the inner function g. g(1) 5 12 4 Substitute g(1) 4 into f. (g(1)) (4) Evaluate (4). (4) 42 7 9 b. Write the desired quantity. (g(2)) Find the value of the inner function g. g(2) 5 (2)2 1 Substitute g(2) 1 into f. (g(2)) (1) Evaluate (1). (1) 12 7 6 c. Write the desired quantity. g(f(3)) Find the value of the inner function f. (3) 32 7 2 Substitute (3) 2 into g. g(f(3)) g(2) Evaluate g(2). g(2) 5 22 1 d. Write the desired quantity. g(f(4)) Find the value of the inner function f. f(4) (4)2 7 9 Substitute (4) 9 into g. g(f(4)) g(9) Evaluate g(9). g(9) 5 92 76 g( f(-4)) = -76 g( f(3)) = 1 f(g(-2)) = -6 f(g(1)) = 9 Answer: (g(1)) 5 and g((1)) 7 Application Problems Recall the example at the beginning of this chapter regarding the clothes that are on sale. Often, real-world applications are modeled with composite functions. In the clothes example, x is the original price of each item. The rst function maps its input (original price) to an output (sale price). The second function maps its input (sale price) to an output (checkout price). Example 7 is another real-world application of composite functions. Three temperature scales are commonly used: The degree Celsius (C) scale This scale was devised by dividing the range between the freezing (0C) and boiling (100C) points of pure water at sea level into 100 equal parts. This scale is used in science and is one of the standards of the metric (SI) system of measurements. c03c.qxd 12/27/11 1:15 PM Page 328 358. 3.4 Operations on Functions and Composition of Functions 329 The Kelvin (K) temperature scale This scale shifts the Celsius scale down so that the zero point is equal to absolute zero (about 273.15C), a hypothetical temperature at which there is a complete absence of heat energy. Temperatures on this scale are called kelvins, not degrees kelvin, and kelvin is not capitalized. The symbol for the kelvin is K. The degree Fahrenheit (F) scale This scale evolved over time and is still widely used mainly in the United States, although Celsius is the preferred metric scale. With respect to pure water at sea level, the degrees Fahrenheit are gauged by the spread from 32F (freezing) to 212F (boiling). The equations that relate these temperature scales are F = 9 5 C + 32 C = K - 273.15 E X AM P LE 7 Applications Involving Composite Functions Determine degrees Fahrenheit as a function of kelvins. Solution: Degrees Fahrenheit is a function of degrees Celsius. Now substitute C K 273.15 into the equation for F. Simplify. F = 9 5 K - 459.67 F = 9 5 K - 491.67 + 32 F = 9 5 (K - 273.15) + 32 F = 9 5 C + 32 SMH S U M MARY Operations on Functions Function Notation Sum ( g)(x) (x) g(x) Difference ( g)(x) (x) g(x) Product ( g)(x) (x) g(x) Quotient The domain of the sum, difference, and product functions is the intersection of the domains, or common domain shared by both f and g. The domain of the quotient function is also the intersection of the domain shared by both f and g with an additional restriction that g(x) Z 0. a f g b(x) = f(x) g(x) g(x) Z 0 ## S E CTI O N 3.4 Composition of Functions The domain restrictions cannot always be determined simply by inspecting the nal form of f(g(x)). Rather, the domain of the composite function is a subset of the domain of g(x). Values of x must be eliminated if their corresponding values of g(x) are not in the domain of f. ( f g)(x) = f(g(x)) c03c.qxd 12/27/11 1:15 PM Page 329 359. 330 C HAP TE R 3 Functions and Their Graphs In Exercises 110, given the functions f and g, nd f g, f g, f . g, and , and state the domain of each. 1. f(x) 2x 1 2. f(x) 3x 2 3. f(x) 2x2 x 4. f(x) 3x 2 5. g(x) 1 x g(x) 2x 4 g(x) x2 4 g(x) x2 25 g(x) x 6. 7. 8. 9. 10. g(x) 2x2 In Exercises 1120, for the given functions f and g, nd the composite functions f g and g f, and state their domains. 11. f(x) 2x 1 12. f(x) x2 1 13. 14. 15. f(x) x g(x) x2 3 g(x) 2 x g(x) x 2 g(x) 2 x 16. f (x) x 1 17. 18. 19. f(x) x3 4 20. g(x) x 5 g(x) x2 2 g(x) (x 4)1/3 g(x) x2/3 1 In Exercises 2138, evaluate the functions for the specied values, if possible. 21. ( f g)(2) 22. ( f g)(10) 23. ( f g)(2) 24. ( f g)(5) 25. ( f g)(4) 26. ( f g)(5) 27. 28. 29. (g(2)) 30. (g(1)) 31. g((3)) 32. g((4)) 33. (g(0)) 34. g((0)) 35. (g(3)) 36. 37. 38. In Exercises 3950, evaluate f(g(1)) and g( f(2)), if possible. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. (x) (x 1)1/3 , g(x) x2 2x 1 50. f(x) (1 x2)1/2 , g(x) (x 3)1/3 f(x) = x 2 - x , g(x) = 4 - x2 f(x) = 1 x2 - 3 , g(x) = 1x - 3f(x) = 3 1x - 3, g(x) = 1 x - 3 f(x) = 1x - 1, g(x) = x2 + 5 f(x) = 1 x , g(x) = 2x - 3f(x) = 1 x - 1 , g(x) = x + 3f(x) = 13 - x, g(x) = x2 + 1 f(x) = 11 - x, g(x) = x2 + 2f(x) = x2 + 1, g(x) = 1 2 - x f(x) = 1 x , g(x) = 2x + 1 (g f )(-3)( f g)(4)gA f A 17B B a f g b(2)a f g b(10) f(x) = x2 + 10 g(x) = 1x - 1 g(x) = 1 x f(x) = 3 2x2 - 1f(x) = 12 - xf(x) = 1x - 1 g(x) = 1 x - 1 f(x) = 2 x - 3 f(x) = 1 x - 1 g(x) = 1 x g(x) = 1x + 3g(x) = 21xg(x) = x - 4 3x + 2 f(x) = 11 - 2xf(x) = 14 - xf(x) = 1x - 1f (x) = 1xf(x) = 2x + 3 x - 4 f(x) = 1 x f g S E CTI O N 3.4 E X E R C I S E S S K I LL S c03c.qxd 11/24/11 4:58 PM Page 330 360. 3.4 Operations on Functions and Composition of Functions 331 In Exercises 5160, show that f(g(x)) x and g( f(x)) x. 51. 52. 53. for x 1 54. for x 2 55. for 56. (x) (5 x)1/3 , g(x) 5 x3 57. 58. 59. 60. In Exercises 6166, write the function as a composite of two functions f and g. (More than one answer is correct.) 61. (g(x)) 2(3x 1)2 5(3x 1) 62. 63. 64. 65. 66. f(g(x)) = 1x 31x + 2 f(g(x)) = 3 1x + 1 - 2 f(g(x)) = 21 - x2 f(g(x)) = 2 x - 3 f(g(x)) = 1 1 + x2 f(x) = 225 - x2 , g(x) = 225 - x2 for 0 x 5f(x) = 1 x - 1 , g(x) = x + 1 x for x Z 0, x Z 1 f(x) = 3 18x - 1, g(x) = x3 + 1 8 f(x) = 4x2 - 9, g(x) = 1x + 9 2 for x 0 x Z 0f(x) = 1 x , g(x) = 1 x f(x) = 2 - x2 , g(x) = 12 - xf(x) = 1x - 1, g(x) = x2 + 1 f(x) = x - 2 3 , g(x) = 3x + 2f(x) = 2x + 1, g(x) = x - 1 2 71. Market Price. Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price p and the number of units for sale x are related by the demand equation: Assume that the cost C(x) of producing x items is governed by the equation and the revenue R(x) generated by selling x units is governed by a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the prot as a function of price p. 72. Market Price. Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price p and the number of units for sale x are related by the demand equation: Assume that the cost C(x) of producing x items is governed by the equation and the revenue R(x) generated by selling x units is governed by a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the prot as a function of price p. R(x) = 1000x C(x) = 30,000 + 5x p = 10,000 - 1 4 x R(x) = 100x C(x) = 2000 + 10x p = 3000 - 1 2 x Exercises 67 and 68 depend on the relationship between degrees Fahrenheit, degrees Celsius, and kelvins: 67. Temperature. Write a composite function that converts kelvins into degrees Fahrenheit. 68. Temperature. Convert the following degrees Fahrenheit to kelvins: 32F and 212F. 69. Dog Run. Suppose that you want to build a square fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? 70. Dog Run. Suppose that you want to build a circular fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? F = 9 5 C + 32 C = K - 273.15 A P P L I C AT I O N S c03c.qxd 11/24/11 4:58 PM Page 331 361. 332 C HAP TE R 3 Functions and Their Graphs In Exercises 8186, for the functions f(x) x 2 and g(x) x2 4, nd the indicated function and state its domain. Explain the mistake that is made in each problem. 81. Solution: Domain: (, ) This is incorrect. What mistake was made? = x - 2 = (x - 2)(x + 2) x + 2 g(x) f(x) = x2 - 4 x + 2 g f 82. Solution: Domain: This is incorrect. What mistake was made? (-, 2) (2, ) = 1 x - 2 = x + 2 (x - 2)(x + 2) = 1 x - 2 f(x) g(x) = x + 2 x2 - 4 f g In Exercises 73 and 74, refer to the following: The cost of manufacturing a product is a function of the number of hours t the assembly line is running per day. The number of products manufactured n is a function of the number of hours t the assembly line is operating and is given by the function n(t). The cost of manufacturing the product C measured in thousands of dollars is a function of the quantity manufactured, that is, the function C(n). 73. Business. If the quantity of a product manufactured during a day is given by and the cost of manufacturing the product is given by a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, b. Find the cost of production on a day when the assembly line was running for 16 hours. Interpret your answer. 74. Business. If the quantity of a product manufactured during a day is given by and the cost of manufacturing the product is given by a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, b. Find the cost of production on a day when the assembly line was running for 24 hours. Interpret your answer. In Exercises 75 and 76, refer to the following: Surveys performed immediately following an accidental oil spill at sea indicate the oil moved outward from the source of the spill in a nearly circular pattern. The radius of the oil spill r measured in miles is a function of time t measured in days from the start of the spill, while the area of the oil spill is a function of radius, that is, the function A(r). C(n(t)). C(n) = 8n + 2375 n(t) = 100t - 4t2 C(n(t)). C(n) = 10n + 1375 n(t) = 50t - t2 75. Environment: Oil Spill. If the radius of the oil spill is given by and the area of the oil spill is given by a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, b. Find the area of the oil spill to the nearest square mile 7 days after the start of the spill. 76. Environment: Oil Spill. If the radius of the oil spill is given by and the area of the oil spill is given by a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, b. Find the area of the oil spill to the nearest square mile 5 days after the start of the spill. 77. Environment: Oil Spill. An oil spill makes a circular pattern around a ship such that the radius in feet grows as a function of time in hours . Find the area of the spill as a function of time. 78. Pool Volume. A 20 foot by 10 foot rectangular pool has been built. If 50 cubic feet of water is pumped into the pool per hour, write the water-level height (feet) as a function of time (hours). 79. Fireworks. A family is watching a reworks display. If the family is 2 miles from where the reworks are being launched and the reworks travel vertically, what is the distance between the family and the reworks as a function of height above ground? 80. Real Estate. A couple are about to put their house up for sale. They bought the house for $172,000 a few years ago, and when they list it with a realtor they will pay a 6% commission. Write a function that represents the amount of money they will make on their home as a function of the asking price p. r(t) = 1501t A(r(t)). A(r) = pr2 r(t) = 8t - 0.1t2 A(r(t)). A(r) = pr2 r(t) = 10t - 0.2t2 C AT C H T H E M I S TA K E c03c.qxd 11/24/11 4:58 PM Page 332 362. 3.4 Operations on Functions and Composition of Functions 333 93. For the functions and g(x) x2 a nd and state its domain. 94. For the functions and , nd and state its domain. Assume a 7 1 and b 7 1. g fg(x) = 1 xb f(x) = 1 xa g f f(x) = 1x + a91. For the functions (x) x a and , nd and state its domain. 92. For the functions (x) ax2 bx c and , nd and state its domain.g f g(x) = 1 x - c g f g(x) = 1 x - a 83. Solution: Domain: (, ) This is incorrect. What mistake was made? 84. Given the function (x) x2 7 and , nd , and state the domain. Solution: Domain: (, ) This is incorrect. What mistake was made? = x - 4 = f(g(x)) = x - 3 + 7 f g = f(g(x)) = A 1x - 3 2 2 + 7 f g g(x) = 1x - 3 = x3 + 2x2 - 4x - 8 = (x + 2)(x2 - 4) f g = f(x)g(x) f g 85. Domain: (, ) This is incorrect. What mistake was made? 86. Domain: (, ) This is incorrect. What mistake was made? = -x2 + x - 2 f(x) - g(x) = x + 2 - x2 - 4 = 2x2 + 2x - 4 = (x2 + x - 2)(2) ( f + g)(2) = (x + 2 + x2 - 4)(2) 89. For any functions f and g, ( f g)(x) exists for all values of x that are in the domain of g(x), provided the range of g is a subset of the domain of f. 90. The domain of a composite function can be found by inspection, without knowledge of the domain of the individual functions. In Exercises 8790, determine whether each statement is true or false. 87. When adding, subtracting, multiplying, or dividing two functions, the domain of the resulting function is the union of the domains of the individual functions. 88. For any functions f and g, (g(x)) g((x)) for all values of x that are in the domain of both f and g. C O N C E P T U A L C HALLE N G E 95. Using a graphing utility, plot and . Plot y3 y1 y2. What is the domain of y3? 96. Using a graphing utility, plot , , and . What is the domain of y3?y3 = y1 y2 y2 = 1 13 - x y1 = 3 1x + 5 y2 = 19 - xy1 = 1x + 7 97. Using a graphing utility, plot , , and . If y1 represents a functiony3 = 1 y2 1 - 14 y2 = 1 x2 - 14 y1 = 2x2 - 3x - 4 f and y2 represents a function g, then y3 represents the composite function g f. The graph of y3 is only dened for the domain of g f. State the domain of g f. 98. Using a graphing utility, plot , and y3 y2 1 2. If y1 represents a function f and y2 represents a function g, then y3 represents the composite function g f. The graph of y3 is only dened for the domain of g f. State the domain of g f. y1 = 11 - x, y2 = x2 + 2 T E C H N O L O G Y c03c.qxd 11/25/11 5:39 PM Page 333 363. Every human being has a blood type, and every human being has a DNA sequence. These are examples of functions, where a person is the input and the output is blood type or DNA sequence. These relationships are classied as functions because each person can have one and only one blood type or DNA strand. The difference between these functions is that many people have the same blood type, but DNA is unique to each individual. Can we map backwards? For instance, if you know the blood type, do you know specically which person it came from? No, but, if you know the DNA sequence, you know exactly to which person it corresponds. When a function has a one-to-one correspondence, like the DNA example, then mapping backwards is possible. The map back is called the inverse function. Determine Whether a Function Is One-to-One In Section 3.1, we dened a function as a relationship that maps an input (contained in the domain) to exactly one output (found in the range). Algebraically, each value for x can correspond to only a single value for y. Recall the square, identity, absolute value, and reciprocal functions from our library of functions in Section 3.3. All of the graphs of these functions satisfy the vertical line test. Although the square function and the absolute value function map each value of x to exactly one value for y, these two functions map two values of x to the same value for y. For example, (1, 1) and (1, 1) lie on both graphs. The identity and reciprocal functions, on the other hand, map each x to a single value for y, and no two x-values map to the same y-value. These two functions are examples of one-to-one functions. C O N C E P TUAL O BJ E CTIVE S Visualize the relationships between the domain and range of a function and the domain and range of its inverse. Understand why functions and their inverses are symmetric about y x. O N E-TO-O N E F U N CTI O N S AN D I NVE R S E F U N CTI O N S S E CTI O N 3.5 S K I LLS O BJ E CTIVE S Determine whether a function is a one-to-one function. Verify that two functions are inverses of one another. Graph the inverse function given the graph of the function. Find the inverse of a function. A function (x) is one-to-one if no two elements in the domain correspond to the same element in the range; that is, if x1 Z x2, then f(x1) Z f(x2). D E F I N I T I O N In other words, it is one-to-one if no two inputs map to the same output. One-to-One Function 334 c03c.qxd 11/24/11 4:58 PM Page 334 364. 3.5 One-to-One Functions and Inverse Functions 335 Just as there is a graphical test for functions, the vertical line test, there is a graphical test for one-to-one functions, the horizontal line test. Note that a horizontal line can be drawn on the square and absolute value functions so that it intersects the graph of each function at two points. The identity and reciprocal functions, however, will intersect a horizontal line in at most only one point. This leads us to the horizontal line test for one-to-one functions. EXAMPLE 1 Determining Whether a Function Dened as a Set of Points Is a One-to-One Function For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. Solution: h = {(-1, -1), (0, 0), (1, 1)} g = {(-1, 1), (0, 0), (1, 1)} f = {(0, 0), (1, 1), (1, -1)} f is not a function. g is a function,but not one-to-one. h is a one-to-one function. If every horizontal line intersects the graph of a function in at most one point, then the function is classied as a one-to-one function. D E F I N I T I O N Horizontal Line Test E X AM P LE 2 Using the Horizontal Line Test to Determine Whether a Function Is One-to-One For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. Assume that x is the independent variable and y is the dependent variable. x = y2 y = x2 y = x3 c03c.qxd 11/24/11 4:58 PM Page 335 365. 336 C HAP TE R 3 Functions and Their Graphs Solution: y = x3 y = x2 x = y2 YO U R TU R N Determine whether each of the functions is a one-to-one function. a. (x) x 2 b. (x) x2 1 Another way of writing the denition of a one-to-one function is: If (x1) (x2), then x1 x2. In the Your Turn following Example 2, we found (using the horizontal line test) that (x) x 2 is a one-to-one function, but that (x) x2 1 is not a one-to-one function. We can also use this alternative definition to determine algebraically whether a function is one-to-one. WORDS MATH State the function. (x) x 2 Let there be two real numbers, x1 and x2, such that (x1) (x2). x1 2 x2 2 Subtract 2 from both sides of the equation. x1 x2 (x) x 2 is a one-to-one function. WORDS MATH State the function. f(x) x2 1 Let there be two real numbers, x1 and x2, such that (x1) (x2). 1 1 Subtract 1 from both sides of the equation. Solve for x1. x1 x2 (x) x2 2 is not a one-to-one function. x2 2x2 1 x2 2x2 1 (fails vertical line test) (passes vertical line test (passes both horizontal but fails horizontal line test) and vertical line tests) One-to-one functionFunction, but not one-to-one Not a function x y 5 5 10 10 x y 5 5 10 x y 10 5 5 Answer: a. yes b. no c03c.qxd 11/24/11 4:58 PM Page 336 366. 3.5 One-to-One Functions and Inverse Functions 337 E X AM P LE 3 Determining Algebraically Whether a Function Is One-to-One Determine algebraically whether the following functions are one-to-one: a. (x) 5x3 2 b. (x) x 1 Solution (a): Find (x1) and f(x2). (x1) 5 2 and f(x2) 5 2 Let (x1) (x2). 5 2 5 2 Add 2 to both sides of the equation. 5 5 Divide both sides of the equation by 5. Take the cube root of both sides of the equation. 1/3 1/3 Simplify. x1 x2 f(x) 5x3 2 is a one-to-one function. Solution (b): Find (x1) and (x2). (x1) x1 1 and (x2) x2 1 Let (x1) (x2). x1 1 x2 1 Solve the absolute value equation. (x1 1) (x2 1) or (x1 1) (x2 1) x1 x2 or x1 x2 2 (x) x 1 is not a one-to-one function. Inverse Functions If a function is one-to-one, then the function maps each x to exactly one y, and no two x-values map to the same y-value. This implies that there is a one-to-one correspondence between the inputs (domain) and outputs (range) of a one-to-one function f (x). In the special case of a one-to-one function, it would be possible to map from the output (range of ) back to the input (domain of ), and this mapping would also be a function. The function that maps the output back to the input of a function f is called the inverse function and is denoted f 1 (x). A one-to-one function f maps every x in the domain to a unique and distinct corresponding y in the range. Therefore, the inverse function f 1 maps every y back to a unique and distinct x. The function notations (x) y and 1 (y) x indicate that if the point (x, y) satises the function, then the point (y, x) satises the inverse function. For example, let the function h(x) {(1, 0), (1, 2), (3, 4)}. 1x3 221x3 12 x3 2x3 1 x3 2x3 1 x3 2x3 1 x3 2x3 1 x f y Range of f1 Domain of f1 Domain of f Range of f f1 c03c.qxd 11/24/11 4:58 PM Page 337 367. 338 C HAP TE R 3 Functions and Their Graphs The inverse function undoes whatever the function does. For example, if f(x) 5x, then the function f maps any value x in the domain to a value 5x in the range. If we want to map backwards or undo the 5x, we develop a function called the inverse function that takes 5x as input and maps back to x as output. The inverse function is . Note that if we input 5x into the inverse function, the output is .x: f -1 (5x) = 1 5(5x) = x f -1 (x) = 1 5 x If f and g denote two one-to-one functions such that (g(x)) x for every x in the domain of g and g((x)) x for every x in the domain of f, then g is the inverse of the function f. The function g is denoted by 1 (read f-inverse). Inverse FunctionD E F I N I T I O N Two properties hold true relating one-to-one functions to their inverses: (1) the range of the function is the domain of the inverse, and the range of the inverse is the domain of the function, and (2) the composite function that results with a function and its inverse (and vice versa) is the identity function x. C A U T I O N f -1 Z 1 f Domain of range of 1 and range of domain of 1 1 ((x)) x and (1 (x)) x x f(x) = 5x f1 (5x) = x 5x x f(x) = 5x 5x Domain of f Range of f ? Note: 1 is used to denote the inverse of f. The 1 is not used as an exponent and, therefore, does not represent the reciprocal of f: . 1 f E X AM P LE 4 Verifying Inverse Functions Verify that is the inverse of (x) 2x 4. Solution: Show that 1 ((x)) x and (1 (x)) x. Write 1 using placeholder notation. f1 (n) 1 2 (n) 2 f -1 (x) = 1 2 x - 2 Substitute (x) 2x 4 into 1 . f1 ( f(x)) 1 2 (2x 4) 2 Simplify. 1 ((x)) x 2 2 x 1 ((x)) x Write f using placeholder notation. Substitute into f. f(f 1 (x)) 2a 1 2 x 2b 4f 1 (x) 1 2 x 2 f(n) 2(n) 4 Simplify. (1 (x)) x 4 4 x (1 (x)) x c03c.qxd 11/24/11 4:58 PM Page 338 368. 3.5 One-to-One Functions and Inverse Functions 339 Note the relationship between the domain and range of f and 1 . E X AM P LE 5 Verifying Inverse Functions with Domain Restrictions Verify that 1 (x) x2 , for x 0, is the inverse of . Solution: Show that 1 ((x)) x and (1 (x)) x. Write 1 using placeholder notation. 1 ( ) ( )2 Substitute into 1 . 1 ((x)) x for x 0 Write f using placeholder notation. Substitute 1 (x) x2 , x 0 into f. (1 (x)) x, x 0 (1 (x)) x for x 0 1x2 f(n) 1(n) f1 ( f(x)) A 1x B2 xf(x) = 1x nn f(x) = 1x DOMAIN RANGE (x) 2x 4 (-, )(-, )f -1 (x) = 1 2x - 2 (-, )(-, ) DOMAIN RANGE [0, ) [0, ) 1 (x) x2 , x 0 [0, ) [0, ) f(x) = 1x Graphical Interpretation of Inverse Functions In Example 4, we showed that is the inverse of (x) 2x 4. Lets now investigate the graphs that correspond to the function f and its inverse f 1 . (x) 1 (x) f 1 (x) 1 2 x 2 x y 3 2 2 0 1 2 0 4 x y 2 3 0 2 2 1 4 0 Note that the point (3, 2) lies on the function and the point (2, 3) lies on the inverse. In fact, every point (a, b) that lies on the function corresponds to a point (b, a) that lies on the inverse. Draw the line y x on the graph. In general, the point (b, a) on the inverse 1 (x) is the reection (about y x) of the point (a, b) on the function (x). In general, if the point (a, b) is on the graph of a function, then the point (b, a) is on the graph of its inverse. x y (2, 0) (1, 2) (0, 4) (3, 2) (2, 3) (0, 2) (2, 1) (4, 0) Study Tip If the point (a, b) is on the function, then the point (b, a) is on the inverse. Notice the interchanging of the x- and y-coordinates. c03c.qxd 11/24/11 4:58 PM Page 339 369. Answer: x y 340 C HAP TE R 3 Functions and Their Graphs YO U R TU R N Given the graph of a function f, plot the inverse function. E X AM P LE 6 Graphing the Inverse Function Given the graph of the function f(x), plot the graph of its inverse f1 (x). Solution: Because the points (3, 2), (2, 0), (0, 2), and (2, 4) lie on the graph of f, then the points (2, 3), (0, 2), (2, 0), and (4, 2) lie on the graph of 1 . We have developed the denition of an inverse function and described properties of inverses. At this point, you should be able to determine whether two functions are inverses of one another. Lets turn our attention to another problem: How do you nd the inverse of a function? Finding the Inverse Function If the point (a, b) lies on the graph of a function, then the point (b, a) lies on the graph of the inverse function. The symmetry about the line y x tells us that the roles of x and y interchange. Therefore, if we start with every point (x, y) that lies on the graph of a function, then every point (y, x) lies on the graph of its inverse. Algebraically, this corresponds to interchanging x and y. Finding the inverse of a nite set of ordered pairs is easy: simply interchange the x- and y-coordinates. Earlier, we found that if h(x) {(1, 0), (1, 2), (3, 4)}, then h1 (x) {(0, 1), (2, 1), (4, 3)}. But how do we nd the inverse of a function dened by an equation? Recall the mapping relationship if f is a one-to-one function. This relationship implies that (x) y and 1 (y) x. Lets use these two identities to nd the inverse. Now consider the x f f1 f(x) Domain of f Range of f f1 (y) y Range of f1 Domain of f1 x f (x)y x y (2, 0) (0, 2) (2, 4) (3, 2) (2, 3) (0, 2) (2, 0) (4, 2) x y c03c.qxd 11/24/11 4:58 PM Page 340 370. 3.5 One-to-One Functions and Inverse Functions 341 function dened by (x) 3x 1. To nd 1 , we let (x) y, which yields y 3x 1. Solve for the variable . Recall that 1 (y) x, so we have found the inverse to be . It is customary to write the independent variable as x, so we write the inverse as . Now that we have found the inverse, lets conrm that the properties 1 ((x)) x and ( f1 (x)) x hold. f -1 ( f(x)) = 1 3 (3x - 1) + 1 3 = x - 1 3 + 1 3 = x f 1 f -1 (x)2 = 3a 1 3 x + 1 3 b - 1 = x + 1 - 1 = x f -1 (x) = 1 3 x + 1 3 f -1 (y) = 1 3 y + 1 3 x:x = 1 3 y + 1 3 Let f be a one-to-one function. Then the following procedure can be used to nd the inverse function f 1 if the inverse exists. STEP PROCEDURE EXAMPLE 1 Let y (x). 2 Solve the resulting equation for x in terms of y (if possible). 3 Let x 1 (y). 4 Let y x (interchange x and y). The same result is found if we rst interchange x and y and then solve for y in terms of x. STEP PROCEDURE EXAMPLE 1 Let y (x). 2 Interchange x and y. 3 Solve for y in terms of x. 4 Let y 1 (x) Note the following: Verify rst that a function is one-to-one prior to nding an inverse (if it is not one-to-one, then the inverse does not exist). State the domain restrictions on the inverse function. The domain of f is the range of 1 and vice versa. To verify that you have found the inverse, show that (1 (x)) x for all x in the domain of 1 and 1 ((x)) x for all x in the domain of f. f -1 (x) = -1 3 x + 5 3 y = -1 3 x + 5 3 3y = -x + 5 x = -3y + 5 y = -3x + 5 f(x) = -3x + 5 f -1 (x) = -1 3 x + 5 3 f -1 (y) = -1 3 y + 5 3 x = -1 3 y + 5 3 3x = -y + 5 y = -3x + 5 f(x) = -3x + 5 FINDING THE INVERSE OF A FUNCTION c03c.qxd 12/27/11 1:15 PM Page 341 371. 342 C HAP TE R 3 Functions and Their Graphs E X AM P LE 7 The Inverse of a Square Root Function Find the inverse of the function . State the domain and range of both f and 1 . Solution: (x) is a one-to-one function because it passes the horizontal line test. STEP 1 Let y (x). STEP 2 Interchange x and y. STEP 3 Solve for y. Square both sides of the equation. x2 y 2 Subtract 2 from both sides. x2 2 y or y x2 2 STEP 4 Let y 1 (x). 1 (x) x2 2 Note any domain restrictions. (State the domain and range of both f and 1 .) x = 1y + 2 y = 1x + 2 f(x) = 1x + 2 x y 5 5 5 5 f(x) f 1(x) Technology Tip Using a graphing utility, plot , y2 1 (x) x2 2, and y3 x. y1 = f(x) = 1x + 2 Note that the function f(x) and its inverse 1 (x) are symmetric about the line y x. YO U R TU R N Find the inverse of the given function. State the domain and range of the inverse function. a. (x) 7x 3 b. g(x) = 1x - 1 x y 5 5 5 5 Answers: a. , Domain:f -1 (x) = x + 3 7 Study Tip Had we ignored the domain and range in Example 7, we would have found the inverse function to be the square function f(x) x2 2, which is not a one-to-one function. It is only when we restrict the domain of the square function that we get a one-to-one function. (, ), Range: (, ) b. g1 (x) x2 1, Domain: [0, ), Range: [1, ) f: Domain: [2, ) Range: [0, ) 1 : Domain: [0, ) Range: [2, ) The inverse of is . Check. 1 ((x)) x for all x in the domain of f. (1 (x)) x for all x in the domain of f 1 . Note that the function and its inverse 1 (x) x2 2 for x 0 are symmetric about the line y x. f(x) = 1x + 2 = x = 2x2 for x 0 f( f -1 (x)) = 21x2 - 22 + 2 = x = x + 2 - 2 for x -2 f -1 ( f(x)) = A1x + 2B 2 - 2 f -1 (x) = x2 - 2 for x 0f(x) = 1x + 2 c03c.qxd 11/24/11 4:58 PM Page 342 372. 3.5 One-to-One Functions and Inverse Functions 343 E X AM P LE 8 A Function That Does Not Have an Inverse Function Find the inverse of the function (x) x if it exists. Solution: The function (x) x fails the horizontal line test and therefore is not a one-to-one function. Because is not a one-to-one function, its inverse function does not exist. x y f(x) = |x| Study Tip The range of the function is equal to the domain of its inverse function. YO U R TU R N The function is a one-to-one function. Find its inverse. f(x) = 4 x - 1 , x Z 1, E X AM P LE 9 Finding the Inverse Function The function , is a one-to-one function. Find its inverse. Solution: STEP 1 Let y (x). STEP 2 Interchange x and y. STEP 3 Solve for y. Multiply the equation by (y 3). x(y 3) 2 Eliminate the parentheses. xy 3x 2 Subtract 3x from both sides. xy 3x 2 Divide the equation by x. STEP 4 Let y 1 (x). Note any domain restrictions on f1 (x). The inverse of the function , is . Check. f 1 f -1 (x)2 = 2 a-3 + 2 x b + 3 = 2 a 2 x b = x, x Z 0 f -1 ( f(x)) = -3 + 2 a 2 x + 3 b = -3 + (x + 3) = x, x Z -3 f -1 (x) = -3 + 2 x , x Z 0f(x) = 2 x + 3 , x Z -3 x Z 0 f-1 (x) = -3 + 2 x y = -3x + 2 x = -3 + 2 x x = 2 y + 3 y = 2 x + 3 f(x) = 2 x + 3 , x Z -3 Answer: f -1 (x) = 1 + 4 x , x Z 0 Technology Tip The graphs of , , and , are shown.x Z 0 + 2 x ,y2 = f -1 (x) = -3x Z -3 y1 = f(x) = 2 x + 3 Note that the function f(x) and its inverse 1 (x) are symmetric about the line y x. Note in Example 9 that the domain of f is and the domain of f1 is . Therefore, we know that the range of f is , and the range of f 1 is .(-, -3)(-3, ) (-, 0) (0, )(-, 0) (0, ) (-, -3)(-3, ) c03c.qxd 11/24/11 4:58 PM Page 343 373. 344 C HAP TE R 3 Functions and Their Graphs E X AM P LE 10 Finding the Inverse of a Piecewise-Dened Function The function , is a one-to-one function. Find its inverse. Solution: From the graph of we can make a table with corresponding domain and range values. From this information we can also list domain and range values for f1 . on nd 1 (x) on STEP 1 Let y (x). y 3x STEP 2 Solve for x in terms of y. x 3y STEP 3 Solve for y. STEP 4 Let y 1 (x). on on nd 1 (x) on STEP 1 Let y (x). y x2 STEP 2 Solve for x in terms of y. x y2 STEP 3 Solve for y. STEP 4 Let y 1 (x). STEP 5 The range of 1 is Combining the two pieces yields a piecewise-dened inverse function. f -1 (x) = c 1 3 x x 6 0 1x x 0 f-1 (x) = 1x[0, ) f-1 (x) = ; 1x y = ; 1x [0, ).[0, );f(x) = x2 (-, 0)f-1 (x) = 1 3 x y = 1 3 x (-, 0).(-, 0);f(x) = 3x f(x) = b 3x x 6 0 x2 x 0 x y 5 5 25 25 f(x) = x2 f(x) = 3x (0, 0) DOMAIN OF f RANGE OF f [0, )[0, ) (-, 0)(-, 0) DOMAIN OF f / RANGE OF f1 RANGE OF f / DOMAIN OF f1 [0, )[0, ) (-, 0)(-, 0) c03c.qxd 12/27/11 1:15 PM Page 344 374. 3.5 One-to-One Functions and Inverse Functions 345 Properties of Inverse Functions 1. If is a one-to-one function, then 1 exists. 2. Domain and range Domain of range of 1 Domain of 1 range of 3. Composition of inverse functions 1 ((x)) x for all x in the domain of f. (1 (x)) x for all x in the domain of 1 . 4. The graphs of and 1 are symmetric with respect to the line y x. Procedure for Finding the Inverse of a Function 1. Let y (x). 2. Interchange x and y. 3. Solve for y. 4. Let y 1 (x). S U M MARY One-to-One Functions Each input in the domain corresponds to exactly one output in the range, and no two inputs map to the same output. There are three ways to test a function to determine whether it is a one-to-one function. 1. Discrete points: For the set of all points (a, b) verify that no y-values are repeated. 2. Algebraic equations: Let (x1) (x2); if it can be shown that x1 x2, then the function is one-to-one. 3. Graphs: Use the horizontal line test; if any horizontal line intersects the graph of the function in more than one point, then the function is not one-to-one. S E CTI O N 3.5 E X E R C I S E S S E CTI O N 3.5 In Exercises 116, determine whether the given relation is a function. If it is a function, determine whether it is a one-to-one function. 1. 2. 3. 4. Domain Range A B COURSE GRADE Carrie Michael Jennifer Sean PERSON Domain Range Chris Alex Morgan SPOUSEPERSON Jordan Pat Tim Domain Range (202) 555-1212 (307) 123-4567 (878) 799-6504 10-DIGIT PHONE # PERSON Mary Jason Chester Domain Range 78F 68F AVERAGE TEMPERATURE MONTH October January April S K I LL S c03c.qxd 11/24/11 4:58 PM Page 345 375. 346 C HAP TE R 3 Functions and Their Graphs 5. {(0, 1), (1, 2), (2, 3), (3, 4)} 6. {(0, 2), (2, 0), (5, 3), (5, 7)} 7. {(0, 0), (9, 3), (4, 2), (4, 2), (9, 3)} 8. {(0, 1), (1, 1), (2, 1), (3, 1)} 9. {(0, 1), (1, 0), (2, 1), (2, 1), (5, 4), (3, 4)} 10. {(0, 0), (1, 1), (2, 8), (1, 1), (2, 8)} 11. 12. 13. 14. 15. 16. In Exercises 1724, determine algebraically and graphically whether the function is one-to-one. 17. (x) x 3 18. (x) (x 2)2 1 19. 20. 21. (x) x2 4 22. 23. f(x) x3 1 24. In Exercises 2534, verify that the function f1 (x) is the inverse of f(x) by showing that f( f1 (x)) x and f1 ( f(x)) x. Graph f(x) and f1 (x) on the same axes to show the symmetry about the line y x. 25. 26. 27. 28. 29. 30. f(x) (5 x)1/3 ; 1 (x) 5 x3 31. 32. 33. 34. f(x) = x - 5 3 - x , x Z 3; -1 (x) = 3x + 5 x + 1 , x Z -1f(x) = x + 3 x + 4 , x Z -4; f -1 (x) = 3 - 4x x - 1 , x Z 1 f(x) = 3 4 - x , x Z 4; -1 (x) = 4 - 3 x , x Z 0f(x) = 1 2x + 6 , x Z -3; f -1 (x) = 1 2x - 3, x Z 0 f(x) = 1 x ; f -1 (x) = 1 x , x Z 0 f(x) = 2 - x2 , x 0; f -1 (x) = 12 - x, x 2f(x) = 1x - 1, x 1; f -1 (x) = x2 + 1, x 0 f(x) = x - 2 3 ; f -1 (x) = 3x + 2f(x) = 2x + 1; f -1 (x) = x - 1 2 f(x) = 1 x + 2 f(x) = 1x + 1 f(x) = 3 1xf(x) = 1 x - 1 x y 10 5 x y 5 5 10 x y (2, 1) (2, 3) x y x y x y (6, 4) (0, 1)(1, 1) (2, 2) (2, 2) c03c.qxd 11/24/11 4:58 PM Page 346 376. 3.5 One-to-One Functions and Inverse Functions 347 In Exercises 3542, graph the inverse of the one-to-one function that is given. 35. 36. 37. 38. 39. 40. 41. 42. In Exercises 4360, the function f is one-to-one. Find its inverse, and check your answer. State the domain and range of both f and f1 . 43. (x) x 1 44. (x) 7x 45. (x) 3x 2 46. (x) 2x 3 47. (x) x3 1 48. (x) x3 1 49. 50. 51. (x) x2 1, x 0 52. (x) 2x2 1, x 0 53. (x) (x 2)2 3, x 2 54. (x) (x 3)2 2, x 3 55. 56. 57. 58. 59. 60. In Exercises 6164, graph the piecewise-dened function to determine whether it is a one-to-one function. If it is a one-to-one function, nd its inverse. 61. 62. 63. 64. (x) = u x + 3 x -2 x -2 6 x 6 2 x2 x 2 (x) = u x x -1 x3 -1 6 x 6 1 x x 1 G(x) = u 1 x x 6 0 1x x 0 G(x) = e 0 x 6 0 1x x 0 f(x) = 2x + 5 7 + x f(x) = 7x + 1 5 - x f(x) = 7 x + 2 f(x) = 2 3 - x f(x) = - 3 x f(x) = 2 x (x) = 13 - x(x) = 1x - 3 x y 10 5 x y 10 5 x y 5 5 10 10 x y 28 5 x y 10 5 5 x y 82 10 x y (3, 1) (3, 3) x y (1, 1) (1, 3) c03c.qxd 11/24/11 4:58 PM Page 347 377. 348 C HAP TE R 3 Functions and Their Graphs Security, write a function E(x) that expresses the students take-home pay each week. Find the inverse function E1 (x). What does the inverse function tell you? 70. Salary. A grocery store pays you $8 per hour for the rst 40 hours per week and time and a half for overtime. Write a piecewise-dened function that represents your weekly earnings E(x) as a function of the number of hours worked x. Find the inverse function E1 (x). What does the inverse function tell you? In Exercises 7174, refer to the following: By analyzing available empirical data it was determined that during an illness a patients body temperature uctuated during one 24-hour period according to the function where T represents that patients temperature in degrees Fahrenheit and t represents the time of day in hours measured from 12:00 A.M. (midnight). 71. Health/Medicine. Find the domain and range of the function T(t). 72. Health/Medicine. Find time as a function of temperature, that is, the inverse function t(T). 73. Health/Medicine. Find the domain and range of the function t(T) found in Exercise 72. 74. Health/Medicine. At what time, to the nearest hour, was the patients temperature 99.5 F? T(t) = 0.0003(t - 24)3 + 101.70 65. Temperature. The equation used to convert from degrees Celsius to degrees Fahrenheit is .f(x) = 9 5 x + 32 In Exercises 7578, explain the mistake that is made. 75. Is x y2 a one-to-one function? Solution: Yes, this graph represents a one-to-one function because it passes the horizontal line test. This is incorrect. What mistake was made? 76. A linear one-to-one function is graphed below. Draw its inverse. Solution: Note that the points (3, 3) and (0, 4) lie on the graph of the function. By symmetry, the points (3, 3) and (0, 4) lie on the graph of the inverse. This is incorrect. What mistake was made? x y x y (3, 3) (0, 4) (0, 4) (3, 3) Determine the inverse function C1 (x). What does the inverse function represent? 67. Budget. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to gure out how much money to raise. The entry fee is $250 per boat for the rst 10 boats and $175 for each additional boat. Find the cost function C(x) as a function of the number of boats the club enters x. Find the inverse function that will yield how many boats the club can enter as a function of how much money it will raise. 68. Long-Distance Calling Plans. A phone company charges $.39 per minute for the rst 10 minutes of a long-distance phone call and $.12 per minute every minute after that. Find the cost function C(x) as a function of the length of the phone call in minutes x. Suppose you buy a prepaid phone card that is planned for a single call. Find the inverse function that determines how many minutes you can talk as a function of how much you prepaid. 69. Salary. A student works at Target making $10 per hour and the weekly number of hours worked per week x varies. If Target withholds 25% of his earnings for taxes and Social Determine the inverse function f 1 (x). What does the inverse function represent? 66. Temperature. The equation used to convert from degrees Fahrenheit to degrees Celsius is .C(x) = 5 9 (x - 32) A P P L I C AT I O N S C AT C H T H E M I S TA K E x y (3, 3) (0, 4) c03c.qxd 12/27/11 1:15 PM Page 348 378. 3.5 One-to-One Functions and Inverse Functions 349 77. Given the function (x) x2 , nd the inverse function 1 (x). Solution: Step 1: Let y (x). y x2 Step 2: Solve for x. Step 3: Interchange x and y. Step 4: Let y 1 (x). Check: and . The inverse of (x) x2 is . This is incorrect. What mistake was made? -1 (x) = 1x -1 ((x)) = 2x2 = x1-1 (x)2 = 11x2 2 = x -1 (x) = 1x y = 1x x = 1y 78. Given the function , nd the inverse function 1 (x), and state the domain restrictions on 1 (x). Solution: Step 1: Let y (x). Step 2: Interchange x and y. Step 3: Solve for y. y x2 2 Step 4: Let 1 (x) y. 1 (x) x2 2 Step 5: Domain restrictions: has the domain restriction that x 2. The inverse of is 1 (x) x2 2. The domain of 1 (x) is x 2. This is incorrect. What mistake was made? f (x) = 1x - 2 f(x) = 1x - 2 x = 1y - 2 y = 1x - 2 (x) = 1x - 2 In Exercises 7982, determine whether each statement is true or false. 79. Every even function is a one-to-one function. 80. Every odd function is a one-to-one function. 81. It is not possible that 1 . 82. A function has an inverse. If the function lies in quadrant II, then its inverse lies in quadrant IV. 83. If (0, b) is the y-intercept of a one-to-one function , what is the x-intercept of the inverse 1 ? 84. If (a, 0) is the x-intercept of a one-to-one function , what is the y-intercept of the inverse 1 ? 85. The unit circle is not a function. If we restrict ourselves to the semicircle that lies in quadrants I and II, the graph repre- sents a function, but it is not a one-to-one function. If we fur- ther restrict ourselves to the quarter circle lying in quadrant I, the graph does represent a one-to-one function. Determine the equations of both the one-to-one function and its inverse. State the domain and range of both. 86. Find the inverse of 87. Under what conditions is the linear function (x) mx b a one-to-one function? 88. Assuming that the conditions found in Exercise 87 are met, determine the inverse of the linear function. f(x) = c x , c Z 0. In Exercises 9396, graph the functions f and g and the line y x in the same screen. Do the two functions appear to be inverses of each other? 93. 94. 95. (x) (x 7)1/3 2; g(x) x3 6x2 12x 1 96. f(x) = 1 3 x + 3 - 2; g(x) = x3 + 6x2 + 12x + 6 f(x) = 14 - 3x; g(x) = 4 3 - x2 3 , x 0 f(x) = 13x - 5; g(x) = x2 3 + 5 3 In Exercises 8992, graph the following functions and determine whether they are one-to-one. 89. (x) 4 x2 90. 91. (x) x1/3 x5 92. (x) = 1 x12 (x) = 3 x3 + 2 C O N C E P T U A L C HALLE N G E T E C H N O L O G Y c03c.qxd 11/24/11 4:58 PM Page 349 379. In 2005, the national average cost of residential electricity was 9.53 /kWh (cents per kilowatt-hour). For example, if a residence used 3400 kWh, then the bill would be $324.02, and if a residence used 2500 kWh, then the bill would be $238.25. In this section we discuss mathematical models for different applications. Two quantities in the real world often vary with respect to one another. Sometimes, they vary directly. For example, the more money we make, the more total dollars of federal income tax we expect to pay. Sometimes, quantities vary inversely. For example, when interest rates on mortgages decrease, we expect the number of homes purchased to increase because a buyer can afford more house with the same mortgage payment when rates are lower. In this section we discuss quantities varying directly, inversely, and jointly. Direct Variation When one quantity is a constant multiple of another quantity, we say that the quantities are directly proportional to one another. Let x and y represent two quantities. The following are equivalent statements: y kx, where k is a nonzero constant. y varies directly with x. y is directly proportional to x. The constant k is called the constant of variation or the constant of proportionality. DIRECT VARIATION C O N C E P TUAL O BJ E CTIV E S Understand the difference between direct variation and inverse variation. Understand the difference between combined variation and joint variation. M O D E LI N G F U N CTI O N S U S I N G VAR IATI O N S K I LLS O BJ E CTIVE S Develop mathematical models using direct variation. Develop mathematical models using inverse variation. Develop mathematical models using combined variation. Develop mathematical models using joint variation. S E CTI O N 3.6 350 c03d.qxd 12/27/11 2:12 PM Page 350 380. 3.6 Modeling Functions Using Variation 351 Substitute the given data x 3098 kWh and y $179.99 into y kx. 179.99 3098k Solve for k. In Tennessee the cost of electricity is 5.81 /kWh . y = 0.0581x k = 179.99 3098 = 0.05810 YO U R TU R N Find a mathematical model that describes the cost of electricity in California if the cost is directly proportional to the number of kWh used and a residence that consumes 4000 kWh is billed $480. Let x and y represent two quantities. The following are equivalent statements: y kxn , where k is a nonzero constant. y varies directly with the nth power of x. y is directly proportional to the nth power of x. DIRECT VARIATION WITH POWERS Answer: y 0.12x; the cost of electricity in California is 12 /kWh. Not all variation we see in nature is direct variation. Isometric growth, where the vari- ous parts of an organism grow in direct proportion to each other, is rare in living organ- isms. If organisms grew isometrically, young children would look just like adults, only smaller. In contrast, most organisms grow nonisometrically; the various parts of organisms do not increase in size in a one-to-one ratio. The relative proportions of a human body change dramatically as the human grows. Children have proportionately larger heads and shorter legs than adults. Allometric growth is the pattern of growth whereby different parts of the body grow at different rates with respect to each other. Some human body charac- teristics vary directly, and others can be mathematically modeled by direct variation with powers. One example of direct variation with powers is height and weight of humans. Weight (in pounds) is directly proportional to the cube of height (feet). W = kH3 E X AM P LE 1 Finding the Constant of Variation In the United States, the cost of electricity is directly proportional to the number of kilowatt hours (kWh) used. If a household in Tennessee on average used 3098 kWh per month and had an average monthly electric bill of $179.99, nd a mathematical model that gives the cost of electricity in Tennessee in terms of the number of kilowatt hours used. Solution: Write the direct variation model. y kx Label the variables and constant. x number of kWh y cost (dollars) k cost per kWh # # c03d.qxd 11/24/11 5:00 PM Page 351 381. 352 C HAP TE R 3 Functions and Their Graphs Substitute the given data W 194 and H 6 into W kH3 . 194 k(6)3 Solve for k. Let H 5.5 ft. W 0.9(5.5)3 149.73 A woman 56 tall with the same height and weight proportionality as the male would weigh 150 lb . W = 0.9H3 k = 194 216 = 0.898148 L 0.90 Let x and y represent two quantities. The following are equivalent statements: , where k is a nonzero constant. y varies inversely with x. y is inversely proportional to x. The constant k is called the constant of variation or the constant of proportionality. y = k x INVERSE VARIATION YO U R TU R N A brother and sister both have weight (pounds) that varies as the cube of height (feet) and they share the same proportionality constant. The sister is 6 feet tall and weighs 170 pounds. Her brother is 6 feet 4 inches. How much does he weigh? Answer: 200 pounds Inverse Variation Two fundamental topics covered in economics are supply and demand. Supply is the quantity that producers are willing to sell at a given price. For example, an artist may be willing to paint and sell 5 portraits if each sells for $50, but that same artist may be willing to sell 100 portraits if each sells for $10,000. Demand is the quantity of a good that consumers are not only willing to purchase but also have the capacity to buy at a given price. For example, consumers may purchase 1 billion Big Macs from McDonalds every year, but perhaps only 1 million let mignons are sold at Outback. There may be 1 billion people who want to buy the let mignon but dont have the nancial means to do so. Economists study the equilibrium between supply and demand. Demand can be modeled with an inverse variation of price: when the price increases, demand decreases, and vice versa. E X AM P LE 2 Direct Variation with Powers The following is a personal ad: Single professional male (6 ft/194 lbs) seeks single professional female for long-term relationship. Must be athletic, smart, like the movies and dogs, and have height and weight similarly proportioned to mine. Find a mathematical equation that describes the height and weight of the male who wrote the ad. How much would a 56 woman weigh who has the same proportionality as the male? Solution: Write the direct variation (cube) model for height versus weight. W kH3 c03d.qxd 12/27/11 2:37 PM Page 352 382. 3.6 Modeling Functions Using Variation 353 YO U R TU R N In New York City, the number of potential buyers in the housing market is inversely proportional to the price of a house. If there are 12,500 potential buyers for a $2 million condominium, how many potential buyers are there for a $5 million condominium? If x and y are related by the equation , then we say that y varies inversely with the nth power of x, or y is inversely proportional to the nth power of x. y = k xn Two quantities can vary inversely with the nth power of x. Answer: 5000 Solution: Write the inverse variation model. Label the variables and constant. x price of house in thousands of dollars y number of buyers Select any point that lies on the curve. (200, 500) Substitute the given data x 200 and y 500 into . Solve for k. k 200 500 100,000 Let x 2000. There are only 50 potential buyers for a $2 million house in this city. y = 100,000 2000 = 50 y = 100,000 x 500 = k 200 y = k x y = k x E X AM P LE 3 Inverse Variation The number of potential buyers of a house decreases as the price of the house increases (see graph on the right). If the number of potential buyers of a house in a particular city is inversely proportional to the price of the house, nd a mathematical equation that describes the demand for houses as it relates to price. How many potential buyers will there be for a $2 million house? Demand(numberof potentialbuyers) Price of the house (in thousands of dollars) 800600400200 200 400 600 800 1000 (100, 1000) (200, 500) (400, 250) (600, 167) Joint Variation and Combined Variation We now discuss combinations of variations. When one quantity is proportional to the product of two or more other quantities, the variation is called joint variation. When direct variation and inverse variation occur at the same time, the variation is called combined variation. c03d.qxd 11/24/11 5:00 PM Page 353 383. 354 C HAP TE R 3 Functions and Their Graphs An example of a joint variation is simple interest (Section 1.2), which is dened as where I is the interest in dollars P is the principal (initial) dollars r is the interest rate (expressed in decimal form) t is time in years The interest earned is proportional to the product of three quantities (principal, interest rate, and time). Note that if the interest rate increases, then the interest earned also increases. Similarly, if either the initial investment (principal) or the time the money is invested increases, then the interest earned also increases. An example of combined variation is the combined gas law in chemistry, where P is pressure T is temperature (kelvins) V is volume k is a gas constant This relation states that the pressure of a gas is directly proportional to the temperature and inversely proportional to the volume containing the gas. For example, as the temperature increases, the pressure increases, but when the volume decreases, pressure increases. As an example, the gas in the headspace of a soda bottle has a xed volume. Therefore, as temperature increases, the pressure increases. Compare the different pressures of opening a twist-off cap on a bottle of soda that is cold versus one that is hot. The hot one feels as though it releases more pressure. P = k T V I = Prt E X AM P LE 4 Combined Variation The gas in the headspace of a soda bottle has a volume of 9.0 ml, pressure of 2 atm (atmospheres), and a temperature of 298 K (standard room temperature of 77F). If the soda bottle is stored in a refrigerator, the temperature drops to approximately 279 K (42F). What is the pressure of the gas in the headspace once the bottle is chilled? Solution: Write the combined gas law. Let P 2 atm, T 298 K, and V 9.0 ml. Solve for k. Let , , and . Since we used the same physical units for both the chilled and room-temperature soda bottles, the pressure is in atmospheres. P = 1.87 atm P = 18 298 # 279 9 L 1.87V = 9.0 in P = k T V T = 279k = 18 298 k = 18 298 2 = k 298 9 P = k T V c03d.qxd 11/24/11 5:00 PM Page 354 384. 3.6 Modeling Functions Using Variation 355 In Exercises 116, write an equation that describes each variation. Use k as the constant of variation. 1. y varies directly with x. 2. s varies directly with t. 3. V varies directly with x3 . 4. A varies directly with x2 . 5. z varies directly with m. 6. h varies directly with . 7. varies inversely with . 8. P varies inversely with r2 . 9. F varies directly with w and inversely with L. 10. V varies directly with T and inversely with P. 11. v varies directly with both g and t. 12. S varies directly with both t and d. 13. R varies inversely with both P and T. 14. y varies inversely with both x and z. 15. y is directly proportional to the square root of x. 16. y is inversely proportional to the cube of t. In Exercises 1736, write an equation that describes each variation. 17. d is directly proportional to t. d r when t 1. 18. F is directly proportional to m. F a when m 1. 19. V is directly proportional to both l and w. V 6h when w 3 and l 2. 20. A is directly proportional to both b and h. A 10 when b 5 and h 4. 21. A varies directly with the square of r. A 9p when r 3. 22. V varies directly with the cube of r. V 36p when r 3. 23. V varies directly with both h and r2 . V 1 when r 2 and . 24. W is directly proportional to both R and the square of I. W 4 when R 100 and I 0.25. 25. V varies inversely with P. V 1000 when P 400. 26. I varies inversely with the square of d. I 42 when d 16. 27. F varies inversely with both l and L. F 20 /m2 when l 1 mm and L 100 kilometers.p h = 4 p 1t E X E R C I S E S S E CTI O N 3.6 Joint variation occurs when one quantity is directly proportional to two or more quantities. Combined variation occurs when one quantity is directly proportional to one or more quantities and inversely proportional to one or more other quantities. S U M MARY Direct, inverse, joint, and combined variation can be used to model the relationship between two quantities. For two quantities x and y, we say that y is directly proportional to x if y 5 kx. y is inversely proportional to x if .y = k x S E CTI O N 3.6 S K I LL S c03d.qxd 1/2/12 3:02 PM Page 355 385. 356 C HAP TE R 3 Functions and Their Graphs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 3 5 8 16 17 18 Exercises 41 and 42 are examples of the golden ratio, or phi, a proportionality constant that appears in nature. The numerical approximate value of phi is 1.618. From www.goldenratio.net. 41. Human Anatomy. The length of your forearm F (wrist to elbow) is directly proportional to the length of your hand H (length from wrist to tip of middle nger). Write the equation that describes this relationship if the length of your forearm is 11 inches and the length of your hand is 6.8 inches. 42. Human Anatomy. Each section of your index nger, from the tip to the base of the wrist, is larger than the preceding one by about the golden (Fibonacci) ratio. Find an equation that represents the ratio of each section of your nger related to the previous one if one section is eight units long and the next section is ve units long. 38. Sales Tax. The sales tax in Orange and Seminole counties in Florida differs by only 0.5%. A new resident knows this but doesnt know which of the counties has the higher tax. The resident lives near the border of the counties and is in the market for a new plasma television and wants to purchase it in the county with the lower tax. If the tax on a pair of $40 sneakers is $2.60 in Orange County and the tax on a $12 T-shirt is $0.84 in Seminole County, write two equations: one for each county that describes the tax T, which is directly proportional to the purchase price P. For Exercises 39 and 40, refer to the following: The ratio of the speed of an object to the speed of sound determines the Mach number. Aircraft traveling at a subsonic speed (less than the speed of sound) have a Mach number less than 1. In other words, the speed of an aircraft is directly proportional to its Mach number. Aircraft traveling at a supersonic speed (greater than the speed of sound) have a Mach number greater than 1. The speed of sound at sea level is approximately 760 miles per hour. 39. Military. The U.S. Navy Blue Angels y F-18 Hornets that are capable of Mach 1.7. How fast can F-18 Hornets y at sea level? 40. Military. The U.S. Air Forces newest ghter aircraft is the F-35, which is capable of Mach 1.9. How fast can an F-35 y at sea level? 37. Wages. Jason and Valerie both work at Panera Bread and have the following paycheck information for a certain week. Find an equation that shows their wages W varying directly with the number of hours worked H. EMPLOYEE HOURS WORKED WAGES Jason 23 $172.50 Valerie 32 $240.00 28. y varies inversely with both x and z. y 32 when x 4 and z 0.05. 29. t varies inversely with s. t 2.4 when s 8. 30. W varies inversely with the square of d. W 180 when d 0.2. 31. R varies inversely with the square of I. R 0.4 when I 3.5. 32. y varies inversely with both x and the square root of z. y 12 when x 0.2 and z 4. 33. R varies directly with L and inversely with A. R 0.5 when L 20 and A 0.4. 34. F varies directly with m and inversely with d. F 32 when m 20 and d 8. 35. F varies directly with both m1 and m2 and inversely with the square of d. F 20 when m1 8, m2 16, and d 0.4. 36. w varies directly with the square root of g and inversely with the square of t. w 20 when g 16 and t 0.5. A P P L I C AT I O N S KimSteele/GettyImages,Inc. c03d.qxd 12/27/11 2:14 PM Page 356 386. 3.6 Modeling Functions Using Variation 357 Equilibrium position 2F 2x F x PLANET DISTANCE TO THE SUN Mercury 58,000 km Earth 150,000 km Mars 228,000 km For Exercises 43 and 44, refer to the following: Hookes law in physics states that if a spring at rest (equilibrium position) has a weight attached to it, then the distance the spring stretches is directly proportional to the force (weight), according to the formula: where F is the force in Newtons (N), x is the distance stretched in meters (m), and k is the spring constant (N/m). F = kx For Exercises 49 and 50, refer to the following: In physics, the inverse square law states that any physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity. In particular, the intensity of light radiating from a point source is inversely proportional to the square of the distance from the source. Below is a table of average distances from the Sun: 43. Physics. A force of 30 N will stretch the spring 10 centimeters. How far will a force of 72 N stretch the spring? 44. Physics. A force of 30 N will stretch the spring 10 centimeters. How much force is required to stretch the spring 18 centimeters? 45. Business. A cell phone company develops a pay-as-you-go cell phone plan in which the monthly cost varies directly as the number of minutes used. If the company charges $17.70 in a month when 236 minutes are used, what should the company charge for a month in which 500 minutes are used? 46. Economics. Demand for a product varies inversely with the price per unit of the product. Demand for the product is 10,000 units when the price is $5.75 per unit. Find the demand for the product (to the nearest hundred units) when the price is $6.50. 47. Sales. Levis makes jeans in a variety of price ranges for juniors. The Flare 519 jeans sell for about $20, whereas the 646 Vintage Flare jeans sell for $300. The demand for Levis jeans is inversely proportional to the price. If 300,000 pairs of the 519 jeans were bought, approximately how many of the Vintage Flare jeans were bought? 48. Sales. Levis makes jeans in a variety of price ranges for men. The Silver Tab Baggy jeans sell for about $30, whereas the Offender jeans sell for about $160. The demand for Levis jeans is inversely proportional to the price. If 400,000 pairs of the Silver Tab Baggy jeans were bought, approximately how many of the Offender jeans were bought? 49. Solar Radiation. The solar radiation on the Earth is approximately 1400 watts per square meter . How much solar radiation is there on Mars? Round to the nearest hundred watts per square meter. 50. Solar Radiation. The solar radiation on the Earth is approximately 1400 watts per square meter. How much solar radiation is there on Mercury? Round to the nearest hundred watts per square meter. 51. Investments. Marilyn receives a $25,000 bonus from her company and decides to put the money toward a new car that she will need in two years. Simple interest is directly proportional to the principal and the time invested. She compares two different banks rates on money market accounts. If she goes with Bank of America, she will earn $750 in interest, but if she goes with the Navy Federal Credit Union, she will earn $1500. What is the interest rate on money market accounts at both banks? 52. Investments. Connie and Alvaro sell their house and buy a xer-upper house. They made $130,000 on the sale of their previous home. They know it will take 6 months before the general contractor will start their renovation, and they want to take advantage of a 6-month CD that pays simple interest. What is the rate of the 6-month CD if they will make $3250 in interest? 53. Chemistry. A gas contained in a 4 milliliter container at a temperature of 300 K has a pressure of 1 atmosphere. If the temperature decreases to 275 K, what is the resulting pressure? 54. Chemistry. A gas contained in a 4 milliliter container at a temperature of 300 K has a pressure of 1 atmosphere. If the container changes to a volume of 3 millileters, what is the resulting pressure? (w/m2 ) c03d.qxd 11/24/11 5:00 PM Page 357 387. 358 C HAP TE R 3 Functions and Their Graphs Exercises 61 and 62 involve the theory governing laser propagation through the Earths atmosphere. The three parameters that help classify the strength of optical turbulence are: , index of refraction structure parameter k, wave number of the laser, which is inversely proportional to the wavelength l of the laser: L, propagation distance The variance of the irradiance of a laser s2 is directly proportional to , k7/6 , and L11/16 .C2 n k = 2p C2 n 61. When 1.0 1013 m2/3 , L 2 km, and l 1.55 mm, the variance of irradiance for a plane wave is 7.1. Find the equation that describes this variation. 62. When 1.0 1013 m2/3 , L 2 km, and l 1.55 mm, the variance of irradiance for a spherical wave is 2.3. Find the equation that describes this variation. s2 sp C2 n s2 pl C2 n In Exercises 55 and 56, explain the mistake that is made. 55. y varies directly with t and indirectly with x. When x 4 and t 2, then y 1. Find an equation that describes this variation. Solution: Write the variation equation. y ktx Let x 4, t 2, and y 1. 1 k(2)(4) Solve for k. Substitute into y ktx. This is incorrect. What mistake was made? y = 1 8 txk = 1 8 k = 1 8 56. y varies directly with t and the square of x. When x 4 and t 1, then y 8. Find an equation that describes this variation. Solution: Write the variation equation. Let x 4, t 1, and y 8. Solve for k. Substitute k 4 into . This is incorrect. What mistake was made? y = 4t1xy = kt1x k = 4 8 = k(1)14 y = kt1x In Exercises 57 and 58, determine whether each statement is true or false. 57. The area of a triangle is directly proportional to both the base and the height of the triangle (joint variation). 58. Average speed is directly proportional to both distance and time (joint variation). In Exercises 59 and 60, match the variation with the graph. 59. Inverse variation 60. Direct variation a. b. 10 10 x y 10 10 x y C AT C H T H E M I S TA K E C O N C E P T U A L C HALLE N G E c03d.qxd 11/24/11 5:00 PM Page 358 388. 3.6 Modeling Functions Using Variation 359 NEW, PRIVATELY 5-YEAR TREASURY OIL PRICE, U.S. DOW JONES OWNED CONSTANT MATURITY JANUARY OF EACH YEAR $ PER BARREL UTILITIES STOCK INDEX HOUSING UNITS RATE 1995 17.99 193.12 1407 7.76 1996 18.88 230.85 1467 5.36 1997 25.17 232.53 1355 6.33 1998 16.71 263.29 1525 5.42 1999 12.47 302.80 1748 4.60 2000 27.18 315.14 1636 6.58 2001 29.58 372.32 1600 4.86 2002 19.67 285.71 1698 4.34 2003 32.94 207.75 1853 3.05 2004 32.27 271.94 1911 3.12 2005 46.84 343.46 2137 3.71 2006 65.51 413.84 2265 4.35 a. Use the calculator commands , (ax b),linRegSTAT For Exercises 6366, refer to the following: Data from 1995 to 2006 for oil prices in dollars per barrel, the U.S. Dow Jones Utilities Stock Index, New Privately Owned Housing, and 5-year Treasury Constant Maturity Rate are given in the table. (Data are from Forecast Centers Historical Economic and Market Home Page at www.neatideas.com/djutil.htm.) Use the calculator commands to enter the table with L1 as the oil price, L2 as the utilities stock index, L3 as number of housing units, and L4 as the 5-year maturity rate. EDITSTAT 63. An increase in oil price in dollars per barrel will drive the U.S. Dow Jones Utilities Stock Index to soar. a. Use the calculator commands , (ax b),linRegSTAT 64. An increase in oil price in dollars per barrel will affect the interest rates across the boardin particular, the 5-year Treasury constant maturity rate. and to model the data using the least squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the oil price in dollars per barrel, then use the calculator commands , , and to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel. c. Use the equations you found in (a) and (b) to predict the stock index when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number. STATPLOTPwrRegSTAT STATPLOT and to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel. b. If the 5-year Treasury constant maturity rate varies inversely as the oil price in dollars per barrel, then use the calculator commands , , and STATPLOTPwrRegSTAT STATPLOT to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel. c. Use the equations you found in (a) and (b) to predict the maturity rate when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual maturity rate at 5.02%? Round all answers to two decimal places. T E C H N O L O G Y c03d.qxd 11/24/11 5:00 PM Page 359 389. 360 C HAP TE R 3 Functions and Their Graphs 66. An increase in the number of new, privately owned housing units will affect the U.S. Dow Jones Utilities Stock Index. a. Use the calculator commands , (ax 1 b), and to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the number of housing units. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the number of new, privately owned housing units, then use the calculator commands , , and to model the data using the power function. Find the variation constant and equation of variation using x as the number of housing units. c. Use the equations you found in (a) and (b) to predict the utilities stock index if there are 1861 new, privately owned housing units in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number. STATPLOT PwrRegSTAT STATPLOT linRegSTAT MARCH OF EACH YEAR 2000 2001 2002 2003 2004 2005 2006 2007 2008 RETAIL GASOLINE PRICE $ PER GALLON 1.517 1.409 1.249 1.693 1.736 2.079 2.425 2.563 3.244 67. a. Use the calculator commands to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the year (x 1 for year 2000) and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2006. Round all answers to three decimal places. Is the answer close to the actual price? c. Use the equation to predict the gasoline price in March 2009. Round all answers to three decimal places. LinRegSTAT 68. a. Use the calculator commands to model the data using the power function. Find the variation constant and equation of variation using x as the year (x 1 for year 2000) and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2006. Round all answers to three decimal places. Is the answer close to the actual price? c. Use the equation to predict the gasoline price in March 2009. Round all answers to three decimal places. PwrRegSTAT 65. An increase in interest ratesin particular, the 5-year Treasury constant maturity ratewill affect the number of new, privately owned housing units. a. Use the calculator commands , (ax 1 b), and to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the 5-year rate. b. If the number of new privately owned housing units varies inversely as the 5-year Treasury constant maturity rate, then use the calculator commands , , and to model the data using the power function. Find the variation constant and equation of variation using x as the 5-year rate. c. Use the equations you found in (a) and (b) to predict the number of housing units when the maturity rate is 5.02% in September 2006. Which answer is closer to the actual number of new, privately owned housing units of 1861? Round all answers to the nearest unit. STATPLOTPwrReg STAT STATPLOT linRegSTAT For Exercises 67 and 68, refer to the following: Data for retail gasoline price in dollars per gallon for the period March 2000 to March 2008 are given in the following table. (Data are from Energy Information Administration, Ofcial Energy Statistics from the U.S. government at http://tonto.eia.doe.gov/oog/info/gdu/gaspump.html.) Use the calculator command to enter the table below with L1 as the year (x 1 for year 2000) and L2 as the gasoline price in dollars per gallon. EDITSTAT c03d.qxd 12/27/11 3:10 PM Page 360 390. Transformations of Functions Being a creature of habit, Dylan usually sets out each morning at 7 AM from his house for a jog. Figure 1 shows the graph of a function, y = d(t), that represents Dylans jog on Friday. a. Use the graph in Figure 1 to ll in the table below. Describe a jogging scenario that ts the graph and table above. b. The graph shown in Figure 2 represents Dylans jog on Saturday. It is a transformation of the function y = d(t) shown in Figure 1. Complete the table of values below for this transformation. You may nd it helpful to refer to the table in part (a). What is the real-world meaning of this transformation? How is Dylans jog on Saturday different from his usual jog? How is it the same? The original function (in Figure 1) is represented by the equation y = d(t). Write an equation, in terms of d(t), that represents the function graphed in Figure 2. Explain. c. The graph shown in Figure 3 represents Dylans jog on Sunday. It is a transformation of the function y = d(t) shown in Figure 1. Complete the table of values below for this transformation. What is the real-world meaning of this transformation? How is Dylans jog on Sunday different from his usual jog? C HAP TE R 3 I N Q U I RY- BAS E D LE AR N I N G P R OJ E CT Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Friday y = d(t) t y = d(t) t y t y Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Saturday Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Sunday 361 Figure 1 Figure 2 Figure 3 c03d.qxd 11/25/11 5:15 PM Page 361 391. The original function (in Figure 1) is represented by the equation y = d(t). Use function notation to represent the function graphed in Figure 3. Explain. d. Suppose Dylans jog on Monday can be represented by the equation y = d(t). Complete the table of values below and sketch a graph at the right for this transformation. What is the real-world meaning of this transformation? How does Dylans jog on Monday differ from his usual jog? How is it the same? e. Suppose Dylan has a goal of cutting his usual jogging time in half, while covering the same distance. Represent this scenario as a transformation of y = d(t) shown in Figure 1. Complete the table, sketch a graph, and write an equation in function notation. Explain why your equation makes sense. Finally, discuss whether you think Dylans goal is realistic. 1 2 362 Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jog on Monday t y t y Time (minutes) 20 40 60 80100 30 50 70 Distancefromhome(miles) 5 4 3 2 1 Dylan's Jogging Goal y = ______________ c03d.qxd 11/25/11 5:15 PM Page 362 392. The U.S. National Oceanic and Atmospheric Association (NOAA) monitors temperature and carbon emissions at its observatory in Mauna Loa, Hawaii. NOAAs goal is to help foster an informed society that uses a comprehensive understanding of the role of the oceans, coasts, and atmosphere in the global ecosystem to make the best social and economic decisions. The data presented in this chapter is from the Mauna Loa Observatory, where historical atmospheric measurements have been recorded for the last 50 years. You will develop linear models based on this data to predict temperature and carbon emissions in the future. The following table summarizes average yearly temperature in degrees Fahrenheit F and carbon dioxide emissions in parts per million (ppm) for Mauna Loa, Hawaii. 1. Plot the temperature data with time on the horizontal axis and temperature on the vertical axis. Let t 0 correspond to 1960. 2. Find a linear function that models the temperature in Mauna Loa. a. Use data from 1965 and 1995. b. Use data from 1960 and 1990. c. Use linear regression and all data given. 3. Predict what the temperature will be in Mauna Loa in 2020. a. Apply the line found in Exercise 2(a). b. Apply the line found in Exercise 2(b). c. Apply the line found in Exercise 2(c). 4. Predict what the temperature will be in Mauna Loa in 2100. a. Apply the line found in Exercise 2(a). b. Apply the line found in Exercise 2(b). c. Apply the line found in Exercise 2(c). 5. Do you think your models support the claim of global warming? Explain. 6. Plot the carbon dioxide emissions data with time on the horizontal axis and carbon dioxide levels on the vertical axis. Let t 0 correspond to 1960. 7. Find a linear function that models the CO2 emissions (ppm) in Mauna Loa. a. Use data from 1965 and 1995. b. Use data from 1960 and 1990. c. Use linear regression and all data given. 8. Predict the expected CO2 levels in Mauna Loa in 2020. a. Apply the line found in Exercise 7(a). b. Apply the line found in Exercise 7(b). c. Apply the line found in Exercise 7(c). 9. Predict the expected CO2 levels in Mauna Loa in 2100. a. Apply the line found in Exercise 7(a). b. Apply the line found in Exercise 7(b). c. Apply the line found in Exercise 7(c). 10. Do you think your models support the claim of the greenhouse effect? Explain. Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Temperature (F) 44.45 43.29 43.61 43.35 46.66 45.71 45.53 47.53 45.86 46.23 CO2 Emissions 316.9 320.0 325.7 331.1 338.7 345.9 354.2 360.6 369.4 379.7 (ppm) 363 M O D E LI N G O U R W O R LD c03d.qxd 12/27/11 2:15 PM Page 363 393. C HAP TE R 3 R EVI EW SECTION CONCEPT KEY IDEAS/FORMULAS 3.1 Functions Relations and functions All functions are relations, but not all relations are functions. Functions dened by equations A vertical line can intersect a function in at most one point. Function notation Placeholder notation: Difference quotient: Domain of a function Are there any restrictions on x? 3.2 Graphs of functions; piecewise-dened functions; increasing and decreasing functions; average rate of change Recognizing and classifying functions Common functions (x) mx b, (x) x, (x) x2 , Even and odd functions Even: Symmetry about y-axis: (x) (x) Odd: Symmetry about origin: (x) (x) Increasing and decreasing Increasing: rises (left to right) functions Decreasing: falls (left to right) Average rate of change Piecewise-dened functions Points of discontinuity 3.3 Graphing techniques: Shift the graph of (x). Transformations Horizontal and vertical shifts (x c) c units to the left, c 0 (x c) c units to the right, c 0 (x) c c units upward, c 0 (x) c c units downward, c 0 Reection about the axes (x) Reection about the x-axis (x) Reection about the y-axis Stretching and compressing c(x) if c 1; stretch vertically c(x) if 0 c 1; compress vertically (cx) if c 1; compress horizontally (cx) if 0 c 1; stretch horizontally x1 Z x2 (x2) - (x1) x2 - x1 (x) = x , (x) = 1 x f(x) = x3 , f(x) = 1x, f(x) = 3 1x, (x + h) - (x) h ; h Z 0 f(x) = 3x2 - 6x + 2 f(n) = 3(n)2 - 6(n) + 2 CHAPTERREVIEW 364 c03d.qxd 11/25/11 5:15 PM Page 364 394. SECTION CONCEPT KEY IDEAS/FORMULAS 3.4 Operations on functions and composition of functions Adding, subtracting, ( g)(x) (x) g(x) multiplying, and ( g)(x) (x) g(x) dividing functions ( g)(x) (x) g(x) Domain of the resulting function is the intersection of the individual domains. Domain of the quotient is the intersection of the domains of and g, and any points when g(x) 0 must be eliminated. Composition of functions The domain of the composite function is a subset of the domain of g(x). Values for x must be eliminated if their corresponding values g(x) are not in the domain of . 3.5 One-to-one functions and inverse functions Determine whether No two x-values map to the same y-value. a function is one-to-one If A horizontal line may intersect a one-to-one function in at most one point. Inverse functions Only one-to-one functions have inverses. 1 ((x)) x and (1 (x)) x. Domain of range of 1 . Range of domain of 1 . Graphical interpretation of The graph of a function and its inverse are inverse functions symmetric about the line y x. If the point (a, b) lies on the graph of a function, then the point (b, a) lies on the graph of its inverse. Finding the inverse function 1. Let y (x). 2. Interchange x and y. 3. Solve for y. 4. Let y 1 (x). 3.6 Modeling functions using variation Direct variation y kx Inverse variation Joint variation and combined Joint: One quantity is directly proportional variation to the product of two or more other quantities. Combined: Direct variation and inverse variation occur at the same time. y = k x f(x1) = f(x2), then x1 = x2. ( f g)(x) = f(g(x)) g(x) Z 0a f g b(x) = f(x) g(x) , CHAPTERREVIEW 365 c03d.qxd 11/24/11 5:00 PM Page 365 395. C HAP TE R 3 R EVI EW E X E R C I S E S 3.1 Functions Determine whether each relation is a function. 1. 2. {(1, 2), (3, 4), (2, 4), (3, 7)} 3. {(2, 3), (1, 3), (0, 4), (2, 6)} 4. {(4, 7), (2, 6), (3, 8), (1, 7)} 5. x2 y2 36 6. x 4 7. y x 2 8. 9. 10. Use the graphs of the functions to nd: 11. 12. x y x y x y x y y = 1x Domain Range NAMES Allie Hannah Danny Ethan Vickie AGES 27 10 4 21 13. 14. Evaluate the given quantities using the following three functions. (x) 4x 7 F(t) t2 4t 3 g(x) x2 2x 4 15. (3) 16. F(4) 17. f(7) g(3) 18. 19. 20. 21. 22. Find the domain of the given function. Express the domain in interval notation. 23. (x) 3x 4 24. g(x) x2 2x 6 25. 26. 27. 28. Challenge 29. If , (4) and (4) are undened, and , nd D. 30. Construct a function that is undened at x 3 and x 2 such that the point (0, 4) lies on the graph of the function. 3.2 Graphs of Functions Determine whether the function is even, odd, or neither. 31. (x) 2x 7 32. g(x) 7x5 4x3 2x 33. h(x) x3 7x 34. (x) x4 3x2 35. (x) x1/4 x 36. 37. 38. f(x) = 1 x2 + 3x4 + xf(x) = 1 x3 + 3x f(x) = 1x + 4 f(5) = 2 f(x) = D x2 - 16 H(x) = 1 12x - 6 G(x) = 1x - 4 F(x) = 7 x2 + 3 h(x) = 1 x + 4 F(t + h) - F(t) h f(3 + h) - f(3) h f(3 + h) f(2) - F(2) g(0) F(0) g(0) x y x y a. (1) b. (1) c. x, where (x) 0 a. (4) b. (0) c. x, where (x) 0 a. (2) b. (4) c. x, where (x) 0 a. (5) b. (0) c. x, where (x) 0 REVIEWEXERCISES 366 c03d.qxd 11/24/11 5:00 PM Page 366 396. Review Exercises 367 Use the graph of the functions to nd: a. Domain b. Range c. Intervals on which the function is increasing, decreasing, or constant. 39. 40. 41. Find the average rate of change of (x) 4 x2 from x 0 to x 2. 42. Find the average rate of change of (x) 2x 1 from x 1 to x 5. Graph the piecewise-dened function. State the domain and range in interval notation. 43. 44. 45. 46. Applications 47. Tutoring Costs. A tutoring company charges $25.00 for the rst hour of tutoring and $10.50 for every 30-minute period after that. Find the cost function C(x) as a function of the length of the tutoring session. Let x number of 30-minute periods. 48. Salary. An employee who makes $30.00 per hour also earns time and a half for overtime (any hours worked above the normal 40-hour work week). Write a function E(x) that describes her weekly earnings as a function of the number of hours worked x. 3.3 Graphing Techniques: Transformations Graph the following functions using graphing aids. 49. y (x 2)2 4 50. y x 5 7 F(x) = u x2 x 6 0 x3 0 6 x 6 1 - x - 1 x 1 (x) = u x2 x 0 - 2x 0 6 x 1 x + 2 x 7 1 (x) = u -2x - 3 x 0 4 0 6 x 1 x2 + 4 x 7 1 F(x) = b x2 x 6 0 2 x 0 x y 10 10 10 10 x y 10 10 10 10 51. 52. 53. 54. y 2x2 3 Use the given graph to graph the following: 55. 56. y (x 2) y 3(x) 57. 58. y 2(x) y (x) 3 Write the function whose graph is the graph of , but is transformed accordingly, and state the domain of the resulting function. 59. Shifted to the left three units 60. Shifted down four units 61. Shifted to the right two units and up three units 62. Reected about the y-axis 63. Stretched by a factor of 5 and shifted down six units 64. Compressed by a factor of 2 and shifted up three units Transform the function into the form (x) c(x h)2 k by completing the square and graph the resulting function using transformations. 65. y x2 4x 8 66. y 2x2 6x 5 y 1x x y x y x y x y y = -1 2 x3 y = 1 x - 2 - 4y = 3 1x - 3 + 2 REVIEWEXERCISES c03d.qxd 11/24/11 5:00 PM Page 367 397. 368 C HAP TE R 3 Functions and Their Graphs 3.4 Operations on Functions and Composition of Functions Given the functions g and h, nd g h, g h, g h, and , and state the domain. 67. g(x) 3x 4 h(x) x 3 68. g(x) 2x 3 h(x) x2 6 69. 70. 71. 72. For the given functions f and g, nd the composite functions and , and state the domains. 73. (x) 3x 4 g(x) 2x 1 74. f(x) x3 2x 1 g(x) x 3 75. 76. 77. 78. Evaluate (g(3)) and g((1)), if possible. 79. (x) 4x2 3x 2 g(x) 6x 3 80. g(x) x2 5 f(x) = 14 - x g(x) = 1 x2 - 4 f(x) = 1 2x g(x) = x2 - 4 f(x) = 1x - 5 g(x) = 1x + 6 f(x) = 22x2 - 5 g(x) = 1 4 - x f(x) = 2 x + 3 g ff g h(x) = x + 2 g(x) = x2 - 4 h(x) = 12x + 1 g(x) = 1x - 4 h(x) = 3x - 1 x - 2 g(x) = x + 3 2x - 4 h(x) = 1x g(x) = 1 x2 g h 81. 82. 83. 84. Write the function as a composite f(g(x)) of two functions f and g. 85. h(x) 3(x 2)2 4(x 2) 7 86. 87. 88. Applications 89. Rain. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time, in minutes, , nd the area of the ripple as a function of time. 90. Geometry. Let the area of a rectangle be given by 42 l w, and let the perimeter be 36 2 l 2 w. Express the perimeter in terms of w. 3.5 One-to-One Functions and Inverse Functions Determine whether the given function is a one-to-one function. 91. Chris Harold Tom Danny Paula Vickie Renee Gabriel BROTHER SISTER Domain Range r(t) = 252t + 2 h(x) = 23x + 4 h(x) = 1 2x2 + 7 h(x) = 3 2x 1 - 3 2x g(x) = 1 x2 - 9 f(x) = 4 x2 - 2 g(x) = 3 1x - 4 f(x) = x2 - x + 10 g(x) = x2 - 1 f(x) = 1 x - 1 g(x) = 5x + 2 f(x) = x 2x - 3 REVIEWEXERCISES c03d.qxd 11/24/11 5:00 PM Page 368 398. Review Exercises 369 118. Sales Tax. The sales tax in two neighboring counties differs by 1%. A new resident knows the difference but doesnt know which county has the higher tax rate. The resident lives near the border of the two counties and wants to buy a new car. If the tax on a $50.00 jacket is $3.50 in County A and the tax on a $20.00 calculator is $1.60 in County B, write two equations (one for each county) that describe the tax (T), which is directly proportional to the purchase price (P). Technology Exercises Section 3.1 119. Use a graphing utility to graph the function and nd the domain. Express the domain in interval notation. 120. Use a graphing utility to graph the function and nd the domain. Express the domain in interval notation. f(x) = x2 - 4x - 5 x2 - 9 f(x) = 1 2x2 - 2x - 3 92. 93. {(2, 3), (1, 2), (3, 3), (3, 4), (2, 1)} 94. {(3, 9), (5, 25), (2, 4), (3, 9)} 95. {(2, 0), (4, 5), (3, 7)} 96. {(8, 6), (4, 2), (0, 3), (2, 8), (7, 4)} 97. 98. y x2 99. (x) x3 100. Verify that the function 1 (x) is the inverse of (x) by showing that (1 (x)) x. Graph (x) and 1 (x) on the same graph and show the symmetry about the line y x. 101. (x) 3x 4; 102. ; 103. ; x 0 104. ; The function is one-to-one. Find its inverse and check your answer. State the domain and range of both and 1 . 105. (x) 2x 1 106. (x) x5 2 107. 108. (x) (x 4)2 3 x 4 109. 110. Applications 111. Salary. A pharmaceutical salesperson makes $22,000 base salary a year plus 8% of the total products sold. Write a function S(x) that represents her yearly salary as a function of the total dollars worth of products sold x. Find S1 (x). What does this inverse function tell you? 112. Volume. Express the volume V of a rectangular box that has a square base of length s and is 3 feet high as a function of the square length. Find V1 . If a certain volume is desired, what does the inverse tell you? f(x) = 2 3 2x - 5 - 8f(x) = x + 6 x + 3 (x) = 2x + 4 -1 (x) = 7x + 2 x - 1 (x) = x + 2 x - 7 -1 (x) = x2 - 4(x) = 2x + 4 -1 (x) = 1 + 7x 4x (x) = 1 4x - 7 -1 (x) = x - 4 3 f(x) = 1 x2 y = 2x Domain Range Function STUDENTS IN PRECALC GRADE IN PRECALCULUS COURSE A C D F Tonja Troy Maria Martin Bill Tracey 3.6 Modeling Functions Using Variation Write an equation that describes each variation. 113. C is directly proportional to r. C 2p when r 1. 114. V is directly proportional to both l and w. V 12h when w 6 and l 2. 115. A varies directly with the square of r. A 25p when r 5. 116. F varies inversely with both l and L. F 20p when l 10 mm and L 10 km. Applications 117. Wages. Cole and Dickson both work at the same museum and have the following paycheck information for a certain week. Find an equation that shows their wages (W) varying directly with the number of hours (H) worked. REVIEWEXERCISES EMPLOYEE HOURS WORKED WAGE Cole 27 $229.50 Dickson 30 $255.00 c03d.qxd 11/24/11 5:00 PM Page 369 399. 370 C HAP TE R 3 Functions and Their Graphs Section 3.2 121. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. 122. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. Section 3.3 123. Use a graphing utility to graph and . Use transforms to describe the relationship between and 124. Use a graphing utility to graph and Use transforms to describe the relationship between and Section 3.4 125. Using a graphing utility, plot , , and What is the domain of 126. Using a graphing utility, plot , , and . If represents a function and represents a function , then represents the composite function . The graph of is only dened for the domain of . State the domain of . Section 3.5 127. Use a graphing utility to graph the function and determine whether it is one-to-one. 128. Use a graphing utility to graph the functions and and the line in the same screen. Are the two functions inverses of each other? f(x) = 14 x - 3 + 1, g(x) = x4 - 4x3 + 6x2 - 4x + 3 y = x gf f(x) = 6 5 2x3 - 1 g fg f y3g f y3g y2fy1y3 = y1 2 - 5 y2 = x2 - 5y1 = 2x2 - 4 y3?y3 = y1 y2 . y2 = 14 - xy1 = 12x + 3 g(x)?f(x) g(x) = -2x2 - x + 6. f(x) = 2x2 - 3x - 5 g(x)?f(x) g(x) = x2 - 5x f(x) = x2 - x - 6 f(x) = x2 - 1 1x - 2 + 4 -2 6 x 6 2 x 7 2 f(x) = L 1 - x [[x]] x + 1 x 6 -1 -1 x 6 2 x 7 2 Section 3.6 From December 1999 to December 2007, data for gold price in dollars per ounce are given in the table below. (Data are from Finfacts Ireland Business & Finance Portal, Irelands Top Business website at www.nfacts.ie/Private/curency/ goldmarketprice.htm.) Use the calculator commands to enter the table below with L1 as the year (x 1 for year 1999) and L2 as the gold price in dollars per ounce. EDITSTAT 129. a. Use the calculator commands to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the year (x 1 for year 1999) and y as the gold price in dollars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2005. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2008. Round all answers to two decimal places. 130. a. Use the calculator commands to model the data using the power function. Find the variation constant and equation of variation using x as the year (x 1 for year 1999) and y as the gold price in dollars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2005. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2008. Round all answers to two decimal places. PwrRegSTAT LinRegSTAT DECEMBER OF GOLD PRICE IN EACH YEAR $ PER OUNCE 1999 287.90 2000 272.15 2001 278.70 2002 346.70 2003 414.80 2004 438.10 2005 517.20 2006 636.30 2007 833.20 REVIEWEXERCISES e c03d.qxd 12/27/11 2:06 PM Page 370 400. C HAP TE R 3 P R ACTI C E TE ST Assuming that x represents the independent variable and y represents the dependent variable, classify the relationships as: a. not a function b. a function, but not one-to-one c. a one-to-one function 1. f(x) 2x 3 2. x y2 2 3. Use and g(x) x2 11, and determine the desired quantity or expression. In the case of an expression, state the domain. 4. (11) 2g(1) 5. 6. 7. 8. ( g)(6) 9. Determine whether the function is odd, even, or neither. 10. (x) x x2 11. (x) 9x3 5x 3 12. Graph the functions. State the domain and range of each function. 13. 14. (x) 2(x 1)2 15. Use the graphs of the function to nd: 16. a. (3) b. (0) c. (4) d. x, where (x) 3 e. x, where (x) 0 17. a. g(3) b. g(0) c. g(4) d. x, where g(x) 0 x y y g(x) x y y f(x) (x) = u -x x 6 -1 1 -1 6 x 6 2 x2 x 2 f(x) = - 1x - 3 + 2 f(x) = 2 x f AgA 27B B g( f(x))a g f b(x) a f g b(x) f(x) 1x 2 y = 3 1x + 1 18. a. p(0) b. x, where p(x) 0 c. p(1) d. p(3) 19. 20. Find the average rate of change of the given functions. 21. 22. Given the function f, nd the inverse if it exists. State the domain and range of both and 1 . 23. 24. f(x) x2 5 25. 26. 27. What domain restriction can be made so that (x) x2 has an inverse? 28. If the point (2, 5) lies on the graph of a function, what point lies on the graph of its inverse function? 29. Discount. Suppose a suit has been marked down 40% off the original price. An advertisement in the newspaper has an additional 30% off the sale price coupon. Write a function that determines the checkout price of the suit. 30. Temperature. Degrees Fahrenheit (F), degrees Celsius (C), and kelvins (K) are related by the two equations: and K C 273.15. Write a function whose input is kelvins and output is degrees Fahrenheit. 31. Circles. If a quarter circle is drawn by tracing the unit circle in quadrant III, what does the inverse of that function look like? Where is it located? 32. Sprinkler. A sprinkler head malfunctions at mideld in an NFL football eld. The puddle of water forms a circular pattern around the sprinkler head with a radius in yards that grows as a function of time, in hours: . When will the puddle reach the sidelines? (A football eld is 30 yards from sideline to sideline.) r (t) = 101t F = 9 5C + 32 (x) = e -x x 0 -x2 x 7 0 f(x) = 2x + 1 5 - x f(x) = 1x - 5 f(x) = 1x - 1 for x = 2 to x = 10 f(x) = 64 - 16x2 for x = 0 to x = 2 f(x) = 5 - 7xf(x) = 3x2 - 4x + 1 Find f(x h) f(x) h for: x y y p(x) PRACTICETEST 371 c03d.qxd 11/24/11 5:00 PM Page 371 401. 372 C HAP TE R 3 Functions and Their Graphs 33. Internet. The cost of airport Internet access is $15 for the rst 30 minutes and $1 per minute for each minute after that. Write a function describing the cost of the service as a function of minutes used. Use variation to nd a model for the given problem. 34. y varies directly with the square of x. y 8 when x 5. 35. F varies directly with m and inversely with p. F 20 when m 2 and p 3. 36. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. 37. Use a graphing utility to graph the function and determine whether it is one-to-one. y = x3 - 12x2 + 48x - 65 f(x) = e 5 -4 x 6 -2 3 + 2x - x2 -2 6 x 4 PRACTICETEST c03d.qxd 11/24/11 5:00 PM Page 372 402. C HAP TE R S 13 C U M U LATIVE TE ST CUMULATIVETEST 1. Simplify . 2. Factor completely: 10x2 29x 21. 3. Simplify and state the domain: . 4. Solve for x: . 5. Perform the operation, simplify, and express in standard form: (8 9i)(8 9i). 6. Solve for x, and give any excluded values: . 7. The original price of a hiking stick is $59.50. The sale price is $35.70. Find the percent of the markdown. 8. Solve by factoring: x(6x 1) 12. 9. Solve by completing the square: . 10. Solve and check: . 11. Solve using substitution: x4 x2 12 0. Solve and express the solution in interval notation. 12. 7 3 2x 5 13. 14. 2.7 3.2x 1.3 15. Calculate the distance and midpoint between the segment joining the points (2.7, 1.4) and (5.2, 6.3). 16. Find the slope of the line passing through the points (0.3, 1.4) and (2.7, 4.3). x x - 5 6 0 3 2x + 2 = -3 x2 2 - x = 1 5 5 x - 10 = 10 3x 1 6 x = - 1 5 x + 11 x3 - 4x x + 2 2 3 - 25 17. Write an equation of a line that passes through the points (1.2, 3) and (0.2, 3). 18. Transform the equation into standard form by completing the square, and state the center and radius of the circle: x2 y2 12x 18y 4 0. 19. Find the equation of a circle with center (2, 1) and passing through the point (4, 3). 20. If a cellular phone tower has a reception radius of 100 miles and you live 85 miles north and 23 miles east of the tower, can you use your cell phone at home? Explain. 21. Use interval notation to express the domain of the function . 22. Find the average rate of change for f (x) 5x2 , from x 2 to x 4. 23. Evaluate g((1)) for (x) 6 x and g(x) x2 3. 24. Find the inverse of the function (x) x2 3 for x 0. 25. Write an equation that describes the variation: r is inversely proportional to t. r 45 when t 3. 26. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. 27. Use a graphing utility to graph the function in the same screen. Find the function h such that .g h = f and g(x) = x2 + x - 2 f(x) = x2 - 3x f(x) = e 1 - x -1 x 6 1 1 - x - 2 1 6 x 3 g(x) = 1 x - 1 373373 c03d.qxd 11/24/11 5:00 PM Page 373 403. 4 Polynomial and Rational Functions I ndoor football stadiums are designed so that football punters will not hit the roof with the football. One of the greatest NFL punters of all time was Ray Guy, who played 14 seasons from 1973 to 1986. In the 1976 Pro Bowl, one of his punts hit the giant TV screen hanging from the rafters in the Louisiana Superdome. Not only did Guy punt high and farthe term hang time came into the NFL lexicon during his tenureonce he even had an opponent take a ball he punted and test it for helium! (www.prokicker.com; Ray Guy Fact Sheet) The path that punts typically follow is called a parabola and is classified as a quadratic function. The distance of a punt is measured in the horizontal direction. The yard line where the punt is kicked from and the yard line where the punt either hits the field or is caught are the zeros of the quadratic function. Zeros are the points where the function value is equal to zero. Ray Guy FocusonSport/GettyImages c04a.qxd 11/25/11 3:23 PM Page 374 404. I N TH I S C HAP TE R we will start by discussing quadratic functions (polynomial functions of degree 2) whose graphs are parabolas. We will find the vertex, which is the maximum or minimum point on the graph. Then we will expand our discussion to higher degree polynomial functions. We will discuss techniques to find zeros of polynomial functions and strategies for graphing polynomial functions. Last we will discuss rational functions, which are ratios of polynomial functions. 375 Graphs of Quadratic Functions: Parabolas Finding the Equation of a Parabola Identifying Polynomial Functions Graphing Polynomial Functions Using Transforma- tions of Power Functions Real Zeros of a Polynomial Function Graphing General Polynomial Functions Long Division of Polynomials Synthetic Division of Polynomials The Remainder Theorem and the Factor Theorem The Rational Zero Theorem and Descartes Rule of Signs Factoring Polynomials The Intermediate Value Theorem Graphing Polynomial Functions Complex Zeros Factoring Polynomials Domain of Rational Functions Vertical, Horizontal, and Slant Asymptotes Graphing Rational Functions POLYNOMIAL AND RATIONAL FUNCTIONS 4.1 Quadratic Functions 4.2 Polynomial Functions of Higher Degree 4.3 Dividing Polynomials: Long Division and Synthetic Division 4.4 The Real Zeros of a Polynomial Function 4.5 Complex Zeros: The Fundamental Theorem of Algebra 4.6 Rational Functions Find the vertex (maximum or minimum) of the graph of a quadratic function. Graph polynomial functions. Divide polynomials using long division and synthetic division. Develop strategies for searching for zeros of a polynomial function. Understand that complex zeros come in conjugate pairs. Graph rational functions. L E A R N I N G O B J E C T I V E S c04a.qxd 11/25/11 3:23 PM Page 375 405. Graphs of Quadratic Functions: Parabolas In Chapter 3, we studied functions in general. In this chapter we will learn about a special group of functions called polynomial functions. Polynomial functions are simple functions; often, more complicated functions are approximated by polynomial functions. Polynomial functions model many real-world applications such as the stock market, football punts, business costs, revenues and prots, and the ight path of NASAs vomit comet. Lets start by dening a polynomial function. Q UAD R ATI C F U N CTI O N S S E CTI O N 4.1 Let n be a nonnegative integer, and let an, an1, . . . , a2, a1, a0 be real numbers with The function is called a polynomial function of x with degree n. The coefcient an is called the leading coefcient, and a0 is the constant. f(x) = anxn + an-1xn-1 + + a2x2 + a1x + a0 an Z 0. Polynomial Function Polynomials of particular degrees have special names. In Chapter 3, the library of functions included the constant function f(x) b, which is a horizontal line; the linear function f(x) mx b, which is a line with slope m and y-intercept (0, b); the square function f(x) x2 ; and the cube function f(x) x3 . These are all special cases of a polynomial function. Here are more examples of polynomial functions of particular degree together with their names: Polynomial Degree Special Name f(x) 3 0 Constant function f(x) 2x 1 1 Linear function f(x) 7x2 5x 19 2 Quadratic function f(x) 4x3 2x 7 3 Cubic function The leading coefcients of these functions are 3, 2, 7, and 4, respectively. The constants of these functions are 3, 1, 19, and 7, respectively. In Section 2.3, we discussed graphs of linear functions, which are rst-degree polynomial functions. In this section we will discuss graphs of quadratic functions, which are second-degree polynomial functions. D E F I N I T I O N 376 C O N C E P TUAL O BJ E CTIVE S Recognize characteristics of graphs of quadratic functions (parabolas): whether the parabola opens up or down whether the vertex is a maximum or minimum the axis of symmetry S K I LLS O BJ E CTIVE S Graph a quadratic function in standard form. Graph a quadratic function given in general form. Find the equation of a parabola. c04a.qxd 11/25/11 3:23 PM Page 376 406. 4.1 Quadratic Functions 377 In Section 3.2, the library of functions that we compiled included the square function f(x) x2 , whose graph is a parabola. See the graph on the right. In Section 3.3, we graphed functions using transformation techniques such as F(x) (x 1)2 2, which can be graphed by starting with the square function y x2 and shifting one unit to the left and down two units. See the graph on the right. Note that if we eliminate the parentheses in F(x) (x 1)2 2 to get the result is a function dened by a second-degree polynomial (a polynomial with x2 as the highest degree term), which is also called a quadratic function. = x2 + 2x - 1 F(x) = x2 + 2x + 1 - 2 x F(x) = (x+1)2 2 y 5 5 8 f(x) = x2 x y (h, k) Vertex x = h Axis of Symmetry h k f(x) = ax2 + bx + c a > 0 x y (h, k) Vertex x = h Axis of Symmetry h k f(x) = ax2 + bx + c a < 0 Let a, b, and c be real numbers with . The function is called a quadratic function. f(x) = ax2 + bx + c a Z 0 Quadratic FunctionD E F I N I T I O N The graph of any quadratic function is a parabola. If the leading coefcient a is positive, then the parabola opens up. If the leading coefcient a is negative, then the parabola opens down. The vertex (or turning point) is the minimum point, or low point, on the graph if the parabola opens up, whereas it is the maximum point, or high point, on the graph if the parabola opens down. The vertical line that intersects the parabola at the vertex is called the axis of symmetry. The axis of symmetry is the line x h, and the vertex is located at the point (h, k), as shown in the following two gures. x y 5 5 10 f(x) = x2 At the start of this section, the function F(x) x2 2x 1, which can also be written as F(x) (x 1)2 2, was shown through graph-shifting techniques to have a graph of a parabola. Looking at the graph on top of the margin, we see that the parabola opens up (a 1 0), has a vertex at the point (1, 2), and has an axis of symmetry of x 1. c04a.qxd 11/25/11 3:23 PM Page 377 407. 378 C HAP TE R 4 Polynomial and Rational Functions Graphing Quadratic Functions Given in Standard Form In general, writing a quadratic function in the form allows the vertex (h, k) and the axis of symmetry x h to be determined by inspection. This form is a convenient way to express a quadratic function in order to quickly determine its corresponding graph. Hence, this form is called standard form. f(x) = a(x - h)2 + k To graph f(x) a(x h)2 k Step 1: Determine whether the parabola opens up or down. a 0 up a 0 down Step 2: Determine the vertex (h, k). Step 3: Find the y-intercept (by setting x 0). Step 4: Find any x-intercepts (by setting f(x) 0 and solving for x). Step 5: Plot the vertex and intercepts and connect them with a smooth curve. GRAPHING QUADRATIC FUNCTIONS x y x y x y The quadratic function is in standard form. The graph of f is a parabola whose vertex is the point (h, k). The parabola is symmetric with respect to the line x h. If a 0, the parabola opens up. If a 0, the parabola opens down. f(x) = a(x - h)2 + k QUADRATIC FUNCTION: STANDARD FORM Recall that graphing linear functions requires nding two points on the line, or a point and the slope of the line. However, for quadratic functions simply knowing two points that lie on its graph is no longer sufcient. Below is a general step-by-step procedure for graphing quadratic functions given in standard form. Note that Step 4 says to nd any x-intercepts. Parabolas opening up or down will always have a y-intercept. However, they can have one, two, or no x-intercepts. The gures above illustrate this for parabolas opening up, and the same can be said about parabolas opening down. c04a.qxd 11/25/11 3:23 PM Page 378 408. E X AM P LE 1 Graphing a Quadratic Function Given in Standard Form Graph the quadratic function f(x) (x 3)2 1. Solution: STEP 1 The parabola opens up. a 1, so a 0 STEP 2 Determine the vertex. (h, k) (3, 1) STEP 3 Find the y-intercept. f(0) (3)2 1 8 (0, 8) corresponds to the y-intercept STEP 4 Find any x-intercepts. f(x) (x 3)2 1 0 (x 3)2 1 Use square-root method. x 3 ;1 Solve. x 2 or x 4 (2, 0) and (4, 0) correspond to the x-intercepts STEP 5 Plot the vertex and intercepts (3, 1), (0, 8), (2, 0), (4, 0). Connect the points with a smooth curve opening up. The graph in Example 1 could also have been found by shifting the square function to the right three units and down one unit. YO U R TU R N Graph the quadratic function f(x) (x 1)2 4. x y (3, 1) (2, 0) (4, 0) (0, 8) x y (3, 1) (2, 0) (4, 0) (0, 8) 4.1 Quadratic Functions 379 Answer: x y (1, 4) (0, 3) (1, 0) (3, 0) Study Tip A quadratic function given in stan- dard form can be graphed using the transformation techniques shown in Section 3.3 for the square function. c04a.qxd 11/25/11 3:23 PM Page 379 409. When graphing quadratic functions (parabolas), have at least 3 points labeled on the graph. When there are x-intercepts (Example 1), label the vertex, y-intercept, and x-intercepts. When there are no x-intercepts (Example 2), label the vertex, y-intercept, and another point. 380 C HAP TE R 4 Polynomial and Rational Functions x y (0, 5) (2, 5) (1, 3) Answer: x y (2, 5) (0, 5) (1, 2) Graphing Quadratic Functions in General Form A quadratic function is often written in one of two forms: Standard form: f(x) a(x h)2 k General form: f(x) ax2 bx c When the quadratic function is expressed in standard form, the graph is easily obtained by identifying the vertex (h, k) and the intercepts and drawing a smooth curve that opens either up or down, depending on the sign of a. E X AM P LE 2 Graphing a Quadratic Function Given in Standard Form with a Negative Leading Coefcient Graph the quadratic function f(x) 2(x 1)2 3. Solution: STEP 1 The parabola opens down. a 2, so a 0 STEP 2 Determine the vertex. (h, k) (1, 3) STEP 3 Find the y-intercept. f(0) 2(1)2 3 2 3 5 (0, 5) corresponds to the y-intercept STEP 4 Find any x-intercepts. f(x) 2(x 1)2 3 0 2(x 1)2 3 The square root of a negative number is not a real number. No real solutions. There are no x-intercepts. STEP 5 Plot the vertex (1, 3) and y-intercept (0, 5). Connect the points with a smooth curve. Note that the axis of symmetry is x 1. Because the point (0, 5) lies on the parabola, then by symmetry with respect to x 1, the point (2, 5) also lies on the graph. YO U R TU R N Graph the quadratic function f(x) 3(x 1)2 2. x - 1 = ; A - 3 2 (x - 1)2 = - 3 2 c04a.qxd 11/25/11 3:23 PM Page 380 410. 4.1 Quadratic Functions 381 E X AM P LE 3 Graphing a Quadratic Function Given in General Form Graph the quadratic function f(x) x2 6x 4. Solution: Express the quadratic function in standard form by completing the square. Write the original function. f(x) x2 6x 4 Group the variable terms together. (x2 6x) 4 Complete the square. Half of 6 is 3; 3 squared is 9. Add and subtract 9 within the parentheses. (x2 6x 9 9) 4 Write the 9 outside the parentheses. (x2 6x 9) 9 4 Write the expression inside the parentheses as a perfect square and simplify. (x 3)2 5 Now that the quadratic function is written in standard form, f(x) (x 3)2 5, we follow our step-by-step procedure for graphing a quadratic function in standard form. STEP 1 The parabola opens up. a 1, so a 0 STEP 2 Determine the vertex. (h, k) (3, 5) STEP 3 Find the y-intercept. f(0) (0)2 6(0) 4 4 (0, 4) corresponds to the y-intercept STEP 4 Find any x-intercepts. f(x) 0 f(x) (x 3)2 5 0 (x 3)2 5 ( , 0) and ( , 0) correspond to the x-intercepts. STEP 5 Plot the vertex and intercepts (3, 5), (0, 4), ( , 0), and ( , 0). Connect the points with a smooth parabolic curve. Note: and YO U R TU R N Graph the quadratic function f(x) x2 8x 14. 3 - 15 L 0.76. 3 + 15 L 5.24 3 - 15 3 + 15 3 - 153 + 15 x = 3 ; 15 x - 3 = ; 15 Study Tip Although either form (standard or general) can be used to nd the inter- cepts, it is often more convenient to use the general form when nding the y-intercept and the standard form when nding the x-intercept. Technology Tip Use a graphing utility to graph the function f(x) x2 6x 4 as y1. x y (3, 5) (0, 4) (0.76, 0) (5.24, 0) Answer: x y (5.4, 0) (2.6, 0) (4, 2) Typically, quadratic functions are expressed in general form and a graph is the ultimate goal, so we must rst express the quadratic function in standard form. One technique for transforming a quadratic function from general form to standard form was introduced in Section 1.3 and is called completing the square. c04a.qxd 11/25/11 3:23 PM Page 381 411. 382 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 4 Graphing a Quadratic Function Given in General Form with a Negative Leading Coefcient Graph the quadratic function f(x) 3x2 6x 2. Solution: Express the function in standard form by completing the square. Write the original function. f(x) 3x2 6x 2 Group the variable terms together. (3x2 6x) 2 Factor out 3 in order to make the coefcient of x2 equal to 1 inside the parentheses. 3(x2 2x) 2 Add and subtract 1 inside the parentheses to create a perfect square. 3(x2 2x 1 1) 2 Regroup the terms. 3(x2 2x 1) 3(1) 2 Write the expression inside the parentheses as a perfect square and simplify. 3(x 1)2 5 Now that the quadratic function is written in standard form, f(x) 3(x 1)2 5, we follow our step-by-step procedure for graphing a quadratic function in standard form. STEP 1 The parabola opens down. a 3, therefore a 0 STEP 2 Determine the vertex. (h, k) (1, 5) STEP 3 Find the y-intercept using f(0) 3(0)2 6(0) 2 2 the general form. (0, 2) corresponds to the y-intercept STEP 4 Find any x-intercepts using the standard form. f(x) 3(x 1)2 5 0 3(x 1)2 5 x - 1 = ; A 5 3 (x - 1)2 = 5 3 x y (1, 5) (0, 2) (0.3, 0) (2.3, 0) Answer: x y 5 5 5 (1, 3) (2.2, 0) (0, 1) (0.2, 0) 5 Technology Tip Use a graphing utility to graph the function f(x) 3x2 6x 2 as y1. x = 1 ; A 5 3 The x-intercepts are and STEP 5 Plot the vertex and intercepts (1, 5), (0, 2), , anda1 + 115 3 , 0b a1 - 115 3 , 0b.a1 + 115 3 , 0b = 1 ; 115 3 . Connect the points with a smooth curve. Note: and . YO U R TU R N Graph the quadratic function f(x) 2x2 4x 1. 1 - 115 3 L -0.3 1 + 115 3 L 2.3 a1 - 115 3 , 0b When the leading coefcient of a quadratic function is not equal to 1, the leading coefcient must be factored out before completing the square. c04a.qxd 11/25/11 3:23 PM Page 382 412. 4.1 Quadratic Functions 383 The graph of a quadratic function f(x) ax2 bx c is a parabola with the vertex located at the point a- b 2a , 4ac - b2 4a ba- b 2a , f a- b 2a b b VERTEX OF A PARABOLA Step 1: Find the vertex. Step 2: Determine whether the parabola opens up or down. If a 0, the parabola opens up. If a 0, the parabola opens down. Step 3: Find additional points near the vertex. Step 4: Sketch the graph with a parabolic curve. GRAPHING A QUADRATIC FUNCTION IN GENERAL FORM x 1 0 1 2 3 f(x) f(1) 1 f(0) 5 f(1) 7 f(2) 5 f(3) 1 x y (0, 5) (2, 5) (1, 7) (3, 1)(1, 1) Answer: x y 2 2 8 8 (1, 1) (2, 4)(0, 4) E X AM P LE 5 Graphing a Quadratic Function Given in General Form Sketch the graph of f(x) 2x2 4x 5. Solution: Let a 2, b 4, and c 5. STEP 1 Find the vertex. x = - b 2a = - 4 2(-2) = 1 In Examples 3 and 4, the quadratic functions were given in general form, and they were transformed into standard form by completing the square. It can be shown (by completing the square) that the vertex of a quadratic function in general form, f(x) ax2 bx c, is located at Another approach to sketching the graphs of quadratic functions is to rst nd the vertex and then nd additional points through point-plotting. x = - b 2a . f(1) 2(1)2 4(1) 5 7 Vertex (1, 7) STEP 2 The parabola opens down. a 2 STEP 3 Find additional points near the vertex. STEP 4 Label the vertex and additional points, then sketch the graph. YO U R TU R N Sketch the graph of f(x) 3x2 6x 4. c04a.qxd 11/25/11 3:23 PM Page 383 413. 384 C HAP TE R 4 Polynomial and Rational Functions Finding the Equation of a Parabola It is important to understand that the equation y x2 is equivalent to the quadratic function f(x) x2 . Both have the same parabolic graph. Thus far, we have been given the equation and then asked to nd characteristics (vertex and intercepts) in order to graph. We now turn our attention to the problem of determining the equation, or function, given certain characteristics. E X AM P LE 6 Finding the Equation of a Parabola Given the Vertex and Another Point Find the equation of a parabola whose graph has a vertex at (3, 4) and that passes through the point (2, 3). Express the quadratic function in both standard and general forms. Solution: Write the standard form of the equation of a parabola. f(x) a(x h)2 k Substitute the coordinates of the vertex (h, k) (3, 4). f(x) a(x 3)2 4 Use the point (2, 3) to nd a. The point (2, 3) implies f(2) 3. f(2) a(2 3)2 4 3 Solve for a. a(2 3)2 4 3 a(1)2 4 3 a 4 3 a 1 Write both forms of the equation of this parabola. Standard form: General form: YO U R T U R N Find the standard form of the equation of a parabola whose graph has a vertex at (3, 5) and that passes through the point (2, 4). f(x) = -x2 + 6x - 5f(x) = -(x - 3)2 + 4 As we have seen in Example 6, once the vertex is known, the leading coefcient a can be found from any point that lies on the parabola. Application Problems That Involve Quadratic Functions What is the minimum distance a driver has to maintain in order to be at a safe distance between her car and the car in front as a function of speed? How many units produced will yield a maximum prot for a company? Given the number of linear feet of fence, what rectangular dimensions will yield a maximum fenced-in area? If a particular stock price has been shown to follow a quadratic trend, when will the stock achieve a maximum value? If a gun is red into the air, where will the bullet land? These are all problems that can be solved using quadratic functions. Because the vertex of a parabola represents either the minimum or maximum value of the quadratic function, in application problems it often sufces simply to nd the vertex. Answer: Standard form: f(x) (x 3)2 5 c04a.qxd 11/25/11 3:23 PM Page 384 414. 4.1 Quadratic Functions 385 In Example 6, the function F(x) x2 6x 5, which can also be written as F(x) (x 3)2 4, was shown to open down and has a vertex at the point (3, 4). Suppose this function represents prot, where x is the number (millions) of units made. The vertex Technology Tip Use a graphing utility to graph the cost function C(x) 2000 15x 0.05x2 as y1. Answer: 250,000 bottles would then represent the maximum prot. Instead of rewriting the function in standard form through completing the square, we use the vertex formula . WORDS MATH Quadratic function. F(x) x2 6x 5 Coefcients. a 1, b 6, c 5 Find the x-coordinate of the vertex. Find the value of the function at x 3. f(3) (3)2 6(3) 5 4 Therefore, the vertex is located at the point (3, 4) . x = - b 2a = - 6 2(-1) = 3 x = - b 2a E X AM P LE 7 Finding the Minimum Cost of Manufacturing a Motorcycle A company that produces motorcycles has a daily production cost of where C is the cost in dollars to manufacture a motorcycle and x is the number of motorcycles produced. How many motorcycles can be produced each day in order to minimize the cost of each motorcycle? What is the corresponding minimum cost? Solution: The graph of the quadratic function is a parabola. Rewrite the quadratic function in general form. C(x) 0.05x2 15x 2000 The parabola opens up, because a is positive. a 0.05 0 Because the parabola opens up, the vertex of the parabola is a minimum. Find the x-coordinate of the vertex. The company keeps costs to a minimum when 150 motorcycles are produced each day. The minimum cost is $875 per motorcycle. C(150) 875 YO U R T U R N The revenue associated with selling vitamins is where R is the revenue in dollars and x is the number of bottles of vitamins sold. Determine how many bottles of vitamins should be sold to maximize the revenue. R(x) = 500x - 0.001x2 x = - b 2a = - (-15) 2(0.05) = 150 C(x) = 2000 - 15x + 0.05x2 c04a.qxd 11/25/11 3:24 PM Page 385 415. 386 C HAP TE R 4 Polynomial and Rational Functions Technology Tip Use a graphing utility to graph the area function A(x) 2x2 100x. E X AM P LE 8 Finding the Dimensions That Yield a Maximum Area You have just bought a puppy and want to fence in an area in the backyard for her. You buy 100 linear feet of fence from Home Depot and have decided to make a rectangular fenced-in area using the back of your house as one side. Determine the dimensions of the rectangular pen that will maximize the area in which your puppy may roam. What is the maximum area of the rectangular pen? Solution: STEP 1 Identify the question. Find the dimensions of the rectangular pen. STEP 2 Draw a picture. STEP 3 Set up a function. If we let x represent the length of one side of the rectangle, then the opposite side is also of length x. Because there are 100 feet of fence, the remaining fence left for the side opposite the house is 100 2x. The area of a rectangle is equal to length times width: STEP 4 Find the maximum value of the function. Find the maximum of the parabola that corresponds to the quadratic function for area A(x) 2x2 100x. a 2 and b 100; therefore, the maximum occurs when Replacing x with 25 in our original diagram: The dimensions of the rectangle are 25 feet by 50 feet . The maximum area A(25) 1250 is 1250 square feet . STEP 5 Check the solution. Two sides are 25 feet and one side is 50 feet, and together they account for all 100 feet of fence. YO U R T U R N Suppose you have 200 linear feet of fence to enclose a rectangular garden. Determine the dimensions of the rectangle that will yield the greatest area. x = - b 2a = - 100 2(-2) = 25 A(x) = x(100 - 2x) = -2x2 + 100x A(x) = x(100 - 2x) The maximum occurs when x 25. The maximum area is y 1250 square feet. A table of values supports the solution. Pen House Pen 100 2x xx House Pen 100 2x = 100 2(25) = 50 x = 25x = 25 House Answer: 50 feet by 50 feet c04a.qxd 11/25/11 3:24 PM Page 386 416. 4.1 Quadratic Functions 387 E X AM P LE 9 Path of a Punted Football The path of a particular punt follows the quadratic function: , where h(x) is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 0 corresponds to mideld (the 50 yard line). For example, x 20 corresponds to the punters own 30 yard line, whereas x 20 corresponds to the other teams 30 yard line. a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. Solution (a): Identify the vertex since it is given in standard form. (h, k) (5, 50) The maximum height of the punt occurs at the other teams 45 yard line, and the height the ball achieves is 50 yards (150 feet) . Solution (b): The height when the ball is kicked or caught is zero. Solve for x. (x 5)2 400 x 5 20 x 15 and x 25 The horizontal distance is the distance between these two points: 25 (15) 40 yards . ; (x - 5) = ;1400 1 8 (x - 5)2 = 50 h(x) = - 1 8 (x - 5)2 + 50 = 0 x = 0x = 20 x = 20 Punter h(x) = - 1 8 (x - 5)2 + 50 When the quadratic function is given in general form, completing the square can be used to rewrite the function in standard form. At least three points are needed to graph a quadratic function. vertex y-intercept x-intercept(s) or other point(s) S U M MARY All quadratic functions f(x) ax2 bx c or f(x) a(x h)2 k have graphs that are parabolas: If a 0, the parabola opens up. If a 0, the parabola opens down. The vertex is at the point a - b 2a , 4ac - b2 4a b(h, k) = a- b 2a , f a- b 2a bb S E CTI O N 4.1 c04a.qxd 11/25/11 3:24 PM Page 387 417. 388 C HAP TE R 4 Polynomial and Rational Functions In Exercises 14, match the quadratic function with its graph. 1. f (x) 3(x 2)2 5 2. f (x) 2(x 1)2 3 3. 4. a. b. c. d. In Exercises 58, match the quadratic function with its graph. 5. f (x) 3x2 5x 2 6. f(x) 3x2 x 2 7. f(x) x2 2x 1 8. f(x) 2x2 x 3 a. b. c. d. In Exercises 922, graph the quadratic function, which is given in standard form. 9. f(x) (x 1)2 2 10. f(x) (x 2)2 1 11. f(x) (x 2)2 3 12. f(x) (x 4)2 2 13. f(x) (x 3)2 9 14. f(x) (x 5)2 4 15. f(x) (x 1)2 3 16. f(x) (x 2)2 6 17. f(x) 2(x 2)2 2 18. f(x) 3(x 2)2 15 19. 20. 21. f(x) 0.5(x 0.25)2 0.75 22. f(x) 0.2(x 0.6)2 0.8 In Exercises 2332, rewrite the quadratic function in standard form by completing the square. 23. f(x) x2 6x 3 24. f(x) x2 8x 2 25. f(x) x2 10x 3 26. f(x) x2 12x 6 27. f(x) 2x2 8x 2 28. f(x) 3x2 9x 11 29. f(x) 4x2 16x 7 30. f(x) 5x2 100x 36 31. 32. In Exercises 3340, graph the quadratic function. 33. f(x) x2 6x 7 34. f(x) x2 3x 10 35. f(x) x2 2x 15 36. f(x) x2 3x 4 37. f(x) 2x2 12x 16 38. f(x) 3x2 12x 12 39. 40. f(x) = - 1 3 x2 + 4 3f(x) = 1 2 x2 - 1 2 f(x) = - 1 3 x2 + 6x + 4f(x) = 1 2 x2 - 4x + 3 f (x) = Ax + 1 4 B 2 - 1 2f(x) = Ax - 1 3 B 2 + 1 9 x y 5 5 6 4 x y 5 5 2 8 x y 5 5 4 6 y 4 6 x 5 5 y x 3 3 10 x y 4 6 4 6 x y 5 2 5 5 x y 8 2 4 6 f (x) = - 1 3 (x - 2)2 + 3f (x) = - 1 2 (x + 3)2 + 2 E X E R C I S E S S E CTI O N 4.1 S K I LL S c04a.qxd 11/25/11 3:24 PM Page 388 418. 4.1 Quadratic Functions 389 In Exercises 4148, nd the vertex of the parabola associated with each quadratic function. 41. f(x) 33x2 2x 15 42. f(x) 17x2 4x 3 43. 44. 45. f(x) 0.002x2 0.3x 1.7 46. f(x) 0.05x2 2.5x 1.5 47. 48. In Exercises 4958, nd the quadratic function that has the given vertex and goes through the given point. 49. vertex: (1, 4) point: (0, 2) 50. vertex: (2, 3) point: (0, 1) 51. vertex: (2, 5) point: (3, 0) 52. vertex: (1, 3) point: (2, 0) 53. vertex: (1, 3) point: (4, 2) 54. vertex: (0, 2) point: (3, 10) 55. vertex: point: 56. vertex: point: (0, 0) 57. vertex: (2.5, 3.5) point: (4.5, 1.5) 58. vertex: (1.8, 2.7) point: (2.2, 2.1) A- 5 6, 2 3 BA3 4, 0BA1 2, - 3 4 B f(x) = - 1 7 x2 - 2 3 x + 1 9f(x) = - 2 5 x2 + 3 7 x + 2 f(x) = - 1 3 x2 + 2 5 x + 4f(x) = 1 2 x2 - 7x + 5 Exercises 63 and 64 concern the path of a punted football. Refer to the diagram in Example 9. 63. Sports. The path of a particular punt follows the quadratic function where h(x) is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 0 corresponds to mideld (the 50 yard line). For example, x 20 corresponds to the punters own 30 yard line, whereas x 20 corresponds to the other teams 30 yard line. a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. 64. Sports. The path of a particular punt follows the quadratic function where h(x) is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 0 corresponds to mideld (the 50 yard line). For example, x 20 corresponds to the punters own 30 yard line, whereas x 20 corresponds to the other teams 30 yard line. a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. h(x) = - 5 40 (x - 30)2 + 50 h(x) = - 8 125 (x + 5)2 + 40 59. Business. The annual prot for a company that manufactures cell phone accessories can be modeled by the function where x is the number of units sold and P is the total prot in dollars. a. What sales level maximizes the companys annual prot? b. Find the maximum annual prot for the company. 60. Business. A manufacturer of ofce supplies has daily production costs of where x is the number of units produced measured in thousands and C is cost in hundreds of dollars. a. What production level will minimize the manufacturers daily production costs? b. Find the minimum daily production costs for the manufacturer. For Exercises 61 and 62, refer to the following: An adult males weight, in kilograms, can be modeled by the function where t measures months (t 1 is January 2010, t 2 is February 2010, etc.) and W is the males weight. 61. Health/Medicine. During which months was the male losing weight and gaining weight? 62. Health/Medicine. Find the maximum weight to the nearest kilogram of the adult male during the 18 months. W(t) = - 2 3 t2 + 13 5 t + 433 5 ; 1 t 18 C(x) = 0.5x2 - 20x + 1600 P(x) = -0.0001x2 + 70x + 12,500 A P P L I C AT I O N S c04a.qxd 11/25/11 3:24 PM Page 389 419. 390 C HAP TE R 4 Polynomial and Rational Functions 65. Ranching. A rancher has 10,000 linear feet of fencing and wants to enclose a rectangular eld and then divide it into two equal pastures with an internal fence parallel to one of the rectangular sides. What is the maximum area of each pasture? Round to the nearest square foot. 66. Ranching. A rancher has 30,000 linear feet of fencing and wants to enclose a rectangular eld and then divide it into four equal pastures with three internal fences parallel to one of the rectangular sides. What is the maximum area of each pasture? 67. Gravity. A person standing near the edge of a cliff 100 feet above a lake throws a rock upward with an initial speed of 32 feet per second. The height of the rock above the lake at the bottom of the cliff is a function of time and is described by a. How many seconds will it take until the rock reaches its maximum height? What is that height? b. At what time will the rock hit the water? 100 feet h(t) = -16t2 + 32t + 100 68. Gravity. A person holds a pistol straight upward and res. The initial velocity of most bullets is around 1200 feet per second. The height of the bullet is a function of time and is described by How long, after the gun is red, does the person have to get out of the way of the bullet falling from the sky? 69. Zero Gravity. As part of their training, astronauts ride the vomit comet, NASAs reduced gravity KC 135A aircraft that performs parabolic ights to simulate weightlessness. The plane starts at an altitude of 20,000 feet and makes a steep climb at 52 with the horizon for 2025 seconds and then dives at that same angle back down, repeatedly. The equation governing the altitude of the ight is where A(x) is altitude and x is horizontal distance in feet. a. What is the maximum altitude the plane attains? b. Over what horizontal distance is the entire maneuver performed? (Assume the starting and ending altitude is 20,000 feet.) 70. Sports. A soccer ball is kicked from the ground at a 45 angle with an initial velocity of 40 feet per second. The height of the soccer ball above the ground is given by H(x) 0.0128x2 x, where x is the horizontal distance the ball travels. a. What is the maximum height the ball reaches? b. What is the horizontal distance the ball travels? 71. Prot. A small company in Virginia Beach manufactures handcrafted surfboards. The prot of selling x boards is given by a. How many boards should be made to maximize the prot? b. What is the maximum prot? P(x) = 20,000 + 80x - 0.4x2 NASAs Vomit Comet CourtesyNASA A(x) = -0.0003x2 + 9.3x - 46,075 h(t) = -16t2 + 1200t c04a.qxd 11/25/11 3:24 PM Page 390 420. 4.1 Quadratic Functions 391 72. Environment: Fuel Economy. Gas mileage (miles per gallon, mpg) can be approximated by a quadratic function of speed. For a particular automobile, assume the vertex occurs when the speed is 50 miles per hour (mph); the mpg will be 30. a. Write a quadratic function that models this relationship, assuming 70 mph corresponds to 25 mpg. b. What gas mileage would you expect for this car driving 90 mph? For Exercises 73 and 74, use the following information: One function of particular interest in economics is the prot function. We denote this function by P(x). It is dened to be the difference between revenue R(x) and cost C(x) so that The total revenue received from the sale of x goods at price p is given by The total cost function relates the cost of production to the level of output x. This includes both xed costs Cf and variable costs Cv (costs per unit produced). The total cost in producing x goods is given by Thus, the prot function is Assume xed costs are $1000, variable costs per unit are $20, and the demand function is 73. Prot. How many units should the company produce to break even? 74. Prot. What is the maximum prot? p = 100 - x P(x) = px - Cf - Cv x C(x) = Cf + Cv x R(x) = px P(x) = R(x) - C(x) Speed (mph) 9010 705030 MPG 5 15 25 35 45 (50, 30) (70, 25) 75. Cell Phones. The number of cell phones in the United States can be approximated by a quadratic function. In 1996 there were approximately 16 million cell phones, and in 2005 there were approximately 100 million. Let t be the number of years since 1996. The number of cell phones in 1996 is represented by (0, 16), and the number in 2005 is (9, 100). Let (0, 16) be the vertex. a. Find a quadratic function that represents the number of cell phones. b. Based on this model, how many cell phones were used in 2010? 76. Underage Smoking. The number of underage cigarette smokers (ages 1017) has declined in the United States. The peak percent was in 1998 at 49%. In 2006 this had dropped to 36%. Let t be time in years after 1998 (t 0 corresponds to 1998). a. Find a quadratic function that models the percent of underage smokers as a function of time. Let (0, 49) be the vertex. b. Based on this model, what was the percent of underage smokers in 2010? 77. Drug Concentration. The concentration of a drug in the bloodstream, measured in parts per million, can be modeled with a quadratic function. In 50 minutes the concentration is 93.75 parts per million. The maximum concentration of the drug in the bloodstream occurs in 225 minutes and is 400 parts per million. a. Find a quadratic function that models the concentration of the drug as a function of time in minutes. b. In how many minutes will the drug be eliminated from the bloodstream? 78. Revenue. Jeff operates a mobile car washing business. When he charged $20 a car, he washed 70 cars a month. He raised the price to $25 a car and his business dropped to 50 cars a month. a. Find a linear function that represents the demand equation (the price per car as a function of the number of cars washed). b. Find the revenue function R(x) xp. c. How many cars should he wash to maximize the revenue? d. What price should he charge to maximize revenue? c04a.qxd 11/25/11 3:24 PM Page 391 421. 392 C HAP TE R 4 Polynomial and Rational Functions In Exercises 7982, explain the mistake that is made. There may be a single mistake, or there may be more than one mistake. 79. Plot the quadratic function f(x) (x 3)2 1. Solution: Step 1: The parabola opens up because a 1 0. Step 2: The vertex is (3, 1). Step 3: The y-intercept is (0, 8). Step 4: The x-intercepts are (2, 0) and (4, 0). Step 5: Plot the vertex and intercepts, and connect the points with a smooth curve. This is incorrect. What mistake(s) was made? 80. Determine the vertex of the quadratic function f(x) 2x2 6x 18. Solution: Step 1: The vertex is given by . In this case, a 2 and b 6. Step 2: The x-coordinate of the vertex is Step 3: The y-coordinate of the vertex is This is incorrect. What mistake(s) was made? = - 63 2 = - 9 2 - 9 - 18 = -2a 9 4 b - 18 2 - 18 f a- 3 2 b = -2a- 3 2 b 2 + 6 a- 3 2 b - 18 x = - 6 2(2) = - 6 4 = - 3 2 (h, k) = a- b 2a , f a- b 2a bb x y (3, 1) (2, 0) (4, 0) 81. Rewrite the following quadratic function in standard form: Solution: Step 1: Group the variables together. (x2 2x) 3 Step 2: Factor out a negative. (x2 2x) 3 Step 3: Add and subtract 1 inside the parentheses. (x2 2x 1 1) 3 Step 4: Factor out the 1. (x2 2x 1) 1 3 Step 5: Simplify. (x 1)2 4 This is incorrect. What mistake(s) was made? 82. Find the quadratic function whose vertex is (2, 3) and whose graph passes through the point (9, 0). Solution: Step 1: Write the quadratic function in standard form. f(x) a(x h)2 k Step 2: Substitute (h, k) (2, 3). f(x) a(x 2)2 3 Step 3: Substitute the point (9, 0) and solve for a. f(0) a(0 2)2 3 9 4a 3 9 4a 12 a 3 The quadratic function sought is: f(x) 3(x 2)2 3. This is incorrect. What mistake(s) was made? f(x) = -x2 + 2x + 3 C AT C H T H E M I S TA K E In Exercises 8386, determine whether each statement is true or false. 83. A quadratic function must have a y-intercept. 84. A quadratic function must have an x-intercept. 85. A quadratic function may have more than one y-intercept. 86. A quadratic function may have more than one x-intercept. C O N C E P T U A L c04a.qxd 11/25/11 3:24 PM Page 392 422. 4.1 Quadratic Functions 393 BENS DATA RICKS DATA x y x y 0 4 0 4 1 14.2 1 8.5 2 28.4 2 10.6 3 30.1 3 13.3 4 35.9 4 16.2 5 37.8 5 17.3 6 41.1 6 19.3 7 38.2 7 19.5 87. For the general quadratic equation, f(x) ax2 bx c, show that the vertex is .(h, k) = a- b 2a , f a- b 2a bb 88. Given the quadratic function f(x) a(x h)2 k, determine the x- and y-intercepts in terms of a, h, and k. 89. A rancher has 1000 feet of fence to enclose a pasture. a. Determine the maximum area if a rectangular fence is used. b. Determine the maximum area if a circular fence is used. 90. A 600-room hotel in Orlando is lled to capacity every night when the rate is $90 per night. For every $5 increase in the rate, 10 fewer rooms are lled. How much should the hotel charge to produce the maximum income? What is the maximum income? 91. On a graphing calculator, plot the quadratic function f(x) 0.002x2 5.7x 23. a. Identify the vertex of this parabola. b. Identify the y-intercept. c. Identify the x-intercepts (if any). d. What is the axis of symmetry? 92. Determine the quadratic function whose vertex is (0.5, 1.7) and whose graph passes through the point (0, 4). a. Write the quadratic function in general form. b. Plot this quadratic function with a graphing calculator. c. Zoom in on the vertex and y-intercept. Do they agree with the given values? In Exercises 93 and 94: (a) use the calculator commands STAT QuadReg to model the data using a quadratic function; (b) write the quadratic function in standard form and identify the vertex; (c) plot this quadratic function with a graphing calculator and use the TRACE key to highlight the given points. Do they agree with the given values? 93. 94. x 2 2 5 y 29.28 21.92 18.32 x 9 2 4 y 2.72 16.18 6.62 For Exercises 95 and 96, refer to the following discussion of quadratic regression: The least-squares criterion used to create a quadratic regression curve that ts a set of n data points (x1, y1), (x2, y2), . . . , (xn , yn) is that the sum of the squares of the vertical distances from the points to the curve be minimum. This means that we need to determine values of a, b, and c for which is as small as possible. Calculus can be used to determine formulas for a, b, and c that do the job, but computing them by hand is tedious and unnecessary because the TI-83 has a built-in program called QuadReg that does this. In fact, this was introduced in Section 2.5*, Problems 3942. The following are application problems that involve experimental data for which the best t curve is a parabola. a n i=1 (yi - (axi 2 + bxi + c))2 y = ax2 + bx + c Projectile Motion It is well known that the trajectory of an object thrown with initial velocity v0 from an initial height s0 is described by the quadratic function . If we have data points obtained in such a context, we can apply the procedure outlined in Section 2.5* with QuadReg in place of LinReg to nd a best t parabola of the form . Each year during Halloween season, it is tradition to hold the Pumpkin Launching Contest where students literally hurl their pumpkins in the hope of throwing them the farthest horizontal distance. Ben claims that it is better to use a steeper trajectory since it will have more air time, while Rick believes in throwing the pumpkin with all of his might, but with less inclination. The following data points that describe the pumpkins horizontal x and vertical y distances (measured in feet) are collected during the ights of their pumpkins: y = ax2 + bx + c (ax+b) s(t) = -16t2 + v0t + s0 95. a. Form a scatterplot for Bens data. b. Determine the equation of the best t parabola and report the value of the associated correlation coefcient. c. Use the best t curve from (b) to answer the following: i. What is the initial height of the pumpkins trajectory and with what initial velocity was it thrown? ii. What is the maximum height of Bens pumpkins trajectory? iii. How much horizontal distance has the pumpkin traveled by the time it lands? 96. Repeat Exercise 95 for Ricks data. C H A L L E N G E T E C H N O L O G Y c04a.qxd 6/6/12 7:48 PM Page 393 423. Identifying Polynomial Functions Polynomial functions model many real-world applications. Often, the input is time, and the output of the function can be many things. For example, the number of active-duty military personnel in the United States, the number of new cases in the spread of an epidemic, and the stock price as a function of time t can all be modeled with polynomial functions. Let n be a nonnegative integer and let an, an 1, . . . , a2, a1, a0 be real numbers with . The function is called a polynomial function of x with degree n. The coefcient an is called the leading coefcient. f(x) = an xn + an-1xn-1 + + a2 x2 + a1x + a0 an Z 0 D E F I N I T I O N Polynomial Function Note: If n is a nonnegative integer, then n 1, n 2, . . . , 2, 1, 0 are also nonnegative integers. E X AM P LE 1 Identifying Polynomials and Their Degree For each of the functions given, determine whether the function is a polynomial function. If it is a polynomial function, then state the degree of the polynomial. If it is not a polynomial function, justify your answer. a. f(x) 3 2x5 b. c. g(x) 2 d. h(x) 3x2 2x 5 e. H(x) 4x5 (2x 3)2 f. G(x) 2x4 5x3 4x2 F(x) = 1x + 1 P O LYN O M IAL F U N CTI O N S O F H I G H E R D E G R E E S E CTI O N 4.2 C O N C E P TUAL O BJ E CTIVE S Understand that real zeros of polynomial functions correspond to x-intercepts. Understand the intermediate value theorem and how it assists in graphing polynomial functions. Realize that end behavior is a result of the leading term dominating. Understand that zeros correspond to factors of the polynomial. S K I LLS O BJ E CTIVE S Identify a polynomial function and determine its degree. Graph polynomial functions using transformations. Identify real zeros of a polynomial function and their multiplicities. Determine the end behavior of a polynomial function. Graph polynomial functions. x-intercepts multiplicity (touch/cross) of each zero end behavior 394 c04a.qxd 11/25/11 3:24 PM Page 394 424. 4.2 Polynomial Functions of Higher Degree 395 How do we graph polynomial functions that are of degree 3 or higher, and why do we care? Polynomial functions model real-world applications, as mentioned earlier. One example is the percentage of fat in our bodies as we age. We can model the weight of a baby after it comes home from the hospital as a function of time. When a baby comes home from the hospital, it usually experiences weight loss. Then typically there is an increase in the percent of body fat when the baby is nursing. When infants start to walk, the increase in exercise is associated with a drop in the percentage of fat. Growth spurts in children are examples of the percent of body fat increasing and decreasing. Later in life, our metabolism slows down, and typically the percent of body fat increases. We will model this with a polynomial function. Other examples are stock prices, the federal funds rate, and yo-yo dieting as functions of time. POLYNOMIAL DEGREE SPECIAL NAME GRAPH f(x) c 0 Constant function Horizontal line f(x) mx b 1 Linear function Line Slope m y-intercept: (0, b) f(x) ax2 bx c 2 Quadratic function Parabola Opens up if a 0. Opens down if a 0. Solution: a. f(x) is a polynomial function of degree 5. b. F(x) is not a polynomial function. The variable x is raised to the power of , which is not an integer. c. g(x) is a polynomial function of degree zero, also known as a constant function. Note that g(x) 2 can also be written as g(x) 2x0 (assuming x 0). d. h(x) is a polynomial function of degree 2. A polynomial function of degree 2 is called a quadratic function. e. H(x) is a polynomial function of degree 7. Note: 4x5 (4x2 12x 9) 16x7 48x6 36x5 . f. G(x) is not a polynomial function. 4x2 has an exponent that is negative. YO U R TU R N For each of the functions given, determine whether the function is a polynomial function. If it is a polynomial function, then state the degree of the polynomial. If it is not a polynomial function, justify your answer. a. b. g(x) = 3x8 (x - 2)2 (x + 1)3 f(x) = 1 x + 2 1 2 Whenever we have discussed a particular polynomial, we have graphed it too. The graph of a constant function (degree 0) is a horizontal line. The graph of a general linear function (degree 1) is a slant line. The graph of a quadratic function (degree 2) is a parabola. These functions are summarized in the table below. Answer: a. f(x) is not a polynomial because x is raised to the power of 1, which is a negative integer. b. g(x) is a polynomial of degree 13. c04a.qxd 11/25/11 3:24 PM Page 395 425. 396 C HAP TE R 4 Polynomial and Rational Functions Let n be a positive integer and the coefcient be a real number. The function is called a power function of degree n. f (x) = axn a Z 0 Power FunctionD E F I N I T I O N Study Tip All polynomial functions have graphs that are both continuous and smooth. Polynomial functions are considered simple functions. Graphs of all polynomial functions are both continuous and smooth. A continuous graph is one you can draw completely without picking up your pencil (the graph has no jumps or holes). A smooth graph has no sharp corners. The following graphs illustrate what it means to be smooth (no sharp corners or cusps) and continuous (no holes or jumps). The graph is not continuous. The graph is not continuous. The graph is continuous but not smooth. The graph is continuous and smooth. All polynomial functions have graphs that are both continuous and smooth. Graphing Polynomial Functions Using Transformations of Power Functions Recall from Chapter 3 that graphs of functions can be drawn by hand using graphing aids such as intercepts and symmetry. The graphs of polynomial functions can be graphed using these same aids. Lets start with the simplest types of polynomial functions called power functions. Power functions are monomial functions of the form f(x) xn , where n is an integer greater than zero. x y x y Corner x y Hole x y Jump Power functions with even powers look similar to the square function. Power functions with odd powers (other than n 1) look similar to the cube function. f(x) = x2 f(x) = x4 f(x) = x6 f(x) = x3 f(x) = x5 f(x) = x7 c04a.qxd 11/25/11 3:24 PM Page 396 426. 4.2 Polynomial Functions of Higher Degree 397 All even power functions have similar characteristics to a quadratic function (parabola), and all odd (n 1) power functions have similar characteristics to a cubic function. For example, all even functions are symmetric with respect to the y-axis, whereas all odd functions are symmetric with respect to the origin. The following table summarizes their characteristics. We now have the tools to graph polynomial functions that are transformations of power functions. We will use the power functions summarized above, combined with our graphing techniques such as horizontal and vertical shifting and reection (Section 3.3). E X AM P LE 2 Graphing Transformations of Power Functions Graph the function f(x) (x 1)3 . Solution: STEP 1 Start with the graph of y x3 . STEP 2 Shift y x3 to the right one unit to yield the graph of f(x) (x 1)3 . YO U R TU R N Graph the function f(x) 1 x4 . x 5 5 y 10 10 x 5 5 y 10 10 n EVEN n ODD Symmetry y-axis Origin Domain (, ) (, ) Range [0, ) (, ) Some key points that lie on the graph (1, 1), (0, 0), and (1, 1) (1, 1), (0, 0), and (1, 1) Increasing (0, ) (, ) Decreasing (, 0) N/A CHARACTERISTICS OF POWER FUNCTIONS: f(x) xn Answer: f(x) 1 x4 x y (0, 1) (1, 0) (1, 0)5 5 5 5 c04b.qxd 11/25/11 3:31 PM Page 397 427. 398 C HAP TE R 4 Polynomial and Rational Functions Real Zeros of a Polynomial Function How do we graph general polynomial functions of degree greater than or equal to 3 if they cannot be written as transformations of power functions? We start by identifying the x-intercepts of the polynomial function. Recall that we determine the x-intercepts by setting the function equal to zero and solving for x. Therefore, an alternative name for an x-intercept of a function is a zero of the function. In our experience, to set a quadratic function equal to zero, the rst step is to factor the quadratic expression into linear factors and then set each factor equal to zero. Therefore, there are four equivalent relationships that are summarized in the following box. If f(x) is a polynomial function and a is a real number, then the following statements are equivalent. Relationship 1: x a is a solution, or root, of the equation f(x) 0. Relationship 2: (a, 0) is an x-intercept of the graph of f(x). Relationship 3: x a is a zero of the function f(x). Relationship 4: (x a) is a factor of f(x). REAL ZEROS OF POLYNOMIAL FUNCTIONS Lets use a simple polynomial function to illustrate these four relationships. Well focus on the quadratic function f(x) x2 1. The graph of this function is a parabola that opens up and has as its vertex the point (0, 1). Study Tip Real zeros correspond to x-intercepts. Illustration of Relationship 1 Set the function equal to zero, f (x) 0. x2 1 0 Factor. (x 1)(x 1) 0 Solve. x 1 or x 1 x 1 and x 1 are solutions, or roots, of the equation x2 1 0. Illustration of Relationship 2 The x-intercepts are (1, 0) and (1, 0). Illustration of Relationship 3 The value of the function at x 1 and x 1 is 0. f(1) = (1)2 - 1 = 0 f(-1) = (-1)2 - 1 = 0 x y (0, 1) (1, 0)(1, 0) f(x) = x2 1 c04b.qxd 11/25/11 8:53 PM Page 398 428. 4.2 Polynomial Functions of Higher Degree 399 Let a and b be real numbers such that a b and let f be a polynomial function. If f(a) and f(b) have opposite signs, then there is at least one zero between a and b. Intermediate Value TheoremD E F I N I T I O N The intermediate value theorem will be used later in this chapter to assist us in nding real zeros of a polynomial function. For now, it tells us that in order to change signs, the polynomial function must pass through the x-axis. In other words, once we know the zeros, then we know that between two consecutive zeros the polynomial is either entirely above the x-axis or entirely below the x-axis. This enables us to break the x-axis down into intervals that we can test, which will assist us in graphing polynomial functions. Keep in mind, though, that the existence of a zero does not imply that the function will change signsas you will see in the subsection on graphing general polynomial functions. E X AM P LE 3 Identifying the Real Zeros of a Polynomial Function Find the zeros of the polynomial function f(x) x3 x2 2x. Solution: Set the function equal to zero. x3 x2 2x 0 Factor out an x common to all three terms. x(x2 x 2) 0 Factor the quadratic expression inside the parentheses. x(x 2)(x 1) 0 Apply the zero product property. x 0 or (x 2) 0 or (x 1) 0 Solve. x 2, x 0, and x 1 The zeros are 2, 0, and 1 . YO U R TU R N Find the zeros of the polynomial function f (x) x3 7x2 12x. Recall that when we were factoring a quadratic equation, if the factor was raised to a power greater than 1, the corresponding root, or zero, was repeated. For example, the quadratic equation x2 2x 1 0 when factored is written as (x 1)2 0. The solution, or root, in this case is x 1, and we say that it is a repeated root. Similarly, when determining zeros of higher order polynomial functions, if a factor is repeated, we say that the zero is a repeated, or multiple, zero of the function. The number of times that a zero repeats is called its multiplicity. Illustration of Relationship 4 x 1 is a zero of the polynomial, so (x 1) is a factor. x 1 is a zero of the polynomial, so (x 1) is a factor. We have a good reason for wanting to know the x-intercepts, or zeros. When the value of a continuous function transitions from negative to positive and vice versa, it must pass through zero. = (x - 1)(x + 1) f(x) = x2 - 1 Technology Tip Graph y1 x3 x2 2x. The zeros of the function 2, 0, and 1 correspond to the x-intercepts (2, 0), (0, 0), and (1, 0). The table supports the real zeros shown by the graph. Answer: The zeros are 0, 3, and 4. c04b.qxd 11/25/11 3:31 PM Page 399 429. 400 C HAP TE R 4 Polynomial and Rational Functions EXAMPLE 4 Finding the Multiplicities of Zeros of a Polynomial Function Find the zeros, and state their multiplicities, of the polynomial function . Solution: Note: Adding the multiplicities yields the degree of the polynomial. The polynomial g(x) is of degree 10, since 2 7 1 10. YO U R TU R N For the polynomial h(x), determine the zeros and state their multi- plicities. h(x) = x2 (x - 2)3 ax + 1 2 b 5 -5 is a zero of multiplicity 1. -3 5 is a zero of multiplicity 7. 1 is a zero of multiplicity 2. g(x) = (x - 1)2 Ax + 3 5 B 7 (x + 5) If (x a)n is a factor of a polynomial f, then a is called a zero of multiplicity n of f. Multiplicity of a ZeroD E F I N I T I O N Answer: 0 is a zero of multiplicity 2. 2 is a zero of multiplicity 3. is a zero of multiplicity 5. -1 2 EXAMPLE 5 Finding a Polynomial from Its Zeros Find a polynomial of degree 7 whose zeros are: 2 (multiplicity 2) 0 (multiplicity 4) 1 (multiplicity 1) Solution: If x a is a zero, then (x a) is a factor. f(x) (x 2)2 (x 0)4 (x 1)1 Simplify. x4 (x 2)2 (x 1) Square the binomial. x4 (x2 4x 4)(x 1) Multiply the two polynomials. x4 (x3 3x2 4) Distribute x4 . x7 3x6 4x4 Graphing General Polynomial Functions Lets develop a strategy for sketching an approximate graph of any polynomial function. First, we determine the x- and y-intercepts. Then we use the x-intercepts, or zeros, to divide the domain into intervals where the polynomial is positive or negative so that we can nd points in those intervals to assist in sketching a smooth and continuous graph. Note: It is not always possible to nd x-intercepts. Sometimes there are no x-intercepts. Study Tip It is not always possible to nd x-intercepts. Sometimes there are no x-intercepts. c04b.qxd 11/25/11 3:31 PM Page 400 430. 4.2 Polynomial Functions of Higher Degree 401 EXAMPLE 6 Using a Strategy for Sketching the Graph of a Polynomial Function Sketch the graph of f(x) (x 2)(x 1)2 . Solution: STEP 1 Find the y-intercept. f(0) (2)(1)2 2 (Let x 0.) (0, 2) is the y-intercept STEP 2 Find any x-intercepts. f(x) (x 2)(x 1)2 0 (Set f(x) 0.) x 2 or x 1 (2, 0) and (1, 0) are the x-intercepts STEP 3 Plot the intercepts. STEP 4 Divide the x-axis into intervals:* ( , 2), (2, 1), and (1, ) *The x-intercepts (2, 0) and (1, 0) divide the x-axis into three intervals similar to those created by zeros when we studied inequalities in Section 1.5. STEP 5 Select a number in each interval and test each interval. The function f(x) either crosses the x-axis at an x-intercept or touches the x-axis at an x-intercept. Therefore, we need to check each of these intervals to determine whether the function is positive (above the x-axis) or negative (below the x-axis). We do so by selecting numbers in the intervals and determining the value of the function at the corresponding points. x y (1, 0)(2, 0) (0, 2) x y (1, 0)(2, 0) x = 2x = 3 x = 1 (1, ) (2, 1) (, 2) x y (2, 4)(1, 4) (3, 16) Interval (, 2) (2, 1) (1, ) Number Selected in Interval 3 1 2 Value of Function f(3) 16 f(1) 4 f(2) 4 Point on Graph (3, 16) (1, 4) (2, 4) Interval Relation to x-Axis Below x-axis Above x-axis Above x-axis Technology Tip The graph of f(x) (x 2)(x 1)2 is shown. Note: The graph crosses the x-axis at the point x 2 and touches the x-axis at the point x 1. A table of values supports the graph. c04b.qxd 11/25/11 3:31 PM Page 401 431. From the table, we nd three additional points on the graph: (3, 16), (1, 4), and (2, 4). The point (2, 0) is an intercept where the function crosses the x-axis, because it is below the x-axis to the left of 2 and above the x-axis to the right of 2. The point (1, 0) is an intercept where the function touches the x-axis, because it is above the x-axis on both sides of x 1. Connecting these points with a smooth curve yields the graph. STEP 6 Sketch a plot of the function. YO U R TU R N Sketch the graph of f(x) x2 (x 3)2 . x y (1, 0) (1, 4) (2, 4) (0, 2) Study Tip We do not know for sure that the points (1, 4) and (1, 0) are turning points. A graphing utility can conrm this, and later in calculus you will learn how to nd relative maximum points and relative minimum points. In Example 6, we found that the function crosses the x-axis at the point (2, 0). Note that 2 is a zero of multiplicity 1. We also found that the function touches the x-axis at the point (1, 0). Note that 1 is a zero of multiplicity 2. In general, zeros with even multiplicity correspond to intercepts where the function touches the x-axis and zeros with odd multiplicity correspond to intercepts where the function crosses the x-axis. x y 5 5 10 Answer: Study Tip If f is a polynomial of degree n, then the graph of f has at most n 1 turning points. x y Turning points Also in Example 6, we know that somewhere in the interval (2, 1) the function must reach a maximum and then turn back toward the x-axis, because both points (2, 0) and (1, 0) correspond to x-intercepts. When we sketch the graph, it appears that the point (1, 4) is a turning point. The point (1, 0) also corresponds to a turning point. In general, if f is a polynomial of degree n, then the graph of f has at most n 1 turning points. The point (1, 4), which we call a turning point, is also a high point on the graph in the vicinity of the point (1, 4). Also note that the point (1, 0), which we call a turning point, is a low point on the graph in the vicinity of the point (1, 0). We call a high point on a graph a local (relative) maximum and a low point on a graph a local (relative) minimum. For quadratic functions we can nd the maximum or minimum point by nding the vertex. However, for higher degree polynomial functions, we rely on graphing utilities to locate such points. Later in calculus, techniques will be developed for nding such points exactly. For now, we use the and features to locate such points on a graph, and we can use the feature of a graphing utility to approximate relative minima or maxima.table tracezoom Study Tip In general, zeros with even multiplicity correspond to intercepts where the function touches the x-axis and zeros with odd multiplicity correspond to intercepts where the function crosses the x-axis. MULTIPLICITY f (x) ON EITHER GRAPH OF FUNCTION OF a SIDE OF x a AT THE INTERCEPT Even Does not change sign Touches the x-axis (turns around) at point (a, 0) Odd Changes sign Crosses the x-axis at point (a, 0) If a is a zero of f(x), then: MULTIPLICITY OF A ZERO AND RELATION TO THE GRAPH OF A POLYNOMIAL 402 C HAP TE R 4 Polynomial and Rational Functions c04b.qxd 11/25/11 3:31 PM Page 402 432. 4.2 Polynomial Functions of Higher Degree 403 As x gets large in the positive and negative directions, the graph of the polynomial has the same behavior as the power function y = anxn f(x) = anxn + an-1xn-1 + + a2x2 + a1x + a0 (x S -)(x S ) END BEHAVIOR Power functions behave much like a quadratic function (parabola) for even-degree polynomial functions and much like a cubic function for odd-degree polynomial functions. There are four possibilities because the leading coefcient can be positive or negative with either an odd or even power. Let y = anxn ; then Let us take the polynomial f(x) x3 2x2 5x 6. Using methods discussed thus far, we can nd that the x-intercepts of its graph are (2, 0), (1, 0), and (3, 0) and the y-intercept is the point (0, 6). We can also nd additional points that lie on the graph such as (1, 8) and (2, 4). Plotting these points we might think that the points (1, 8) and (2, 4) might be turning points, but a graphing utility reveals an approximate relative maximum at the point (0.7863, 8.2088207) and an approximate relative minimum at the point (2.1196331, 4.060673). Intercepts and turning points assist us in sketching graphs of polynomial functions. Another piece of information that will assist us in graphing polynomial functions is knowledge of the end behavior. All polynomials eventually rise or fall without bound as x gets large in both the positive and negative directions. The highest degree monomial within the polynomial dominates the end behavior. In other words, the highest power term is eventually going to overwhelm the other terms as x grows without bound. (x S -)(x S ) n Even Even Odd Odd an Positive Negative Negative Positive The graph of the The graph of the The graph of the The graph of the (Left) function rises function falls function rises function falls The graph of the The graph of the The graph of the The graph of the (Right) function rises function falls function falls function rises Graph x y an > 0 x y an < 0 x y an < 0 x y an > 0 x S x S - Technology Tip Use TI to graph the function as Y1.f(x) = x3 - 2x2 - 5x + 6 c04b.qxd 11/25/11 3:32 PM Page 403 433. 404 C HAP TE R 4 Polynomial and Rational Functions EXAMPLE 7 Graphing a Polynomial Function Sketch a graph of the polynomial function f(x) 2x4 8x2 . Solution: STEP 1 Determine the y-intercept: (x 0). f(0) 0 The y-intercept corresponds to the point (0, 0). STEP 2 Find the zeros of the polynomial. f(x) 2x4 8x2 Factor out the common 2x2 . 2x2 (x2 4) Factor the quadratic binomial. 2x2 (x 2)(x 2) Set f(x) 0. 2x2 (x 2)(x 2) 0 0 is a zero of multiplicity 2. The graph will touch the x-axis. 2 is a zero of multiplicity 1. The graph will cross the x-axis. 2 is a zero of multiplicity 1. The graph will cross the x-axis. STEP 3 Determine the end behavior. f(x) 2x4 8x2 behaves like y 2x4 . y 2x4 is of even degree, and the leading coefcient is positive, so the graph rises without bound as x gets large in both the positive and negative directions. STEP 4 Sketch the intercepts and end behavior. x y (2, 0) (0, 0) (2, 0) STEP 5 Find additional points. STEP 6 Sketch the graph. Estimate additional points. Connect with a smooth curve. Note the symmetry about the y-axis. This function is an even function: f(x) f(x). It is important to note that the local minimums occur at but at this time can only be illustrated using a graphing utility. YO U R TU R N Sketch a graph of the polynomial function f(x) x5 4x3 . x = ; 12 L ;1.14 x y (2, 0)(2, 0) 10 10 x 1 1 f(x) 6 6-15 8-15 8 1 2-1 2 Answer: x y 5 5 10 10 Technology Tip The graph of is shown. Note: The graph crosses the x-axis at the points x 2 and x 2. The graph touches the x-axis at the point x 0. A table of values supports the graph. f (x) = 2x4 - 8x2 c04b.qxd 12/23/11 6:27 PM Page 404 434. 4.2 Polynomial Functions of Higher Degree 405 3. x-intercepts (real zeros) divide the x-axis into intervals. Test points in the intervals to determine whether the graph is above or below the x-axis. 4. Determine the end behavior by investigating the end behavior of the highest degree monomial. 5. Sketch the graph with a smooth curve. In general, polynomials can be graphed in one of two ways: Use graph-shifting techniques with power functions. General polynomial function. 1. Identify intercepts. 2. Determine each real zero and its multiplicity, and ascertain whether the graph crosses or touches the x-axis there. In Exercises 110, determine which functions are polynomials, and for those that are, state their degree. 1. f(x) 3x2 15x 7 2. f(x) 2x5 x2 13 3. 4. g(x) x4 (x 1)2 (x 2.5)3 5. 6. h(x) (x 1)1/2 5x 7. F(x) x1/3 7x2 2 8. 9. 10. In Exercises 1118, match the polynomial function with its graph. 11. f(x) 3x 1 12. f(x) 3x2 x 13. f(x) x2 x 14. f(x) 2x3 4x2 6x 15. f(x) x3 x2 16. f(x) 2x4 18x2 17. f(x) x4 5x3 18. f(x) x5 5x3 4x a. b. c. d. e. f. g. h. x y 5 5 5 5 x y 5 5 5 5 x y 5 5 10 10 x y 5 5 5 5 x y 10 10 50 50 x y 10 10 90 x y 3 2 4 1 x y 5 5 5 5 H(x) = x2 + 1 2 G(x) = x + 1 x2 F(x) = 3x2 + 7x - 2 3x h(x) = 1x + 1 g(x) = (x + 2)3 Ax - 3 5 B2 E X E R C I S E S S E CTI O N 4.2 S E CTI O N 4.2 S U M MARY S K I LL S c04b.qxd 12/23/11 6:27 PM Page 405 435. 406 C HAP TE R 4 Polynomial and Rational Functions In Exercises 1926, graph each function by transforming a power function y xn . 19. f(x) x5 20. f(x) x4 21. f(x) (x 2)4 22. f(x) (x 2)5 23. f(x) x5 3 24. f(x) x4 3 25. f(x) 3 (x 1)4 26. f(x) (x 3)5 2 In Exercises 2738, nd all the real zeros (and state their multiplicities) of each polynomial function. 27. f(x) 2(x 3)(x 4)3 28. f(x) 3(x 2)3 (x 1)2 29. f(x) 4x2 (x 7)2 (x 4) 30. f(x) 5x3 (x 1)4 (x 6) 31. f(x) 4x2 (x 1)2 (x2 4) 32. f(x) 4x2 (x2 1)(x2 9) 33. f(x) 8x3 6x2 27x 34. f(x) 2x4 5x3 3x2 35. f(x) 2.7x3 8.1x2 36. f(x) 1.2x6 4.6x4 37. 38. In Exercises 3952, nd a polynomial (there are many) of minimum degree that has the given zeros. 39. 3, 0, 1, 2 40. 2, 0, 2 41. 5, 3, 0, 2, 6 42. 0, 1, 3, 5, 10 43. 44. 45. 46. 47. 2 (multiplicity 3), 0 (multiplicity 2) 48. 4 (multiplicity 2), 5 (multiplicity 3) 49. 3 (multiplicity 2), 7 (multiplicity 5) 50. 0 (multiplicity 1), 10 (multiplicity 3) 51. (multiplicity 2), 1 (multiplicity 1), 0 (multiplicity 2), (multiplicity 2) 52. (multiplicity 2), 0 (multiplicity 1), 1 (multiplicity 2), (multiplicity 2) In Exercises 5372, for each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each x-intercept; (c) nd the y-intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. 53. f(x) x2 6x 9 54. f(x) x2 4x 4 55. f(x) (x 2)3 56. f(x) (x 3)3 57. f(x) x3 9x 58. f(x) x3 4x2 59. f(x) x3 x2 2x 60. f(x) x3 6x2 9x 61. f(x) x4 3x3 62. f(x) x5 x3 63. f(x) 12x6 36x5 48x4 64. f(x) 7x5 14x4 21x3 65. f(x) 2x5 6x4 8x3 66. f(x) 5x4 10x3 5x2 67. f(x) x3 x2 4x 4 68. f(x) x3 x2 x 1 69. f(x) (x 2)2 (x 1)2 70. f(x) (x 2)3 (x 1)3 71. f(x) x2 (x 2)3 (x 3)2 72. f(x) x3 (x 4)2 (x 2)2 In Exercises 7376, for each graph given: (a) list each real zero and its smallest possible multiplicity; (b) determine whether the degree of the polynomial is even or odd; (c) determine whether the leading coefcient of the polynomial is positive or negative; (d) nd the y-intercept; and (e) write an equation for the polynomial function (assume the least degree possible). 73. 74. 75. 76. 5 5 x y 32 8 5 5 x y 10 10 5 5 x y 10 10 5 5 x y 4 16 15- 15 13- 13 1 - 13, 1 + 131 - 12, 1 + 12-3 4, -1 3, 0, 1 2-1 2, 2 3, 3 4 f(x) = 2 7 x5 - 3 4 x4 + 1 2 x3 f(x) = 1 3 x6 + 2 5 x4 c04b.qxd 11/25/11 3:32 PM Page 406 436. 4.2 Polynomial Functions of Higher Degree 407 A P P L I C AT I O N S 81. Weight. Jennifer has joined a gym to lose weight and feel better. She still likes to cheat a little and will enjoy the occasional bad meal with an ice cream dream dessert and then miss the gym for a couple of days. Given in the table is Jennifers weight for a period of 8 months. Her weight can be modeled as a polynomial. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest degree polynomial that can represent Jennifers weight? 82. Stock Value. A day trader checks the stock price of Coca-Cola during a 4-hour period (given below). The price of Coca-Cola stock during this 4-hour period can be modeled as a polynomial function. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest degree polynomial that can represent the Coca-Cola stock price? 83. Stock Value. The price of Tommy Hilger stock during a 4-hour period is given in the following table. If a third- degree polynomial models this stock, do you expect the stock to go up or down in the fth period? PERIOD WATCHING STOCK MARKET PRICE 1 $15.10 2 $14.76 3 $15.50 4 $14.85 PERIOD WATCHING STOCK MARKET PRICE 1 $53.00 2 $56.00 3 $52.70 4 $51.50 MONTH WEIGHT 1 169 2 158 3 150 4 161 5 154 6 159 7 148 8 153 For Exercises 77 and 78, refer to the following: The relationship between a companys total revenue R (in millions of dollars) is related to its advertising costs x (in thousands of dollars). The relationship between revenue R and advertising costs x is illustrated in the graph. 77. Business. Analyze the graph of the revenue function. a. Determine the intervals on which revenue is increasing and decreasing. b. Identify the zeros of the function. Interpret the meaning of zeros for this function. 78. Business. Use the graph to identify the maximum revenue for the company and the corresponding advertising costs that produce maximum revenue. For Exercises 79 and 80, refer to the following: During a cough, the velocity v (in meters per second) of air in the trachea may be modeled by the function where r is the radius of the trachea (in centimeters) during the cough. 79. Health/Medicine. Graph the velocity function and estimate the intervals on which the velocity of air in the trachea is increasing and decreasing. 80. Health/Medicine. Estimate the radius of the trachea that produces the maximum velocity of air in the trachea. Use this radius to estimate the maximum velocity of air in the trachea. v(r) = -120r3 + 80r2 100 200 300 400 600500 50 45 40 35 30 25 20 15 10 5 x R Advertising Costs (in thousands of dollars) Revenue(inmillionsofdollars) c04b.qxd 11/25/11 3:32 PM Page 407 437. 408 C HAP TE R 4 Polynomial and Rational Functions In Exercises 8790, explain the mistake that is made. 87. Find a fourth-degree polynomial function with zeros 2, 1, 3, 4. Solution: f(x) (x 2)(x 1)(x 3)(x 4) This is incorrect. What mistake was made? 88. Determine the end behavior of the polynomial function f(x) x(x 2)3 . Solution: This polynomial has similar end behavior to the graph of y x3 . End behavior falls to the left and rises to the right. This is incorrect. What mistake was made? x y 5 5 10 10 x y 5 5 5 5 89. Graph the polynomial function f(x) (x 1)2 (x 2)3 . Solution: The zeros are 2 and 1, and therefore, the x-intercepts are (2, 0) and (1, 0). The y-intercept is (0, 8). Plotting these points and connecting with a smooth curve yields the graph on the right. This graph is incorrect. What did we forget to do? 90. Graph the polynomial function f(x) (x 1)2 (x 1)2 . Solution: The zeros are 1 and 1, so the x-intercepts are (1, 0) and (1, 0). The y-intercept is (0, 1). Plotting these points and connecting with a smooth curve yields the graph on the right. This graph is incorrect. What did we forget to do? 84. Stock Value. The stock prices for Coca-Cola during a 4-hour period on another day yield the following results. If a third-degree polynomial models this stock, do you expect the stock to go up or down in the fth period? PERIOD WATCHING STOCK MARKET PRICE 1 $52.80 2 $53.00 3 $56.00 4 $52.70 For Exercises 85 and 86, the following graph illustrates the average federal funds rate in the month of January (2000 to 2008). 85. Finance. If a polynomial function is used to model the federal funds rate data shown in the graph, determine the degree of the lowest degree polynomial that can be used to model those data. 86. Finance. Should the leading coefcient in the polynomial found in Exercise 85 be positive or negative? Explain. Year 2001 2003 2005 2007 2009 FederalFundsRate 10.00% 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 C AT C H T H E M I S TA K E c04b.qxd 12/22/11 7:14 PM Page 408 438. 4.2 Polynomial Functions of Higher Degree 409 In Exercises 9194, determine if each statement is true or false. 94. The range of all polynomial functions is ( , ). 95. What is the maximum number of zeros that a polynomial of degree n can have? 96. What is the maximum number of turning points a graph of an nth-degree polynomial can have? 91. The graph of a polynomial function might not have any y-intercepts. 92. The graph of a polynomial function might not have any x-intercepts. 93. The domain of all polynomial functions is ( , ). 99. Determine the zeros of the polynomial f(x) x3 (b a)x2 abx for the positive real numbers a and b. 100. Graph the function f(x) x2 (x a)2 (x b)2 for the positive real numbers a, b, where b a. 97. Find a seventh-degree polynomial that has the following graph characteristics: The graph touches the x-axis at x 1, and the graph crosses the x-axis at x 3. Plot this polynomial function. 98. Find a fth-degree polynomial that has the following graph characteristics: The graph touches the x-axis at x 0 and crosses the x-axis at x 4. Plot the polynomial function. In Exercises 101 and 102, use a graphing calculator or computer to graph each polynomial. From that graph, estimate the x-intercepts (if any). Set the function equal to zero, and solve for the zeros of the polynomial. Compare the zeros with the x-intercepts. 101. f(x) x4 2x2 1 102. f(x) 1.1x3 2.4x2 5.2x For each polynomial in Exercises 103 and 104, determine the power function that has similar end behavior. Plot this power function and the polynomial. Do they have similar end behavior? 103. f(x) 2x5 5x4 3x3 104. f(x) x4 6x2 9 In Exercises 105 and 106, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the x-intercepts and state the zeros of the function and their multiplicities. 105. f (x) x4 15.9x3 1.31x2 292.905x 445.7025 106. f (x) x5 2.2x4 18.49x3 29.878x2 76.5x 100.8 In Exercises 107 and 108, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the coordinates of the relative maximum and minimum points. Round your answers to two decimal places. 107. 108. f (x) = 2x5 - 4x4 - 12x3 + 18x2 + 16x - 7 f (x) = 2x4 + 5x3 - 10x2 - 15x + 8 C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y c04b.qxd 11/25/11 3:32 PM Page 409 439. C O N C E P TUAL O BJ E CTIVE S Extend long division of real numbers to polynomials. Understand when synthetic division can be used. D IVI D I N G P O LYN O M I ALS: LO N G D IVI S I O N AN D SYNTH ETI C D IVI S I O N S E CTI O N 4.3 S K I LLS O BJ E CTIVE S Divide polynomials with long division. Divide polynomials with synthetic division. Long Division of Polynomials Lets start with an example whose answer we already know. We know that a quadratic expression can be factored into the product of two linear factors: x2 4x 5 (x 5)(x 1). Therefore, if we divide both sides of the equation by (x 1), we get We can state this by saying x2 4x 5 divided by x 1 is equal to x 5. Conrm this statement by long division: Note that although this is standard division notation, the dividend and the divisor are both polynomials that consist of multiple terms. The leading terms of each algebraic expression will guide us. WORDS MATH Q: x times what quantity gives x2 ? A: x x x Multiply x(x 1) x2 x. x2 x Subtract (x2 x) from x2 4x 5. x Note: (x2 x) x2 x. Bring down the 5. 5x - 5- -x2 + x x - 1x2 + 4x - 5 x - 1x2 + 4x - 5 x - 1x2 + 4x - 5 x - 1x2 + 4x - 5 x - 1x2 + 4x - 5 x2 + 4x - 5 x - 1 = x + 5 To divide polynomials, we rely on the technique we use for dividing real numbers. For example, if you were asked to divide 6542 by 21, the long division method used is illustrated in the margin. This solution can be written two ways: . In this example, the dividend is 6542, the divisor is 21, the quotient is 311, and the remainder is 11. We employ a similar technique (dividing the leading terms) when dividing polynomials. 311 R11 or 311 + 11 21 311 63 24 21 32 21 11 216542 410 c04b.qxd 11/25/11 3:32 PM Page 410 440. 4.3 Dividing Polynomials: Long Division and Synthetic Division 411 WORDS MATH Q: x times what quantity is 5x? A: 5 Multiply 5(x 1) 5x 5. Subtract (5x 5). Note: (5x 5) 5x 5. 0 As expected, the remainder is 0. By long division we have shown that Check: Multiplying the equation by x 1 yields x2 4x 5 (x 5)(x 1), which we knew to be true. x2 + 4x - 5 x - 1 = x + 5 -5x + 5 5x - 5 -x2 + x x - 1x2 + 4x - 5 x + 5 5x 5 5x - 5-- -x2 + x x - 1x2 + 4x - 5 x + 5 Answer: x2 2x 3, remainder 0. Technology Tip A graphing utility can be used to check (x2 5x 6)(2x 1) 2x3 9x2 7x 6 using their graphs. Notice that the graphs are the same. E X AM P LE 1 Dividing Polynomials Using Long Division; Zero Remainder Divide 2x3 9x2 7x 6 by 2x 1. Solution: Multiply: x2 (2x 1). Subtract: Bring down the 7x. Multiply: 5x(2x 1). Subtract: Bring down the 6. Multiply: 6(2x 1). Subtract. 0 Quotient: Check: (2x 1)(x2 5x 6) 2x3 9x2 7x 6. Note: Since the divisor cannot be equal to zero, 2x 1 Z 0, then we say x Z . YO U R TU R N Divide 4x3 13x2 2x 15 by 4x 5. 1 2 x2 - 5x + 6 -(12x 6) 12x + 6 -(10x2 5x) -10x2 + 7x -(2x3 x2 ) 2x + 12x3 - 9x2 + 7x + 6 x2 - 5x + 6 Why are we interested in dividing polynomials? Because it helps us nd zeros of polynomials. In Example 1, using long division, we found that 2x3 9x2 7x 6 (2x 1)(x2 5x 6) c04b.qxd 11/25/11 3:32 PM Page 411 441. 412 C HAP TE R 4 Polynomial and Rational Functions Factoring the quadratic expression enables us to write the cubic polynomial as a product of three linear factors: 2x3 9x2 7x 6 (2x 1)(x2 5x 6) (2x 1)(x 3)(x 2) Set the value of the polynomial equal to zero, (2x 1)(x 3)(x 2) 0, and solve for x. The zeros of the polynomial are , 2, and 3. In Example 1 and in the Your Turn, the remainder was 0. Sometimes there is a nonzero remainder (Example 2). 1 2 E X AM P LE 2 Dividing Polynomials Using Long Division; Nonzero Remainder Divide 6x2 x 2 by x 1. Solution: Multiply 6x(x 1). Subtract and bring down 2. Multiply 7(x 1). Subtract and identify the remainder. Dividend Quotient Remainder Divisor Divisor Check: Multiply the quotient and (6x 7)(x 1) remainder by x 1. 6x2 x 7 5 The result is the dividend. 6x2 x 2 YO U R TU R N Divide 2x3 x2 4x 3 by x 1. 5 (x + 1) # (x + 1) x Z -1 6x2 x 2 x 1 = 6x 7 + 5 x 1 In general, when a polynomial is divided by another polynomial, we express the result in the following form: where P(x) is the dividend, is the divisor, Q(x) is the quotient, and r(x) is the remainder. Multiplying this equation by the divisor, d(x), leads us to the division algorithm. d(x) Z 0 P(x) d(x) = Q(x) + r(x) d(x) + 5 -(-7x - 7) -7x - 2 -(6x2 + 6x) x + 16x2 - x - 2 6x - 7 Answer: 2x2 3x 1 R: 4 or 2x2 + 3x - 1 - 4 x - 1 If P(x) and d(x) are polynomials with , and if the degree of P(x) is greater than or equal to the degree of d(x), then unique polynomials Q(x) and r(x) exist such that If the remainder r(x) 0, then we say that d(x) divides P(x) and that d(x) and Q(x) are factors of P(x). P(x) = d(x) # Q(x) + r(x) d(x) Z 0 THE DIVISION ALGORITHM c04b.qxd 11/25/11 3:32 PM Page 412 442. 4.3 Dividing Polynomials: Long Division and Synthetic Division 413 E X AM P LE 3 Long Division of Polynomials with Missing Terms Divide x3 8 by x 2. Solution: Insert 0x2 0x for placeholders. Multiply x2 (x 2) x3 2x2 . Subtract and bring down 0x. Multiply 2x(x 2) 2x2 4x. Subtract and bring down 8. Multiply 4(x 2) 4x 8. Subtract and get remainder 0. Since the remainder is 0, x 2 is a factor of x3 8. Check: x3 8 (x2 2x 4)(x 2) x3 2x2 4x 2x2 4x 8 x3 8 YO U R TU R N Divide x3 1 by x 1. x3 - 8 x - 2 = x2 + 2x + 4, x Z 2 x2 + 2x + 4 -(x3 - 2x2 ) 2x2 + 0x -(2x2 - 4x) 4x - 8 - (4x - 8) 0 x - 2x3 + 0x2 0x - 8 E X AM P LE 4 Long Division of Polynomials Divide 3x4 2x3 x2 4 by x2 1. Solution: Insert 0x as a placeholder in both the divisor and the dividend. Multiply 3x2 (x2 0x 1). Subtract and bring down 0x. Multiply 2x(x2 0x 1). Subtract and bring down 4. Multiply 2(x2 2x 1). Subtract and get remainder 2x 6. 3x2 + 2x - 2 x2 + 0x + 1 3x4 + 2x3 + x2 + 0x + 4 - A3x4 + 0x3 + 3x2 B 2x3 - 2x2 + 0x -(2x3 + 0x2 + 2x) -2x2 - 2x + 4 -(-2x2 + 0x - 2) -2x + 6 YO U R TU R N Divide 2x5 3x2 12 by x3 3x 4. 3x4 + 2x3 + x2 + 4 x2 + 1 = 3x2 + 2x - 2 + -2x + 6 x2 + 1 Answer: x2 x 1 Answer: 2x2 + 6 + 11x2 + 18x + 36 x3 - 3x - 4 c04b.qxd 11/25/11 3:32 PM Page 413 443. 414 C HAP TE R 4 Polynomial and Rational Functions Answer: 5x2 - 3 2 x + 5 2 + 7 2 x - 3 2 2x2 - 1 In Examples 1 through 4 the dividends, divisors, and quotients were all polynomials with integer coefcients. In Example 5, however, the resulting quotient has rational (noninteger) coefcients. E X AM P LE 5 Long Division of Polynomials Resulting in Quotients with Rational Coefcients Divide Solution: Insert as a placeholder in the dividend and as a placeholder in the divisor. Multiply Subtract and bring down remaining terms. Multiply Subtract and bring down remaining terms. Multiply Subtract and bring down the remainder YO U R TU R N Divide 10x4 - 3x3 + 5x - 4 by 2x2 - 1. 8x4 - 5x3 + 7x - 2 2x2 + 1 = 4x2 - 5 2 x - 2 + 19 2 x 2x2 + 1 19 2 x. -2(2x2 + 0x + 1). -5 2 x(2x2 + 0x + 1). 4x2 (2x2 + 0x + 1). 0x 0x2 8x4 - 5x3 + 7x - 2 by 2x2 + 1. 19 2 x -(-4x2 + 0x - 2) - 4x2 + 19 2 x - 2 -(-5x3 + 0x2 - 5 2 x) - 5x3 - 4x2 + 7x -(8x4 + 0x3 + 4x2 ) 2x2 + 0x + 1 8x4 - 5x3 + 0x2 + 7x - 2 4x2 - 5 2x - 2 Synthetic Division of Polynomials In the special case when the divisor is a linear factor of the form x a or x a, there is another, more efcient way to divide polynomials. This method is called synthetic division. It is called synthetic because it is a contrived shorthand way of dividing a polynomial by a linear factor. A detailed step-by-step procedure is given below for synthetic division. Lets divide x4 x3 2x 2 by x 1 using synthetic division. STEP 1 Write the division in synthetic form. List the coefcients of the dividend. Remember to use 0 for a placeholder. The divisor is x 1. Since x 1 0 x 1 is used. STEP 2 Bring down the rst term (1) in the dividend. STEP 3 Multiply the 1 by this leading coefcient (1), and place the product up and to the right in the second column. Coefcients of Dividend 1 1 1 0 2 2 1 1 1 0 2 2 Bring down the 1 1 1 1 1 0 2 2 1 1 Study Tip If (x a) is the divisor, then a is the number used in synthetic division. c04b.qxd 12/22/11 7:14 PM Page 414 444. STEP 4 Add the values in the second column. STEP 5 Repeat Steps 3 and 4 until all columns are lled. STEP 6 Identify the quotient by assigning powers of x in descending order, beginning with xn1 x41 x3 . The last term is the remainder. We know that the degree of the rst term of the quotient is 3 because a fourth-degree polynomial was divided by a rst-degree polynomial. Lets compare dividing x4 x3 2x 2 by x 1 using both long division and synthetic division. Long Division Synthetic Division x 1x4 x3 0x2 2x 2 x4 + x3 -2x3 + 0x2 -(-2x3 - 2x2 ) 2x2 - 2x -(2x2 + 2x) -4x + 2 -(-4x - 4) 6 x3 2x2 2x 4 1 1 1 0 2 2 1 ADD 1 2 1 1 1 0 2 2 1 2 2 4 1 2 2 4 6 1 1 1 0 2 2 1 2 2 4 1 2 2 4 6 x3 2x2 2x 4 f 1 1 1 0 2 2 1 2 2 4 1 2 2 4 6 Quotient Coefcients Remainder x3 2x2 2x 4 f Study Tip Synthetic division can only be used when the divisor is of the form x a or x a. Both long division and synthetic division yield the same answer. x4 - x3 - 2x + 2 x + 1 = x3 - 2x2 + 2x - 4 + 6 x + 1 E X AM P LE 6 Synthetic Division Use synthetic division to divide 3x5 2x3 x2 7 by x 2. Solution: STEP 1 Write the division in synthetic form. List the coefcients of the dividend. Remember to use 0 for a placeholder. The divisor of the original problem is x 2. If we set x 2 0 we nd that x 2, so 2 is the divisor for synthetic division. 2 3 0 2 1 0 7 4.3 Dividing Polynomials: Long Division and Synthetic Division 415 c04b.qxd 11/25/11 3:32 PM Page 415 445. 416 C HAP TE R 4 Polynomial and Rational Functions 2 3 0 2 1 0 7 6 12 20 38 76 3 6 10 19 38 83 2 3 0 2 1 0 7 6 12 20 38 76 3 6 10 19 38 83 3x4 6x3 10x2 19x 38 f Answer: 2x2 + 2x + 1 + 4 x - 1 Expressing Results Dividend (quotient)(divisor) remainder When Remainder Is Zero Dividend (quotient)(divisor) Quotient and divisor are factors of the dividend. Dividend Divisor = quotient + remainder divisor Division of Polynomials Long division can always be used. Synthetic division is restricted to when the divisor is of the form x a or x a. STEP 2 Perform the synthetic division steps. STEP 3 Identify the quotient and remainder. YO U R TU R N Use synthetic division to divide 2x3 x 3 by x 1. 3x5 - 2x3 + x2 - 7 x + 2 = 3x4 - 6x3 + 10x2 - 19x + 38 - 83 x + 2 E X E R C I S E S S E C TI O N 4.3 S E CTI O N 4.3 S U M MARY S K I LL S In Exercises 130, divide the polynomials using long division. Use exact values. Express the answer in the form Q(x) ?, r(x) ?. 1. (2x2 5x 3) (x 3) 2. (2x2 5x 3) (x 3) 3. (x2 5x 6) (x 2) 4. (2x2 3x 1) (x 1) 5. (3x2 9x 5) (x 2) 6. (x2 4x 3) (x 1) 7. (3x2 13x 10) (x 5) 8. (3x2 13x 10) (x 5) 9. (x2 4) (x 4) 10. (x2 9) (x 2) 11. (9x2 25) (3x 5) 12. (5x2 3) (x 1) 13. (4x2 9) (2x 3) 14. (8x3 27) (2x 3) 15. (11x 20x2 12x3 2) (3x 2) 16. (12x3 2 11x 20x2 ) (2x 1) 17. (4x3 2x 7) (2x 1) 18. (6x4 2x2 5) (3x 2) 19. 20. 21. (2x5 3x4 2x2 ) (x3 3x2 1) 22. (9x6 7x4 2x3 5) (3x4 2x 1) (12x3 + 1 + 7x + 16x2 ) , Ax + 1 3 B(4x3 - 12x2 - x + 3) , Ax - 1 2 B c04b.qxd 11/25/11 3:32 PM Page 416 446. 4.3 Dividing Polynomials: Long Division and Synthetic Division 417 63. Travel. If a car travels a distance of x3 60x2 x 60 miles at an average speed of x 60 miles per hour, how long does the trip take? 64. Sports. If a quarterback throws a ball x2 5x 50 yards in 5 x seconds, how fast is the football traveling? 61. Geometry. The area of a rectangle is 6x4 4x3 x2 2x 1 square feet. If the length of the rectangle is 2x2 1 feet, what is the width of the rectangle? 62. Geometry. If the rectangle in Exercise 61 is the base of a rectangular box with volume 18x5 18x4 x3 7x2 5x 1 cubic feet, what is the height of the box? 23. 24. 25. 26. 27. 28. 29. (x4 0.8x3 0.26x2 0.168x 0.0441) (x2 1.4x 0.49) 30. (x5 2.8x4 1.34x3 0.688x2 0.2919x 0.0882) (x2 0.6x 0.09) In Exercises 3150, divide the polynomial by the linear factor with synthetic division. Indicate the quotient Q(x) and the remainder r(x). 31. (3x2 7x 2) (x 2) 32. (2x2 7x 15) (x 5) 33. (7x2 3x 5) (x 1) 34. (4x2 x 1) (x 2) 35. (3x2 4x x4 2x3 4) (x 2) 36. (3x2 4 x3 ) (x 1) 37. (x4 1) (x 1) 38. (x4 9) (x 3) 39. (x4 16) (x 2) 40. (x4 81) (x 3) 41. 42. 43. 44. 45. (2x4 9x3 9x2 81x 81) (x 1.5) 46. (5x3 x2 6x 8) (x 0.8) 47. 48. 49. 50. In Exercises 5160, divide the polynomials by either long division or synthetic division. 51. (6x2 23x 7) (3x 1) 52. (6x2 x 2) (2x 1) 53. (x3 x2 9x 9) (x 1) 54. (x3 2x2 6x 12) (x 2) 55. (x5 4x3 2x2 1) (x 2) 56. (x4 x2 3x 10) (x 5) 57. (x4 25) (x2 1) 58. (x3 8) (x2 2) 59. (x7 1) (x 1) 60. (x6 27) (x 3) (x6 - 4x4 - 9x2 + 36) , Ax - 13B(x6 - 49x4 - 25x2 + 1225) , Ax - 15B x6 + 4x5 - 2x3 + 7 x + 1 x7 - 8x4 + 3x2 + 1 x - 1 (3x4 + x3 + 2x - 3) , Ax - 3 4 B(2x4 - 3x3 + 7x2 - 4) , Ax - 2 3 B (3x3 - 8x2 + 1) , Ax + 1 3 B(2x3 - 5x2 - x + 1) , Ax + 1 2 B 2x5 - 4x3 + 3x2 + 5 x - 0.9 -3x4 + 7x3 - 2x + 1 x - 0.6 -13x2 + 4x4 + 9 4x2 - 9 40 - 22x + 7x3 + 6x4 6x2 + x - 2 x4 - 9 x2 + 3 x4 - 1 x2 - 1 A P P L I C AT I O N S c04b.qxd 11/25/11 3:32 PM Page 417 447. 418 C HAP TE R 4 Polynomial and Rational Functions In Exercises 6568, explain the mistake that is made. 65. Divide x3 4x2 x 6 by x2 x 1. Solution: This is incorrect. What mistake was made? x - 3 x2 + x + 1x3 - 4x2 + x + 6 x3 + x2 + x -3x2 + 2x + 6 -3x2 - 3x - 3 -x + 3 67. Divide x3 4x 12 by x 3. Solution: This is incorrect. What mistake was made? 2 1 3 5 2 2 10 30 1 5 15 28 x2 5x 15 1 1 3 2 1 1 2 4 1 2 4 5 x2 2x 4 3 1 4 12 3 21 1 7 9 x 7 c 69. A fth-degree polynomial divided by a third-degree polynomial will yield a quadratic quotient. 70. A third-degree polynomial divided by a linear polynomial will yield a linear quotient. 71. Synthetic division can be used whenever the degree of the dividend is exactly one more than the degree of the divisor. 72. When the remainder is zero, the divisor is a factor of the dividend. d d 66. Divide x4 3x2 5x 2 by x 2. Solution: This is incorrect. What mistake was made? 68. Divide x3 3x2 2x 1 by x2 1. Solution: This is incorrect. What mistake was made? 77. Plot . What type of function is it? Perform this division using long division, and conrm that the graph corresponds to the quotient. 78. Plot . What type of function is it? Perform this division using synthetic division, and conrm that the graph corresponds to the quotient. 79. Plot . What type of function is it? Perform this division using synthetic division, and conrm that the graph corresponds to the quotient. x4 + 2x3 - x - 2 x + 2 x3 - 3x2 + 4x - 12 x - 3 2x3 - x2 + 10x - 5 x2 + 5 80. Plot . What type of function is it? Perform this division using long division, and conrm that the graph corresponds to the quotient. 81. Plot . What type of function is it? Perform this division using long division, and conrm that the graph corresponds to the quotient. 82. Plot . What type of function is it? Perform this division using long division, and conrm that the graph corresponds to the quotient. -3x5 - 4x4 + 29x3 + 36x2 - 18x 3x2 + 4x - 2 -6x3 + 7x2 + 14x - 15 2x + 3 x5 - 9x4 + 18x3 + 2x2 - 5x - 3 x4 - 6x3 + 2x + 1 75. Divide x3n x2n xn 1 by xn 1. 76. Divide x3n 5x2n 8xn 4 by xn 1. 73. Is x b a factor of x3 (2b a)x2 (b2 2ab)x ab2 ? 74. Is x b a factor of x4 (b2 a2 )x2 a2 b2 ? C AT C H T H E M I S TA K E C O N C E P T U A L In Exercises 6972, determine whether each statement is true or false. C H A L L E N G E T E C H N O L O G Y c04c.qxd 11/25/11 3:43 PM Page 418 448. C O N C E P TUAL O BJ E CTIVE S Understand that a polynomial of degree n has at most n real zeros. Understand that a real zero can be either rational or irrational and that irrational zeros will not be listed as possible zeros through the rational zero test. Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated. TH E R EAL Z E R O S O F A P O LYN O M IAL F U N CTI O N S E C TI O N 4.4 S K I LLS O BJ E CTIVE S Apply the remainder theorem to evaluate a polynomial function. Apply the factor theorem. Use the rational zero (root) theorem to list possible rational zeros. Apply Descartes rule of signs to determine the possible combination of positive and negative real zeros. Utilize the upper and lower bound theorems to narrow the search for real zeros. Find the real zeros of a polynomial function. Factor a polynomial function. Employ the intermediate value theorem to approximate a real zero. *There are complicated formulas for nding the zeros of polynomial functions of degree 3 and 4, but there are no such formulas for degree 5 and higher polynomials (according to the AbelRufni theorem). The Remainder Theorem and the Factor Theorem The zeros of a polynomial function assist us in nding the x-intercepts of the graph of a polynomial function. How do we nd the zeros of a polynomial function? For polynomial functions of degree 2, we have the quadratic formula, which allows us to nd the two zeros. For polynomial functions whose degree is greater than 2, much more work is required.* In this section we focus our attention on nding the real zeros of a polynomial function. Later, in Section 4.5, we expand our discussion to complex zeros of polynomial functions. In this section we start by listing possible rational zeros.As you will see there are sometimes many possibilities. We can then narrow the search using Descartes rule of signs, which tells us possible combinations of positive and negative real zeros. We can narrow the search even further with the upper and lower bound rules. Once we have tested possible values and determined a zero, we will employ synthetic division to divide the polynomial by the linear factor associated with the zero. We will continue the process until we have factored the polynomial function into a product of either linear factors or irreducible quadratic factors. Last, we will discuss how to nd irrational real zeros using the intermediate value theorem. If we divide the polynomial function f(x) x3 2x2 x 3 by x 2 using synthetic division, we nd the remainder is 1. Notice that if we evaluate the function at x 2, the result is 1. f(2) 1 This leads us to the remainder theorem. 2 1 2 1 3 2 0 2 1 0 1 1 419 c04c.qxd 11/25/11 3:43 PM Page 419 449. 420 C HAP TE R 4 Polynomial and Rational Functions If a polynomial P(x) is divided by x a, then the remainder is r P(a). REMAINDER THEOREM The remainder theorem tells you that polynomial division can be used to evaluate a polynomial function at a particular point. E X AM P LE 1 Two Methods for Evaluating Polynomials Let P(x) 4x5 3x4 2x3 7x2 9x 5 and evaluate P(2) by a. Evaluating P(2) directly b. The remainder theorem and synthetic division Solution: a. P(2) 4(2)5 3(2)4 2(2)3 7(2)2 9(2) 5 4(32) 3(16) 2(8) 7(4) 9(2) 5 128 48 16 28 18 5 81 Recall that when a polynomial is divided by x a, if the remainder is zero, we say that x a is a factor of the polynomial. Through the remainder theorem, we now know that the remainder is related to evaluation of the polynomial at the point x a. We are then led to the factor theorem. If P(a) 0, then x a is a factor of P(x). Conversely, if x a is a factor of P(x), then P(a) 0. FACTOR THEOREM 2 4 3 2 7 9 5 8 10 24 34 86 4 5 12 17 43 81 YO U R TU R N Let P(x) x3 2x2 5x 2 and evaluate P(2) using the remainder theorem and synthetic division. b. WORDS MATH Recall the Division Algorithm. Let for any real number . The degree of the remainder is always less than the degree of the divisor: therefore the remainder must be a constant (Call it . Let Simplify. 0 P(a) = r P(a) = (a - a) # Q(x) + rx = a. P(x) = (x - a) # Q(x) + rr, r(x) = r) P(x) = (x - a) # Q(x) + r(x)ad(x) = x - a P(x) = d(x) # Q(x) + r(x) c Technology Tip A graphing utility can be used to evaluate P(2). Enter P(x) 4x5 3x4 2x3 7x2 9x 5 as Y1. To evaluate P(2), press VARS Y-VARS 1:Function 1:Y1 ( 2 ) ENTER Answer: P(2) 28 c04c.qxd 11/25/11 3:43 PM Page 420 450. 4.4 The Real Zeros of a Polynomial Function 421 STEP 1 Divide P(x) x3 2x2 5x 6 by x 2 using synthetic division. Since the remainder is zero, P(2) 0, x 2 is a factor of P(x) x3 2x2 5x 6. STEP 2 Write P(x) as a product. P(x) (x 2)(x2 4x 3) STEP 3 Factor the quadratic polynomial. P(x) (x 2)(x 3)(x 1) YO U R TU R N Determine whether x 1 is a factor of P(x) x3 4x2 7x 10. If so, factor P(x) completely. 1 6 8 1 3 0 x2 - 4x + 3 -4 -6-2 -5-2-2 3 1 0 0 36 3 9 12 36 1 3 4 12 0 x3 + 3x2 - 4x - 12 -13 Solution: STEP 1 With synthetic division divide P(x) x4 13x2 36 by x 3. Because the remainder is 0, x 3 is a factor , and we can write the polynomial as STEP 2 With synthetic division divide the remaining cubic polynomial (x3 3x2 4x 12) by x 2. Because the remainder is 0, x 2 is a factor , and we can now write the polynomial as STEP 3 Factor the quadratic polynomial: x2 x 6 (x 3)(x 2). STEP 4 Write P(x) as a product of linear factors: YO U R TU R N Determine whether x 3 and x 2 are factors of P(x) x4 x3 7x2 x 6. If so, factor P(x) completely. P(x) = (x - 3)(x - 2)(x + 2)(x + 3) P(x) = (x - 3)(x + 2)(x2 + x - 6) P(x) = (x - 3)(x3 + 3x2 - 4x - 12) E X AM P LE 3 Using the Factor Theorem to Factor a Polynomial Determine whether x 3 and x 2 are factors of P(x) x4 13x2 36. If so, factor P(x) completely. 2 1 3 4 12 2 2 12 1 1 6 0 x2 + x - 6 E X AM P LE 2 Using the Factor Theorem to Factor a Polynomial Determine whether x 2 is a factor of P(x) x3 2x2 5x 6. If so, factor P(x) completely. Solution: By the factor theorem, x 2 is a factor of P(x) x3 2x2 5x 6 if P(2) 0. By the remainder theorem, if we divide P(x) x3 2x2 5x 6 by x 2, then the remainder is equal to P(2). Answer: (x 1) is a factor; P(x) (x 5)(x 1) (x 2) The zeros of the function at x 3, x 2, x 2, and x 3 are shown in the graph. A table of values supports the zeros of the graph. Technology Tip Answer: are factors; P(x) (x 3)(x 2) (x 1)(x 1) (x - 3) and (x + 2) Technology Tip The three zeros of the function give the three factors x 2, x 1, and x 3. A table of values supports the zeros of the graph. c04c.qxd 11/25/11 8:56 PM Page 421 451. 422 C HAP TE R 4 Polynomial and Rational Functions The Search for Real Zeros In all of the examples thus far, the polynomial function and one or more real zeros (or linear factors) were given. Now, we will not be given any real zeros to start with. Instead, we will develop methods to search for them. Each real zero corresponds to a linear factor, and each linear factor is of degree 1. Therefore, the largest number of real zeros a polynomial function can have is equal to the degree of the polynomial. A polynomial function cannot have more real zeros than its degree. THE NUMBER OF REAL ZEROS The following functions illustrate that a polynomial function of degree n can have at most n real zeros. Study Tip The largest number of zeros a polynomial can have is equal to the degree of the polynomial. POLYNOMIAL FUNCTION DEGREE REAL ZEROS COMMENTS f(x) x2 9 2 x ;3 Two real zeros f(x) x2 4 2 None No real zeros f(x) x3 1 3 x 1 One real zero f(x) x3 x2 6x 3 x 2, 0, 3 Three real zeros Now that we know the maximum number of real zeros a polynomial function can have, let us discuss how to nd these zeros. The Rational Zero Theorem and Descartes Rule of Signs When the coefcients of a polynomial are integers, then the rational zero theorem (rational root test) gives us a list of possible rational zeros. We can then test these possible values to determine whether they really do correspond to actual zeros. Descartes rule of signs tells us the possible combinations of positive real zeros and negative real zeros. Using Descartes rule of signs will assist us in narrowing down the large list of possible zeros generated through the rational zero theorem to a (hopefully) shorter list of possible zeros. First, lets look at the rational zero theorem; then well turn to Descartes rule of signs. If the polynomial function has integer coefcients, then every rational zero of P(x) has the form: = ; positive integer factors of constant term positive integer factors of leading coefficient Rational zero = integer factors of a0 integer factors of an = integer factors of constant term integer factors of leading coefficient P(x) = anxn + an-1xn-1 + + a2x2 + a1x + a0 THE RATIONAL ZERO THEOREM (RATIONAL ROOT TEST) To use this theorem, simply list all combinations of integer factors of both the constant term a0 and the leading coefcient term an and take all appropriate combinations of ratios. This procedure is illustrated in Example 4. Notice that when the leading coefcient is 1, then the possible rational zeros will simply be the possible integer factors of the constant term. c04c.qxd 11/25/11 3:43 PM Page 422 452. 4.4 The Real Zeros of a Polynomial Function 423 2 is not a zero: 2 1 1 5 1 6 2 2 6 14 1 1 3 7 20 2 is a zero: 2 1 1 5 1 6 2 6 2 6 1 3 1 3 0 Since 2 is a zero, then x 2 is a factor of P(x), and the remaining quotient is x3 3x2 x 3. Therefore, if there are any other real roots remaining, we can now use the simpler x3 3x2 x 3 for the dividend. Also note that the rational zero theorem can be applied to the new dividend and possibly shorten the list of possible rational zeros. In this case, the possible rational zeros of F(x) x3 3x2 x 3 are 3 is a zero: 3 1 3 1 3 3 0 3 1 0 1 0 We now know that 2 and 3 are conrmed zeros. If we continue testing, we will nd that the other possible zeros fail. This is a fourth-degree polynomial, and we have found two rational real zeros. We see in the graph on the right that these two real zeros correspond to the x-intercepts. YO U R TU R N List the possible rational zeros of the polynomial P(x) x4 2x3 2x2 2x 3, and determine rational real zeros. ; 1 and ; 3. E X AM P LE 4 Using the Rational Zero Theorem Determine possible rational zeros for the polynomial P(x) x4 x3 5x2 x 6 by the rational zero theorem. Test each one to nd all rational zeros. Solution: STEP 1 List factors of the constant a0 6 ;1, ;2, ;3, ;6 and leading coefcient terms. an 1 ;1 STEP 2 List possible rational zeros . There are three ways to test whether any of these are zeros: Substitute these values into the polynomial to see which ones yield zero; use either polynomial division or synthetic division to divide the polynomial by these possible zeros; and look for a zero remainder. STEP 3 Test possible zeros by looking for zero remainders. 1 is not a zero: P(1) (1)4 (1)3 5(1)2 (1) 6 12 1 is not a zero: P(1) (1)4 (1)3 5(1)2 (1) 6 8 We could continue testing with direct substitution, but let us now use synthetic division as an alternative. ;1 ;1 , ;2 ;1 , ;3 ;1 , ;6 ;1 = ;1, ;2, ;3, ;6 a0 an x y 20 55 20 Study Tip The remainder can be found by evaluating the function or synthetic division. For simple values like x ;1, it is easier to evaluate the polynomial function. For other values, it is often easier to use synthetic division. Study Tip Notice in Step 3 that the polynomial can be factored by grouping: F(x) = (x - 3)(x2 + 1). F(x) = x3 - 3x2 + x - 3 Answer: Possible rational zeros: ;1 and ;3. Rational real zeros: 1 and 3. c04c.qxd 11/25/11 3:43 PM Page 423 453. 424 C HAP TE R 4 Polynomial and Rational Functions Notice in Example 4 that the polynomial function P(x) x4 x3 5x2 x 6 had two rational real zeros, x 2 and x 3. This implies that x 2 and x 3 are factors of P(x). Also note in the last step when we divided by the zero x 3, the quotient was x2 1. Therefore, we can write the polynomial in factored form as linear linear irreducible factor factor quadratic factor Notice that the first two factors are of degree 1, so we call them linear factors. The third expression, x2 1, is of degree 2 and cannot be factored in terms of real numbers. We will discuss complex zeros in the next section. For now, we say that a quadratic expression, ax2 bx c, is called irreducible if it cannot be factored over the real numbers. P(x) = (x + 2) (x - 3)(x2 + 1) E X AM P LE 5 Factoring a Polynomial Function Write the following polynomial function as a product of linear and/or irreducible quadratic factors: . Solution: Use the rational zero theorem to list possible rational roots. x ;1, ;3, ;5, ;9, ;15, ;45 Test possible zeros by evaluating the function or by utilizing synthetic division. x 1 is not a zero. P(1) 80 x 1 is a zero. P(1) 0 Divide P(x) by x 1. 1 1 4 4 36 45 1 5 9 45 1 5 9 45 0 x 5 is a zero. 5 1 5 9 45 5 0 45 1 0 9 0 x2 9 The factor x2 9 is irreducible. Write the polynomial as a product of linear and/or irreducible quadratic factors. Notice that the graph of this polynomial function has x-intercepts at x 1 and x 5. YO U R TU R N Write the following polynomial function as a product of linear and/or irreducible quadratic factors. P(x) = x4 - 2x3 - x2 - 4x - 6 P(x) = (x - 5)(x + 1)(x2 + 9) P(x) = x4 - 4x3 + 4x2 - 36x - 45 Technology Tip The real zeros of the function at x 1 and x 5 give the linear factors x 1 and x 5. Use the synthetic division to nd the irreducible quadratic factors. A table of values supports the real zeros of the graph. Answer: P(x) (x 1)(x 3) x2 2BA c04c.qxd 11/25/11 3:43 PM Page 424 454. 4.4 The Real Zeros of a Polynomial Function 425 d d d d d d If the polynomial function has real coefcients and then: The number of positive real zeros of the polynomial is either equal to the number of variations of sign of P(x) or less than that number by an even integer. The number of negative real zeros of the polynomial is either equal to the number of variations of sign of P(x) or less than that number by an even integer. a0 Z 0, P(x) = an xn + an-1xn-1 + + a2 x2 + a1x + a0 DESCARTES RULE OF SIGNS The rational zero theorem lists possible zeros. It would be helpful if we could narrow that list. Descartesrule of signs determines the possible combinations of positive real zeros and negative real zeros through variations of sign.Avariation in sign is a sign difference seen between consecutive coefcients. Sign Change to Sign Change Sign Change to to This polynomial experiences three sign changes or variations in sign. P(x) = 2x6 - 5x5 - 3x4 + 2x3 - x2 - x - 1 E X AM P LE 6 Using Descartes Rule of Signs Determine the possible combinations of zeros for P(x) x3 2x2 5x 6. Solution: Determine the number of variations of sign in P(x). P(x) has 2 variations in sign. Apply Descartes rule of signs. P(x) has either 2 or 0 positive real zeros. Determine the number of (x)3 2(x)2 5(x) 6 variations of sign in P(x). x3 2x2 5x 6 P(x) has 1 variation in sign. P(x) x3 2x2 5x 6 Apply Descartes rule of signs. P(x) must have 1 negative real zero. Since P(x) x3 2x2 5x 6 is a third-degree polynomial, there are at most 3 real zeros. One zero is a negative real number, and there can be either 2 positive real zeros or 0 positive real zeros. Now look back at Example 2 and see that in fact there were 1 negative real zero and 2 positive real zeros. d d d d Sign Change P(x) x3 2x2 5x 6 Sign Change Descartesrule of signs narrows our search for real zeros because we dont have to test all of the possible rational zeros. For example, if we know there is one positive real zero, then if we nd a positive rational zero we no longer need to continue to test possible positive zeros. d d Sign Change c04c.qxd 11/25/11 3:43 PM Page 425 455. 426 C HAP TE R 4 Polynomial and Rational Functions Technology Tip There are 1 negative real zero and 1 positive real zero. The number of negative real zeros is 1 and the number of positive real zeros is 1. So, it has 2 complex conjugate zeros. A table of values supports the zeros of the function. E X AM P LE 7 Using Descartes Rule of Signs to Find Possible Combinations of Real Zeros Determine the possible combinations of real zeros for P(x) x4 2x3 x2 2x 2. Solution: P(x) has 3 variations in sign. Apply Descartes rule of signs. P(x) has either 3 or 1 positive real zero. Find P(x). P(x) (x)4 2(x)3 (x)2 2(x) 2 x4 2x3 x2 2x 2 P(x) has 1 variation in sign. P(x) x4 2x3 x2 2x 2 Sign Change Apply Descartes rule of signs. P(x) has 1 negative real zero. Since P(x) x4 2x3 x2 2x 2 is a fourth-degree polynomial, there are at most 4 real zeros. P(x) has 1 negative real zero and could have 3 or 1 positive real zeros. The Technology Tip in the margin conrms 1 negative real zero and 1 positive real zero. YO U R TU R N Determine the possible combinations of zeros for: P(x) x4 2x3 x2 8x 12 Factoring Polynomials Now lets draw on the tests discussed in this section thus far to help us in nding all real zeros of a polynomial function. Doing so will enable us to factor polynomials. Answer: Positive real zeros: 1 Negative real zeros: 3 or 1 d d Sign Change P(x) x4 2x3 x2 2x 2 Sign Change Sign Change d d d d d d c04c.qxd 11/25/11 3:43 PM Page 426 456. 4.4 The Real Zeros of a Polynomial Function 427 Positive Real Zeros: 1 Negative Real Zeros: 2 or 0 1 1 3 3 3 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 2 0 2 1 0 1 0 x2 1 E X AM P LE 8 Factoring a Polynomial Write the polynomial P(x) x5 2x4 x 2 as a product of linear and/or irreducible quadratic factors. Solution: STEP 1 Determine variations in sign. P(x) has 1 sign change. P(x) x5 2x4 x 2 P(x) has 2 sign changes. P(x) x5 2x4 x 2 STEP 2 Apply Descartes rule of signs. STEP 3 Use the rational zero theorem to determine the possible rational zeros. ;1, ;2 We know (Step 2) that there is one positive real zero, so test the possible positive rational zeros rst. STEP 4 Test possible rational zeros. 1 1 2 0 0 1 2 1 3 3 3 2 1 is a zero: 1 3 3 3 2 0 Now that we have found the positive zero, we can test the other two possible negative zerosbecause either they both are zeros or neither is a zero. 1 is a zero: At this point, from Descartes rule of signs we know that 2 must also be a zero, since there are either 2 or 0 negative zeros. Lets conrm this: 2 is a zero: STEP 5 Three of the ve zeros have been found to be zeros: 1, 2, and 1. STEP 6 Write the fth-degree polynomial as a product of 3 linear factors and an irreducible quadratic factor. YO U R TU R N Write the polynomial P(x) x5 2x4 x3 2x2 2x 4 as a product of linear and/or irreducible quadratic factors. The rational zero theorem gives us possible rational zeros of a polynomial, and Descartes rule of signs gives us possible combinations of positive and negative real zeros. Additional aids that help eliminate possible zeros are the upper and lower bound rules. These rules can give you an upper and lower bound on the real zeros of a polynomial function. If f(x) has a common monomial factor, you should factor it out rst, and then follow the upper and lower bound rules. P(x) = (x - 1)(x + 1)(x + 2)Ax2 + 1B Answer: P(x) (x 2)(x 1)(x 1)(x2 2) Study Tip If f(x) has a common monomial factor, it should be factored out rst before applying the bound rules. c04c.qxd 12/22/11 7:17 PM Page 427 457. 428 C HAP TE R 4 Polynomial and Rational Functions Let f(x) be a polynomial with real coefcients and a positive leading coefcient. Suppose f(x) is divided by x c using synthetic division. 1. If c 0 and each number in the bottom row is either positive or zero, c is an upper bound for the real zeros of f. 2. If c 0 and the numbers in the bottom row are alternately positive and negative (zero entries count as either positive or negative), c is a lower bound for the real zeros of f. UPPER AND LOWER BOUND RULES E X AM P LE 9 Using Upper and Lower Bounds to Eliminate Possible Zeros Find the real zeros of f(x) 4x3 x2 36x 9. Solution: STEP 1 The rational zero theorem gives possible rational zeros. STEP 2 Apply Descartes rule of signs: f(x) has 3 sign variations. 3 or 1 positive real zeros f(x) has no sign variations. no negative real zeros STEP 3 Try x 1. 1 4 1 36 9 4 3 39 4 3 39 30 x 1 is not a zero, but because the last row contains all positive entries, x 1 is an upper bound. Since we know there are no negative real zeros, we restrict our search to between 0 and 1. STEP 4 Try . 4 1 36 9 1 0 9 4 0 36 0 is a zero and the quotient 4x2 36 has all positive coefcients; therefore, is the only real zero . Note: If f(x) has a common monomial factor, it should be factored out rst before applying the bound rules. 1 4 1 4 1 4x = 1 4 = ;1, ; 1 2 , ; 1 4 , ; 3 4 , ; 3 2 , ; 9 4 , ;3, ; 9 2 , ;9 Factors of 9 Factors of 4 = ;1, ;3, ;9 ;1, ;2, ;4 c04c.qxd 11/25/11 3:43 PM Page 428 458. 4.4 The Real Zeros of a Polynomial Function 429 The Intermediate Value Theorem In our search for zeros, we sometimes encounter irrational zeros, as in, for example, the polynomial Descartes rule of signs tells us there is exactly one real positive zero. However, the rational zero test yields only x ;1, neither of which are zeros. So if we know there is a real positive zero and we know its not rational, it must be irrational. Notice that f(1) 1 and f(2) 15. Since polynomial functions are continuous and the function goes from negative to positive between x 1 and x 2, we expect a zero somewhere in that interval. Generating a graph with a graphing utility, we nd that there is a zero around x 1.3. The intermediate value theorem is based on the fact that polynomial functions are continuous. f(x) = x5 - x4 - 1 Let a and b be real numbers such that a b and f(x) be a polynomial function. If f(a) and f(b) have opposite signs, then there is at least one real zero between a and b. INTERMEDIATE VALUE THEOREM x zero f(x) y a b f (b) f (a) *In calculus you will learn Newtons method, which is a more efcient approximation technique for nding zeros. If the intermediate value theorem tells us that there is a real zero in the interval (a, b), how do we approximate that zero? The bisection method* is a root-nding algorithm that approximates the solution to the equation f(x) 0. In the bisection method the interval is divided in half, and then the subinterval that contains the zero is selected. This is repeated until the bisection method converges to an approximate root of f. c04c.qxd 11/25/11 3:43 PM Page 429 459. 430 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 10 Approximating Real Zeros of a Polynomial Function Approximate the real zero of f(x) x5 x4 1. Note: Descartes rule of signs tells us that there are no real negative zeros and there is exactly one real positive zero. Solution: Find two consecutive integer values for x that have corresponding function values opposite in sign. Note that a graphing utility would have shown an x-intercept between x 1 and x 2. Apply the bisection method, with a 1 and b 2. Evaluate the function at x c. Compare the values of f at the endpoints and midpoint. f(1) 1, , f(2) 15 Select the subinterval corresponding to the opposite signs of f. (1, 1.5) Apply the bisection method again (repeat the algorithm). Evaluate the function at x 1.25. Compare the values of f at the endpoints and midpoint. f(1) 1, , Select the subinterval corresponding to the opposite signs of f. (1.25, 1.5) Apply the bisection method again (repeat the algorithm). Evaluate the function at x 1.375. Compare the values of f at the endpoints and midpoint. , , Select the subinterval corresponding to the opposite signs of f. (1.25, 1.375) We can continue this procedure (applying the bisection method) to nd that the zero is somewhere between and . We nd that to three signicant digits, 1.32 is an approximation to the real zero. f(1.33) L 0.0326f(1.32) L -0.285 f(1.5) L 1.53f(1.375) L 0.3404f(1.25) L -0.38965 f(1.375) L 0.3404 1.25 + 1.5 2 = 1.375 f(1.5) L 1.53f(1.25) L -0.38965 f(1.25) L -0.38965 1 + 1.5 2 = 1.25 f(1.5) L 1.53 f(1.5) L 1.53 c = a + b 2 = 1 + 2 2 = 3 2 x f (x) 1 1 2 15 Technology Tip The graph of f(x) x5 x4 1 is shown. To nd the zero of the function, press 2nd TRACE 2 Zero 1 ENTER 2 ENTER ENTER A table of values supports this zero of the function. Graphing Polynomial Functions In Section 4.2, we graphed simple polynomial functions that were easily factored. Now that we have procedures for nding real zeros of polynomial functions (rational zero theorem, Descartes rule of signs, and upper and lower bound rules for rational zeros, and the c04c.qxd 11/25/11 3:43 PM Page 430 460. 4.4 The Real Zeros of a Polynomial Function 431 E X AM P LE 11 Graphing a Polynomial Function Graph the function f(x) 2x4 2x3 5x2 17x 22. Solution: STEP 1 Find the y-intercept. f(0) 22 STEP 2 Find any x-intercepts (real zeros). Apply Descartesrule of signs. 3 sign changes correspond to 3 or 1 positive real zeros. f(x) 2x4 2x3 5x2 17x 22 1 sign change corresponds to 1 negative real zero. f(x) 2x4 2x3 5x2 17x 22 Apply the rational zero theorem. Let a0 22 and an 2. Test the possible zeros. x 1 is a zero. f(1) 0 There are no other rational zeros. Apply the upper bound rule. 1 2 2 5 17 22 2 0 5 22 2 0 5 22 0 Since x 1 is positive and all of the numbers in the bottom row are positive (or zero), x 1 is an upper bound for the real zeros. We know there is exactly one negative real zero, but none of the possible zeros from the rational zero theorem is a zero. Therefore, the negative real zero is irrational. Apply the intermediate value theorem and the bisection method. f is positive at x 2. f(2) 12 f is negative at x 1. f(1) 30 Use the bisection method to nd the negative real zero between 2 and 1. STEP 3 Determine the end behavior. y 2x4 x y x L - 1.85 Factors of a0 Factors of an = ; 1 2 , ;1, ;2, ; 11 2 , ;11, ;22 intermediate value theorem and the bisection method for irrational zeros), let us return to the topic of graphing polynomial functions. Since a real zero of a polynomial function corresponds to an x-intercept of its graph, we now have methods for nding (or estimating) any x-intercepts of the graph of any polynomial function. Technology Tip The graph of f(x) 2x4 2x3 5x2 17x 22 is shown. A table of values supports this zero of the function and the graph. To nd the zero of the function, press: 2nd TRACE 2 Zero - 3 ENTER - 1 ENTER ENTER c04c.qxd 11/25/11 3:43 PM Page 431 461. 432 C HAP TE R 4 Polynomial and Rational Functions STEP 5 Sketch the graph. x y (1, 0) (2, 48) (1, 30) (0, 22) (1.85, 0) (2, 12) 48 22 32 STEP 4 Find additional points. x 2 1.85 1 0 1 2 f(x) 12 0 30 22 0 48 Point (2, 12) (1.85, 0) (1, 30) (0, 22) (1, 0) (2, 48) Irrational zeros: Approximate zeros by determining when the polynomial function changes sign (intermediate value theorem). Procedure for Factoring a Polynomial Function List possible rational zeros (rational zero theorem). List possible combinations of positive and negative real zeros (Descartes rule of signs). Test possible values until a zero is found.* Once a real zero is found, repeat testing on the quotient until linear and/or irreducible quadratic factors remain. If there is a real zero but all possible rational roots have failed, then approximate the zero using the intermediate value theorem and the bisection method. *Depending on the form of the quotient, upper and lower bounds may eliminate possible zeros. In this section we discussed how to nd the real zeros of a polynomial function. Once real zeros are known, it is possible to write the polynomial function as a product of linear and/or irreducible quadratic factors. The Number of Zeros A polynomial of degree n has at most n real zeros. Descartesrule of signs determines the possible combinations of positive and negative real zeros. Upper and lower bounds help narrow the search for zeros. How to Find Zeros Rational zero theorem: List possible rational zeros: Factors of constant, a0 Factors of leading coefficient, an S E CTI O N 4.4 S U M MARY E X E R C I S E S S E CTI O N 4.4 In Exercises 16, nd the following values by using synthetic division. Check by substituting the value into the function. f(x) 3x4 2x3 7x2 8 g(x) 2x3 x2 1 1. f(1) 2. f(1) 3. g(1) 4. g(1) 5. f(2) 6. g(2) In Exercises 710, determine whether the number given is a zero of the polynomial. 7. 7, P(x) x3 2x2 29x 42 8. 2, P(x) x3 2x2 29x 42 9. 3, P(x) x3 x2 8x 12 10. 1, P(x) x3 x2 8x 12 S K I LL S c04c.qxd 11/25/11 3:43 PM Page 432 462. 4.4 The Real Zeros of a Polynomial Function 433 In Exercises 1120, given a real zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear and/or irreducible quadratic factors. Polynomial Zero Polynomial Zero 11. P(x) x3 13x 12 1 12. P(x) x3 3x2 10x 24 3 13. P(x) 2x3 x2 13x 6 14. P(x) 3x3 14x2 7x 4 15. P(x) x4 2x3 11x2 8x 60 3, 5 16. P(x) x4 x3 7x2 9x 18 1, 2 17. P(x) x4 5x2 10x 6 1, 3 18. P(x) x4 4x3 x2 6x 40 4, 2 19. P(x) x4 6x3 13x2 12x 4 2 (multiplicity 2) 20. P(x) x4 4x3 2x2 12x 9 1 (multiplicity 2) In Exercises 2128, use the rational zero theorem to list the possible rational zeros. 21. P(x) x4 3x2 8x 4 22. P(x) x4 2x3 5x 4 23. P(x) x5 14x3 x2 15x 12 24. P(x) x5 x3 x2 4x 9 25. P(x) 2x6 7x4 x3 2x 8 26. P(x) 3x5 2x4 5x3 x 10 27. P(x) 5x5 3x4 x3 x 20 28. P(x) 4x6 7x4 4x3 x 21 In Exercises 2932, list the possible rational zeros, and test to determine all rational zeros. 29. P(x) x4 2x3 9x2 2x 8 30. P(x) x4 2x3 4x2 2x 3 31. P(x) 2x3 9x2 10x 3 32. P(x) 3x3 5x2 26x 8 In Exercises 3344, use Descartes rule of signs to determine the possible number of positive real zeros and negative real zeros. 33. P(x) x4 32 34. P(x) x4 32 35. P(x) x5 1 36. P(x) x5 1 37. P(x) x5 3x3 x 2 38. P(x) x4 2x2 9 39. P(x) 9x7 2x5 x3 x 40. P(x) 16x7 3x4 2x 1 41. P(x) x6 16x4 2x2 7 42. P(x) 7x6 5x4 x2 2x 1 43. P(x) 3x4 2x3 4x2 x 11 44. P(x) 2x4 3x3 7x2 3x 2 For each polynomial in Exercises 4562: (a) use Descartes rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) test for rational zeros; and (d) factor as a product of linear and/or irreducible quadratic factors. 45. P(x) x3 6x2 11x 6 46. P(x) x3 6x2 11x 6 47. P(x) x3 7x2 x 7 48. P(x) x3 5x2 4x 20 49. P(x) x4 6x3 3x2 10x 50. P(x) x4 x3 14x2 24x 51. P(x) x4 7x3 27x2 47x 26 52. P(x) x4 5x3 5x2 25x 26 53. P(x) 10x3 7x2 4x 1 54. P(x) 12x3 13x2 2x 1 55. P(x) 6x3 17x2 x 10 56. P(x) 6x3 x2 5x 2 57. P(x) x4 2x3 5x2 8x 4 58. P(x) x4 2x3 10x2 18x 9 59. P(x) x6 12x4 23x2 36 60. P(x) x4 x2 16x2 16 61. P(x) 4x4 20x3 37x2 24x 5 62. P(x) 4x4 8x3 7x2 30x 50 In Exercises 6366, use the information found in Exercises 47, 51, 55, and 61 to assist in sketching a graph of each polynomial function. 63. Exercise 47. 64. Exercise 51. 65. Exercise 55. 66. Exercise 61. - 1 3 1 2 c04c.qxd 11/25/11 3:43 PM Page 433 463. 434 C HAP TE R 4 Polynomial and Rational Functions In Exercises 6772, use the intermediate value theorem and the bisection method to approximate the real zero in the indicated interval. Approximate to two decimal places. 67. f(x) x4 3x3 4 [1, 2] 68. f(x) x5 3x3 1 [0, 1] 69. f(x) 7x5 2x2 5x 1 [0, 1] 70. f(x) 2x3 3x2 6x 7 [2, 1] 71. f(x) x3 2x2 8x 3 [1, 0] 72. f(x) x4 4x2 7x 13 [2, 1] 73. Prot. A mathematics honorary society wants to sell magazine subscriptions to Math Weekly. If there are x hundred subscribers, its monthly revenue and cost are given by: a. Determine the prot function. Hint: P R C. b. Determine the number of subscribers needed in order to break even. 74. Prot. Using the prot equation P(x) x3 5x2 3x 6, when will the company break even if x represents the units sold? For Exercises 75 and 76, refer to the following: The demand function for a product is where p is the unit price (in dollars) of the product and x is the number of units produced and sold. The cost function for the product is where C is the total cost (in dollars) and x is the number of units produced. The total prot obtained by producing and selling x units is P(x) = xp(x) - C(x) C(x) = 20x + 1500 p(x) = 28 - 0.0002x R(x) = 46 - 3x2 and C(x) = 20 + 2x 75. Business. Find the total prot function when x units are produced and sold. Use Descartes rule of signs to determine possible combinations of positive zeros for the prot function. 76. Business. Find the break-even point(s) for the product to the nearest unit. Discuss the signicance of the break-even point(s) for the product. 77. Health/Medicine. During the course of treatment of an illness the concentration of a dose of a drug (in mcg/mL) in the bloodstream uctuates according to the model where t 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream? 78. Health/Medicine. During the course of treatment of an illness, the concentration of a dose of a drug (in mcg/mL) in the bloodstream uctuates according to the model where t 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream? C(t) = 60 - 0.75t2 C(t) = 15.4 - 0.05t2 80. Determine whether x 2 is a factor of P(x) x3 2x2 5x 6. Solution: 2 1 2 5 6 2 8 6 1 4 3 0 Yes, x 2 is a factor of P(x). This is incorrect. What mistake was made? In Exercises 79 and 80, explain the mistake that is made. 79. Use Descartes rule of signs to determine the possible combinations of zeros of Solution: No sign changes, so no positive real zeros. Five sign changes, so ve negative real zeros. This is incorrect. What mistake was made? P(- x) = - 2x5 + 7x4 - 9x3 + 9x2 - 7x + 2 P(x) = 2x5 + 7x4 + 9x3 + 9x2 + 7x + 2 P(x) = 2x5 + 7x4 + 9x3 + 9x2 + 7x + 2 C AT C H T H E M I S TA K E A P P L I C AT I O N S c04c.qxd 11/25/11 3:43 PM Page 434 464. In Exercises 8184 determine whether each statement is true or false. 81. All real zeros of a polynomial correspond to x-intercepts. 82. A polynomial of degree n, n 0, must have at least one zero. 83. A polynomial of degree n, n 0, can be written as a product of n linear factors over real numbers. 84. The number of sign changes in a polynomial is equal to the number of positive real zeros of that polynomial. 85. Given that x a is a zero of P(x) x3 (a b c)x2 (ab ac bc)x abc, nd the other two zeros, given that a, b, c are real numbers and a b c. 86. Given that x a is a zero of p(x) x3 (a b c)x2 (ab bc ac)x abc, nd the other two real zeros, given that a, b, c are real positive numbers. In Exercises 87 and 88, determine all possible rational zeros of the polynomial. There are many possibilities. Instead of trying them all, use a graphing calculator or software to graph P(x) to help nd a zero to test. 87. P(x) x3 2x2 16x 32 88. P(x) x3 3x2 16x 48 In Exercises 89 and 90: (a) determine all possible rational zeros of the polynomial, use a graphing calculator or software to graph P(x) to help nd the zeros; and (b) factor as a product of linear and/or irreducible quadratic factors. 89. 90. P(x) = 3x3 - x2 - 7x - 49 P(x) = 12x4 + 25x3 + 56x2 - 7x - 30 C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y C O N C E P TUAL O BJ E CTIVE S Extend the domain of polynomial functions to complex numbers. Understand how the fundamental theorem of algebra guarantees at least one zero. Understand why complex zeros occur in conjugate pairs. C O M P LE X Z E R O S: TH E F U N DAM E NTAL TH E O R E M O F ALG E B R A S K I LLS O BJ E CTIVE S Find the complex zeros of a polynomial function. Use the complex conjugate zeros theorem. Factor polynomial functions. S E CTI O N 4.5 Complex Zeros In Section 4.4, we found the real zeros of a polynomial function. In this section we nd the complex zeros of a polynomial function. In this chapter we assume the coefcients of polynomial functions are real numbers. The domain of polynomial functions thus far has been the set of all real numbers. Now, we consider a more general case. In this section, the coefcients of a polynomial function and the domain of a polynomial function are complex numbers. Note that the set of real numbers is a subset of the complex numbers. (Choose the imaginary part to be zero.) It is important to note, however, that when we are discussing graphs of polynomial functions, we restrict the domain to the set of real numbers. Study Tip The zeros of a polynomial can be complex numbers. Only when the zeros are real numbers do we interpret zeros as x-intercepts. 4.5 Complex Zeros: The Fundamental Theorem of Algebra 435 c04c.qxd 11/25/11 3:43 PM Page 435 465. 436 C HAP TE R 4 Polynomial and Rational Functions Every polynomial P(x) of degree n 0 has at least one zero in the complex number system. THE FUNDAMENTAL THEOREM OF ALGEBRA The fundamental theorem of algebra and the factor theorem are used to prove the following n zeros theorem. Study Tip The largest number of zeros a polynomial can have is equal to the degree of the polynomial. A zero of a polynomial P(x) is the solution or root of the equation P(x) 0. The zeros of a polynomial can be complex numbers. However, since the xy-plane represents real numbers, we interpret zeros as x-intercepts only when the zeros are real numbers. We can illustrate the relationship between real and complex zeros of polynomial functions and their graphs with two similar examples. Lets take the two quadratic functions f(x) x2 4 and g(x) x2 4. The graphs of these two functions are parabolas that open upward with f(x) shifted down four units and g(x) shifted up four units as shown on the left. Setting each function equal to zero and solving for x, we nd that the zeros for f(x) are 2 and 2 and the zeros for g(x) are 2i and 2i. Notice that the x-intercepts for f(x) are (2, 0) and (2, 0) and g(x) has no x-intercepts. The Fundamental Theorem of Algebra In Section 4.4, we were able to write a polynomial function as a product of linear and/or irreducible quadratic factors. Now, we consider factors over complex numbers. Therefore, what were irreducible quadratic factors over real numbers will now be a product of two linear factors over the complex numbers. What are the minimum and maximum number of zeros a polynomial can have? Every polynomial has at least one zero (provided the degree is greater than zero). The largest number of zeros a polynomial can have is equal to the degree of the polynomial. x y (0, 4) (2, 0)(2, 0) (0, 4) 8 8 Every polynomial P(x) of degree n 0 can be expressed as the product of n linear factors. Hence, P(x) has exactly n zeros, not necessarily distinct. n ZEROS THEOREM These two theorems are illustrated with ve polynomials below. a. The rst-degree polynomial f(x) x 3 has exactly one zero: x 3. b. The second-degree polynomial f(x) x2 10x 25 (x 5)(x 5) has exactly two zeros: x 5 and x 5. It is customary to write this as a single zero of multiplicity 2 or refer to it as a repeated root. c. The third-degree polynomial f(x) x3 16x x(x2 16) x(x 4i)(x 4i) has exactly three zeros: x 0, x 4i, and x 4i. d. The fourth-degree polynomial f(x) x4 1 (x2 1)(x2 1) (x 1)(x 1)(x i)(x i) has exactly four zeros: x 1, x 1, x i, and x i. e. The fth-degree polynomial f(x) x5 x x x x x has exactly ve zeros: x 0, which has multiplicity 5. #### The fundamental theorem of algebra and the n zeros theorem only tell you that the zeros existnot how to nd them. We must rely on techniques discussed in Section 4.4 and additional strategies discussed in this section to determine the zeros. c04c.qxd 11/25/11 3:43 PM Page 436 466. 4.5 Complex Zeros: The Fundamental Theorem of Algebra 437 Study Tip If we restrict the coefcients of a polynomial to real numbers, complex zeros always come in conjugate pairs. If a polynomial P(x) has real coefcients, and if a bi is a zero of P(x), then its complex conjugate a bi is also a zero of P(x). COMPLEX CONJUGATE ZEROS THEOREM E X AM P LE 1 Zeros That Appear as Complex Conjugates Find the zeros of the polynomial P(x) x2 4x 13. Solution: Set the polynomial equal to zero. P(x) x2 4x 13 0 Use the quadratic formula to solve for x. Simplify. x 2 ; 3i The zeros are the complex conjugates 2 3i and 2 3i. Check: This is a second-degree polynomial, so we expect two zeros. x = -(-4) ; 2(-4)2 - 4(1)(13) 2(1) Complex Conjugate Pairs Often, at a grocery store or a drugstore, we see signs for special offersbuy one, get one free. A similar phenomenon occurs for complex zeros of a polynomial function with real coefcients. If we restrict the coefcients of a polynomial to real numbers, complex zeros always come in conjugate pairs. In other words, if a zero of a polynomial function is a complex number, then another zero will always be its complex conjugate. Look at the third-degree polynomial in the above illustration, part (c), where two of the zeros were 4i and 4i, and in part (d), where two of the zeros were i and i. In general, if we restrict the coefcients of a polynomial to real numbers, complex zeros always come in conjugate pairs. E X AM P LE 2 Finding a Polynomial Given Its Zeros Find a polynomial of minimum degree that has the zeros: 2, 1 i, 1 i. Solution: Write the factors corresponding to each zero: 2: (x 2) 1 i: [x (1 i)] 1 i: [x (1 i)] Express the polynomial as the product of the three factors. Regroup inner parentheses. Use the difference of two squares formula (a b)(a b) a2 b2 for the product of the latter two factors. (x 1)2 i2 Simplify. x2 2x + 1 1 YO U R TU R N Find a polynomial of minimum degree that has the zeros: 1, 2 i, 2 i. P(x) = x3 - 2x + 4 P(x) = x3 - 2x2 + 2x + 2x2 - 4x + 4 P(x) = (x + 2)(x2 - 2x + 2) P(x) = (x + 2)[(x - 1)2 - i2 ] P(x) = (x + 2)[(x - 1) - i][(x - 1) + i] P(x) = (x + 2)[(x - 1) - i][(x - 1) + i] P(x) = (x + 2)[x - (1 + i)][x - (1 - i)] y r r Answer: P(x) = x3 - 5x2 + 9x - 5 c04c.qxd 11/25/11 3:43 PM Page 437 467. 438 C HAP TE R 4 Polynomial and Rational Functions 0 -(-6x2 + 0x - 6) -6x2 + 0x - 6 -(-x3 + 0x2 - x) -x3 - 6x2 - x -(x4 + 0x3 + x2 ) x2 + 0x + 1 x4 - x3 - 5x2 - x - 6 x2 - x - 6 Technology Tip The graph of P(x) x4 x3 5x2 x 6 is shown. The real zeros of the function at x 2 and x 3 give the factors of x 2 and x 3. Use synthetic division to nd the other factors. A table of values supports the real zeros of the function and its factors. Answer: P(x) (x 2i)(x 2i)(x 1)(x 2) Note: The zeros of P(x) are 1, 2, 2i, and 2i. E X AM P LE 3 Factoring a Polynomial with Complex Zeros Factor the polynomial P(x) x4 x3 5x2 x 6 given that i is a zero of P(x). Since P(x) is a fourth-degree polynomial we expect four zeros. The goal in this problem is to write P(x) as a product of four linear factors: P(x) (x a)(x b)(x c)(x d ), where a, b, c, and d are complex numbers and represent the zeros of the polynomial. Solution: Write known zeros and linear factors. Since i is a zero, we know that i is a zero. x i and x i We now know two linear factors of P(x). (x i) and (x i) Write P(x) as a product of four factors. P(x) (x i)(x i)(x c)(x d) Multiply the two known factors. (x i)(x i) x2 i2 x2 (1) x2 1 Rewrite the polynomial. P(x) (x2 1)(x c)(x d) Divide both sides of the equation by x2 1. Divide P(x) by x2 1 using long division. Since the remainder is 0, x2 x 6 is a factor. P(x) (x2 1)(x2 x 6) Factor the quotient x2 x 6. x2 x 6 (x 3)(x 2) Write P(x) as a product of four linear factors. Check: P(x) is a fourth-degree polynomial and we found four zeros, two of which are complex conjugates. YO U R TU R N Factor the polynomial P(x) x4 3x3 6x2 12x 8 given that x 2i is a factor. P(x) = (x - i)(x + i)(x - 3)(x + 2) P(x) x2 + 1 = (x - c)(x - d) c04c.qxd 11/25/11 3:43 PM Page 438 468. 4.5 Complex Zeros: The Fundamental Theorem of Algebra 439 Answer: P(x) [x (1 2i)] [x (1 2i)](x 1)(x 3) Note: The zeros of P(x) are 1, 3, 1 2i, and 1 2i. # E X AM P LE 4 Factoring a Polynomial with Complex Zeros Factor the polynomial P(x) x4 2x3 x2 2x 2 given that 1 i is a zero of P(x). Since P(x) is a fourth-degree polynomial, we expect four zeros. The goal in this problem is to write P(x) as a product of four linear factors: P(x) (x a)(x b)(x c)(x d ), where a, b, c, and d are complex numbers and represent the zeros of the polynomial. Solution: STEP 1 Write known zeros and linear factors. Since 1 i is a zero, we know that 1 i is a zero. x 1 i and x 1 i We now know two linear factors of P(x). [x (1 i)] and [x (1 i)] STEP 2 Write P(x) as a product of four factors. P(x) [x (1 i)][x (1 i)](x c)(x d) STEP 3 Multiply the rst two terms. [x (1 i)][x (1 i)] First group the real parts together in each bracket. [(x 1) i][(x 1) i] Use the special product (x 1)2 i2 (a b)(a b) a2 b2 , (x2 2x 1) (1) where a is (x 1) and b is i. x2 2x 2 STEP 4 Rewrite the polynomial. P(x) (x2 2x 2)(x c)(x d) STEP 5 Divide both sides of the equation by x2 2x 2, and substitute in the original polynomial (x c)(x d) P(x) x4 2x3 x2 2x 2. STEP 6 Divide the left side of the equation using long division. STEP 7 Factor x2 1. (x 1)(x 1) STEP 8 Write P(x) as a product of four linear factors. P(x) [x (1 i)][x (1 i)][x 1][x 1] YO U R TU R N Factor the polynomial P(x) x4 2x2 16x 15 given that 1 2i is a zero. P(x) = (x - 1 - i)(x - 1 + i)(x - 1)(x + 1) x4 - 2x3 + x2 + 2x - 2 x2 - 2x + 2 = x2 - 1 x4 - 2x3 + x2 + 2x - 2 x2 - 2x + 2 = Because an n-degree polynomial function has exactly n zeros and since complex zeros always come in conjugate pairs, if the degree of the polynomial is odd, there is guaranteed to be at least one zero that is a real number. If the degree of the polynomial is even, there is no guarantee that a zero will be realall the zeros could be complex. Study Tip Odd-degree polynomials have at least one real zero. or c04c.qxd 11/25/11 3:43 PM Page 439 469. 440 C HAP TE R 4 Polynomial and Rational Functions Applying Descartes rule of signs, we nd that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros. b. Because this is a fourth-degree polynomial, there are four zeros. Since complex zeros come in conjugate pairs, the table describes the possible four zeros. Applying Descartes rule of signs we nd that there are 2 or 0 positive real zeros and 2 or 0 negative real zeros. YO U R TU R N List the possible combinations of real and complex zeros for: P(x) = x6 - 7x5 + 8x3 - 2x + 1 Answer: REAL COMPLEX ZEROS ZEROS 0 6 2 4 4 2 6 0 REAL ZEROS COMPLEX ZEROS 0 4 2 2 4 0 POSITIVE NEGATIVE COMPLEX REAL ZEROS REAL ZEROS ZEROS 1 0 4 3 0 2 1 2 2 3 2 0 POSITIVE NEGATIVE COMPLEX REAL ZEROS REAL ZEROS ZEROS 0 0 4 2 0 2 0 2 2 2 2 0 Factoring Polynomials Now lets draw on the tests discussed in this chapter to help us nd all the zeros of a polynomial. Doing so will enable us to write polynomials as a product of linear factors. Before reading Example 5, reread Section 4.4, Example 8. E X AM P LE 5 Finding Possible Combinations of Real and Complex Zeros List the possible combinations of real and complex zeros for the given polynomials. a. 17x5 2x4 3x3 x2 5 b. 5x4 2x3 x 2 Solution: a. Since this is a fth-degree polynomial, there are ve zeros. Because complex zeros come in conjugate pairs, the table describes the possible ve zeros. REAL ZEROS COMPLEX ZEROS 1 4 3 2 5 0 c04c.qxd 11/25/11 3:43 PM Page 440 470. 4.5 Complex Zeros: The Fundamental Theorem of Algebra 441 STEP 3 Utilize the rational zero theorem to determine the possible rational zeros. ;1, ;2 STEP 4 Test possible rational zeros. 1 is a zero: 1 1 2 0 0 1 2 1 3 3 3 2 1 3 3 3 2 0 1 is a zero: 1 1 3 3 3 2 1 2 1 2 1 2 1 2 0 2 is a zero: 2 1 2 1 2 2 0 2 1 0 1 0 STEP 5 Write P(x) as a product of linear factors. P(x) (x 1)(x 1)(x 2)(x i)(x i) = (x - i)(x + i)x2 + 1 Technology Tip The graph of P(x) x5 2x4 x 2 is shown. The real zeros of the function at x 2, x 1, and x 1 give the factors of x 2, x 1, and x 1. Use synthetic division to nd the other factors. A table of values supports the real zeros of the function and its factors. Study Tip From Step 2 we know there is one positive real zero, so test the positive possible rational zeros rst in Step 4. Study Tip P(x) is a fth-degree polynomial, so we expect ve zeros. S U M MARY S E CTI O N 4.5 P(x) has at least one zero and no more than n zeros. If a bi is a zero, then a bi is also a zero. The polynomial can be written as a product of linear factors, not necessarily distinct. In this section we discussed complex zeros of polynomial functions. A polynomial function, P(x), of degree n with real coefcients has the following properties: t E X AM P LE 6 Factoring a Polynomial Factor the polynomial P(x) x5 2x4 x 2. Solution: STEP 1 Determine variations in sign. P(x) has 1 sign change. P(x) x5 2x4 x 2 P(x) has 2 sign changes. P(x) x5 2x4 x 2 STEP 2 Apply Descartes rule of signs and summarize the results in a table. POSITIVE NEGATIVE COMPLEX REAL ZEROS REAL ZEROS ZEROS 1 2 2 1 0 4 c04c.qxd 11/25/11 3:43 PM Page 441 471. 442 C HAP TE R 4 Polynomial and Rational Functions In Exercises 18, nd all zeros (real and complex). Factor the polynomial as a product of linear factors. 1. P(x) x2 4 2. P(x) x2 9 3. P(x) x2 2x 2 4. P(x) x2 4x 5 5. P(x) x4 16 6. P(x) x4 81 7. P(x) x4 25 8. P(x) x4 9 In Exercises 916, a polynomial function with real coefcients is described. Find all remaining zeros. 9. Degree: 3 Zeros: 1, i 10. Degree: 3 Zeros: 1, i 11. Degree: 4 Zeros: 2i, 3 i 12. Degree: 4 Zeros: 3i, 2 i 13. Degree: 6 Zeros: 2 (multiplicity 2), 1 3i, 2 5i 14. Degree: 6 Zeros: 2 (multiplicity 2), 1 5i, 2 3i 15. Degree: 6 Zeros: i, 1 i (multiplicity 2) 16. Degree: 6 Zeros: 2i, 1 i (multiplicity 2) In Exercises 1722, nd a polynomial of minimum degree that has the given zeros. 17. 18. 19. 20. 21. 22. In Exercises 2334, given a zero of the polynomial, determine all other zeros (real and complex) and write the polynomial in terms of a product of linear factors. Polynomial Zero Polynomial Zero 23. P(x) x4 2x3 11x2 8x 60 2i 24. P(x) x4 x3 7x2 9x 18 3i 25. P(x) x4 4x3 4x2 4x 3 i 26. P(x) x4 x3 2x2 4x 8 2i 27. P(x) x4 2x3 10x2 18x 9 3i 28. P(x) x4 3x3 21x2 75x 100 5i 29. P(x) x4 9x2 18x 14 1 i 30. P(x) x4 4x3 x2 6x 40 1 2i 31. P(x) x4 6x3 6x2 24x 40 3 i 32. P(x) x4 4x3 4x2 4x 5 2 i 33. P(x) x4 9x3 29x2 41x 20 2 i 34. P(x) x4 7x3 14x2 2x 20 3 i In Exercises 3558, factor each polynomial as a product of linear factors. 35. P(x) x3 x2 9x 9 36. P(x) x3 2x2 4x 8 37. P(x) x3 5x2 x 5 38. P(x) x3 7x2 x 7 39. P(x) x3 x2 4x 4 40. P(x) x3 x2 2 41. P(x) x3 x2 18 42. P(x) x4 2x3 2x2 2x 3 43. P(x) x4 2x3 11x2 8x 60 44. P(x) x4 x3 7x2 9x 18 45. P(x) x4 4x3 x2 16x 20 46. P(x) x4 3x3 11x2 27x 18 47. P(x) x4 7x3 27x2 47x 26 47. P(x) x4 5x3 5x2 25x 26 49. P(x) x4 3x3 x2 13x 10 50. P(x) x4 x3 12x2 26x 24 -i, i, 1 - 2i, 1 + 2i1 - i, 1 + i, -3i, 3i2, 4 - i, 4 + i 1, 1 - 5i, 1 + 5i0, 2 - i, 2 + i0, 1 - 2i, 1 + 2i E X E R C I S E S S E CTI O N 4.5 S K I LL S c04c.qxd 12/22/11 7:17 PM Page 442 472. 4.5 Complex Zeros: The Fundamental Theorem of Algebra 443 In Exercises 5962, assume the prot model is given by a polynomial function P(x) where x is the number of units sold by the company per year. A P P L I C AT I O N S 59. Prot. If the prot function of a given company has all imaginary zeros and the leading coefcient is positive, would you invest in this company? Explain. 60. Prot. If the prot function of a given company has all imaginary zeros and the leading coefcient is negative, would you invest in this company? Explain. 61. Prot. If the prot function of a company is modeled by a third-degree polynomial with a negative leading coefcient and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? Explain. 62. Prot. If the prot function of a company is modeled by a third-degree polynomial with a positive leading coefcient and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? Explain. For Exercises 63 and 64, refer to the following: The following graph models the prot P of a company where t is months and t 0. 1 2 3 4 65 15 12 9 6 3 t (months) P (profit in millions of dollars) 63. Business. If the prot function pictured is a third-degree polynomial, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 64. Business. If the prot function pictured is a fourth-degree polynomial with a negative leading coefcient, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. For Exercises 65 and 66, refer to the following: The following graph models the concentration, C (in g/mL) of a drug in the bloodstream; and t is time in hours after the drug is administered where t 0. 65. Health/Medicine. If the concentration function pictured is a third-degree polynomial, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 66. Health/Medicine. If the concentration function pictured is a fourth-degree polynomial with a negative leading coefcient, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 1 2 3 4 8765 50 40 30 20 10 45 35 25 15 5 t (hours) C (mcg/mL) 51. P(x) x4 2x3 5x2 8x 4 52. P(x) x4 2x3 10x2 18x 9 53. P(x) x6 12x4 23x2 36 54. P(x) x6 2x5 9x4 16x3 24x2 32x 16 55. P(x) 4x4 20x3 37x2 24x 5 56. P(x) 4x4 44x3 145x2 114x 26 57. P(x) 3x5 2x4 9x3 6x2 12x 8 58. P(x) 2x5 5x4 4x3 26x2 50x 25 c04c.qxd 11/25/11 3:43 PM Page 443 473. 444 C HAP TE R 4 Polynomial and Rational Functions 72. A polynomial function of degree n, n 0 can be written as a product of n linear factors. 73. Is it possible for an odd-degree polynomial to have all imaginary complex zeros? Explain. 74. Is it possible for an even-degree polynomial to have all imaginary zeros? Explain. In Exercises 6972, determine whether each statement is true or false. 69. If x 1 is a zero of a polynomial function, then x 1 is also a zero of the polynomial function. 70. All zeros of a polynomial function correspond to x-intercepts. 71. A polynomial function of degree n, n 0 must have at least one zero. In Exercises 75 and 76, assume a and b are nonzero real numbers. 75. Find a polynomial function that has degree 6, and for which bi is a zero of multiplicity 3. 76. Find a polynomial function that has degree 4, and for which a bi is a zero of multiplicity 2. For Exercises 77 and 78, determine possible combinations of real and complex zeros. Plot P(x) and identify any real zeros with a graphing calculator or software. Does this agree with your list? For Exercises 79 and 80, nd all zeros (real and complex). Factor the polynomial as a product of linear factors. In Exercises 67 and 68, explain the mistake that is made. 67. Given that 1 is a zero of P(x) x3 2x2 7x 6, nd all other zeros. Solution: Step 1: P(x) is a third-degree polynomial, so we expect three zeros. Step 2: Because 1 is a zero, 1 is a zero, so two linear factors are (x 1) and (x 1). Step 3: Write the polynomial as a product of three linear factors. P(x) (x 1)(x 1)(x c) P(x) (x2 1)(x c) Step 4: To nd the remaining linear factor, we divide P(x) by x2 1. Which has a nonzero remainder? What went wrong? x3 - 2x2 + 7x - 6 x2 - 1 = x - 2 + 6x - 8 x2 - 1 68. Factor the polynomial P(x) 2x3 x2 2x 1. Solution: Step 1: Since P(x) is an odd-degree polynomial, we are guaranteed one real zero (since complex zeros come in conjugate pairs). Step 2: Apply the rational zero test to develop a list of potential rational zeros. Possible zeros: ;1 Step 3: Test possible zeros. 1 is not a zero: P(x) 2(1)3 (1)2 2(1) 1 6 1 is not a zero: P(x) 2(1)3 (1)2 2(1) 1 2 Note: is the real zero. Why did we not nd it?-1 2 C AT C H T H E M I S TA K E C O N C E P T U A L C H A L L E N G E T E C H N O L O G Y 77. P(x) x4 13x2 36 78. P(x) x6 2x4 7x2 130x 288 79. P(x) 5x5 3x4 25x3 15x2 20x 12 80. P(x) x5 2.1x4 5x3 5.592x2 9.792x 3.456 c04c.qxd 11/25/11 3:43 PM Page 444 474. C O N C E P TUAL O BJ E CTIVE S Understand arrow notation. Interpret the behavior of the graph of a rational function near an asymptote. S K I LLS O BJ E CTIVE S Find the domain of a rational function. Determine vertical, horizontal, and slant asymptotes of rational functions. Graph rational functions. Domain of Rational Functions So far in this chapter we have discussed polynomial functions. We now turn our attention to rational functions, which are ratios of polynomial functions. Ratios of integers are called rational numbers. Similarly, ratios of polynomial functions are called rational functions. A function f(x) is a rational function if where the numerator, n(x), and the denominator, d(x), are polynomial functions. The domain of f (x) is the set of all real numbers x such that . Note: If d(x) is a constant, then f(x) is a polynomial function. d(x) Z 0 f(x) = n(x) d(x) d(x) Z 0 Rational FunctionD E F I N I T I O N The domain of any polynomial function is the set of all real numbers. When we divide two polynomial functions, the result is a rational function, and we must exclude any values of x that make the denominator equal to zero. R ATI O NAL F U N CTI O N S S E C TI O N 4.6 445 c04c.qxd 11/25/11 3:43 PM Page 445 475. E X AM P LE 1 Finding the Domain of a Rational Function Find the domain of the rational function Express the domain in interval notation. Solution: Set the denominator equal to zero. x2 x 6 0 Factor. (x 2)(x 3) 0 Solve for x. x 2 or x 3 Eliminate these values from the domain. or State the domain in interval notation. YO U R TU R N Find the domain of the rational function Express the domain in interval notation. It is important to note that there are not always restrictions on the domain. For example, if the denominator is never equal to zero, the domain is the set of all real numbers. f(x) = x - 2 x2 - 3x - 4 . (-, -2)(-2, 3)(3, ) x Z 3x Z -2 f(x) = x + 1 x2 - x - 6 . Answer: The domain is the set of all real numbers such that or Interval notation: (-, -1)(-1, 4)(4, ) x Z 4. x Z -1 446 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 2 When the Domain of a Rational Function Is the Set of All Real Numbers Find the domain of the rational function Express the domain in interval notation. Solution: Set the denominator equal to zero. x2 9 0 Subtract 9 from both sides. x2 9 Solve for x. x 3i or x 3i There are no real solutions; therefore the domain has no restrictions. R, the set of all real numbers State the domain in interval notation. YO U R TU R N Find the domain of the rational function Express the domain in interval notation. g(x) = 5x x2 + 4 . (-, ) g(x) = 3x x2 + 9 . Answer: The domain is the set of all real numbers. Interval notation: (, ) It is important to note that , where , and are not the same function. Although f(x) can be written in the factored form g(x) = x - 2x Z -2f(x) = x2 - 4 x + 2 , its domain is different. The domain of g(x) is the set off(x) = (x - 2)(x + 2) x + 2 = x - 2 all real numbers, whereas the domain of f(x) is the set of all real numbers such that If we were to plot f(x) and g(x), they would both look like the line y x 2. However, f(x) would have a hole, or discontinuity, at the point x 2. x Z -2. x y 5 55 5 hole c04d.qxd 11/25/11 4:04 PM Page 446 476. 4.6 Rational Functions 447 x y (1, 1) (1, 1) x 1f(x) = x approaching 0 from the left x approaching 0 from the right We cannot let x 0 because that point is not in the domain of the function. We should, however, ask the question, how does f(x) behave as x approaches zero? Let us take values that get closer and closer to x 0, such as , , , . . . (See the table above.) We use an arrow to represent the word approach, a positive superscript to represent from the right, and a negative superscript to represent from the left. A plot of this function can be generated using point-plotting techniques. The following are observations of the graph f(x) = 1 x . 1 1000 1 100 1 10 x 0 10 100 1000 undened 1000 100 10f(x) 1 x 1 10 1 100 1 1000 - 1 1000 - 1 100 - 1 10 x f(x) 10 1 1 1 1 10 1 10 1 10 1 x Vertical, Horizontal, and Slant Asymptotes If a function is not dened at a point, then it is still useful to know how the function behaves near that point. Lets start with a simple rational function, the reciprocal function This function is dened everywhere except at x 0.f(x) = 1 x . WORDS MATH As x approaches zero from the right, the function f(x) increases without bound. As x approaches zero from the left, the function f(x) decreases without bound. As x approaches innity (increases without bound), the function f(x) approaches zero from above. As x approaches negative innity (decreases without bound), the function f(x) approaches zero from below. The symbol does not represent an actual real number. This symbol represents growing without bound. 1. Notice that the function is not dened at x 0. The y-axis, or the vertical line x 0, represents the vertical asymptote. 2. Notice that the value of the function is never equal to zero. The x-axis is never touched by the function. The x-axis, or y 0, is a horizontal asymptote. 1 x S 0 x S 1 x S 0 x S 1 x S x S 0 1 x S x S 0 x y (1, 1) (1, 1) x 1f(x) = c04d.qxd 11/25/11 4:04 PM Page 447 477. 448 C HAP TE R 4 Polynomial and Rational Functions The line x a is a vertical asymptote for the graph of a function if f(x) either increases or decreases without bound as x approaches a from either the left or the right. Vertical AsymptotesD E F I N I T I O N x y x y x y x y x = a f(x) x = a f(x)x = a f(x) x = a f(x) Asymptotes are lines that the graph of a function approaches. Suppose a football teams defense is its own 8 yard line and the team gets an offsides penalty that results in loss of half the distance to the goal. Then the offense would get the ball on the 4 yard line. Suppose the defense gets another penalty on the next play that results in half the distance to the goal. The offense would then get the ball on the 2 yard line. If the defense received 10 more penalties all resulting in half the distance to the goal, would the referees give the offense a touchdown? No, because although the offense may appear to be snapping the ball from the goal line, technically it has not actually reached the goal line. Asymptotes utilize the same concept. We will start with vertical asymptotes. Although the function had one vertical asymptote, in general, rational functions can have none, one, or several vertical asymptotes. We will rst formally dene what a vertical asymptote is and then discuss how to nd it. f(x) = 1 x Vertical asymptotes assist us in graphing rational functions since they essentially steer the function in the vertical direction. How do we locate the vertical asymptotes of a rational function? Set the denominator equal to zero. If the numerator and denominator have no common factors, then any numbers that are excluded from the domain of a rational function locate vertical asymptotes. A rational function is said to be in lowest terms if the numerator n(x) and denominator d(x) have no common factors. Let be a rational function in lowest terms; then any zeros of the numerator n(x) correspond to x-intercepts of the graph of f, and any zeros of the denominator d(x) correspond to vertical asymptotes of the graph of f. If a rational function does have a common factor (is not in lowest terms), then the common factor(s) should be canceled, resulting in an equivalent rational function R(x) in lowest f(x) = n(x) d(x) f(x) = n(x) d(x) c04d.qxd 11/25/11 4:04 PM Page 448 478. 4.6 Rational Functions 449 Let be a rational function in lowest terms (that is, assume n(x) and d(x) are polynomials with no common factors); then the graph of f has a vertical asymptote at any real zero of the denominator d(x). That is, if d(a) 0, then x a corresponds to a vertical asymptote on the graph of f. Note: If f is a rational function that is not in lowest terms, then divide out the common factors, resulting in a rational function R that is in lowest terms. Any common factor x a of the function f corresponds to a hole in the graph of f at x a provided the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. f(x) = n(x) d(x) LOCATING VERTICAL ASYMPTOTES Study Tip The vertical asymptotes of a rational function in lowest terms occur at x-values that make the denominator equal to zero. E X AM P LE 3 Determining Vertical Asymptotes Locate any vertical asymptotes of the rational function Solution: Factor the denominator. The numerator and denominator have no common factors. Set the denominator equal to zero. and Solve for x. and The vertical asymptotes are and . YO U R TU R N Locate any vertical asymptotes of the following rational function: f(x) = 3x - 1 2x2 - x - 15 x = 2 3x = - 1 2 x = 2 3 x = - 1 2 3x - 2 = 02x + 1 = 0 f(x) = 5x + 2 (2x + 1)(3x - 2) f(x) = 5x + 2 6x2 - x - 2 . Answer: and x 3x = - 5 2 terms. If (x a)p is a factor of the numerator and (x a)q is a factor of the denominator, then there is a hole in the graph at x a provided p q and x a is a vertical asymptote if p q. c04d.qxd 11/25/11 4:04 PM Page 449 479. 450 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 4 Determining Vertical Asymptotes When the Rational Function Is Not in Lowest Terms Locate any vertical asymptotes of the rational function Solution: Factor the denominator. x3 3x2 10x x(x2 3x 10) f(x) = x + 2 x3 - 3x2 - 10x . We now turn our attention to horizontal asymptotes. As we have seen, rational functions can have several vertical asymptotes. However, rational functions can have at most one horizontal asymptote. Horizontal asymptotes imply that a function approaches a constant value as x becomes large in the positive or negative direction. Another difference between vertical and horizontal asymptotes is that the graph of a function never touches a vertical asymptote but, as you will see in the next box, the graph of a function may cross a horizontal asymptote, just not at the ends (x S ;). The line y b is a horizontal asymptote of the graph of a function if f(x) approaches b as x increases or decreases without bound. The following are three examples: As x S , f(x) S b Horizontal AsymptoteD E F I N I T I O N x y x y y = b f(x) y = b f(x) x y y = b f(x) Answer: x 3 Note: A horizontal asymptote steers a function as x gets large. Therefore, when x is not large, the function may cross the asymptote. x(x 5)(x 2) Write the rational function in factored form. Cancel (divide out) the common factor (x 2). Find the values when the denominator of R is equal to zero. x 0 and x 5 The vertical asymptotes are and . Note: x 2 is not in the domain of f(x), even though there is no vertical asymptote there. There is a hole in the graph at x 2. Graphing calculators do not always show such holes. YO U R TU R N Locate any vertical asymptotes of the following rational function: f(x) = x2 - 4x x2 - 7x + 12 x = 5x = 0 x Z -2R(x) = 1 x(x - 5) f(x) = (x + 2) x(x - 5)(x + 2) c04d.qxd 11/25/11 4:04 PM Page 450 480. How do we determine whether a horizontal asymptote exists? And, if it does, how do we locate it? We investigate the value of the rational function as or as . One of two things will happen: either the rational function will increase or decrease without bound or the rational function will approach a constant value. We say that a rational function is proper if the degree of the numerator is less than the degree of the denominator. Proper rational functions, like , approach zero as x gets large. Therefore, all proper rational functions have the specic horizontal asymptote, y 0 (see Example 5a). We say that a rational function is improper if the degree of the numerator is greater than or equal to the degree of the denominator. In this case, we can divide the numerator by the denominator and determine how the quotient behaves as x increases without bound. If the quotient is a constant (resulting when the degrees of the numerator and denominator are equal), then as or as , the rational function approaches the constant quotient (see Example 5b). If the quotient is a polynomial function of degree 1 or higher, then the quotient depends on x and does not approach a constant value as x increases (see Example 5c). In this case, we say that there is no horizontal asymptote. We nd horizontal asymptotes by comparing the degree of the numerator and the degree of the denominator. There are three cases to consider: 1. The degree of the numerator is less than the degree of the denominator. 2. The degree of the numerator is equal to the degree of the denominator. 3. The degree of the numerator is greater than the degree of the denominator. x S -x S f(x) = 1 x x S -x S Let f be a rational function given by where n(x) and d(x) are polynomials. 1. When n m, the x-axis (y 0) is the horizontal asymptote. 2. When n m, the line (ratio of leading coefcients) is the horizontal asymptote. 3. When n m, there is no horizontal asymptote. y = an bm f(x) = n(x) d(x) = an xn + an-1xn-1 + + a1x + a0 bm xm + bm-1xm-1 + + b1x + b0 LOCATING HORIZONTAL ASYMPTOTES In other words, 1. When the degree of the numerator is less than the degree of the denominator, then y 0 is the horizontal asymptote. 2. When the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefcients. 3. If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. 4.6 Rational Functions 451 c04d.qxd 11/25/11 4:04 PM Page 451 481. Thus far we have discussed linear asymptotes: vertical and horizontal. There are three types of lines: horizontal (slope is zero), vertical (slope is undened), and slant (nonzero slope). Similarly, there are three types of linear asymptotes: horizontal, vertical, and slant. 452 C HAP TE R 4 Polynomial and Rational Functions Answer: is the horizontal asymptote. y = -7 4 Study Tip There are three types of linear asymptotes: horizontal, vertical, and slant. b. Graph .g(x) = 8x2 + 3 4x2 + 1 c. Graph .h(x) = 8x3 + 3 4x2 + 1 E X AM P LE 5 Finding Horizontal Asymptotes Determine whether a horizontal asymptote exists for the graph of each of the given rational functions. If it does, locate the horizontal asymptote. a. b. c. Solution (a): The degree of the numerator 8x 3 is 1. n 1 The degree of the denominator 4x2 1 is 2. m 2 The degree of the numerator is less than the degree of the denominator. n m The x-axis is the horizontal asymptote for the graph of f(x). y 0 The line is the horizontal asymptote for the graph of f(x). Solution (b): The degree of the numerator 8x2 3 is 2. n 2 The degree of the denominator 4x2 1 is 2. m 2 The degree of the numerator is equal to the degree of the denominator. n m The ratio of the leading coefcients is the horizontal asymptote for the graph of g(x). The line is the horizontal asymptote for the graph of g(x). If we divide the numerator by the denominator, the resulting quotient is the constant 2. Solution (c): The degree of the numerator 8x3 3 is 3. n 3 The degree of the denominator 4x2 1 is 2. m 2 The degree of the numerator is greater than the degree of the denominator. n m The graph of the rational function h(x) has . If we divide the numerator by the denominator, the resulting quotient is a linear function. YO U R TU R N Find the horizontal asymptote (if one exists) for the graph of the rational function f(x) = 7x3 + x - 2 -4x3 + 1 . h(x) = 8x3 + 3 4x2 + 1 = 2x + -2x + 3 4x2 + 1 no horizontal asymptote g(x) = 8x2 + 3 4x2 + 1 = 2 + 1 4x2 + 1 y = 2 y = 8 4 = 2 y = 0 h(x) = 8x3 + 3 4x2 + 1 g(x) = 8x2 + 3 4x2 + 1 f(x) = 8x + 3 4x2 + 1 Technology Tip The following graphs correspond to the rational functions given in Example 5. The horizontal asymptotes are apparent, but are not drawn in the graph. a. Graph .f(x) = 8x + 3 4x2 + 1 c04d.qxd 11/25/11 4:04 PM Page 452 482. E X AM P LE 6 Finding Slant Asymptotes Determine the slant asymptote of the rational function . Solution: Divide the numerator by the denominator with long division. f(x) = 4x3 + x2 + 3 x2 - x + 1 Answer: y x 5 Let f be a rational function given by , where n(x) and d(x) are polynomials and the degree of n(x) is one more than the degree of d(x). On dividing n(x) by d(x), the rational function can be expressed as where the degree of the remainder r(x) is less than the degree of d(x) and the line y mx b is a slant asymptote for the graph of f. Note that as or , .f (x) S mx + bx S x S - f(x) = mx + b + r(x) d(x) f(x) = n(x) d(x) SLANT ASYMPTOTES 4x + 5 x2 - x + 14x3 + x2 + 0x + 3 -(4x3 - 4x2 + 4x) 5x2 - 4x + 3 -(5x2 - 5x + 5) x - 2 Note that as the rational expression approaches 0. The quotient is the slant asymptote. YO U R TU R N Find the slant asymptote of the rational function .f(x) = x2 + 3x + 2 x - 2 y = 4x + 5 x S ; S 0 as f(x) = 4x 5 + x - 2 x2 - x + 1 x S ; Recall that when dividing polynomials the degree of the quotient is always the difference between the degree of the numerator and the degree of the denominator. For example, a cubic (third-degree) polynomial divided by a quadratic (second-degree) polynomial results in a linear (rst-degree) polynomial. A fth-degree polynomial divided by a fourth-degree polynomial results in a rst-degree (linear) polynomial. When the degree of the numerator is exactly one more than the degree of the denominator, the quotient is linear and represents a slant asymptote. 4.6 Rational Functions 453 Technology Tip The graph of has a slant asymptote of y = 4x + 5 f(x) = 4x3 + x2 + 3 x2 - x + 1 c04d.qxd 12/22/11 7:24 PM Page 453 483. 454 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 7 Graphing a Rational Function Graph the rational function . Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 4 0 Solve for x. x ;2 State the domain. STEP 2 Find the intercepts. y-intercept: y 0 x-intercepts: x 0 The only intercept is at the point .(0, 0) f(x) = x x2 - 4 = 0 f(0) = 0 -4 = 0 (-, -2)(-2, 2)(2, ) f(x) = x x2 - 4 Study Tip Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole on its graph. It is important to note that any real number eliminated from the domain of a rational function corresponds to either a vertical asymptote or a hole on its graph. Let f be a rational function given by . Step 1: Find the domain of the rational function f. Step 2: Find the intercept(s). y-intercept: evaluate f(0). x-intercept: solve the equation n(x) 0 for x in the domain of f. Step 3: Find any holes. Factor the numerator and denominator. Divide out common factors. A common factor x a corresponds to a hole in the graph of f at x a if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. The result is an equivalent rational function in lowest terms. Step 4: Find any asymptotes. Vertical asymptotes: solve q(x) 0. Compare the degree of the numerator and the degree of the denominator to determine whether either a horizontal or slant asymptote exists. If one exists, nd it. Step 5: Find additional points on the graph of fparticularly near asymptotes. Step 6: Sketch the graph; draw the asymptotes, label the intercept(s) and additional points, and complete the graph with a smooth curve between and beyond the vertical asymptotes. R(x) = p(x) q(x) f(x) = n(x) d(x) GRAPHING RATIONAL FUNCTIONS Study Tip Common factors need to be divided out rst; then the remaining x-values corresponding to a denominator value of 0 are vertical asymptotes. Graphing Rational Functions We can now graph rational functions using asymptotes as graphing aids. The following box summarizes the ve-step procedure for graphing rational functions. c04d.qxd 11/25/11 4:04 PM Page 454 484. 4.6 Rational Functions 455 x 3 1 1 3 f(x) 3 5-1 3 1 3-3 5 x y 5 5 5 5 Answer: E X AM P LE 8 Graphing a Rational Function with No Horizontal or Slant Asymptotes State the asymptotes (if there are any) and graph the rational function . Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 1 0 Solve for x. x ;1 State the domain. STEP 2 Find the intercepts. y-intercept: x-intercepts: n(x) x4 x3 6x2 0 Factor. x2 (x 3)(x 2) 0 Solve. x 0, x 3, and x 2 The intercepts are the points (0, 0), (3, 0), and (2, 0). f(0) = 0 -1 = 0 (-, -1)(-1, 1)(1, ) f(x) = x4 - x3 - 6x2 x2 - 1 Technology Tip The behavior of each function as x approaches or can be shown using tables of values. Graph f(x) = x4 - x3 - 6x2 x2 - 1 . - x y 5 5 5 5 STEP 3 Find any holes. There are no common factors, so f is in lowest terms. Since there are no common factors, there are no holes on the graph of f. STEP 4 Find any asymptotes. Vertical asymptotes: d(x) (x 2)(x 2) 0 and Horizontal asymptote: Degree of numerator Degree of denominator y 0 STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching the asymptotes. YO U R TU R N Graph the rational function .f(x) = x x2 - 1 Degree of numerator = 1 Degree of denominator = 2 x = 2x = -2 f(x) = x (x + 2)(x - 2) c04d.qxd 11/25/11 4:04 PM Page 455 485. 456 C HAP TE R 4 Polynomial and Rational Functions STEP 3 Find any holes. There are no common factors, so f is in lowest terms. Since there are no common factors, there are no holes on the graph of f. STEP 4 Find the asymptotes. Vertical asymptote: d(x) x2 1 0 Factor. (x 1)( x 1) 0 Solve. x 1 and x 1 No horizontal asymptote: degree of n(x) degree of d(x) [4 2] No slant asymptote: degree of n(x) degree of d(x) 1 [4 2 2 1] The asymptotes are x 1 and x 1. STEP 5 Find additional points on the graph. f(x) = x2 (x - 3)(x + 2) (x - 1)(x + 1) The graph of f(x) shows that the vertical asymptotes are at and there is no horizontal asymptote or slant asymptote. x = ;1 x 3 0.5 0.5 2 4 f(x) 6.75 1.75 2.08 5.33 6.4 STEP 6 Sketch the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote. 5 5 10 10 x y (3, 0)(2, 0) x = 1 x = 1 YO U R TU R N State the asymptotes (if there are any) and graph the rational function f(x) = x3 - 2x2 - 3x x + 2 . 5 10 25 50 x y (3, 0) (1, 0) x = 2 Answer: Vertical asymptote: x 2. No horizontal or slant asymptotes. E X AM P LE 9 Graphing a Rational Function with a Horizontal Asymptote State the asymptotes (if there are any) and graph the rational function Solution: STEP 1 Find the domain. Set the denominator equal to zero. 8 x3 0 Solve for x. x 2 State the domain. (-, 2)(2, ) f(x) = 4x3 + 10x2 - 6x 8 - x3 c04d.qxd 11/25/11 4:04 PM Page 456 486. Technology Tip The behavior of each function as x approaches or can be shown using tables of values. Graph f(x) = 4x3 + 10x2 - 6x 8 - x3 . - 4.6 Rational Functions 457 The graph of f(x) shows that the vertical asymptote is at and the horizontal asymptote is at .y = -4 x = 2 STEP 2 Find the intercepts. y-intercept: x-intercepts: n(x) 4x3 10x2 6x 0 Factor. 2x(2x 1)(x 3) 0 Solve. x 0, and x 3 The intercepts are the points (0, 0), , and (3, 0). STEP 3 Find the holes. There are no common factors, so f is in lowest terms (no holes). STEP 4 Find the asymptotes. Vertical asymptote: d(x) 8 x3 0 Solve. x 2 Horizontal asymptote: degree of n(x) degree of d(x) Use leading coefcients. The asymptotes are x 2 and y 4. STEP 5 Find additional points on the graph. y = 4 -1 = -4 f(x) = 2x(2x - 1)(x + 3) (2 - x)(x2 + 2x + 4) A1 2, 0B x = 1 2 , f(0) = 0 8 = 0 STEP 6 Sketch the graph; label the intercepts and asymptotes and complete with a smooth curve. YO U R TU R N Graph the rational function Give equations of the vertical and horizontal asymptotes and state the intercepts. f(x) = 2x2 - 7x + 6 x2 - 3x - 4 . 10 10 10 10 x y (3, 0) x = 2 y = 4 ( , 0)1 2 Answer: Vertical asymptotes: x 4, x 1 Horizontal asymptote: y 2 Intercepts: A0, - 3 2 B, A3 2, 0B, (2, 0) 4 6 x y 5 5 y = 2 x = 1 x = 4 x 4 1 1 3 f(x) 1 1.33 0.10 1.14 9.47 1 4 c04d.qxd 12/23/11 7:16 PM Page 457 487. 458 C HAP TE R 4 Polynomial and Rational Functions E X AM P LE 10 Graphing a Rational Function with a Slant Asymptote Graph the rational function Solution: STEP 1 Find the domain. Set the denominator equal to zero. x 2 0 Solve for x. x 2 State the domain. STEP 2 Find the intercepts. y-intercept: f(0) = - 4 2 = -2 (-,-2)(-2, ) f(x) = x2 - 3x - 4 x + 2 .Technology Tip The behavior of each function as x approaches or can be shown using tables of values. Graph f(x) = x2 - 3x - 4 x + 2 . - The graph of f(x) shows that the vertical asymptote is at and the slant asymptote is at y = x - 5. x = -2 20 20 20 20 x y (4, 0)(0, 2) (1, 0) x = 2 y = x 5 x 6 5 3 5 6 f(x) 12.5 12 14 0.86 1.75 Answer: Horizontal asymptote: x 3 Slant asymptote: y x 4 10 10 10 20 x y x = 3 y = x + 4 (2, 0) (1, 0) (0, )2 3 x-intercepts: n(x) x2 3x 4 0 Factor. (x 1)(x 4) 0 Solve. x 1 and x 4 The intercepts are the points (0, 2), (1, 0), and (4, 0). STEP 3 Find any holes. There are no common factors, so f is in lowest terms. Since there are no common factors, there are no holes on the graph of f. STEP 4 Find the asymptotes. Vertical asymptote: d(x) x 2 0 Solve. x 2 Slant asymptote: degree of n(x) degree of d(x) 1 Divide n(x) by d(x). Write the equation of the asymptote. y x 5 The asymptotes are x 2 and y x 5. STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote. YO U R TU R N For the function , state the asymptotes (if any exist) and graph the function. f(x) = x2 + x - 2 x - 3 f(x) = x2 - 3x - 4 x + 2 = x - 5 + 6 x + 2 f(x) = (x - 4)(x + 1) (x + 2) c04d.qxd 12/23/11 6:29 PM Page 458 488. 4.6 Rational Functions 459 E X AM P LE 11 Graphing a RationalFunction with a Hole in the Graph Graph the rational function . Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 x 2 0 Solve for x. (x 2)(x 1) 0 x 1 or x 2 State the domain. STEP 2 Find the intercepts. y-intercept: y 3 x-intercepts: n(x) x2 x 6 0 (x 3)(x 2) 0 x 3 or x 2 The intercepts correspond to the points and . The point (2, 0) appears to be an x-intercept; however, x 2 is not in the domain of the function. STEP 3 Find any holes. Since x 2 is a common factor, there is a hole in the graph of f at x 2. Dividing out the common factor generates an equivalent rational function in lowest terms. STEP 4 Find the asymptotes. Vertical asymptotes: x 1 0 Horizontal asymptote: Since the degree of the numerator equals the degree of the denominator, use the leading coefcients. STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching asymptotes. Recall the hole at . Note that so the open hole is located at the point (2, 5/3). YO U R TU R N Graph the rational function .f(x) = x2 - x - 2 x2 + x - 6 R(2) = 5 3 x = 2 y = 1 1 = 1 Degree of numerator of f Degree of denominator of f 2 and Degree of numerator of R Degree of denominator of R 1 x = -1 R(x) = (x + 3) (x + 1) f(x) = (x - 2)(x + 3) (x - 2)(x + 1) (-3, 0)(0, 3) f(0) = -6 -2 = 3 (-, -1)(-1, 2)(2, ) f(x) = x2 + x - 6 x2 - x - 2 Technology Tip The behavior of each function as x approaches or can be shown using tables of values. Graph f(x) = x2 + x - 6 x2 - x - 2 . - The graph of f(x) shows that the vertical asymptote is at and the horizontal asymptote is at .y = 1 x = -1 Notice that the hole at is not apparent in the graph. A table of values supports the graph. x = 2 x 4 2 1 3 f(x) or R(x) 1 5 2 3 2 1 3 -1 2 x y 5 5 5 5 Answer: x 7 3 y 5 5 c04d.qxd 11/25/11 4:04 PM Page 459 489. 460 C HAP TE R 4 Polynomial and Rational Functions 1. If degree of the numerator degree of the denominator 1, then there is a slant asymptote. 2. Divide the numerator by the denominator. The quotient corresponds to the equation of the line (slant asymptote). Procedure for Graphing Rational Functions 1. Find the domain of the function. 2. Find the intercept(s). y-intercept x-intercepts (if any) 3. Find any holes. If x a is a common factor of the numerator and denominator, then x a corresponds to a hole in the graph of the rational function if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. The result after the common factor is canceled is an equivalent rational function in lowest terms (no common factor). 4. Find any asymptotes. Vertical asymptotes Horizontal/slant asymptotes 5. Find additional points on the graph. 6. Sketch the graph: draw the asymptotes and label the intercepts and points and connect with a smooth curve. S U M MARY In this section, rational functions were discussed. Domain: All real numbers except the x-values that make the denominator equal to zero, d(x) 0. Vertical Asymptotes: Vertical lines, x a, where d(a) a, after all common factors have been divided out. Vertical asymptotes steer the graph and are never touched. Horizontal Asymptotes: Horizontal lines, y b, that steer the graph as . 1. If degree of the numerator degree of the denominator, then y 0 is a horizontal asymptote. 2. If degree of the numerator degree of the denominator, then y c is a horizontal asymptote where c is the ratio of the leading coefcients of the numerator and denominator, respectively. 3. If degree of the numerator degree of the denominator, then there is no horizontal asymptote. Slant Asymptotes: Slant lines, y mx b, that steer the graph as .x S ; x S ; f(x) = n(x) d(x) S E CTI O N 4.6 In Exercises 110, nd the domain of each rational function. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. In Exercises 1120, nd all vertical asymptotes and horizontal asymptotes (if there are any). 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. f(x) = 0.8x4 - 1 x2 - 0.25 f(x) = (0.2x - 3.1)(1.2x + 4.5) 0.7(x - 0.5)(0.2x + 0.3) f(x) = 1 10 Ax2 - 2x + 3 10 B 2x - 1 f(x) = 1 3x2 + 1 3x - 1 4 x2 + 1 9 f(x) = 6x2 + 3x + 1 3x2 - 5x - 2 f(x) = 6x5 - 4x2 + 5 6x2 + 5x - 4 f(x) = 2 - x3 2x - 7 f(x) = 7x3 + 1 x + 5 f(x) = 1 5 - x f(x) = 1 x + 2 f(x) = 5(x2 - 2x - 3) (x2 - x - 6) f(x) = - 3(x2 + x - 2) 2(x2 - x - 6) f(x) = - 2x x2 + 9 f(x) = 7x x2 + 16 f(x) = x - 1 x2 + 2x - 3 f(x) = x + 4 x2 + x - 12 f(x) = 5 - 3x (2 - 3x)(x - 7) f(x) = 2x + 1 (3x + 1)(2x - 1) f(x) = 3 4 - x f(x) = 1 x + 3 S K I LL S E X E R C I S E S S E CTI O N 4.6 c04d.qxd 11/25/11 4:04 PM Page 460 490. 4.6 Rational Functions 461 In Exercises 2126, nd the slant asymptote corresponding to the graph of each rational function. 21. 22. 23. 24. 25. 26. In Exercises 2732, match the function to the graph. 27. 28. 29. 30. 31. 32. a. b. c. d. e. f. In Exercises 3358, use the graphing strategy outlined in this section to graph the rational functions. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. f(x) = -2x(x - 3) x(x2 + 1) f(x) = 3x(x - 1) x(x2 - 4) f(x) = (x - 1)(x2 - 9) (x - 3)(x2 + 1) f(x) = (x - 1)(x2 - 4) (x - 2)(x2 + 1) f(x) = (x + 1)2 (x2 - 1) f(x) = (x - 1)2 (x2 - 1) f(x) = x - 4 x f(x) = 3x + 4 x f (x) = 25x2 - 1 (16x2 - 1)2 f(x) = 1 - 9x2 (1 - 4x2 )3 f(x) = 12x4 (3x + 1)4 f(x) = 7x2 (2x + 1)2 f(x) = 1 - x2 x2 + 1 f(x) = x2 + 1 x2 - 1 f(x) = 3x3 + 5x2 - 2x x2 + 4 f(x) = 2x3 - x2 - x x2 - 4 f(x) = x2 - 9 x + 2 f(x) = x2 x + 1 f(x) = 3(x2 - 1) x2 - 3x f(x) = 2(x2 - 2x - 3) x2 + 2x f(x) = 2 + x x - 1 f(x) = x - 1 x f(x) = 4x x + 2 f(x) = 2x x - 1 f(x) = 4 x - 2 f(x) = 2 x + 1 5 5 5 5 x y 5 5 5 5 x y 10 10 10 10 x y 10 10 150 150 x y 10 10 10 10 x y 10 10 10 10 x y f(x) = 3x2 x + 4 f(x) = 3x2 4 - x2 f(x) = - 3x2 x2 + 4 f(x) = 3x2 x2 - 4 f(x) = 3x x - 4 f(x) = 3 x - 4 f(x) = 2x6 + 1 x5 - 1 f(x) = 8x4 + 7x3 + 2x - 5 2x3 - x2 + 3x - 1 f(x) = 3x3 + 4x2 - 6x + 1 x2 - x - 30 f(x) = 2x2 + 14x + 7 x - 5 f(x) = x2 + 9x + 20 x - 3 f(x) = x2 + 10x + 25 x + 4 c04d.qxd 11/25/11 4:04 PM Page 461 491. 462 C HAP TE R 4 Polynomial and Rational Functions In Exercises 5962, for each graph of the rational function given determine: (a) all intercepts, (b) all asymptotes, and (c) equation of the rational function. 59. 60. 61. 62. x y 10 10 10 10 x y 10 10 10 10 x y 10 10 10 10 x y 5 5 5 5 66. Memorization. A professor teaching a large lecture course tries to learn students names. The number of names she can remember N(t) increases with each week in the semester t and is given by the rational function: How many students names does she know by the third week in the semester? How many students names should she know by the end of the semester (16 weeks)? According to this function, what are the most names she can remember? 67. Food. The amount of food that cats typically eat increases as their weight increases. A rational function that describes this is , where the amount of food F(x) is given in ounces and the weight of the cat x is given in pounds. Calculate the horizontal asymptote. How many ounces of food will most adult cats eat? 10 10 x y F(x) = 10x2 x2 + 4 N(t) = 600t t + 20 63. Medicine. The concentration C of a particular drug in a persons bloodstream t minutes after injection is given by a. What is the concentration in the bloodstream after 1 minute? b. What is the concentration in the bloodstream after 1 hour? c. What is the concentration in the bloodstream after 5 hours? d. Find the horizontal asymptote of C(t). What do you expect the concentration to be after several days? 64. Medicine. The concentration C of aspirin in the bloodstream t hours after consumption is given by a. What is the concentration in the bloodstream after hour? b. What is the concentration in the bloodstream after 1 hour? c. What is the concentration in the bloodstream after 4 hours? d. Find the horizontal asymptote for C(t). What do you expect the concentration to be after several days? 65. Typing. An administrative assistant is hired after graduating from high school and learns to type on the job. The number of words he can type per minute is given by where t is the number of months he has been on the job. a. How many words per minute can he type the day he starts? b. How many words per minute can he type after 12 months? c. How many words per minute can he type after 3 years? d. How many words per minute would you expect him to type if he worked there until he retired? N(t) = 130t + 260 t + 5 t 0 1 2 C(t) = t t2 + 40 . C(t) = 2t t2 + 100 A P P L I C AT I O N S c04d.qxd 11/25/11 4:04 PM Page 462 492. 4.6 Rational Functions 463 68. Memorization. The Guinness Book of World Records, 2004 states that Dominic OBrien (England) memorized on a single sighting a random sequence of 54 separate packs of cards all shufed together (2808 cards in total) at Simpsons-In-The-Strand, London, England, on May 1, 2002. He memorized the cards in 11 hours 42 minutes, and then recited them in exact sequence in a time of 3 hours 30 minutes. With only a 0.5% margin of error allowed (no more than 14 errors), he broke the record with just 8 errors. If we let x represent the time (hours) it takes to memorize the cards and y represent the number of cards memorized, then a rational function that models this event is given by According to this model, how many cards could be memorized in an hour? What is the greatest number of cards that can be memorized? 69. Gardening. A 500-square-foot rectangular garden will be enclosed with fenc