CORE CONNECTIONS ALGEBRA 2 Parent Guide with Extra Practice

Transcription

1 CORE CONNECTIONS ALGEBRA 2 Parent Guide with Etra Practice Managing Editors / Authors Jud Ksh (First Edition) San Francisco State Universit San Francisco, CA Michael Kassarjian (Second Edition) CPM Educational Program Kensington, CA Evra Baldinger (First Edition) Universit of California, Berkele Berkele, CA Contributing Authors Scott Coner, P.E. Christian Brothers High School Sacramento, CA Bob Petersen CPM Educational Program Sacramento, CA Brian Hoe CPM Educational Program Sacramento, CA Karen Wootton CPM Educational Program Odenton, MD Sarah Maile CPM Educational Program Sacramento, CA Technical Assistants Madeline Kimball Sarah Maile Aubrie Maze Cover Art Kevin Coffe San Francisco, CA Program Directors Elizabeth Coner CPM Educational Program Sacramento, CA Brian Hoe CPM Educational Program Sacramento, CA Tom Sallee, Ph.D. Department of Mathematics Universit of California, Davis Leslie Dietiker, Ph.D. Boston Universit Boston, MA Michael Kassarjian CPM Educational Program Kensington, CA Karen Wootton CPM Educational Program Odenton, MD Lori Hamada CPM Educational Program Fresno, CA Jud Ksh, Ph.D. Departments of Education and Mathematics San Francisco State Universit, CA

2 Copright 2013 b CPM Educational Program. All rights reserved. No part of this publication ma be reproduced or transmitted in an form or b an means, electronic or mechanical, including photocop, recording, or an information storage and retrieval sstem, without permission in writing from the publisher. Requests for permission should be made in writing to: CPM Educational Program, 9498 Little Rapids Wa, Elk Grove, CA cpm@cpm.org Printed in the United States of America ISBN:

3 Introduction to the Parent Guide with Etra Practice Welcome to the Core Connections Algebra 2 Parent Guide with Etra Practice. The purpose of this guide is to assist ou should our child need help with homework or the ideas in the course. We believe all students can be successful in mathematics as long as the are willing to work and ask for help when the need it. We encourage ou to contact our child s teacher if our student has additional questions that this guide or other resources do not answer. This guide was written to address the major topics in each chapter of the tetbook. Each section begins with a title bar and the lesson(s) in the book that it addresses. In man cases the eplanation bo at the beginning of the section refers ou to one or more Math Notes boes in the student tet for additional information about the fundamentals of the idea. Detailed eamples follow a summar of the concept or skill and include complete solutions. The eamples are similar to the work our child has done in class. Additional problems, with answers, are provided for our child to practice. There will be some topics that our child understands quickl and some concepts that ma take longer to master. The big ideas of the course take time to learn. This means that students are not necessaril epected to master a concept when it is first introduced. When a topic is first introduced in the tetbook, there will be several problems to do for practice. Subsequent lessons and homework assignments will continue to practice the concept or skill over weeks and months so that master will develop over time. Practice and discussion are required to understand mathematics. When our child comes to ou with a question about a homework problem, often ou ma simpl need to ask our child to read the problem and then ask her/him what the problem is asking. Reading the problem aloud is often more effective than reading it silentl. When ou are working problems together, have our child talk about the problems. Then have our child practice on his/her own. Below is a list of additional questions to use when working with our child. These questions do not refer to an particular concept or topic. Some questions ma or ma not be appropriate for some problems. What have ou been doing in class or during this chapter that might be related to this problem? Let's look at our notebook, class notes, and Learning Log. Do ou have them? Were the other members of our team having difficult with this as well? Can ou call our stud partner or someone from our stud team? Have ou checked the online homework help (in the tetbook resources section at cpm.org)? What have ou tried? What steps did ou take? What did not work? Wh did it not work? Which words are most important? Wh? What does this word/phrase tell ou? What do ou know about this part of the problem? Eplain what ou know right now. What is unknown? What do ou need to know to solve the problem? How did the members of our stud team eplain this problem in class? What important eamples or ideas were highlighted b our teacher? How did ou organize our information? Do ou have a record of our work? Can ou draw a diagram or sketch to help ou? Have ou tried making a list, looking for a pattern, etc.? What is our estimate/prediction? Is there a simpler, similar problem we can do first? i

4 If our student has made a start at the problem, tr these questions: What do ou think comes net? Wh? What is still left to be done? Is that the onl possible answer? Is that answer reasonable? Are the units correct? How could ou check our work and our answer? If ou do not seem to be making an progress, ou might tr these questions. Let s look at our notebook, class notes, and Toolkit. Do ou have them? Were ou listening to our team members and teacher in class? What did the sa? Did ou use the class time working on the assignment? Show me what ou did. Were the other members of our team having difficult with this as well? Can ou call our stud partner or someone from our stud team? This is certainl not a complete list; ou will probabl come up with some of our own questions as ou work through the problems with our child. Ask an question at all, even if it seems too simple to ou. To be successful in mathematics, students need to develop the abilit to reason mathematicall. To do so, students need to think about what the alread know and then connect this knowledge to the new ideas the are learning. Man students are not used to the idea that what the learned esterda or last week will be connected to toda s lesson. Too often students do not have to do much thinking in school because the are usuall just told what to do. When students understand that connecting prior learning to new ideas is a normal part of their education, the will be more successful in this mathematics course (and an other course, for that matter). The student s responsibilities for learning mathematics include the following: Activel contributing in whole class and stud team work and discussion. Completing (or at least attempting) all assigned problems and turning in assignments in a timel manner. Checking and correcting problems on assignments (usuall with their stud partner or stud team) based on answers and solutions provided in class and online. Asking for help when needed from his or her stud partner, stud team, and/or teacher. Attempting to provide help when asked b other students. Taking notes and using his/her Toolkit when recommended b the teacher or the tet. Keeping a well-organized notebook. Not distracting other students from the opportunit to learn. Assisting our child to understand and accept these responsibilities will help him or her to be successful in this course, develop mathematical reasoning, and form habits that will help her/him become a life-long learner. Additional support for students and parents is provided at the CPM website (search Homework Help) cpm.org. The website provides a variet of complete solutions, hints, and answers. Some problems refer back to other similar problems. The homework help is designed to assist students to be able to do the problems but not necessaril do the problems for them. ii

5 CALCULATORS IN CCA2 Which calculator will Algebra students need when? Man lessons in Core Connections Algebra 2, particularl in the second half of the book, also assume that students have dail access to a graphing calculator or other graphing technolog in the classroom. Tpical graphing calculators are those in the Teas Instruments TI-83+ and TI-84+ families. A graphing calculator makes some of the homework problems more efficient, but with few eceptions, a graphing calculator is not required for homework. The statistical functions and statistical graphing capabilit of a graphing calculator first become a necessar part of the classwork lessons in Core Connections Algebra Chapter 6, Modeling Two-Variable Data. Then the abilit of a graphing calculator to efficientl create multiple representations of a function is leveraged in the investigations of eponential and quadratic functions in Chapters 7 through 10. Chapter 11 contains challenging, culminating investigations in which students use the skills the learned throughout the course. Strategicall choosing a graphing calculator when it is an appropriate tool is an important objective of these investigations. ADDITIONAL SUPPORT Consider these additional resources for assisting students with the CPM Educational Program: This Core Connections Algebra 2 Parent Guide with Etra Practice. This booklet can be printed for no charge (or a bound cop ordered for $20) from the tetbook resources section at cpm.org. CPM Homework Help site in the tetbook resources section at cpm.org. A variet of complete solutions, hints, and answers are provided. Some problems refer back to other similar problems. The homework help is designed to assist students to be able to do the problems but not necessaril do the problems for them. Checkpoints. The back of the student tet has Checkpoint materials to assist students with skills the should master. The checkpoints are numbered to align with the chapter in the tet. For eample, the topics in Checkpoint 5A and Checkpoint 5B should be mastered while students complete Chapter 5. Resource Pages. The resource pages referred to in the student tet can be found in the tetbook resources section of cpm.org. Previous tests. Man teachers allow students to eamine their own tests from previous chapters in the course. Even if the are not allowed to bring these tests home, a student can learn much b analzing errors on past tests. Math Notes in the student tet. The Closure section at the end of each chapter in the student tet has a list of the Math Notes in that chapter. Note that relevant Math Notes are sometimes found in other chapters than the one currentl being studied. Glossar and Inde at the back of the student tet. Answers and Support table. The What Have I Learned questions (in the closure section a the end of each chapter) are followed b an Answers and Support table that indicates where students can get more help with problems. After-school assistance. Some schools have after-school or at-lunch support programs for students. Ask the teacher. iii

6 Other students. Consider asking our child to obtain the contact information for a couple other students in class. Parent Guides from previous courses. If our student needs help with the concepts from previous courses that are necessar preparation for this class, the Core Connections Algebra and Core Connections Geometr Parent Guides with Etra Practice are available for no charge at in the tetbook resources section at cpm.org. iv

7 Table of Contents Chapter 1 Investigations and Functions Investigations and Functions 1 Investigating a Function 6 SAT Prep 11 Chapter 2 Transformations of Parent Graphs Transformations of f () = 2 13 Transforming Parent Graphs 18 More on Completing the Square 23 SAT Prep 25 Chapter 3 Equivalent Forms Equivalence 27 Simplifing Epressions 32 Multiplication and Division of Rational Epressions 34 Addition and Subtraction of Rational Epressions 36 SAT Prep 38 Chapter 4 Solving and Intersections Solving Sstems of Equations 40 Inequalities 47 SAT Prep 53 Chapter 5 Inverses and Logarithms Inverses 55 Logarithms 59 SAT Prep 62 Chapter 6 3-D Graphing and Logarithms Sstems of Three Equations 64 Solving With Logarithms 70 SAT Prep 75 v

8 Chapter 7 Trigonometric Functions Trigonometric Functions 77 Transforming Trig Functions 83 SAT Prep 89 Chapter 8 Polnomials Polnomials 91 Comple Numbers 96 Factoring Polnomial Functions 100 SAT Prep 104 Chapter 9 Randomization and Normal Distributions Surve Design, Sampling, and Bias 105 Testing Cause and Effect with Eperiments and 107 Conclusions From Observational Studies Relative Frequenc Histograms 109 The Normal Probabilit Densit Function and Percentiles 111 SAT Prep 114 Chapter 10 Series Series 116 Geometric Series 119 Pascal s Triangle 122 SAT Prep 124 Chapter 11 Simulating Sampling Variabilit Simulation 126 Sample-to-Sample Variabilit, Margin of Error, and 129 Statistical Hpothesis Testing Counterintuitive Probabilit 133 SAT Prep 136 Chapter 12 Analtic Trigonometr Analtic Trigonometr 138 Trigonometric Identities 142 SAT Prep 146 vi

9 Appendi A Sequences Introduction to Sequences 147 Equations for Sequences 151 Patterns of Growth in Tables and Graphs 160 Appendi B Eponential Functions Eponential Functions 163 Fractional Eponents 168 Curve Fitting 171 Appendi C Comparing Single-Variable Data Comparing and Representing Data 174 vii

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11 Chapter 1 INVESTIGATIONS AND FUNCTIONS This opening section introduces the students to man of the big ideas of Algebra 2, as well as different was of thinking and various problem solving strategies. Students are also introduced to their graphing calculators. The also learn the appropriate use of the graphing tool, so that time is not wasted using it when a problem can be solved more efficientl b hand. Not onl are students working on challenging, interesting problems, the are also reviewing topics from earlier math courses such as graphing, trigonometric ratios, and solving equations. The also practice algebraic manipulations b inputting values into function machines and calculating each output. For further information see the Math Notes boes in Lessons 1.1.1, 1.1.2, and Eample 1 Talula s function machine at right shows its inner workings, written in function notation. Note that = 10 2 is an equivalent form. What will the output be if: a. 2 is dropped in? b. 2 is dropped in? c. 10 is dropped in? d is dropped in? Solution: The number dropped in, that is, substituted for, takes the place of in the equation in the machine. Follow the Order of Operations to simplif the epression to determine the value of f(). a. f (2) = 10 (2) 2 = 10 4 = 6 c. f ( 10) = 10 ( 10) 2 = = 0 b. f ( 2) = 10 ( 2) 2 = 10 4 = 6 d. f ( 3.45) = 10 ( 3.45) 2 = = Parent Guide with Etra Practice 1

12 Eample 2 Consider the functions f () = 3 and g() = ( + 5 )2. a. What is f (4)? b. What is g(4)? c. What is the domain of f ()? d. What is the domain of g()? e. What is the range of f ()? f. What is the range of g()? Solution: Substitute the values of in the functions for parts (a) and (b): f (4) = = 2 1 = 2 g(4) = ( 4 + 5) 2 = (9) 2 = 81 The domain of f () is the set of -values that are allowable, and this function has some restrictions. First, we cannot take the square root of a negative number, so cannot be less than zero. Additionall, the denominator of a fraction cannot be zero, so 3. Therefore, the domain of f () is 0, 3. For g(), we could substitute an number for, add five, and then square the result. This function has no restrictions so the domain of g() is all real numbers. The range of these functions is the set of all possible values that result when substituting the domain, or -values. We need to decide if there are an values that the functions could never reach, or are impossible to produce. Consider the range of g() first. Since the function g squares the amount in the final step, the output will alwas be positive. It could equal zero (when = 5), but it will never be negative. Therefore, the range of g() is 0. The range of f () is more complicated. Tr finding some possibilities first. Can this function ever equal zero? Yes, when = 0, then f () = 0. Can the function ever equal a ver large positive number? Yes, this happens when < 3, but ver close to 3. (For eample, let = , then f () is approimatel equal to 17,320.) Can f () become an etremel negative number? Yes, when > 3, but ver close to 3. (Here, tr = , and then f () is approimatel equal to 17,320.) There does not seem to be an restrictions on the range of f (), therefore we can sa that the range is all real numbers. 2 Core Connections Algebra 2

13 Chapter 1 Eample 3 For each problem below, first decide how ou will answer the question: b using a graphing tool, our algebra skills, or a combination of the two. Use the most efficient method. Show our work, including a justification of the method ou chose for each problem. a. What is the -intercept of the graph of = ? b. Does the graph of = cross the -ais? If so, how man times? c. Where do the graphs of = and = intersect? d. What are the domain and range of = ? Solutions: Part (a): The -intercept is the point (0, b). Therefore, the -intercept can be found b substituting 0 for. In this case, the -intercept can be found b calculating = 2 3 (0)+19 and hence the -intercept is the point (0, 19). Part (b): The most efficient method is to use a graphing calculator to see if the graph does or does not cross the -ais. The graph at right shows the complete graph, meaning we see everthing that is important about the graph, and the rest of the graph is predictable based on what we see. This graph shows us that the graph intersects the -ais, twice. Part (c): It is best to use algebra and solve this sstem of two equations with two unknowns to find where the graphs intersect. This can be done b using the Equal Values method = = (multipl all terms b 5) 26 = 130 (add and 100 to both sides) = 5 (divide both sides b 26) = 5(5)+ 20 = 45 Therefore the graphs intersect at the point (5, 45). (substitute 5 for in the first equation) Part (d): We need to find the acceptable values for the input, and the possible outputs. The equation offers no restrictions, like dividing b zero, or taking the square root of a negative number, so the domain is all real numbers. Since this equation is a parabola when graphed, there will be restrictions on the range. Just as the equation = 2 can never be negative, the ranges of quadratic functions (parabolas) will have a lowest point, or a highest point. If we graph this parabola, 8 we can see that the lowest point, the verte, is at (6, 10). The graph onl has values of 10, therefore the range is Parent Guide with Etra Practice 3

14 Most of the earl homework assignments review skills developed in Algebra 1 and geometr. The problems below include these tpes of problems. Problems Solve the following equations for and/or. 1. 5( + 7) = = 12 = = b( a) = c Find the error and show the correct solution = 2( 3) 5 9 = = 14 = = = 3 2( + 1) = 3 2 = 3 or + 1 = 3 = 3 2 or = 2 Sketch a complete graph of each of the following equations. Be sure to label the graph carefull so that all ke points are identified. What are the domain and range of each function? 7. = = 3 6 If f () = 3 2 6, find: 9. f (1) 10. f ( 3) 11. f (2.75) 4 Core Connections Algebra 2

15 Chapter 1 Use the graph at right to complete the following problems. 12. Briefl, what does the graph represent? 13. Based on the graph, give all the information ou can about the Ponda Concord. 14. Based on the graph, give all the information ou can about the Neo Brism. 15. Does it make sense to etend these lines into the second and fourth quadrants? Eplain. Gas in tank (gallons) Neo Brism Ponda Concord Distance traveled (miles) Answers 1. = = 2, = 6 3. = 7, 3 4. = c+ab = c b b + a 5. When distributing, ( 2)( 3) = 6. = Must set the equation equal to zero before factoring. = 1, The graph shows how much gas is in the tank of a Ponda Concord or a Neo Brism as the car is driven. 13. Ponda Concord s gas tank holds 16 gallons of gas, and the car has a driving range of about 350 miles on one tank of gas. It gets 22 miles per gallon. 14. The Neo Brism s gas tank holds onl 10 gallons of gas, but it has a driving range of 400 miles on a tank of gas. It gets 40 miles per gallon. 15. No, the car doesn t travel negative miles, nor can the gas tank hold negative gallons. Parent Guide with Etra Practice 5

16 INVESTIGATING A FUNCTION A goal of this course is for students to make connections among mathematical ideas. To reach this goal, we develop the idea of investigating a function. We want students to find out everthing the can about a function or a situation, b asking questions about it, and drawing conclusions, so that the have a complete picture of the function or situation. In this section we consider the different questions to ask about the functions, and the different components to consider in order for students to understand the function completel. For further information see the Math Notes bo in Lesson 1.2.2, as well as the bo contained within problem Eample 1 Investigate the function f () = Solution: We want to find out all we can about this function, and although the graph of f ()will be helpful in our understanding, graphs can be unclear or incomplete. To investigate this function completel, answer the following questions: Is this function linear? If not, can ou classif it? Does the function have an -intercepts? Does it have an -intercepts? If so, what are the? What is the domain of the function? What is the range of the function? Does the function have an asmptotes? If so, what are the and wh do the occur? Are there an important points on the graph of this function? (High points, low points, turning points, etc.) What makes these points important? What is the shape of the graph? An eas first step is to graph this function on a graphing calculator or using some other graphing tool. However, take care when interpreting the graph. With complicated functions such as this one, the graphing tool ma not clearl displa important points or trends. It is important to be thinking while graphing! In particular, notice that this function has two restrictions. First, we cannot have zero in the denominator. Second, we cannot have a negative quantit under the radical sign With a quick sketch and some preliminar thinking, we can proceed. This function is not linear, but curved. Some students might know it as a hperbola, but that is not essential at this point. It is not obvious from this graph that there are an -intercepts, but there are. -intercepts occur when f () = 0. For a rational epression (a fraction) to equal zero, the numerator (the top) must equal to zero. Eample continues on net page 6 Core Connections Algebra 2

17 Chapter 1 Eample continued from previous page. +4 Therefore, f () = 0 = = = 0 when + 4 = 0, or when = 4. Therefore, the -intercept is ( 4, 0). The -intercept occurs when we substitute = 0 into the equation. f (0) = 0+4 3(0) = 4 0 This result is undefined because we cannot divide b zero. Therefore, there are no -intercepts. The work we just did will help us determine the domain and range. Here, the domain is restricted because and the denominator cannot equal zero. Therefore, the domain is the set of all numbers 4, 0. The range is all real numbers (that is, f() can take on all values). When we have the graph, we can also think of the domain and range as shadows of the graphs on the aes. For instance, if we could shine a light from above and below the graph, casting a shadow on the -ais, the shadow would be the domain. Here the shadow would not appear to the left of = 4, nor will it appear at zero. Similarl, a light casting a shadow onto the -ais gives us the range. Here all of the -ais will be covered in shadow. Asmptotes occur when the graph approaches a value, but never quite reaches it. Here, as the -values become larger and larger, the f()-values move closer and closer to zero. This happens because the denominator will grow larger more quickl than the numerator, making smaller and smaller fractions. Therefore = 0 is a horizontal asmptote. Similarl, the line = 0 is also a vertical asmptote. You can convince ourself that the graph becomes ver close to this line as ou substitute values for that are ver close to zero. We covered the last two questions in the other answers therefore we have covered all the ke points for investigating this function. Eample 2 Suppose a ardstick leans against the wall, forming a right triangle with the wall and the floor when viewed from the side. We define a function with inputs,, as the height of this triangle (i.e., the height on the wall where the ardstick touches it), and outputs as the area of the triangle. Note that the ardstick is the hpotenuse of the triangle. Investigate this function and write summar statements for what ou know about this function. To investigate a function we must be able to answer the questions we listed in Eample 1. Before we can answer an of these questions, we must first understand this geometric relationship and translate it into something algebraic. To do this, we will tr some specific eamples. First, suppose the ardstick touches at a point 20 inches up the wall. This means we are considering an input value of = 20. Eample continues on net page Parent Guide with Etra Practice 7

18 Eample continued from previous page. The output will be the area of the triangle. To find the area, we need to know the length of the base of the triangle and for that we use the Pthagorean Theorem. Now that we know the length of the base, we can find the area of this triangle. A = 1 2 bh 1 2 (29.93)(20) A square inches b 2 = b 2 = 1296 b 2 = 896 b = This gives us one input and its corresponding output. Tr this again using an input (height) of 10 inches. You should get a corresponding output (area) of approimatel square inches. Each student needs to do as man eamples as s/he needs in order to understand the general case. In the general case, we let the height (input) be, solve for b, and then find the area of the triangle, all in terms of. b = 36 2 b = 1296 b 2 = b = A = 1 2 bh ( ) () A = in. b in. Now we have an equation to aid us with investigating the problem further. Note that this is just an aid. We will answer man of the questions b knowing the contet of the problem. To begin with, this is a function because ever input has one and onl one output. Net, we can determine the domain. In this situation, the acceptable inputs are the various heights along the wall that the ardstick can lean. The height could never be smaller than zero, and it could never be larger than 36 inches (because the ardstick will onl reach that high when it is flat against the wall, and no triangle is reall visible). Therefore the domain is 0 < < 36. Given this domain, the range is the values that the equation can take on within this restriction on. B looking at the graph of the equation representing this situation, we are looking for the -values this graph takes on. The area of the triangle can be zero (or ver close to zero) and has a maimum value. B using the zoom and trace buttons on our graphing calculator, ou can find the maimum value to be 324. (Note: At this point, we epect students to find the maimum value this wa. It will be some time before the can find the maimum value of a function algebraicall.) Therefore the range is 0 < < 324. If we allow the triangle to have sides with length 0 inches, then the function has two -intercepts at (0, 0) and (36, 0), and the one -intercept (0, 0). This function is also continuous (no breaks), has no asmptotes, and is not linear. At this point, we do not know what tpe of function it is. 8 Core Connections Algebra 2

19 Chapter 1 Problems Solve the following equations for and/or = = 10 = = = 0 If g() = , find: 5. g( 2) 6. g(0.4) 7. g(18) Sketch a complete graph of each of the following equations. Be sure to label the graph carefull so that all ke points are identified. What are the domain and range of each function? 8. = = Investigate the function = Investigate the function = 1 6. Parent Guide with Etra Practice 9

20 Answers = 22, = = 1 4, = ± This is a line with a slope of intercept: (300, 0) -intercept: (0, 30) This is a parabola opening upward. -intercepts: ( 40.88, 0) and (10.88, 0) -intercept: (0, 445) verte: ( 15, 670) Line of smmetr: = The graph of this function is curved. There are no - or -intercepts, it is a function, the domain is 6, the range is 1. The starting point is (6, 1). 11. This graph is a curve and it has two unconnected parts. It has no -intercepts, and the -intercept is ( 0, 1 6 ). The domain is all real values of ecept 6, and the range is all real values of ecept 0. The line = 0 is a horizontal asmptote, and = 6 is a vertical asmptote. This is a function. 10 Core Connections Algebra 2

21 Chapter 1 SAT PREP These problems are ver similar to actual SAT test questions. Use a calculator whenever ou need one. On multiple choice questions, choose the best answer from the ones provided. When a picture has a diagram, assume the diagram is drawn accuratel ecept when a problem sas it is not. These questions address more topics than ou have done in class so far. 1. If + 9 is an even integer, then which of the following could be the value of? a. 4 b. 2 c. 0 d. 1 e If (m + 5)(11 7) = 24, then m =? a. 1 b. 4 c. 8 d. 11 e The fractions 3 d, 4 d, and 5 d be the value of d? are in simplest reduced form. Which of the following could a. 20 b. 21 c. 22 d. 23 e A group of three numbers is called a j-triple for some number j, if ( 4 j, j, 5 4 j ). Which of the following is a j-triple? a. (0, 4, 5) b. ( 5 3 4,6,61 4 ) c. (6, 2, 10) d. (750, 1000, 1250) e. (575, 600, 625) 5. A ball is thrown straight up. The height of the ball can be modeled with the equation h = 38t 16t 2 where h is the height in feet and t is the number of second since the ball was thrown. How high is the ball two seconds after it is thrown? a. 12 b. 16 c. 22 d. 32 e In the figure at right, AC is a line segment with a length of 4 units. What is the value of k? 5k 3k A B C 7. Let the operation be defined as a b is the sum of all integers between a and b. For eample, 4 10 = = 35. What is the value of ( ) ( )? Parent Guide with Etra Practice 11

22 8. An isosceles triangle has a base of length 15. The length of each the other two equal sides is an integer. What is the shortest possible length of these other two sides? 9. Assume that 1 4 quart of cranberr concentrate is mied with quarts of apple juice to make cranapple juice for four people. How man quarts of cranberr concentrate are needed to make a cranapple drink at the same strength for 15 people? 10. A stack of five cards is labeled with a different integer ranging from 0 to 4. If two cards are selected at random without replacement, what is the probabilit that the sum will be 2? Answers 1. D 2. A 3. D 4. D 5. A Core Connections Algebra 2

23 Chapter 2 TRANSFORMATIONS OF f() = In order for the students to be proficient in modeling data or contetual relationships, the must easil recognize and manipulate graphs of various functions. Students investigate the general equation for a famil of quadratic functions, discovering was to shift and change the graphs. Additionall, the learn how to quickl graph a quadratic function when it is written in graphing form. For further information see the Math Notes bo in Lesson Eample 1 The graph of f () = 2 is shown at right. For each new function listed below, eplain how the new graph would differ from this original graph. g() = 2 2 h() = ( + 3) 2 j() = 2 6 k() = l() = 3( 2) Ever function listed above has something in common: the all have 2 as the highest power of. This means that all of these functions are quadratic functions, and all will form a parabola when graphed. The onl differences will be in the direction of opening (up or down), the size (compressed or stretched), and/or the location of the verte. The 2 in g() = 2 2 does two things to the parabola. The negative sign changes the parabola s direction so that it will open downward. The 2 stretches the graph making it appear skinnier. The graph of h() = ( + 3) 2 will have the same shape as f () = 2, open upward, and have a new location: it will move to the left 3 units. The graph of j() = 2 6 will also have the same shape as f () = 2, open upward and be shifted down 6 units. The function k() = does not move, still opens upward, but the 1 4 will compress the parabola, making it appear fatter. The last function, l() = 3( 2) 2 + 7, combines all of these changes into one graph. The 3 causes the graph to be skinnier and open upward, the 2 causes it to shift to the right 2 units, and the + 7 causes the graph to shift up 7 units. All these graphs are shown above. Match the equation with the correct parabola. Parent Guide and Etra Practice 13

24 Eample 2 For each of the quadratic equations below, where is the verte? f () = 2( + 4) g() = 5( 8) 2 h() = For a quadratic equation, the verte is the locator point. It gives ou a starting point for graphing the parabola quickl. The verte for the quadratic equation f () = a( h) 2 + k is the point (h, k). For f () = 2( + 4) the verte is ( 4, 7). Since g() = 5( 8) 2 can also be written g() = 5( 8) 2 + 0, the verte is (8, 0). We can rewrite h() = as h() = 5 3 ( 0)2 2 5 to see that its verte is ( 0, 2 5 ). Eample 3 In a neighborhood water balloon battle, Dudle has developed a winning strateg. He has his home base situated five feet behind an eight-foot fence. 25 feet awa on the other side of the fence is his nemesis camp. Dudle uses a water balloon launcher, and shoots his balloons so that the just miss the fence and land in his opponent s camp. Write an equation that, when graphed, will model the trajector (path) of the water balloon. As with man problems, it is most helpful to first draw a sketch of the situation. The parabola shows the path the balloon will take, starting five feet awa from the fence (point A) and landing 25 feet past the fence (point B). A B There are different was to set up aes for this problem, and depending where ou put them, our answer might be different. Here, the -ais will be at the fence. With the aes in place, we label an coordinates we know. This now shows all of the information we have from the problem description. If we can find the coordinates of the verte (highest point) of this parabola, we will be able to write the equation of it in graphing form. Parabolas are smmetric, therefore the verte will be half-wa between the two -intercepts. The total distance between points A and B is 30 units, so half is 15. Fifteen units from point A is the point (10, 0). We know the equation will be in the form = a( 10) 2 + k, with a < 0. Also, k must be greater than eight since the verte is higher than the -intercept of (0, 8). The parabola passes through the points (0, 8), ( 5, 0) and (25, 0). We will use these points in the equation we have so far and see what else we can find. Using the point (0, 8) we can substitute the - and -values in to the equation and write: 8 = a(0 10) 2 + k or 8 = 100a + k (0, 8) ( 5, 0) (25, 0) 14 Core Connections Algebra 2

25 Chapter 2 This equation has two variables, which means we need another (different) equation with a and k to be able to solve for them. Using the point ( 5, 0): 0 = a( 5 10) 2 + k or 0 = 225a + k Begin solving b subtracting the second equation from the first: 8 = 100a + k (0 = 225a + k) 8 = 125a a = Substitute this value for the variable a back into one of the two equations above to find k. 8 = 100 ( )+ k 8 = k k = 72 5 The equation for the path of a water balloon is = 8 ( ) You should graph this 5 on our graphing calculator to check. Problems Find the - and -intercepts of each of the following quadratic equations. 1. = = = = = = 6 12 Find the error in each of the following solutions. Then find the correct solution to the problem. 7. Solve for if = 0. a = 3, b = 6, c = 1 = 6± 62 4(3)(1) 2(3) 6± = = 6 6± 24 6 = 6±2 6 6 = 3± Solve for if = 0 a = 2, b = 7, c = 5 = 7± 72 4( 2)(5) 2( 2) Parent Guide and Etra Practice 15 = 7± = 7±3 4 = 4 4 = 1 or = 10 4 = 2.5

26 Find the verte of each of the following parabolas b averaging the -intercepts or completing the square. Then write each equation in graphing form, and sketch the graph. 9. = = = ( + 7)( 3) 12. = 2( + 6) 2 1 For each situation, write an appropriate equation that will model the situation effectivel. 13. When Twinkle Toes Ton kicked a football, it landed 100 feet from where he kicked it. It also reached a height of 125 feet. Write an equation that, when graphed, will model the path of the ball from the moment it was kicked until it first touched the ground. 14. When some software companies develop software, the do it with planned obsolescence in mind. This means that the plan on the sale of the software to rise, hit a point of maimum sales, then drop and eventuall stop when the release a newer version of the software. Suppose the curve showing the number of sales over time is parabolic and that the compan plans on the life span of its product to be si months, with maimum sales reaching 1.5 million units. Write an equation that best fits this data. 15. A new skateboarder s ramp just arrived at Bunge s Famil Fun Center. A cross-sectional view shows that the shape is parabolic. The sides are 12 feet high and 15 feet apart. Write an equation that, when graphed, will show the cross section of this ramp. 16 Core Connections Algebra 2

27 Chapter 2 Answers 1. -intercepts: ( 1, 0), ( 3, 0), -intercept: (0, 3) 2. -intercepts: ( 6, 0), (1, 0), -intercept: (0, 6) 3. -intercepts: ( 0.5, 0), (4, 0), -intercept: (0, 4) intercepts: ( 4, 0), ( 3, 0 ), -intercept: (0, 8) 5. -intercepts: ( 5 4, 0 5 ), ( 4, 0 ), -intercept: (0, 25 ) 6. -intercept: (2, 0), -intercept: (0, 12) 7. The formula starts with b; the negative sign is left off. = 3± Under the radical, 4ac should equal = 7± = 2( 1) ± 6 -intercepts: ( 2, 0) verte is (1, 3) = ( + 5) 2 6 ( ) -intercepts: 5 ± 6, 0 verte is ( 5, 6) = ( + 2) intercepts: (3, 0), ( 7, 0) verte is ( 2, 25) 12. = 2( + 6) ± 2 -intercepts:, 0 2 verte is ( 6, 1) ( ) Placing the start of the kick at the origin gives = 0.05( 100). 14. Let the -ais be the number of months, and the -ais be the number of sales in millions. Placing the origin at the beginning of sales, we can use = 1 6 ( 3) Placing the lowest point of the ramp at the origin gives = Parent Guide and Etra Practice 17

28 TRANSFORMING PARENT GRAPHS and Students will generalize what the have learned about transforming the graph of f () = 2 to change the shape and position of the graphs of several other functions. The students start with the simplest form of each function s graph, which is called the parent graph. Students use = 3, = 1, =, and = b as the equations for parent graphs, and what the learn while studing these graphs will appl to all functions. The also learn to appl their knowledge to non-functions. For further information see the Math Notes boes in Lessons and Eample 1 For each of the following equations, state the parent equation, and use it to graph each equation as a transformation of its parent equation. = ( + 4) 3 1 = 1 = 3 2 = 3 6 For each of these equations, we will graph both it and its parent on the same set of aes to help displa the change and movement. The first equation is a cubic (the term given to a polnomial with 3 as the highest power of ), thus its parent is = 3. The given equation will have the same shape as = 3, but it will be shifted to the left 4 units (from the + 4 within the parentheses), and down one unit (from the 1 ). The new graph is the darker curve shown on the graph at right. Notice that the point (0, 0) on the original graph has shifted left 4, and down 1, and now is at ( 4, 1). This point is known as a locator point. It is a ke point of the graph, and graphing its position helps us to graph the rest of the curve. The second curve, = 1, has had onl one change from the parent graph = 1 : the negative sign. Just as a negative at the front of f () = 2 would flip this graph upside down, the negative sign here flips the parts of the parent graph across the -ais. The lighter graph shown at right is the parent = 1, and the darker graph is = Core Connections Algebra 2

29 Chapter 2 In the equation = 3 2 the radical is multiplied b 3, hence the transformed graph will grow verticall more quickl than the parent graph =. It is also shifted to the right 2 units because of the 2 under the radical sign. The new graph is the darker curve on the graph shown at right. Notice that the point (0, 0) on the original graph (the locator point) has shifted right 2 units. This last graph is an eponential function. The parent graph, = b, changes in steepness as b changes. The larger b is, the quicker the graph rises, making a steeper graph. With = 3 6, the graph is a bit steeper than = 2, often thought of as the simplest eponential function, but also shifted down 6 units. The lighter graph is = 2, while the darker graph is = 3 6. Eample 2 Suppose f () is shown at right. From all ou have learned about changing the graphs of functions: a. Graph f () + 3. b. Graph f () 2. c. Graph f ( 1). d. Graph f ( + 3). e. Graph 3 f (). f. Graph 1 2 f (). Each time we alter the equation slightl, the graph is changed. Even though we have no idea what the equation of this function is, we can still shift it on the coordinate grid. Remember that f () represents the range or -values. Therefore, in part (a), f () + 3 sas the -values, plus 3. Adding three to all the -values will shift the graph up three units. This is shown at left. Notice that the shape of the graph is identical to the original, just shifted up three units. Check this b comparing the -intercepts. If f () + 3 shifts the graph up three units, then f () 2 will shift the graph down two units. This graph is shown at right. Again, compare the -intercept on the graph at left to the original. (Note: Using the -intercept or the -intercepts to help ou graph is an effective wa to create a graph.) Parent Guide and Etra Practice 19

30 What happens when the change is made within the parentheses as in parts (c) and (d)? Here the shift is with the -coordinates, thus the graph will move left or right. In part (c), the graph of f ( 1) The graph in part (d), f ( + 3), is shifted to the right 1 unit. is shifted to the left 3 units. When multipling f () b a number as in parts (e) and (f), look at some ke points. In particular, consider the -intercepts. Since the -value is zero at these points, multipling b an number will not change the -value. Therefore, the -intercepts do not change at all, but the -intercept will. In the original graph, the -intercept is 1, so f (0) = 1. The larger the constant b which ou multipl, the more stretched out the graph becomes. A smaller number flattens the graph. Multipling b 3 will raise that point Multipling b 1 2 changes three times as high, to the point (0, 3). the -intercept to ( 0, 1 2 ). Eample 3 Appl our knowledge of parent graphs and transformations to graph the following two nonfunctions. a. = b. ( 2) 2 + ( + 3) 2 = 36 Solution follows on net page 20 Core Connections Algebra 2

31 Chapter 2 Not ever equation is a function, and the two non-functions students consider are = 2 and = r 2. The first is the equation of a sleeping parabola, or a parabola ling on its side. The second equation is the general form of a circle with center (0, 0), and radius of length r. As written, neither of these equations can be entered into a graphing calculator. Students need to solve each of these as = to use the calculator. But rather than doing that, the students can use what the have alread learned to make accurate graphs of each equation. The parent of the equation in part (a) is = 2. The + 3 tells us the graph will shift 3 units, but is it up, down, left, or right? Rewriting the equation as ± 3 = helps us see that this graph is shifted to the right 3 units. At right, the gre curve is the graph of = 2, and the darker curve is the graph of = Graphing the equation of a circle is straightforward: a circle with center (h, k) and radius r has the equation ( h) 2 + ( + k) 2 = r 2. Therefore the graph of the equation in part (b) is a circle with a center at (2, 3), and a radius of 6. The graph of the circle is shown at right. Problems For each of the following equations, state the parent equation and then sketch its graph. Be sure to include an ke and/or locator points. 1. = ( 5) 2 2. = 1 3 ( + 4) ( 2) 2 + ( + 1) 2 = 9 4. = = = = ( 2) = = =± 9 For each of the following problems, state whether or not it is a function. If it is not a function, eplain wh not = 7 ± = 3( 4) 2 Parent Guide and Etra Practice 21

32 Answers 1. Parent graph f () = 2, verte (5, 0) 2. Parent graph f () = 2, verte ( 4, 7) 3. Parent graph ( h) 2 + ( k) 2 = r 2, center (2, 1), radius 3 4. Parent graph f () =, verte ( 5, 2) 5. Parent graph f () = 1, asmptotes = 1, = Parent graph f () = 2, asmptote = 8 7. Parent graph f () = 3, locator point (2, 1) 8. Parent graph f () =, verte ( 7, 0) 9. Parent graph f () =, verte (5, 0) 10. Parent graph 2 =, verte (9, 0) 11. Yes. 12. No, on the left part of the graph, for each -value there are two possible -values. You can see this b drawing a vertical line through the graph. If a vertical line passes through the graph more than once, it is not a function. 13. No, because the equation has ±, for each value substituted for, there will be two -values produced. A function can have onl one output for each input. 14. Yes. 22 Core Connections Algebra 2

33 Chapter 2 MORE ON COMPLETING THE SQUARE Although students can find the verte of a parabola b averaging the -intercepts, the also can use the algebraic method known as completing the square. This allows students to go directl from standard (or non-graphing) form to graphing form without the intermediate step of finding the -intercepts. Completing the square is also used when the equation of a circle is written in an epanded form. When the students first looked at how to complete the square, the used tiles so that the could see how the method works. When the tried to create a square (complete it) b arranging the tiles, there were either too man or missing parts. This visual representation helps students see how to rewrite the equation algebraicall. Eample 1 The function f () = is written in standard form. Complete the square to write it in graphing form. Then state the verte of the parabola and sketch the graph. The general equation of a parabola in graphing form is f () = a( h) 2 + k, where (h, k) is the verte. The original equation needs to be changed into a set of parentheses squared, with a constant either added to or subtracted from it. To do this, we must know that ( h) 2 = 2 2h + h 2. We will use this form of a perfect square to complete the square of the given equation or function. f () = = = = ( + 3) 2 6 The first bo holds a space for the number we have to add to complete the square. The second bo is to subtract that same number so as not to change the balance of the equation. To determine the missing number, take half the coefficient of (half of 6), and then square it and place the result in both boes. With the equation in graphing form, we know the verte is ( 3, 6). The graph is shown below. Parent Guide and Etra Practice 23

34 Eample 2 The equation = 41 is the equation of a circle. Complete the square to determine the coordinates of its center and the length of the radius. As with the last eample, we will fill in the blanks to create perfect squares. We need to do this twice: once for, and again for = = = 41 ( 4) ( + 8) 2 64 = 41 ( 4) 2 + ( + 8) 2 = ( 4) 2 + ( + 8) 2 = 121 This is a circle with center (4, 8) and a radius of 121 = 11. Problems Write each of the following equations in graphing form. Then state the verte and the direction the parabola opens. 1. = = = = = = Find the center and radius of each circle. 7. ( + 2) 2 + ( + 7) 2 = ( 9) 2 + 3( + 1) 2 = = = = = 496 Answers 1. = ( 4) 2 + 2, verte (4, 2), up 2. = ( + 1) 2 6, verte ( 1, 6), down 3. = 3( 4) 2 6, verte (4, 6), up 4. = 2( 0) 2 6, verte (0, 6), up 5. = 1 2 ( 3)2 4, verte (3, 4), up 6. = ( + 9) , verte ( 9, 16), up 7. Center: ( 2, 7), radius: 5 8. Center: (9, 1), radius: 2 9. Center: ( 3, 0), radius: Center: (5, 7), radius: Center: ( 25, 1), radius: 24 = Center: (4, 8), radius: Core Connections Algebra 2

35 Chapter 2 SAT PREP 1. If m is an integer, which of the following could not equal m 3? a. 27 b. 0 c. 1 d. 16 e If n is divided b 7 the remainder is 3. What is the remainder if 3n is divided b 7? a. 2 b. 3 c. 4 d. 5 e What is the slope of the line passing through the point ( 3, 1) and the origin? a. 3 b. 1 3 c. 0 d. 1 3 e If = and 3 = 7 4, what does equal? a. 5 b. 1 5 c d. 2 3 e A bag contains a number of marbles of which 35 are blue, 16 are red and the rest are ellow. If the probabilit of selecting a ellow marble from the bag at random is 1 4, how man ellow marbles are in the bag? a. 4 b. 17 c. 19 d. 41 e If n > 0 and k + 25 = (4 + n) 2 for all values of, what does k n equal? a. 0 b. 5 c. 35 d. 40 e A rectangular solid has two faces congruent to figure I at right and four faces congruent to figure II at right. What is the volume of the solid? I 5 II 8. In the figure at right, PQ = QR. What is the -coordinate of point Q? O 9. The time t, in hours, needed to produce u units of a product is given b the formula t = ku + c, where k and c are constants. If it takes 430 hours to produce 100 units and 840 hours to produce 200 units, what is the value of c? 10. In the figure at right, a square is inscribed in a circle. If the sides of the square measure 3 and the area of the circle is cπ, what is the eact value of c? P Q R (1, 3) (, ) (9, 3) Parent Guide and Etra Practice 25

36 Answers 1. D 2. A 3. D 4. C 5. B 6. C Core Connections Algebra 2

37 Chapter 3 EQUIVALENCE Solving equations is a skill that algebra students practice a great deal. In Algebra 2, the equations become increasingl more comple. Whenever possible, it is beneficial for students to rewrite equations in a simpler form, or as equations the alread know how to solve. This is done b recognizing equivalent epressions and developing algebraic strategies for demonstrating equivalence. Eample 1 Emma and Rueben have been given a sequence the have never seen before. It does not seem to be an arithmetic or a geometric sequence. n t(n) After a great deal of brain strain, Emma eclaims He! I see a pattern! If we look at the differences between the t(n) s, we can list the whole sequence! Rueben agrees, Oh, I see it! But what a pain to list it all out. We should be able to find a formula. Now that we see a pattern, Emma sas, Let s each spend some time thinking of a formula. After a few minutes, both Rueben and Emma have formulas. Wait a minute! sas Rueben. That s not the formula I got. M formula is t(n) = n 2 7n 30, but our formula is t(n) = (n + 3)(n 10). Which one of us is correct? a. Who is correct? Justif our answer completel. b. Later Tess, another team member, sas Ha! I have the right equation! It is ( ) t(n) = n Rueben comments You are reall off, Tess. That is nowhere near the right answer! Is Rueben correct, or has Tess found another equation? Justif our answer. We can show that both Rueben and Emma s equations produce the values in the table b substituting different values for n, but that would onl show that the are equivalent for those specific values. We must show that the two equations are equivalent algebraicall in order to verif that the are the same. To do this, we can use algebra to rewrite one equation, and hopefull get the other one. t(n) = (n + 3)(n 10) = n 2 10n + 3n 30 = n 2 7n 30 We started with Rueben s equation, and through algebraic manipulation, the result was Emma s equation. Therefore the equations are equivalent. Parent Guide and Etra Practice 27

38 Similarl, we can manipulate Tess equation to see if we can get either of the other two ( ) 2. equations. If we can, then it too is equivalent. Start b epanding n 7 2 ( ) ( ) n t(n) = n 7 2 = n = n 2 7n = n 2 7n 30 ( ) Tess has an equivalent equation as well. Therefore all three are equivalent equations. Eample 2 Solve the following equation b rewriting it as a simpler equivalent equation: = 1. This equation would be much simpler if it did not have an fractions, so multipl everthing b 32 to eliminate the denominators = 1 32 ( )= 32 ( 1) 32 ( ) 32 ( 1 8 )= = 32 To solve a quadratic equation, set it equal to zero, and solve b either factoring or using the Quadratic Formula. Since the equation appears to be easil factorable, use that method. 2 4 = = 0 8 ( )= 0 8 = 0, + 4 = 0 = 8, = 4 ( ) Core Connections Algebra 2

39 Chapter 3 Eample 3 Decide whether or not the following pairs of epressions or equations are equivalent for all values of the variables. Justif our answer completel. a. a + b and a + b 12 b. +4 and c. 12 and d = and = 0 In part (a), choose different values for a and b to check. For instance, if a = 1 and b = 2, then we would have = , and = Therefore the two epressions are not equal. (Note: These epressions are onl equal when both a and b are equal to zero.) Tr an value for in part (b), and the two epressions will not be equal. For eample, if = 1, then = = 12 5 and = = = Note that ou onl need to find one eample the does not work to demonstrate that the two epressions of equations are not equivalent. This strateg is known as a countereample. The epressions in part (c) demonstrate how we add fractions with common denominators: b adding the numerators. You ma wish to tr some values of in the two equations of part (d), but the equations are fairl mess. In addition, using a few values would not show that the equations are equivalent for all values of. It is easier to simplif the first equation to see if it results in the second equation = = = = 0 The result is the second equation. Therefore, these two equations are equivalent. Parent Guide and Etra Practice 29

40 Problems Rewrite the following epressions in a simpler form. 1. (3 2 ) 5 (4 3 ) z ( z) 2 0 Decide whether or not the following pairs of epressions or equations are equivalent for an values of the variables. Justif our answer completel = and + 3 = = 12 and = ( )(0.5 2) = 0 and = = 2 3 ( 2) and 3 = 2 3 ( 8) For each sequence below there are two equations. Decide whether or not the equations represent the sequence, and whether or not the equations are equivalent. Justif our answer. 8. n t(n) A: t(n) = 3n + 14 B: t(n) = 3(n 4) n t(n) A: t(n) = (2n + 1) 2 B: t(n) = 4 n A: t(n) = 8 ( ( ) 2 3 ) n ( ) n Simplif, and then solve the following equations. 9. n t(n) A: t(n) = 2 8 n ( ) n 1 B: t(n) = n t(n) 8 B: t(n) = = = 30 2 = = = Core Connections Algebra 2

41 Chapter 3 Answers No, in the first equation, = 1. The second equation has no solution. 5. No, (0, 3) works in the first equation but not in the second. The standard form of the second equation is 3 4 = Equivalent. Multipl the two binomials in the first equation go get = 0. Then multipl all of the terms b No, rewriting the equations in slope-intercept form produces different -intercepts. 8. A and B both represent the sequence, and are equivalent. 9. A and B both represent the sequence, and are equivalent. 10. A and B both represent the sequence, and are equivalent. 11. Onl B represents the sequence. The are not equivalent = 0, = 2, = (5, 5) 14. = = 0, 1 Parent Guide and Etra Practice 31

42 SIMPLIFYING EXPRESSIONS To simplif rational epressions, find factors in the numerator and denominator that are the same and then write them as fractions equal to 1. For eample, 6 6 = = 1 ( + 2) ( + 2) = 1 (3 2) (3 2) = 1 Notice that the last two eamples involved binomial sums and differences. Onl when sums or differences are eactl the same does the fraction equal 1. The rational epressions below cannot be simplified: (6 + 5) As shown in the eamples below, most problems that involve simplifing rational epressions will require that ou factor the numerator and denominator. Note that in all cases we assume the denominator does not equal zero, so in Eample 4 below the simplification is onl valid provided 6 or 2. For more information, see eamples 1 and 2 in the Math Notes bo in Lesson One other special situation is shown in the following eamples: 2 2 = 1 = 1 2 ( + 2) ( 5) Again assume the denominator does not equal zero. Eample = = 2 9 since 2 2 = 3 3 = 1 Eample 3 12( 1) 3 ( + 2) 3( 1) 2 ( + 2) 2 = 4 3( 1)2 ( 1)( + 2) 3( 1) 2 ( + 2)( + 2) Eample = = Eample ( 4)( 2) 2 = ( + 6)( 2) = 4( 1) ( + 2) since 3 ( 1)2, 3 ( 1) 2, and = 1 = ( 4) ( + 6) since ( 2) ( 2) = 1 32 Core Connections Algebra 2

43 Chapter 3 Problems Simplif each of the following epressions completel. Assume the denominator does not equal zero Answers ( ) ( ) ( )( 5) ( )( + 2) ( )( 6) ( )( 6 ) ( ) ( ) ( )( 4 ) ( )( 4) ( )( 16) ( )( 16 ) ( )( 2) ( )( + 2) ( )( 7 ) ( )( 7) ( + 3) 2 ( 2) 4 ( + 3) 4 ( 2) 3 (5 ) 2 ( 2) 2 ( + 5) 4 ( 2) (5 ) 4 (3 1) 2 ( 5) 4 (3 2) ( 7)( + 2) 4 20( 7) 2 ( + 2) (3 7)( +1) 6 20(3 7) 3 ( +1) ( +1)( 2) ( + 3) 2 2 ( ) 2( 5) 3( + 2) ( ) 2( + 6) ( 5 ) 2 ( + 5) 4 ( 2) ( 3) ( ) ( ) 5( + 3) ( ) 8( + 16) ( +1) 2 ( 2) ( + 3) ( ) ( ) 5( + 1) (3 1) 2 (3 2) ( +1) 5(3 7) ( + 5) ( 2) ( + 3) ( ) + 2 ( ) ( +1) Parent Guide and Etra Practice 33

44 MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS Multiplication or division of rational epressions follows the same procedure used with numerical fractions. However, it is often necessar to factor the polnomials in order to simplif them. When dividing rational epressions, change the problem to multiplication b inverting (flipping) the second epression (or an epression that follows a division sign) and completing the process as ou do for multiplication. As in the previous section, remember that simplification assumes that the denominator is not equal to zero. For addition information, see Eamples 3 and 4 in the Math Notes bo in Lesson Eample 1 Multipl ( + 6) After factoring, the epression becomes: and simplif the result. ( + 6) + 6 ( )( + 6) ( + 6)( + 1) ( + 1)( 1) After multipling, reorder the factors: Since ( + 6) + 6 ( ) = 1 and ( + 1) = 1, simplif: ( + 1) ( + 6) + 6 ( ) ( + 6) ( + 6) ( 1) ( + 1) + 1 ( ) for 6, 1, or 1 1 Eample 2 Divide and simplif the result. First change to a multiplication epression b inverting (flipping) the second fraction: After factoring, the epression is: Reorder the factors (if ou need to): Since ( 5) ( 5) = 1 and ( 2) ( 2) = 1, simplif: ( 5) ( + 1) 2 5 ( )( 2) ( + 6) ( 2) ( )( + 3) ( 5) ( 5) ( 2) ( 2) ( + 1) ( 2) ( + 6) ( + 3) ( + 1) ( 2) ( + 6) ( + 3) Thus, = +1 2 ( ) ( ) ( + 6) ( + 3) or for 3, 2, or Core Connections Algebra 2

45 Chapter 3 Problems Multipl or divide each pair of rational epressions. Simplif the result. Assume the denominator is not equal to zero ( 4) w Answers ( + 3) ( ) ( ) w ( + 3)( 6) Parent Guide and Etra Practice 35

46 ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS Addition and Subtraction of Rational Epressions uses the same process as simple numerical fractions. First, find a common denominator (if necessar). Second, convert the original fractions to equivalent ones with the common denominator. Third, add (or subtract) the new numerators over the common denominator. Finall, factor the numerator and denominator and reduce (if possible). For additional information, see the Math Notes bo in Lesson Note that these steps are onl valid provided that the denominator is not zero. Eample 1 The least common multiple of 2(n + 2) and n(n + 2) is 2n(n + 2). 3 2(n + 2) + 3 n(n + 2) To get a common denominator in the first fraction, 3 multipl the fraction b n =, a form of the number 1. n Multipl the second fraction b (n + 2) n n + 3 n(n + 2) 2 2 Multipl the numerator and denominator of each term. It ma be necessar to distribute the numerator. Add, factor, and simplif the result. (Note: n 0 or 2) = 3n 2n(n + 2) + 6 2n(n + 2) = 3n + 6 3(n + 2) 2n(n + 2) 2n(n + 2) 3 2n Eample 2 ( ) Eample ( )( 2) 4 ( 1) ( 2) 36 Core Connections Algebra 2

47 Chapter 3 Problems Add or subtract each epression and simplif the result. In each case assume the denominator does not equal zero ( + 2)( + 3) + 2 ( + 2)( + 3) b 2 + 2b b 2 + 2b a 5a 2 + a a 1 5a 2 + a 6 ( + 3) + 2 ( + 3) a a 2 + 6a 3 3a b a a 2 + 2a a 2 + 2a +1 a + 2b a + b + 2a + b a + b ( 7) Answers a ( + 2) = ( 4)( +1) b 3 b a ( 7)( + 7) = (5 + 6) ( 4)( + 4) ( 3)( 2) = a ( 7) = ( 4) = Parent Guide and Etra Practice 37

48 SAT PREP 1. If 1 < t < 0, which of the following statements must be true? a. t 3 < t < t 2 b. t 2 < t 3 < t c. t 2 < t < t 3 d. t < t 3 < t 2 e. t < t 2 < t 3 2. Without taking a single break, Mercedes hiked for 10 hours, up a mountain and back down b the same path. While hiking, she averaged 2 miles per hour on the wa up and 3 miles per hour on the wa down. How man miles was it from the base of the mountain to the top? a. 4 b. 6 c. 9 d. 12 e When a certain rectangle is folded in half, it forms two squares. If the perimeter of one of these squares is 28, what is the perimeter of the original rectangle? a. 30 b. 42 c. 49 d. 56 e. Cannot be determined from the information given. 4. A class of 50 girls and 60 bos sponsored a road rall race. If 60% of the girls and 50% of the bos participated in the road rall, what percent of the class participated in the road rall? a. 54.5% b. 55% c. 57.5% d. 88% e. 110% 5. The sum of four consecutive integers is s. In terms of s, which of the following is the smallest of these four integers? a. s 6 4 b. s 4 4 c. s 3 4 d. s 2 4 e. s 4 6. On a certain map, 30 miles is represented b one-half inch. On the same map, how man miles are represented b 2.5 inches? 7. How man of the first one hundred positive integers contain the digit 9? 8. The sum of n and n + 1 is greater than five but less than 15. If n is an integer, what is one possible value of n? 38 Core Connections Algebra 2

49 Chapter 3 9. In the figure at right, ΔABC is a right triangle and 6 = 6. What is the value of? 10 8 B 6 A D 10 C 10. For three numbers a, b, and c, the average (arithmetic mean) is twice the median. If a < b < c, a = 0, and c = kb, what is the value of k? Answers 1. D 2. D 3. B 4. A 5. A miles integers 8. n can be 3, 4, 5, or 6 9. = k = 5 Parent Guide and Etra Practice 39

50 SOLVING SYSTEMS OF EQUATIONS Students have been solving equations even before Algebra 1. Now the focus on what a solution means, both algebraicall and graphicall. B understanding the nature of solutions, students are able to solve equations in new and different was. Their understanding also provides opportunities to solve some challenging applications. In this section the will etend their knowledge about solving one and two variable equations to solve sstems with three variables. Eample 1 The graph of = ( 5) 2 4 is shown below right. Solve each of the following equations. Eplain how the graph can be used to solve the equations. a. ( 5) 2 4 = 12 b. ( 5) 2 4 = 3 c. ( 5) 2 = 4 Students can determine the correct answers through a variet of different was. For part (a), most students would do the following: ( 5) 2 4 = 12 ( 5) 2 = 16 5 =±4 = 5 ± 4 = 9, 1 This is correct and a standard procedure. However, with the graph of the parabola provided, the student can find the solution b inspecting the graph. Since we alread have the graph of = ( 5) 2 4, we can add the graph of = 12 which is a horizontal line. These two graphs cross at two points, and the -coordinates of these points are the solutions. The intersection points are (1, 12) and (9, 12). Therefore the solutions to the equation are = 1 and = 9. We can use this method for part (b) as well. Draw the graph of = 3 to find that the graphs intersect at (4, 3) and (6, 3). Therefore the solutions to part (b) are = 4 and = 6. The equation in part (c) might look as if we cannot solve it with the graph, but we can. Granted, this is eas to solve algebraicall: ( 5) 2 = 4 5 =±2 = 7, 3 But, we want students to be looking for alternative approaches to solving problems. B looking for and eploring alternative solutions, students are epanding their repertoire for solving problems. This ear the will encounter equations that can onl be solved through alternative methods. B recognizing the equation in part (c) as equivalent to ( 5) 2 4 = 0 (subtract four from both sides), we can use the graph to find where the parabola crosses the line = 0 (the -ais). The graph tells us the solutions are = 7 and = Core Connections Algebra 2

51 Chapter 4 Eample 2 Solve the equation + 2 = using at least two different approaches. Eplain our methods and the implications of the solution(s). Most students will begin b using algebra to solve this equation. This includes squaring both sides and solving a quadratic equation as shown at right. A problem arises if the students do not check their work. If the students substitute each -value back into the original equation, onl one -value checks: = 1 4. This is the onl solution. We can see wh the other solution does not work if we use a graph to solve the equation. The graphs of = + 2 and = are shown at right. Notice that the graphs onl intersect at one point, namel = 1. This is the onl point where the are 4 equal. The solution to the equation is this one value; the other value is called an etraneous solution. Remember that a solution is a value that makes the equation true. In our original equation, this would mean that both sides of the equation would be equal for certain values of. Using the graphs, the solution is the -value that has the same -value for both graphs, or the point(s) at which the graphs intersect. + 2 = ( + 2 ) 2 = (2 + 1) = = 0 (4 1)( + 1) = 0 = 1 4, 1 Eample 3 Solve each sstem of equations below without graphing. For each one, eplain what the solution (or lack thereof) tells ou about the graph of the sstem. a. = b. = 2( 2) = = c. = = 25 The two equations in part (a) are written in = form, which makes the Substitution method the most efficient method for solving. Since both epressions in are equal to, we set the epressions equal to each other, and solve for. = = = = ( )= ( ) = = 25 = 5 Parent Guide and Etra Practice 41

52 We substitute this value for back into either one of the original equations to determine the value of. Finall, we must check that the solution satisfies both equations. Check Solution to check: (5, 1) Therefore the solution is the point (5, 1), which means that the graphs of these two lines intersect in one point, the point (5, 1). The equations in part (b) are written in the same form so we will solve this sstem the same wa we did in part (a). We substitute each -value into either equation to find the corresponding -value. Here we will use the simpler equation. = 6, = = 2(6) + 15 = = 3 Solution: (6, 3) = 1, = = 2( 1) + 15 = = 17 Solution: ( 1, 17) Lastl, we need to check each point in both equations to make sure we do not have an etraneous solutions. = 2( 2) = ( 2) 2 = ( ) = = = = 0 ( 6)( +1) = 0 = 6, = 1 (6, 3) : = 2( 2) = 2(6 2) = 2(16) + 35 Check (6, 3) : = = 2(6)+15 Check ( 1, 17) : = 2( 2) = 2( 1 2) = 2(9)+ 35 Check ( 1, 17) : = = 2( 1) +15 Check In solving these two equations with two unknowns, we found two solutions, both of which check in the original equations. This means that the graphs of the equations, a parabola and a line, intersect in eactl two distinct points. 42 Core Connections Algebra 2

53 Chapter 4 Part (c) requires substitution to solve. We can replace in the second equation with what the first equation tells us it equals, but that will require us to solve an equation of degree four (an eponent of 4). Instead, we will first rewrite the first equation without fractions in an effort to simplif. We do this b multipling both sides of the equation b 6. = = = 2 Now we can replace the 2 in the second equation with Net we substitute this value back into either equation to find the corresponding -value = 25 (6 + 34) + 2 = = = 0 ( + 3)( + 3) = 0 = 3 = 3: = 2 6( 3) + 34 = = 2 16 = 2 =±4 This gives us two possible solutions: (4, 3) and ( 4, 3). Be sure to check these points for etraneous solutions! (4, 3) : = , 3 = 1 6 (4) = = 18 6, check. (4, 3) : = 25, (4) 2 + ( 3) 2 = = 25, check. ( 4, 3) : = , 3 = 1 6 ( 4) = = 18 6, check. ( 4, 3) : = 25, ( 4) 2 + ( 3) 2 = = 25, check. Since there are two points that make this sstem true, the graphs of this parabola and this circle intersect in onl two points, (4, 3) and ( 4, 3). Parent Guide and Etra Practice 43

54 Eample 4 Jo has small containers of lemonade and lime soda. She once mied one lemonade container with three containers of lime soda to make 17 ounces of a tast drink. Another time, she combined five containers of lemonade with si containers of lime soda to produce 58 ounces of another splendid beverage. Given this information, how man ounces are in each small container of lemonade and lime soda? We can solve this problem b using a sstem of equations. To start, we let equal the number of ounces of lemonade in each small container, and equal the number of ounces of lime soda in each of its small containers. We can write an equation that describes each miture Jo created. The first miture used one container of ounces of lemonade and three containers of ounces of lime soda to produce 17 ounces. This can be represented as = 17. The second miture used five containers of ounces of lemonade and si containers of ounces of lime soda to produce 58 ounces. This can be represented b the equation = 58. We can solve this sstem to find the values for and. + 3 = = = = (3) = = 17 = 8 9 = 27 = 3 (Note: ou should check these values!) Therefore each container of Jo s lemonade has 8 ounces, and each container of her lime soda has onl 3 ounces. 44 Core Connections Algebra 2

55 Chapter 4 Problems Solve each of the following sstems for and. Then eplain what the answer tells ou about the graphs of the equations. Be sure to check our work = 11 3 = = = = = = 24 = = = = = 24 The graph of = 1 2 ( 4)2 + 3 is shown at right. Use the graph to solve each of the following equations. Eplain how ou get our answer ( 4)2 + 3 = ( 4)2 + 3 = ( 4)2 + 3 = ( 4)2 = 8 Solve each equation below. Think about rewriting, looking inside, or undoing to simplif the process ( 4) = = ( )= = 10 Solve each of the following sstems of equations algebraicall. What does the solution tell ou about the graph of the sstem? 15. = = = ( + 1) = = 3( 4) 2 2 = = 0 = ( 4) Adult tickets for the Mr. Moose s Fantas Show on Ice are $6.50 while a child s ticket is onl $2.50. At Tuesda night s performance, 435 people were in attendance. The bo office brought in $ for that evening. How man of each tpe of ticket were sold? Parent Guide and Etra Practice 45

56 20. The net math test will contain 50 questions. Some will be worth three points while the rest will be worth si points. If the test is worth 195 points, how man three-point questions are there, and how man si-point questions are there? 21. Reread Eample 3 from Chapter 4 about Dudle s water balloon fight. If ou did this problem, ou found that Dudle s water balloons followed the path described b the equation = ( 10) Suppose Dudle s nemesis, in a mad dash to save his 5 base from total water balloon bombardment, ran to the wall and set up his launcher at its base. Dudle s nemesis launches his balloons to follow the path = ( ) in an effort to knock Dudle s water bombs out of the air. Is Dudle s nemesis successful? Eplain. Answers 1. (4, 7) 2. ( 2, 5) 3. no solution 4. ( 1 2,11 ) 5. All real numbers 6. (12, 3) 7. = 4. The horizontal line = 3 crosses the parabola in onl one point, at the verte. 9. No solution. The horizontal line = 1 does not cross the parabola. 8. = 2, = = 0, = 8. Add three to both sides to rewrite the equation as 1 2 ( 4)2 + 3 = 11. The horizontal line = 11 crosses at these two points. 11. = 7, = no solution 13. = no solution (A square root must have a positive result.) 15. All real numbers. When graphed, these equations give the same line. 17. No solution. This parabola and this line do not intersect adult tickets were sold, while 290 child tickets were sold. 21. B graphing we see that the nemesis balloon when launched at the base of the wall (the -ais), hits the path of the Dudle s water balloon. Therefore, if timed correctl, the nemesis is successful. 16. (0, 4). The parabola and the line intersect onl once. 18. (2, 2) and (5, 5). The line and the parabola intersect twice. 20. There are 35 three-point questions and 15 si-point questions on the test. 46 Core Connections Algebra 2

57 Chapter 4 INEQUALITIES Once the students understand the notion of a solution, the can etend their understanding to inequalities and sstems of inequalities. Inequalities tpicall have infinitel man solutions, and students learn was to represent such solutions. For further information see the Math Notes boes in Lessons 4.2.2, 4.2.3, and Eample 1 Solve each equation or inequalit below. Eplain what the solution for each one represents. Then eplain how the equation and inequalities are related to each other = < Students have man was to solve the equation, including graphing, factoring, or using the Quadratic Formula. Most students will factor and use the zero-product propert to solve as shown below = 0 ( 5)( + 1) = 0 = 5, = 1 Check: = 5 : (5) 2 4(5) 5 = = 0 = 1: ( 1) 2 4( 1) 5 = = 0 These are the onl two values for that make this equation true, = 5 and = 1. The second quadratic is an inequalit. To solve this we will utilize a number line to emphasize what the solution represents. When solving the equation, we found that the quadratic epression can be factored = ( 5)( + 1) Using the factored form, we want to find all -values so that ( 5)( + 1) < 0, or rephrasing it, where the product is negative. We begin b recording on a number line the places where the product equals zero. We found those two points in the previous part: = 5 and = 1. B placing these two points on the number line, the act as boundar points, dividing the number line into three sections. We choose an number in each of the sections to see if the number will make the inequalit true or false. Solutions will make the inequalit true. Note: We onl need to check one point from each section. As one point goes, so goes the section! To start, choose the point = 2. Substituting this into the inequalit gives: ( 2) 2 4( 2) 5 <? <? 0 7 <? False! Seven is NOT less than zero, so this section is NOT part of the solution. Parent Guide and Etra Practice 47

58 Net we choose a point in the middle section. An eas value to tr is = 0. True! This section is part of the solution. Finall, we check to see if an points in the last section make the inequalit true. Tr = 7. False! Therefore the solution is onl the middle section, the numbers that lie between 1 and 5. (0) 2 4(0) 5 <? <? 0 5 <? 0 (7) 2 4(7) 5? < ? < 0 16 <? 0 We can represent this in a couple of was. We can use smbols to write 1 < < 5. We can also represent the solution on the number line b shading the section of the number line that represents the solution of the inequalit. An point in the shaded section of the number line will make the inequalit true. The last inequalit of the eample has added a. We want to find all -values greater than or equal to the quadratic epression. Having both and means we need to use an -coordinate graph. The graph of the parabola at right divides the plane into two regions: the part within the bowl of the parabola the interior and the region outside the parabola. The points on the parabola represent where = We use a test point from one of the regions to check whether it will make the inequalit true or false. As before, we are looking for the true region. The point (0, 0) is an eas point to use. 0? (0) 2 4(0) 5 0? ? 5 True! Therefore the region containing the point (0, 0) is the solution. This means an point chosen in this region, the bowl of the parabola, will make the inequalit true. To illustrate that this region is the solution, we shade this region of the graph. Note: Since the inequalit was greater than or equal to, the parabola itself is included in the solution. If the inequalit had been strictl greater than, we would have made the curve dashed to illustrate that the parabola itself is not part of the solution. To see how these equations and inequalities are related, eamine the graph of the parabola. Where are the -values of the parabola negative? Where are the equal to zero? The graph is negative when it dips below the -ais, and this happens when is between 1 and 5. Solving the first inequalit told ou that as well. It equals zero at the points = 1 and = 5, which ou found b solving the equation. Therefore, if we had the graph initiall, we could have answered the first two parts quickl b looking at the graph. 48 Core Connections Algebra 2

59 Chapter 4 Eample 2 Han and Lea have been building jet roamers and pod racers. Each jet roamer requires one jet pack and three crstallic fuel tanks, while each pod racer requires two jet packs and four crstallic fuel tanks. Han and Lea s suppliers can onl produce 100 jet packs, and 270 fuel tanks each week, and due to manufacturing conditions, Han and Lea can build no more than 30 pod racers each week. Each jet roamer makes a profit of 1 tig (their form of currenc) while each pod racer makes a profit of four tigs. a. If Han and Lea receive an order for twelve jet roamers and twent-two pod racers, how man of each part will the need to fill this order? If the can fill this order, how man tigs will the make? b. Write a list of constraints, an inequalit for each constraint, and sketch a graph showing all inequalities with the points of intersection labeled. How man jet roamers and pod racers should Han and Lea build to maimize their profits? This problem is an eample of a linear programming problem, and although the name might conjure up images of computer programming, these problems are not done on a computer. We solve this problem b creating a sstem of inequalities that, when graphed, creates a feasibilit region. This region contains the solution for the number of jet roamers and pod racers Han and Lea should make to maimize their profit. We begin b defining the variables. Let represent the number of jet roamers Han and Lea will make, while represents the number of pod racers. We know that 0, and 0. A jet roamer requires one jet pack while a pod racer requires two. There are onl 100 jet packs available each week, so we can write as one of our inequalities. Each jet roamer requires three crstallic fuel tanks and each pod racer requires four. This translates into the inequalit since the have onl 270 fuel tanks available each week. Lastl, since Han and Lea cannot make more than 30 pod racers, we can write 30. To find the number of parts needed to fill the order in part (a), we can use these inequalities with = 12 and = 22. Jet packs: Crstallic fuel tank: 1(12) + 2(22) = = 56 3(12) + 4(22) = = 124 Each result is within the constraints, so it is possible for Han and Lea to fill this order. If the do, the will make 1(12) + 4(22) = = 100 tigs. Part (b) has us generalize this information to determine how Han and Lea can maimize their profits. Given their limited suppl of parts, should the use all of them to make jet roamers? The bring in more mone per vehicle. Or, should the make some combination of the two vehicle tpes to ensure the use their parts and still bring in as much mone as possible? We include all these inequalities on the graph of this sstem at right. The region common to all constraints is shaded. This is the feasibilit region because choosing a point in this shaded area gives ou a combination of jet roamers and pod racers that Han and Lea can produce under the given restraints. Jet Roamers Parent Guide and Etra Practice 49 Pod Racers

60 To maimize profits, we will test all the vertices of this region in our profit equation, profit = + 4, to find the greatest profit. These points are: (0, 0), (0, 30), (40, 30), (70, 15), and (90, 0). profit: + 4 (0,0): 0 + 4(0) = 0 (40, 30): (30) = 160 (0,30): 0 + 4(30) = 120 (70,15): (15) = 130 (90,0): (0) = 90 The greatest profit is 160 tigs when Han and Lea build 40 jet roamers and 30 pod racers. Problems Graph the following sstem of inequalities. Shade the solution region (the region containing the points that satisf all of the inequalities). 1. < > < < 10 + > 4 < 2 > < 12 > ( + 1) < 3 4 ( 1)2 + 6 > 7 6. < ( + 2) 3 > Write a sstem of inequalities that when graphed will produce these regions Core Connections Algebra 2

61 Chapter 4 9. Ramon and Thea are considering opening their own business. The wish to make and sell alien dolls the call Hauteans and Zotions. Each Hautean sells for $1.00 while each Zotion sells for $1.50. The can make up to 20 Hauteans and 40 Zotions, but no more than 50 dolls total. When Ramon and Thea go to cit hall to get a business license, the find there are a few more restrictions on their production. The number of Zotions (the more epensive item) can be no more than three times the number of Hauteans (the cheaper item). How man of each doll should Ramon and Thea make to maimize their profit? What will the profit be? 10. Sam and Emma are plant managers for the Stick Chew Cand Compan that specializes in delectable gourmet candies. Their two most popular candies are Chocolate Chews and Peanut and Jell Jimmies. Each batch of Chocolate Chews takes 1 teaspoon of vanilla while each batch of the Peanut and Jell Jimmies uses two teaspoons of vanilla. The have at most 20 teaspoons of vanilla on hand as the use onl the freshest of ingredients. The Chocolate Chews use two teaspoons of baking soda while the Peanut and Jell Jimmies use three teaspoons of baking soda. The onl have 36 teaspoons of baking soda on hand. Because of production restrictions, the can make no more than 15 batches of Chocolate Chews and no more than 7 batches of Peanut and Jell Jimmies. Sam and Emma have been given the task of determining how man batches of each cand the should produce if the make $3.00 profit for each batch of Chocolate Chews and $2.00 for each batch of Peanut and Jell Jimmies. Help them out b writing the inequalities described here, graphing the feasibilit region, and determining their maimum profit. Answers Parent Guide and Etra Practice 51

62 ( 6) The graph of the feasibilit region is shown at right. The inequalities are 0, 0, + 50, 20, 40, and 3, where is the number of Hauteans and is the number of Zotions. The profit is given b P = Maimum profit seems to occur at point A (12.5, 37.5), but there is a problem with this point. Ramon and Thea cannot make a half of a doll (or at least that does not seem possible). Tr these nearb points: (12, 37), (12, 38), (13, 37), and (13, 38). The point that gives maimum profit and is still in the feasibilit region is (13, 37). The should make 13 Hautean and 37 Zotion dolls for a profit of $ A The graph of the feasibilit region is shown at right. The inequalities are 0, 0, 7, 15, , and , where = number of Chocolate Chews and = number of Peanut and Jell Jimmies. The profit is given b P = The point that seems to give the maimum profit is (15, 2.5) but this onl works if half batches can be made. Instead, choose the point (15, 2) which means Sam and Emma should make 15 batches of Chocolate Chews and 2 batches of Peanut and Jell Jimmies. Their profit will be $ Core Connections Algebra 2

63 Chapter 4 SAT PREP 1. If = 4, then equals: 3 a. 3 b. 6 c. 8 d. 10 e What is the least of three consecutive integers whose sum is 21? a. 5 b. 6 c. 7 d. 8 e Juanita has stocks, bonds, and t-bills for investments. The number of t-bills she has is one more than the number of stocks, and the number of bonds is three times the number of t-bills. Which of the following could be the total number of investments? a. 16 b. 17 c. 18 d. 19 e Through how man degrees would the minute hand of a clock turn from 5:20 p.m. to 5:35 p.m. the same da? a. 15º b. 30º c. 45º d. 60º e. 90º 5. The length of a rectangle is si times its width. If the perimeter of the rectangle is 56, what is the width of the rectangle? a. 4 b. 7 c. 8.5 d. 18 e. 24 Y 6. If m > 1 and m n m 5 = m 15, then what does n equal? 40º b c 7. In the triangle at right, what is the value of a + b + c + d? X a d Z 8. If and are positive integers, + < 12, and > 4, what is the greatest possible value for? 9. If ( )(2 + 4) = a 3 + b 2 + c + d for all values of, what does c equal? 10. Four lines intersect in one point creating eight congruent adjacent angles. What is the measure of one of these angles? Parent Guide and Etra Practice 53

64 Answers 1. E 2. B 3. D 4. E 5. A º º 54 Core Connections Algebra 2

65 Chapter 5 INVERSES Students eplore inverses, that is, equations that undo the actions of functions. Although the ma not be aware of it, students have alread seen inverses without calling them that. When solving equations, students reverse operations to undo the equation leaving just, and that is what inverses do. Students further eplore composition of functions b considering what happens when inverses are combined. For further information see the Math Notes boes in Lessons and Eample 1 Find the inverse (undo) rule for the functions below. Use function notation and give the inverse rule a name different from the original function. a. f () = 6 3 b. g() = ( + 4)2 + 1 The function in part (a) subtracts 6 from the input then divides b 3. The undo rule, or inverse, reverses this process. Therefore, the inverse first multiplies b 3 then adds 6. If we call this inverse h(), we can write h() = The function g() adds 4 to the input, squares that value, then adds 1. The inverse will first subtract 1, take the square root then subtract 4. Calling this rule j() we can write j() =± 1 4. Rather than give the inverse a new name, we can use the notation for inverses. The inverse of f () is written as f 1 (). Note: The inverses h() and j(), are fundamentall different. h() = is the equation of a non-vertical line, therefore h() is a function. j() =± 1 4, however, is not a function. B taking the square root, we created a positive value and a negative value. This gives two outputs for each input, and so b definition it is not a function. Although we computed the inverses through a verbal description of what each function does, the students learn an algorithm for finding an inverse. The switch the and, and then solve for. Using this algorithm on the equations above: f () = 6 3 = 6 3 = = = g() = ( + 4) 2 +1 = ( + 4) 2 +1 = ( + 4) = ( + 4) 2 ± 1 = ± 1 = Parent Guide with Etra Practice 55

66 Eample 2 The graph of f () = is shown at right. Graph the inverse of this function. Following the algorithm for determining the equation for an inverse, as we did above, would be difficult here. The students do not have a method for solving cubic equations. Nevertheless, students can graph the inverse because the know a special propert about the graphs of functions and their inverses: the are smmetrical about the line =. If we add the line = to the graph, the inverse is the reflection across this line. Here we can fold the paper along the line =, and trace the result to create the reflection. Eample 3 Consider the function f () = 2 1. Determine the inverse of f () and label it g(). Verif that 7 these two functions are inverses b calculating f( g() ) and g( f() ). Using the algorithm, we can determine the inverse. = = = = 2 = g() = Composing the two functions f () and g() gives a method for checking whether or not functions are inverses of each other. Since one function undoes the other, when the functions are composed, the output should be. f () = 2 1 g() = f ( g() )= f 7+1 ( ( )= ) 1 g( f() )= g ( 2 1 )= = = ( ) +1 = = = 2 = Since f( g() )= g( f() )=, the functions are inverses. 56 Core Connections Algebra 2

67 Chapter 5 Problems Find the inverse of each of the following functions. 1. f () = 8( 13) 2. = = 5(+2) 3 4. f () = f () = g() = 5 7. g() = ( + 1) = ( + 2) 3 9. = g() = Sketch the graph of the inverse of each of the following functions. 11. = f () = g() = 14. = 1 5 For each of the following pairs of functions, determine f g() to decide whether or not f () and g() are inverses of each other. 16. f () = g() = f () = 8 g() = f () = + 5 g() = f () = 2 3 g() = 3 2 ( ) and g f() ( ), then use the result 20. f () = g() = 3( 6) f () = g() = ( 9 3 ) 2 Parent Guide with Etra Practice 57

68 Answers 1. = = 5 2. = = 1 ± = = =± 6 9. = ( 3) 2 + 4, for 3 5. = = = 6 12 = = f( g() )= g( f() )=. The are inverses. 17. f( g() )= g( f() )=. The are inverses. 18. f ( g() )= 1 + 5, g ( f() +5 )= 1. No, the are not inverses f ( g() )= 4 9, g( f() )= 1. No, the are not inverses. 20. f( g() )= g( f() )=. The are inverses. 21. f ( g() )= 3( 9)2 + 9, g f() 3 ( )= 2. No, the are not inverses. 58 Core Connections Algebra 2

69 Chapter 5 LOGARITHMS The earlier sections of this chapter gave students man opportunities to find the inverses of various functions. Here, students eplore the inverse of an eponential function. Although the can graph the inverse b reflecting the graph of an eponential function across the line =, the cannot write the equation of this new function. Writing the equation requires the introduction of a new function, the logarithm. Students eplore the properties and graphs of logarithms, and in a later chapter use them to solve equations of this tpe. For further information see the Math Notes bo in Lesson Eample 1 Find each of the values below and then justif our answer b writing the equivalent eponential form. a. log 5 (25) =? b. log 7 (?) = 3 c. log 1 2 ( 8 )=? A logarithm is reall just an eponent, so an epression like the one in part (a), log 5 (25), is asking What eponent can I raise the base 5 to, to get 25? We can translate this question into an equation: 5? = 25. B phrasing it this wa, the answer is more apparent: 2. This is true because 5 2 = 25. Part (b) can be rephrased as 7 3 =?. The answer is 343. Part (c) asks 2 to what eponent gives 1 8? or 2? = 1 8. The answer is 3 because 2 3 = = 1 8. Eample 2 The graph of = log() is shown at right. Use this parent graph to graph each of the following equations. Eplain how ou get our new graphs. = log( 4) = 6 log() + 3 = log() Parent Guide with Etra Practice 59

70 The logarithm function follows the same rules for transforming its graphs as other functions we have used. The first equation shifts the original graph to the right four units. The graph of the second equation is shifted up three units (because of the + 3 ) but is also stretched because it is multiplied b si. The third function is flipped across the -ais. All three of these graphs are shown at right. The original function = log() is also there, in light gra. Note: When a logarithm is written without a base, as in = log() and the log ke used on a calculator, the base is 10. Problems Rewrite each logarithmic equation as an eponential equation and vice versa. 1. = log 4 () 2. 3 = log 2 () 3. = log 5 (30) 4. 4 = ( 2 ) = = = log (32) = ( ) = = log 1 16 What is the value of in each equation below? If necessar, rewrite the epression in the equivalent eponential equation to verif our answer = log 5 () = log 9 () = log() = ( 3 ) = = = log 2 (32) = 19. log 11 () = log 1 5 ( 125 )= Graph each of the following equations. 21. = log( + 2) 22. = 5 + log() 23. = log( 4) 24. = log( 7 60 Core Connections Algebra 2

71 Chapter 5 Answers 1. 4 = = 3. 5 = log 4 (80) = 5. log 1/2 (64) = 6. log (343) = 3 7. log 1 5 ( 125 )= 8. = log 11 () = = = = = 1,000,000, = = = = = = = Parent Guide with Etra Practice 61

72 SAT PREP 1. In the figure at right, ΔDAI is isosceles with DI = 13 and base 24. If ΔVAI ΔDAI what is the area of quadrilateral DAVI? a. 60 b. 75 c. 120 d. 156 e. 240 V I A D 2. An eperimental jet flies at a speed of 5280 miles per hour. How man miles can this jet cover in 10 seconds? a b c d e If the angle (not shown) where a and b intersect is three times as large as the angle (not shown) where e and b intersect, what is the value of p? a. 70º b. 85º c. 140º d. 160º e. Cannot determine p a b º 90º e c d 4. Let ζ ζ be defined for all positive integer values of as the product of all even factors of 4. For eample, ζ 3ζ = = 576. What is the value of ζ 5ζ? a b c d e The chart at right shows the distribution of topics covered in a particular business tet, in chapters and pages per chapter. According to the chart, how man total pages are in this tet? Topic No. of Chapters No. of pages Development 3 12 Marketing 4 8 Public Relations 1 11 a. 31 b. 39 c. 48 d. 65 e Core Connections Algebra 2

73 Chapter 5 6. In the figure at right, what is the sum of and? Note: The figure is not drawn to scale. 7. If 2 q = 8 q 1, then q =? 120º 130º 120º 130º 8. If a is 40 percent of 300, b is 40 percent of a, and c is 25 percent of b, what is a + b + c? 9. If 4 = 11, what is the value of? If 3 5 of 1 3 is added to 5, what is the answer? Answers 1. C 2. D 3. C 4. A 5. E º 7. q = = Parent Guide with Etra Practice 63

74 SYSTEMS OF THREE EQUATIONS This chapter begins with students using technolog to eplore graphing in three dimensions. B using strategies that the used for graphing in two dimensions, students etend their skills to plotting points as well as graphing planes represented b equations with three variables. This leads to multiple planes intersecting, and using algebra to find either the equation of the line of the intersecting planes, or the point of intersection for the sstem of three equations and three unknowns. For more information, see the Math Notes boes in Lessons and Eample 1 Graph the 3-D point and equation below. (3, 4, 5) z = 24 Although we live in a three-dimensional world, visualizing threedimensional objects on a two-dimensional piece of paper can be difficult. In class, students used the computer to help them visualize the graphs. (You an access this software at home at Click on the desired item from Chapter 6.) Students begin b plotting points on aes as shown at right. As with plotting points in two dimensions, each number of the coordinate tells us how far to move along the -ais, then the -ais, and finall the z-ais. Here we are onl showing the positive direction for each ais; these aes etend in the negative direction as well. z To plot the point (3, 4, 5), we move along the -ais three units, along the direction of the -ais four units, and then five units along the z-ais. To help illustrate this, the point is marked with a circle. The path to the point is shown with a dotted line along the and directions, and a solid line to show the rise in the z direction. It might help students to think of the point as the corner on a bo, farthest from the origin. This imaginar bo is lightl shaded in above. To graph the equation with three variables on the three dimensional graph, it is helpful to find where it crosses each ais. We do this b letting the different variables equal zero, which allows us to find the -, -, and z-intercepts. z z = 12 = 0, = z = 12, z = 3 = 0, z = = 12, = 4 = 0, z = = 12, = 6 B plotting these intercepts we see how the plane slices through this quadrant of space. The shaded plane continues; it does not stop at the edges of the triangle. 64 Core Connections Algebra 2

75 Chapter 6 Eample 2 Solve the following sstem of three equations and three unknowns. Eplain what our solution sas about the graphs of each equation z = z = z = 7 Before beginning, it is helpful to recall how students solved two equations with two unknowns. If an equation was written in = form, students substituted the epression for into the other equation. That method will not work as easil with three equations. Other times with two equations and two unknowns, students would add or subtract two equations to make a variable disappear. Sometimes the needed to multipl an equation b some number before adding or subtracting. In either procedure, the goal was the same: to eliminate a variable z = z = z = 6 We will use the same approach here. B adding the first and third equation above, we eliminate the. The problem is, we still have two variables. The goal now is to find another pair of equations from which to eliminate. There are different was to do this. Here, we will multipl the second equation b two, then add the result to the third equation. 2( 3 + z) = 2( 21) z = z = z = 49 Now that we have two equations with two unknowns, we will use this simpler problem to solve for and z. 6 (2 + z = 6) z = 36 ( 5 + 6z = 49) 17 = 85 = 5 = z = 6 2(5)+ z = z = 6 z = 4 Now that we know what and z are, we can substitute them back into an one of our original equations to determine the value of. = 5, z = z = 21 3(5) + ( 4) = = 21 = 2 Therefore the solution to this sstem is ( 2, 5, 4) which tells us all three planes intersect in one point. Parent Guide with Etra Practice 65

76 Eample 3 Pizza Planet sells three sizes of combination pizzas. Small (8" diameter) $8.50 Medium (10" diameter) $11.50 Large (13" diameter) $17.50 Assume that the price of the pizza can be modeled with a quadratic function, with the price dependent on the diameter of the pizza. Use the information to write three data points, and determine an equation representing the data points. If Pizza Planet is considering selling an Etra Large combination pizza, with an 18" diameter, what should such a pizza cost? If the wanted to sell a combination pizza for $50.00, how big would it have to be to fit with the rest of the price data for the pizzas? If we let represent the diameter of the pizza in inches, and represent the cost of the pizza in dollars, the three data points are (8, 8.50), (10, 11.50), and (13, 17.50). We use these three points in the general equation for a quadratic, = a 2 + b + c. Our goal is to determine the appropriate values for a, b, and c so that the graph of the quadratic equation passes through the three data points. To be able to do this, we will need to solve three equations with three unknowns. First, we substitute the data points into the general equation. (8, 8.50) = a 2 + b + c 8.50 = a(8) 2 + b(8) + c 8.50 = 64a + 8b + c (10,11.50) = a 2 + b + c = a(10) 2 + b(10) + c = 100a + 10b + c (13,17.50) = a 2 + b + c = a(13) 2 + b(13) + c = 169a + 13b + c This gives us the three equations shown at right. (For further reference, the equations are numbered.) This is now similar to the last eample; we must solve three equations with three unknowns. Our unknowns here are a, b, and c, rather than,, and z = 64a + 8b + c (1) = 100a + 10b + c (2) = 169a + 13b + c (3) To begin we will eliminate c b subtracting pairs of equations. Equation (2) minus equation (1) gives 3 = 36a + 2b; equation (3) minus equation (2) gives 6 = 69a + 3b. Now we are back to the more familiar two equations with two unknowns. To solve them for a and b, we will multipl the first b 3 and the second b 2 then add the results. 3 = 36a + 2b ( 3) 9 = 108a 6b 6 = 69a + 3b 2 12 = 138a + 6b 3 = 30a a = Core Connections Algebra 2

77 Chapter 6 Now that we know the value of a, we substitute it back into the first (or second) equation to find the value of b. Lastl, we use the values for a and b to find c. We can use an of the three equations (1), (2), or (3). To keep it simple, we will use (1). a = 1 10, b = = 64a + 8b + c = 64 ( 1 10 )+ 8 ( 3 10 )+ c 8.50 = c 8.50 = 4 + c c = 4.50 a = = 36a + 2b 3 = 36 ( 10 1 )+ 2b 3 = b 0.6 = 2b b = 0.3 = 10 3 Note: An of the three original equations would have worked, and in fact, equation (2) would have eliminated the fractions and decimals from our work. Now that we have found a, b, and c, we can write the equation that models this data: = We use this equation to determine the cost of a combination pizza with 10 an 18-inch diameter. = 18 = = 1 10 (18)2 3 (18) = = $31.50 Therefore an 18-inch combination pizza should cost $ How large should a $50.00 pizza be to fit with this data? To answer this we must let = 50, and solve for. The solution will require solving a quadratic equation. Although the students know several was to solve quadratics, the best approach here is to use the Quadratic Formula. To begin, we multipl everthing b 10 to eliminate the fractions and the decimals. 50 = = = 0 = 3± ( 3)2 4(1)( 455) 2(1) 3± = 2 3± Therefore, to sell a pizza for $50.00, Pizza Planet should make the diameter of the pizza approimatel inches. A 23-inch diameter would surel suffice! Parent Guide with Etra Practice 67

78 Problems Solve each of the following sstems of equations for,, and z. Eplain what the answer tells ou about the graphs of the equations z = z = z = z = z = z = z = z = = z = z = z = z 2 = z 4 = z 3 = z = z = z = Find the equation of the parabola passing through the three points ( 2, 32), (0, 10), and (2, 12). 8. Find the equation of the parabola passing through the three points (2, 81), (7, 6), and (10, 33). 9. While reading a recent stud done on people of various ages, ou notice a trend. The stud counted the number of times in a 24 hour period that the people misunderstood or misinterpreted a statement, comment, or question. The stud offers the numbers shown in the table below. Misunderstandings or Age (ears) Misinterpretations You believe that the number of misunderstandings should reach a minimum at some age then go up again for ver old people. Therefore, ou assume that a quadratic function will best model this data. Find the equation that best fits this data. Use our equation to predict how man times an 80-ear old person will misunderstand or misinterpret a statement, comment, or question. What about a one-ear old? Who understands the most? 68 Core Connections Algebra 2

79 Chapter In archer, the arrow appears to travel in a straight line when it is released. However, the arrow will actuall travel upward slightl before curving back down toward the earth. For a particular archer, the arrow starts at 5.4 feet above the ground. After 0.3 seconds, the arrow is 5.5 feet above the ground. The arrow hits the target after a total of 2 seconds at a height of 5 feet above the ground. Find the particular equation that models this data. Answers 1. (3, 1, 4), these three planes intersect in a point. 2. (1, 5, 1), these three planes intersect in a point. 3. (3, 1, 2.5), these three planes intersect in a point. 4. No solution or inconsistent. Two of these planes are parallel. 5. (36, 24, 12), these three planes intersect in a point. 6. Infinitel man solutions. All three of these equations represent the same plane. 7. = = The equation that fits this data is = According to this model, an 80-ear old person would make 68 misunderstandings in a 24 hour period. A one ear old would make approimatel 96. The age that understands the most would be the age at which the number of misinterpretations is the lowest. This is at the verte of this function. The verte is at (45, 19) so 45 ear olds have the lowest number with onl 19 mistakes. 10. With rounding, = Parent Guide with Etra Practice 69

80 SOLVING WITH LOGARITHMS and Students turn their attention back to logarithms. Using Guess and Check, Pattern Recognition, and other problem solving strategies, students develop several properties of logs that enable them to solve equations that have been, until now, ver cumbersome to solve. These properties are listed in the Math Notes bo in Lesson Eample 1 Solve each of the following equations for. a. 5 = 67 b. 3(7 ) + 4 = 124 Each of these problems has the variable as the eponent, which makes them different from others that students have been solving. So far, students have been solving problems similar to these b Guess and Check. This approach has been time consuming and difficult to find an accurate answer. Now students can use the log propert, log(b ) = log(b), to solve these equations for. As with other equations, however, students must isolate the variable on one side of the equation. Note: The decimal answer is an approimation. The eact answer is the fraction log(67) log(5). 5 = 67 log(5 ) = log(67) log(5) = log(67) = log(67) log(5) Some work must be done to the second equation before we can incorporate logs. We will move everthing we can to one side of the equation so that the variable is as isolated as possible (steps 1 through 3). 3(7 ) + 4 = 124 3(7 ) = = 40 log(7 ) = log(40) log(7) = log(40) = log(40) log(7) Core Connections Algebra 2

81 Chapter 6 Eample 2 Using the properties of logs of products and quotients, rewrite each product as a sum, each quotient as a difference, and vice versa. a. log 3 (16) = b. log 6 (32) + log 6 (243) = c. log 3 8 ( 7 )= d. log 12 (276) log 12 (23) = The two properties we will use are log(ab) = log(a) + log(b) and log ( b a ) = log(a) log(b). These properties are true for an base, so we can use the first one to rewrite part (a) as log 3 (16) = log 3 (). This new form is not necessaril better or simpler, it is just another wa to represent the epression. In part (b), we can use the first propert to write log 6 (32) + log 6 (243) = log 6 (32 243) = log 6 (7776). Although it is not necessar, this can be simplified further. Since 6 5 = 7776, log 6 (7776) = 5. We will rewrite parts (c) and (d) using the second propert listed above. Therefore, log 3 8 ( 7 )= log 8 (3) log 8 (7). Note: We could use the first propert to epand this further b writing log 8 (3) as log 8 (3) + log 8 (). Working in the opposite direction on part (d), we write log 12 (276) log 12 (23) = log ( 23 ). Simplifing further, log ( 23 ) = log 12 (12) = 1. Eample 3 Fall came earl in Pine Orchard, and the communit swimming pool was still full when the first frost froze the leaves. The outside temperature hovered at 30º. Maintenance quickl turned off the heat so that energ would not be wasted heating a pool that nobod would be swimming in for at least si months. As Tess walked b the pool each da on her wa to school, she would peer through the fence at the slowl cooling pool. She could just make out the thermometer across the deck that displaed the water s temperature. On the first da, she noted that the water temperature was 68º. Four das later, the temperature reading was 58º. Write an equation that models this data. If the outside temperature remains at 30º, and the pool is allowed to cool, how long before it freezes? Heating and cooling problems are tpical application problems that use eponential equations. In class, students solved such a problem, The Case of the Cooling Corpse. The equation that will model this problem is an eponential equation of the form = km + b. In the problem description, we are given two data points: (0, 68º) and (4, 58º). We also have another piece of important information. The outside temperature is hovering at 30. This is the temperature the water will approach, that is, = 30 is the horizontal asmptote for this equation. Knowing this fact allows us to write the equation as = km + b. To determine k and m, we will substitute our values into the equation and solve for k and m. Parent Guide with Etra Practice 71

82 (0, 68) = km = km (4, 58) = km = km This gives us two equations with two unknowns that we can solve. Simplifing first makes our work a lot easier. The first equation simplifies to 38 = k since m 0 = 1. Since k = 38 we can substitute this value into the second equation to determine m. 58 = km = 38m = 38m 4 m 4 = m Therefore the equation is = 38(0.9265) To determine when the pool will freeze, we want to find when the water s temperature reaches 32º. 32 = 38(0.9265) = 38(0.9265) 2 38 = ( )= log( ) ( )= log(0.9265) log 2 38 log 2 38 ( ) = log 38 2 log(0.9265) In approimatel 38.5 das, the water in the pool will freeze if the outside temperature remains at 30º for those das. In realit, the pool would be drained to prevent damage from freezing. 72 Core Connections Algebra 2

83 Chapter 6 Problems Solve each of the following equations for. 1. (2.3) = = 6 3. log 7 (49) = 4. log 3 () = (3.14) = = log (100) = 4 8. log 5 (45) = 9. 2(6.5) + 7 = (14) + 6 = 9.1 Rewrite each log of a product as a sum of logs, each difference of logs as a log of a quotient, and vice versa. ( ) 11. log(23 3) 12. log log 60 2 ( 7 ) 14. log 8 (12) log 8 (2) 15. log 5 (25) + log 5 (25) 16. log(10 10) 17. log 13 (15 2 ) 18. log(123) + log(456) 19. log(10 8 ) log(10 3 ) 20. log(5 4) Simplif. 21. log 2 (64) 22. log 17 (17 1/8 ) log 8 (1.3) log 2.3 (1) 25. Climbing Mt. Everest is not an eas task! Not onl is it a difficult hike, but the Earth s atmosphere decreases eponentiall as ou climb above the Earth s surface, and this makes it harder to breathe. The air pressure at the Earth s surface (sea level) is approimatel 14.7 pounds per square inch (or 14.7 psi). In Denver, Colorado, elevation 5280 feet, the air pressure is approimatel psi. Write the particular equation representing this data epressing air pressure as a function of altitude. What is the air pressure in Meico Cit, elevation 7300 feet? At the top of Mt. Everest, elevation 29,000 feet? (Note: You will need to carr out the decimal values several places to get an accurate equation and air pressures.) Parent Guide with Etra Practice 73

84 Answers 1. = log(7) log(6) log(2.3) = log(12) = 2 4. = = log(3.6) log(3.14) = log(45) log(5) = log(7) log(6.5) = log(30.2) log(14) log(23) + log(3) 12. log(3) log(8) 13. log 2 (60) log 2 (7) 14. log 12 8 ( 2 )= log 8 (6) 15. log 5 (625) 16. log(10) + log(10) 17. log 13 (15) + log 13 ( 2 ) 18. log(56,088) 19. log ( ) = log Alread simplified The particular equation is = 1.47( ) where is the elevation, and is the number of pounds per square inch (psi). The air pressure in Meico Cit is approimatel 11.3 psi, and at the top of Mt. Everest, the air pressure is approimatel psi. 74 Core Connections Algebra 2

85 Chapter 6 SAT PREP 1. If b + 4 = 11, then (b 2) 2 = a. 16 b. 25 c. 36 d. 49 e Let P and Q represent digits in the addition problem shown at right. What must the digit Q be? 25P +P4 32Q a. 0 b. 1 c. 2 d. 3 e If 3 4 = 9, then = a. 2 b. 3 c. 5 d. 8 e When a positive number n is divided b 7 the remainder is 6. Which of the following epressions will ield a remainder of 1 when it is divided b 7? a. n + 1 b. n + 2 c. n + 3 d. n + 4 e. n How man 4-digit numbers have the thousands digit equal to 2 and the units digit equal to 7? a. 100 b. 199 c. 200 d. 500 e In the figure at right, where < 6, what is the value of ? a. 10 b. 50 c. 100 d. 600 e The measures of the angles of a triangle in degrees can be epressed b the ratio 5:6:7. What is the sum of the measures of the two larger angles? a. 110 b. 120 c. 130 d. 160 e If r 3 = 7, what is the value of r? 10 Parent Guide with Etra Practice 75

86 9. If p and q are two different prime numbers greater than 2, and n = pq, how man positive factors, including 1 and n, does n have? 10. If 1 2 ( )= a 3 + b 2 + c + d, for all values of where a, b, c, and d are all constants, what is the value of a + b + c + d? Answers 1. B 2. A 3. A 4. B 5. A 6. B 7. C 8. r = Core Connections Algebra 2

87 Chapter 7 TRIGONOMETRIC FUNCTIONS This chapter etends the students knowledge of trigonometr. Students have alread studied right triangle trigonometr, using sine, cosine and tangent with their calculators to find the lengths of unknown sides of triangles. Now students eplore these same three trigonometr terms as functions. The are introduced to the unit circle, and the eplore how the trigonometric functions are found within the unit circle. In addition, the learn a new wa to measure angles using radian measure. For further information see the Math Notes boes in Lessons 7.1.2, 7.1.5, 7.1.6, and Eample 1 As Daring Davis stands in line waiting to ride the huge Ferris wheel, he notices that this Ferris wheel is not like an of the Board here others he has ridden. First, this Ferris wheel does not board the passengers at the lowest point of the ride; rather, the board after climbing several flights of stairs, at the level of that wheel s horizontal ais. Also, if Davis thinks of the boarding point as a height of zero above that ais, then the maimum height above the boarding point that a person rides is 25 feet, and the minimum height below the boarding point is 25 feet. Use this information to create a graph that shows how a passenger s height on the Ferris wheel depends on the number of degrees of rotation from the boarding point of the Ferris wheel. As the Ferris wheel rotates counterclockwise, a passenger s height above the horizontal ais increases, and reaches its maimum of 25 feet above the ais after 90º of rotation. Then the passenger s height decreases as measured from the horizontal ais, reaching zero feet after 180º of rotation, and continues to decrease as measured from the horizontal ais. The minimum height, 25 feet, occurs when the passenger has rotated 270º. After rotating 360º, the passenger is back where he started, and the ride continues To create this graph, we calculate the height of the passenger at various points along the rotation. These heights are shown using the gre line segments drawn from the passenger s location on the wheel perpendicular to the horizontal ais of the Ferris wheel. Note: Some of these values are easil filled in. At 0º, the height above the ais is zero feet. At 90º, the height is 25 feet. Rotation, Degrees 0º 30º 45º 60º 90º 135º 180º 210º 225º 270º 315º 360º Height, Feet To complete the rest of the table we calculate the heights using right triangle trigonometr. We will demonstrate three of these values, 30º, 135º, and 225º, and allow ou to verif the rest. Parent Guide with Etra Practice 77

88 Each of these calculations involves focusing on the portion of the picture that makes a right triangle. For the 30º point, we look at the right triangle with a hpotenuse of 25 feet. (The radius of the circle is 25 feet because it is the maimum and minimum height the passenger reaches.) In this right triangle, we can use the sine function: 25 ft 30º h At the 135º mark, we use the right triangle on the outside of the curve. Since the angles are supplementar, the angle we use measures 45º. h 25 ft 45º 135º At 225º (225 = ), the triangle we use drops below the horizontal ais. We will use the 45º angle that is within the right triangle, so h 17.68, using the previous calculation and changing the sign to represent that the rider is below the starting point. Now we can fill in all the values of the table. h 180º 45º 25 ft Rotation, Degrees 0º 30º 45º 60º 90º 135º 180º 210º 225º 270º 315º 360º Height, Feet Plot these points and connect them with a smooth curve; our graph should look like the one at right. Note: This curve shows two revolutions of the Ferris wheel. This curve continues, repeating the ccle for each revolution of the Ferris wheel. It also represents a particular sine function: = 25sin() Core Connections Algebra 2

89 Chapter 7 Eample 2 On a unit circle, represent and then calculate cos(60º), cos(150º), and cos(315º). Then graph = cos(). The trigonometric functions ( trig functions) arise naturall in circles as we saw with the first eample. The simplest circle is a unit circle, that is, a circle of radius 1 unit, and it is this circle we often use with the trig functions. On the unit circle at right, several points are labeled. Point P corresponds to a 60º rotation, point Q corresponds to 150º, and Q 30º 1 60º 45º P R R corresponds to 315º. We measure rotations from the point (1, 0) counter-clockwise to determine the angle. If we create right triangles at each of these points, we can use the right triangle trig we learned in geometr to determine the lengths of the legs of the triangle. In the previous eample, the height of the triangle was found using the sine. Here, the cosine will give us the length of the other leg of the triangle. 1 60º cos(60º) 1 30º cos(30º) cos(45º) 45º 1 To full understand wh the length of the horizontal leg is labeled with cosine, consider the triangle below. In the first triangle, if we labeled the short leg, we would write: ( 1, 0) Q 30º (0, 1) (0, 1) 1 60º 45º P R (1, 0) 1 60º Therefore the length of the horizontal leg of the first triangle is cos(60º). Note: The second triangle representing 150º, lies in the second quadrant where the -values are negative. Therefore cos(150º) = cos(30º). Check this on our calculator. 1 θ P It is important to note what this means. On a unit circle, we can find a point P b rotating θ degrees. If we create a right triangle b dropping a height from point P to the -ais, the length of this height is alwas sin(θ ). The length of the horizontal leg is alwas cos(θ ). Additionall, this means that the coordinates of point P are (cos(θ ), sin(θ )). This is the power of using a unit circle: the coordinates of an point on the circle are found b taking the sine and cosine of the angle. The graph at right shows the cosine curve for two rotations around the unit circle Parent Guide with Etra Practice 79

90 Eample 3 On a unit circle, find the points corresponding to the following radians. Then convert each angle given in radians to degrees. a. π 6 b. 11π 12 c. 5π 4 d. 5π 3 One radian is about 57, but that is not the wa to remember how to convert from degrees to radians. Instead, think of the unit circle, and remember that one rotation would be the same as traveling around the unit circle one circumference. The circumference of the unit circle is C = 2πr = 2π(1) = 2π. Therefore, one rotation around the circle, 360º, is the same as traveling 2π radians around the circle. Radians do not just appl to unit circles. A circle with an size radius still has 2π radians in a 360º rotation. We can place these points around the unit circle in appropriate places without converting them. First, remember that 2π radians is the same point as a 360º rotation. That makes half of that, 180º, corresponds to π radians. Half of that, 90º, is 2 π radians. With similar reasoning, 270 corresponds to 3π 2 radians. Using what we know about fractions allows us to place the other radian measures around the circle. For eample, 6 π is one-sith the distance to π. π Sometimes we want to be able to convert from radians to degrees and back. To do so, we can use a ratio of degrees radians. To convert 6 π radians to degrees we create a ratio, and solve for. We will use π 2π as a simpler form of Therefore 6 π radians is equivalent to 30º. Similarl, we can convert the other angles above to degrees: π 180º = π /6 ( ) π = π π = 30π = 30º π 180º = 11π /12 ( ) π = 180º 11π 12 π = 165º π = 165º π 180º = 5π /4 ( ) π = 180º 5π 4 π = 225º π = 225º π 180º = 5π /3 ( ) π = 180º 5π 3 π = 300º π = 300º 80 Core Connections Algebra 2

91 Chapter 7 Eample 4 Graph T(θ ) = tan(θ ). Eplain what happens at the points θ = 2 π, 3π 2, 5π 2, 7π 2 happen?,.... Wh does this As with the graphs of S(θ ) = sin(θ ) and C(θ ) = cos(θ ), T(θ ) = tan(θ ) repeats, that is, it is cclic. The graph does not, however, have the familiar hills and valles the other two trig functions displa. This graph, shown at right, resembles in part the graph of a cubic such as f () = 3. However, it is not a cubic, which is clear from the fact that it has asmptotes and repeats. At θ = 2 π, the graph approaches a vertical asmptote. This also occurs at θ = 2 π, and because the graph is cclic, it happens repeatedl at θ = 3π 2, 5π 2, 7π 2,. In fact, it happens at all values of θ of the form (2k 1)π 2 for all integer values of k (odd values). The real question is, wh does this asmptote occur at these points? Recall that tan(θ ) = sin(θ ) cos(θ ). Ever point where cos(θ ) = 0, this function is undefined (we cannot have zero in the denominator). So at each point where cos(θ ) = 0, the function T(θ ) = tan(θ ) is also undefined. Eamining the graph of C(θ ) = cos(θ ), we can see that this graph is zero (crosses the -ais) at the same tpe of points as above: (2k 1)π 2 for all integer values of k. Parent Guide with Etra Practice 81

92 Problems Graph each of the following trig equations. 1. = sin() 2. = cos() 3. = tan() Find each of the following values without using a calculator, but b using what ou know about right triangle trigonometr, the unit circle, and special right triangles. 4. cos(180º) 5. sin(360º) 6. tan(45º) 7. cos( 90º) 8. sin(150º) 9. tan(240º) Convert each of the angle measures º to radians º to radians º to radians 13. π 15 radians to degrees π 8 radians to degrees 15. 3π 4 radians to degrees Answers π 3 radians π 18 radians 12. 7π 4 radians º º º 82 Core Connections Algebra 2

93 Chapter 7 TRANSFORMING TRIG FUNCTIONS Students appl their knowledge of transforming parent graphs to the trigonometric functions. The will generate general equations for the famil of sine, cosine and tangent functions, and learn about a new propert specific to cclic functions called the period. The Math Notes bo in Lesson illustrates the different transformations of these functions. Eample 1 For each of the following equations, state the amplitude, number of ccles in 2π, horizontal shift, and vertical shift of the graph. Then graph each equation on a separate set aes. = 3cos 2( 3 π ) 2 = sin 1 4 ( + π ) +1 The general form of a sine function is = a sin[b( h)] + k. Some of the transformations of trig functions are standard ones that students learned in Chapter 2. The a will determine the orientation, in this case, whether it is in the standard form, or if it has been reflected across the -ais. With trigonometric functions, a also represents the amplitude of the function: half of the distance the function stretches from the maimum and minimum points verticall. As before, h is the horizontal shift, and k is the vertical shift. This leaves just b, which tells us about the period of the function. The graphs of = sin(θ ) and = cos(θ ) each have a period of 2π, which means that one ccle (before it repeats) has a length of 2π. However, b affects this length since b tells us the number of ccles that occur in the length 2π. The first function, then, has an amplitude of 3, and since this is positive, it is not reflected across the -ais. The graph is shifted horizontall to the right 3 π units, and shifted down (verticall) 2 units. The 2 before the parentheses tells us it does two ccles in 2π units. If the graph does two ccles in 2π units, then the length of the period is π units. The graph of this function is shown at right. = 3cos 2( 3 π ) 2 The second function has an amplitude of 1, but it is reflected across the -ais. It is shifted to the left π units, and shifted up 1 unit. Here we see that within a 2π span, onl one fourth of a ccle appears. This means the period is four times as long as normal, or is 8π. The graph is shown at right. = sin 1 4 ( + π ) +1 Parent Guide with Etra Practice 83

94 Eample 2 For the Fourth of Jul parade, Vicki decorated her triccle with streamers and balloons. She stuck one balloon on the outside rim of one of her back tires. The balloon starts at ground level. As she rides, the height of balloon rises up and down, sinusoidall (that is, a sine curve). The diameter of her tire is 10 inches. a. Sketch a graph showing the height of the balloon above the ground as Vicki rolls along. b. What is the period of this graph? c. Write the equation of this function. d. Use our equation to predict the height of the balloon after Vicki has traveled 42 inches. This problem is similar to the Ferris Wheel eample at the beginning of this chapter. The balloon is rising up and down just as a sine or cosine curve rises up and down. A simple sketch is shown at right. The balloon begins net to the ground and as the triccle wheel rolls, the balloon rises to the top of the wheel, then comes back down. If we let the ground represent the -ais, the balloon is at its highest point when it is at the top of the wheel, a distance of one wheel's height or diameter, 10 inches. So now we know that the distance from the highest point to the lowest point is 10. The amplitude is half of this distance, 5. To determine the period, we need to think about the problem. The balloon starts at ground level, rises as the wheel rolls and comes down again to the ground. What has happened when the balloon returns to the ground? The wheel has made one complete revolution. How far has the wheel traveled then? It has traveled the distance of one circumference. The circumference of a circle with diameter 10 inches is 10π inches. Therefore the period of this graph is 10π. To get the equation for this graph we need to make some decisions. The graphs of sine and cosine are similar. In fact, one is just the other shifted 90º (or 2 π radians). At this point, we need to decide if we want to use sine or cosine to model this data. Either one will work but the answers will look different. Since the graph starts at the lowest point and not in the middle, this suggests that we use cosine. (Yes, cosine starts at the highest point but we can multipl b a negative to flip the graph over and start at the lowest point.) We also know the amplitude is 5 and there is no horizontal shift. All of this information can be written in the equation as = 5cos(b) + k. We can determine k b remembering that we set the -ais as the ground. This implies the graph is shifted up 5 units. To determine the number of ccles in 2π (that is, b), recall that we found that the period of this graph is 10π. Therefore 10 2π = 1 5 of the curve appears within the 2π span. Finall, pulling everthing together we can write = 5cos , and is shown in the following graph. 84 Core Connections Algebra 2

95 Chapter 7 = 5cos Note: If ou decided to use the sine function for this data, ou must realize that the graph is shifted to the right 10π 4 units. One equation that gives this graph is = 5sin 1 5 ( 10π 4 ) + 5. There are other equations that work, so if ou do not get the same equation as shown here, graph ours and compare. To find the height of the balloon after Vicki rides 42 inches, we substitute 42 for in the equation. If ou do not get this answer, make sure our calculator is in radians! = 5cos cos(8.4) inches Problems Eamine each graph below. For each one, draw a sketch of one ccle, then give the amplitude and the period Parent Guide with Etra Practice 85

96 For each equation listed below, state the amplitude and period. 5. = 2cos(3) = 1 2 sin() 6 7. f () = 3sin(4) 8. = sin f () = cos() + 2π 10. f () = 5cos( 1) 1 4 Below are the graphs of = sin() and = cos(). = sin() = cos() Use them to sketch the graphs of each of the following equations and functions b hand. Use our graphing calculator to check our answer. 11. = 2sin( + π ) 12. f () = 1 2 sin(3) 13. f () = cos ( 4 ( π 4 )) 14. = 3cos( + π 4 ) f () = 7sin( 1 4 ) A wooden water wheel is net to an old stone mill. The water wheel makes ten revolutions ever minute, dips down two feet below the surface of the water, and at its highest point is 18 feet above the water. A snail attaches to the edge of the wheel when the wheel is at its lowest point and rides the wheel as it goes round and around. As time passes, the snail rises up and down, and in fact, the height of the snail above the surface of the water varies sinusoidall with time. Use this information to write the particular equation that gives the height of the snail over time. 17. To keep bab Cristina entertained, her mother often puts her in a Johnn Jump Up. It is a seat on the end of a strong spring that attaches in a doorwa. When Mom puts Cristina in, she notices that the seat drops to just 8 inches above the floor. One and a half seconds later (1.5 seconds), the seat reaches its highest point of 20 inches above the ground. The seat continues to bounce up and down as time passes. Use this information to write the particular equation that gives the height of bab Cristina s Johnn Jump Up seat over time. (Note: You can start the graph at the point where the seat is at its lowest.) 86 Core Connections Algebra 2

97 Chapter 7 Answers 1. Amplitude is 2, period is π. 2. Amplitude 0.5, period 2π. 3. The graph shows one ccle alread. Amplitude is 3 and period is 4π. 4. Amplitude is 2.5, period is π Amplitude: 2, period: 2π Amplitude: 1 2, period: 2π. 7. Amplitude: 3, period: π Amplitude: 1, period: 6π. 9. Amplitude: 1, period: 2π. 10. Amplitude: 5, period: 2π Surprised? The negative flips it over, but the + π shifts it right back to how it looks originall! Parent Guide with Etra Practice 87

98 = 10 cos 1 ( 10 )+ 8, and there are other possible equations which will work. 17. = 6cos( 2π 3 ) works if we let the graph be smmetric about the -ais. The -ais does not have to represent the ground. If ou let the -ais represent the ground, ou equation might look like = 6cos( 2π 3 ) Core Connections Algebra 2

99 Chapter 7 SAT PREP 1. If one pentaminute is the same as five minutes of time, how man pentaminutes are equivalent to four hours of time? a b. 240 c. 60 d. 48 e If a = 12 and b = 4, what is the value of 4a 3b? a. 60 b. 36 c. 16 d. 9 e The average (arithmetic mean) of 4 and s is equal to the average of 3, 8 and s. What does s equal? a. 3 b. 5.5 c. 9 d. 10 e. No such s eists. 4. In the figure at right, AB = CD. What does k equal? C( 3, 6) a. 6 b. 5 c. 4 d. 3 e. 2 A( 8, 4) B(2, 4) 5. The initial term of a sequence is 36. Each term after that is half of the term before it if that term is even. If the preceding term is odd, the net term is one half that term, plus one half. What is the sith term of this sequence? D( 3, k) a. 1 b c. 2 d. 3.5 e At a spa, the customer is offered a choice of five different massages and eight different pedicures. How man different combinations are there of one massage and one pedicure? a. 3 b. 13 c. 16 d. 28 e A rectangular bo is 12 cm long, 20 cm wide, and 15 cm high. If eactl 60 smaller identical rectangular boes can be stored perfectl in this larger bo, which of the following could be the dimensions, in cm, of these smaller boes? a. 5 b 6 b 12 b. 4 b 5 b 6 c. 3 b 5 b 6 d. 3 b 4 b 6 e. 2 b 5 b 6 Parent Guide with Etra Practice 89

100 8. When Harr returned his book to the librar, Madam Pince told him he owed a fine of $6.45. This included $3.00 for three weeks, plus a fine of $0.15 per da for ever da he was late in returning the book. How man overdue das did Harr have the book? 9. What is the slope of the line that passes through the points (0, 2) and ( 10, 2)? 10. At right is the complete graph of the function f(). For how man positive values of does f() = 3? 1 Answers 1. D 2. A 3. D 4. C 5. C 6. E 7. E das Core Connections Algebra 2

101 Chapter 8 POLYNOMIALS The chapter eplores polnomial functions in greater depth. Students will learn how to sketch polnomial functions without using their graphing tool b using the factored form of the polnomial. In addition, the learn the reverse process: finding the polnomial equation from the graph. For further information see the Math Notes boes in Lessons 8.1.1, 8.1.2, and Eample 1 State whether or not each of the following epressions is a polnomial. If it is not, eplain wh not. If it is a polnomial, state the degree of the polnomial. a b π c d. ( 3 + 2)( 4 4) A polnomial is an epression that can be written as the sum or difference of terms. The terms are in the form a n where a is an number called the coefficient of, and n, the eponent, must be a whole number. The epression in part (a) is a polnomial. A coefficient that is a fraction ( ) is acceptable. The degree of the polnomial is the largest eponent on the variable, so in 2 3 this case the degree is four. The epression in part (b) is also a polnomial, and its degree is ten. The epression in part (c) is not a polnomial for two reasons. First, the 1 is not allowed because the eponents of the variable cannot be negative. The second reason is because of the 7. The variable cannot be a power in a polnomial. Although the epression in part (d) is not the sum or difference of terms, it can be written as the sum or difference of terms b multipling the epression and simplifing. Doing this gives , which is a polnomial of degree 8. Eample 2 Without using our graphing tool, make a sketch of each of the following polnomials b using the orientation, roots, and degree. a. f () = ( + 1)( 3)( 4) b. = ( 6) 2 ( + 1) c. p() = ( + 1) 2 ( 4) 2 d. f () = ( + 1) 3 ( 1) 2 Parent Guide with Etra Practice 91

102 Through investigations, students learn a number of things about the graphs of polnomial functions. The roots of the polnomial are the -intercepts, which are easil found when the polnomial is in factored form, as are all the polnomials above. Ask ourself the question: what values of will make this epression equal to 0? The answer will give ou the roots. In part (a), the roots of this third degree polnomial are = 1, 3, and 4. In part (b), the roots of this third degree polnomial are 6 and 1. The degree of a polnomial tells ou the maimum number of roots possible, and since this third degree polnomial has just two roots, ou might ask where is the third root? = 6 is called a double root, since that epression is squared and is thus equivalent to ( 6)( 6). The graph will just touch the -ais at = 6, and bounce off. The fifth degree polnomial in part (c) has three roots, 0, 1, and 4 with both 1 and 4 being double roots. The fifth degree polnomial in part (d) has two roots, 1 and 1, with 1 being a double root, and 1 being a triple root. The triple root flattens out the graph at the -ais. With the roots, we can sketch the graphs of each of these polnomials. a. b. c. d. Check that the roots fit the graphs. In addition, the graph in part (d) was the onl one whose orientation was flipped. Normall, a polnomial with an odd degree, starts off negative (as we move left of the graph) and ends up positive (as we move to the right). Because the polnomial in part (d) has a negative leading coefficient, its graph does the opposite. 92 Core Connections Algebra 2

103 Chapter 8 Eample 3 Write the eact equation of the graph shown at right. From the graph we can write a general equation based on the orientation and the roots of the polnomial. Since the -intercepts are 3, 3, and 8, we know ( + 3), ( 3), and ( 8) are factors. Also, since the graph touches at 3 and bounces off, ( + 3) is a double root, so we can write this function as f () = a( + 3) 2 ( 3)( 8). We need to determine the value of a to have the eact equation. ( 3, 0) (3, 0) (0, 2) (8, 0) Using the fact that the graph passes through the point (0, 2), we can write: 2 = a(0 + 3) 2 (0 3)(0 8) 2 = a(9)( 3)( 8) 2 = 216a a = = Therefore the eact equation is f () = ( + 3)2 ( 3)( 8). Problems State whether or not each of the following is a polnomial function. If it is, give the degree. If it is not, eplain wh not π ( + 2) ( ) Sketch the graph of each of the following polnomials. 4. = ( + 5)( 1) 2 ( 7) 5. = ( + 3)( 2 + 2)( + 5) 2 6. f () = ( + 8)( + 1) 7. = ( + 4)( 2 1)( 4) Parent Guide with Etra Practice 93

104 Below are the complete graphs of some polnomial functions. Based on the shape and location of the graph, describe all the roots of the polnomial function, its degree, and orientation. Be sure to include information such as whether or not a root is a double or triple root Using the graphs below and the given information, write the specific equation for each polnomial function intercept: (0, 12) 12. -intercept: (0, 15) 13. -intercept: (0, 3) Answers 1. Yes, degree No. You cannot have in the denominator. 3. No. When ou multipl this out, ou will still have in the denominator. 4. The roots are = 5, 1, and 7 with = 1 being a double root. Remember a double root is where the graph is tangent. This graph has a positive orientation. 94 Core Connections Algebra 2

105 Chapter 8 5. The roots are = 3 and = 5, which is a double root. The term does not produce an real roots since this epression cannot equal zero. The orientation is negative. The graph crosses the -ais at = This graph has negative orientation and the roots are = 8, 1, and 0. Be sure to include = 0 as a root gives us two roots. Since it factors to ( + 1)( 1), the five roots are: = 4, 1, 0, 1, and 4. The graph has a positive orientation. 8. A third degree polnomial (cubic) with one root at = 0, and one double root at = 4. It has a positive orientation. 9. A fourth degree polnomial with real roots at = 5 and 3, and a double root at = 5. It has a negative orientation. 10. A fifth degree polnomial with five real roots: = 5, 1, 2, 4, and 6. It has a positive orientation. 11. = ( + 3)( 1)( 4) 12. = 0.1( + 5)( + 2)( 3)( 5) 13. = 1 12 ( + 3)2 ( 1)( 4) Parent Guide with Etra Practice 95

106 COMPLEX NUMBERS and Students are introduced to the comple number sstem. Comple numbers arise naturall when tring to solve some equations such as = 0, which, until now, students thought had no solution. The see how the solution to this equation relates to its graph, its roots, and how imaginar and comple numbers arise in other polnomial equations as well. For further information see the Math Notes boes in Lessons 8.2.1, 8.2.2, and Eample 1 Solve the equation below using the Quadratic Formula. Eplain what the solution tells ou about the graph of the function = 0 As a quick review, the Quadratic Formula sas: If a 2 + b + c = 0 then = b± b2 4ac. Here, 2a a = 2, b = 20, and c = 53. Therefore, = ( 20)± ( 20)2 4(2)(53) 2(2) = = 20± ± 24 4 We now have an epression with a negative under the radical. Until now, students would claim this equation has no solution. In fact, it has no real solution, but it does have a comple solution. We define i = 1 as an imaginar number. When we combine an imaginar number with a real number, we call it a comple number. Comple numbers are written in the form a + bi. Using i, we can simplif the answer above ± = 4 = 20±2i 6 4 = 20± ( ) = 210±i 6 4 = 10±i Core Connections Algebra 2

107 Chapter 8 Because this equation has no real solutions, if we were to graph = we would see a parabola that does not cross the -ais. If we completed the square and put this into graphing form, we would get = 2( 5) The verte of this parabola is at (5, 3), and since it open upwards, it will never cross the -ais. You should verif this with our graphing tool. Therefore, the graph of the function = has no -intercepts, but it does have two comple roots, = 10±i 6 2. Recall that we said the degree of a polnomial function tells us the maimum number of roots. In fact the degree tells us the eact number of roots; some (or all) might be comple. Eample 2 Simplif each of the following epressions. a b. (3+ 4i) + ( 2 6i) c. (4i)( 5i) d. (8 3i)(8 + 3i) Remember that i = 1. Therefore, the epression in (a) can be written as = = 3 + 4i. This is the simplest form; we cannot combine real and imaginar parts of the comple number. But, as is the case in part (b), we can combine real parts with real parts, and imaginar parts with imaginar parts: (3 + 4i) + ( 2 6i) = 1 2i. In part (c), we can use the commutative rule to rearrange this epression: (4i)( 5i) = (4 5)(i i) = 20i 2. However, remember that i = 1, so i 2 = ( 1) 2 = 1. Therefore, 20i 2 = 20( 1) = 20. Finall in part (d), we will multipl using methods we have used previousl for multipling binomials. You can use the Distributive Propert or generic rectangles to compute this product. (8 3i)(8 + 3i) = 8(8) + 8(3i) 3i(8) 3i(3i) = i 24i + 9 = i 64 24i 3i 24i 9 The two epressions in part (d) are similar. In fact the are the same ecept for the middle sign. These two epressions are called comple conjugates, and the are useful when working with comple numbers. As ou can see, multipling a comple number b its conjugate produces a real number! This will alwas happen. Also, whenever a function with real coefficients has a comple root, it alwas has the conjugate as a root as well. Parent Guide with Etra Practice 97

108 Eample 3 Make a sketch of a graph of a polnomial function p() so that p() = 0 would have onl four real solutions. Change the graph so that it has two real and two comple solutions. If p() = 0 is to have onl four real solutions, then p() will have four real roots. This will be a fourth degree polnomial that crosses the -ais in eactl four different places. One such graph is shown at right. In order for the graph to have onl two real and two comple roots, we must change it so one of the dips does not reach the -ais. One eample is shown at right. Problems Simplif the following epressions. 1. (6 + 4i) (2 i) 2. 8i ( 3)(4i)(7i) 4. (5 7i)( 2 + 3i) 5. (3 + 2i)(3 2i) 6. ( 3 5i)( 3 + 5i) Below are the complete graphs of some polnomial functions. Based on the shape and location of the graph, describe all the roots of the polnomial function. Be sure to include information such as whether roots are double or triple, real or comple, etc Write the specific equation for the polnomial function passing through the point (0, 5), and with roots = 5, = 2 and = 3i. 98 Core Connections Algebra 2

109 Chapter 8 Answers i 2. 4i i A third degree polnomial with negative orientation and with one real root at = 5 and two comple roots. 8. A fifth degree polnomial with negative orientation and with one real root at = 4 and four comple roots. 9. = 1 18 ( ) ( 2 + 9) Parent Guide with Etra Practice 99

110 FACTORING POLYNOMIAL FUNCTIONS and Students learn to divide polnomials as one method for factoring polnomials of degrees higher than two. Through division and with two theorems, students are able to rewrite polnomials in a form more suitable for graphing. The can also easil find a polnomial s roots, both real and comple. For further information see the Math Notes boes in Lessons 8.3.1, 8.3.2, and Eample 1 Divide b + 1. Students have learned to multipl polnomials using several methods, one of which is with generic rectangles. The generic rectangle is a method that works for polnomial division as well. To find the product of two polnomials, students draw a rectangle and label the dimensions with the two polnomials. The area of the rectangle is the product of the two polnomials. For division, we start with the area and one dimension of the rectangle, and use the model to find the other dimension. To review, consider the product ( + 2)( ). We use the two epressions as the dimensions of a rectangle and calculate the area of each smaller part of the rectangle. In this case, the upper left rectangle has an area of 3. The net rectangle to the right has an area of 3 2. We continue to calculate each smaller rectangle s area, and sum the collection to find the total area. The total area represents the product Here, the total area is , or once it is simplified. Now we will do the reverse of this process for our eample. We will set up a rectangle that has a width of + 1 and an area of We have to move slowl, however, as we do not know what the length will be. We will add information to the figure graduall, adjusting as we go. The top left rectangle has an area equal to the highest-powered term: 3. Now 1 3 work backwards: If the area of this rectangle is 3 and the side has a length of, what does the length of the other side have to be? It would be an 2. If we fill this piece of information above the upper left small rectangle we can use it to compute the area of the lower left rectangle. 100 Core Connections Algebra 2

111 Chapter The total area is , but we onl have 1 3 and 1 2 so far. We will need to add 3 2 more to the total area (plus some other terms, but remember we are taking this one step at a time). Once we have filled in the remaining 2 area, we can figure out the length of the top side. Remember that part of the left side has a length of. This means that part of the top must have a length of Use this new piece of information to compute the area of the rectangle that is to the right of the 3 rectangle, and then the small rectangle below that result Our total area has a total of 7, but we have onl 3 so far. This means we will need to add 10 more. Place this amount of area in the rectangle to the right of 3 2. With this new piece of area added, we can compute the top piece s length and use it to calculate the area of the rectangle below the 10. Note that our constant term in the total area is 10, which is what our rectangle has as well. 3 Therefore we can write = , or = ( + 1)( ). Now that one of the terms is a quadratic, students can to factor it. Therefore, = ( + 1)( + 5)( 2) Parent Guide with Etra Practice 101

112 Eample 2 Factor the polnomial and find all its roots. P() = Students learn the Integral Zero Theorem, which sas that zeros, or roots of this polnomial, must be factors of the constant term. This means the possible real roots of this polnomial are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, or ±48. In this case there are 20 possible roots to check! We can check them in a number of was. One method is to divide the polnomial b the corresponding binomial epression (for instance, if 1 is a root, we divide the polnomial b ( + 1) to see if it is a factor. Another method is to substitute each zero into the polnomial to see which of them, if an, make the polnomial equal to zero. We will still have to divide b the corresponding epression once we have the root, but it will mean less division in the long run. Substituting values for P(), we get: P(1) = (1) 4 + (1) 2 14(1) 48 = = 60 P(2) = (2) 4 + (2) 2 14(2) 48 = = 56 P( 1) = ( 1) 4 + ( 1) 2 14( 1) 48 = = 32 P( 2) = ( 2) 4 + ( 2) 2 14( 2) 48 = = 0 We can keep going, but we just found a root, = 2. Therefore, + 2 is a factor of the polnomial. Now we can divide the polnomial b this factor to find the other factors This other factor, however, is degree three, still too high to use easier methods of factoring. Therefore we must use the Integral Zero Theorem again, and find another zero from the list ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24. We can start where we left off, but now using Q() = , a simpler polnomial to evaluate. 102 Core Connections Algebra 2

113 Chapter 8 Q(3) = (3) 3 2(3) 2 + 5(3) 24 = = It is certainl helpful to find a root so quickl! Since = 3 is a root, 3 is a factor. Therefore, we can divide again = 0 Now we have P() = = ( + 2)( 3)( ). Finall! The last polnomial is a quadratic (degree 2) so we can factor or use the Quadratic Formula. If ou tr factoring, ou will not be successful, as this quadratic does not factor with integers. Therefore, we must use the Quadratic Formula to find the roots as shown at right. Therefore, the original polnomial factors as: P() = = ( + 2)( 3) 1+i 31 ( 2 ) 1 i 31 ( 2 ) = 1± 12 4(1)(8) 2(1) = = = 1± ± ±i 31 2 Problems 1. Divide b Divide b Divide b 2 1. Factor the polnomials, keeping the factors real. 4. f () = g() = Find all roots for each of the following polnomials. 6. P() = Q() = Answers f () = ( + 4)( ) 5. g() = ( + 1)( 4)( ) 6. = 1, 3, 2i, 2i 7. = 6, 4 + i, 4 i Parent Guide with Etra Practice 103

114 SAT PREP 1. (5 + 6) 2 =? a. (2 5) + (2 6) b c d. 61 e If 6 7 = 12, what is the value of 2(6 7)? a. 24 b. 13 c. 1 d. 420 e The average (arithmetic mean) of three numbers is 25. If two numbers are 25 and 30, what is the third number? a. 35 b. 30 c. 25 d. 20 e People from the countr of Turpa measure with different units. Each curd is 7 garlongs long and each garlong is made up of 15 bleebs. How man complete curds are there in 510 bleebs? a. 105 b. 15 c. 5 d. 4 e If 2 2 = 12 and = 2, what is the value of +? 6. Five consecutive integers sum to 25. What is the largest of these consecutive numbers? 7. For all positive integers m and n, we define m n to be the whole number remainder when m is divided b n. If 11 k = 3, what does k equal? 8. At Pies R Us, each pie is cut into a slice as shown in the figure at right. Each slice of pie has a central angle of 30º. The sell the pies b the slice. If the weight of each pie is uniforml distributed, weighing 108 grams, how much does each slice weigh in grams? 30º 9. In the figure at right, what is the area of the shaded region if that region is a square? What is the sith term in the sequence beginning 432, 72, 12,? 6 Answers 1. C 2. A 3. D 4. D g Core Connections Algebra 2

115 Chapter 9 SURVEY DESIGN, SAMPLING, AND BIAS Students learn that in man situations the require data from samples to estimate characteristics (parameters) of large populations. The reason ma be that the population is too large to permit the student to gather data from ever subject. For eample a national opinion poll about a political issue. Sometimes the act of collecting the data ruins the item being studied, as when crash-testing automobiles. Understandabl, to get reliable results, one must minimize the sources of bias. Students discuss the potential for bias in the wording of surve questions. Later in the section the do an activit which shows the need for the random selection of subjects to reduce bias in statistical studies. For additional information, see the Math Notes bo in Lesson Additionall, problem 9-3 eplains tpes of bias found within surve questions. Eamples Determine whether each situation describes a sample or census. If the situation describes a sample, discuss the sampling technique and potential sources of bias. a. The Chief of Police calls in his five newest officers to get their opinion on new requirements for department-wide promotions. Answer: The Chief is using a sample, which is not likel representative of the population affected b the new requirements. The newest officers would likel want to please the Chief and would echo his support or dislike of the new requirements. b. A manager compares the annual sales totals for 12 store locations to rank their performance. Answer: This is a census. The manager has information from all 12 of the stores he is interested in comparing. c. Shoppers are invited to fill out a questionnaire about their shopping eperience at the cash register. The store manager uses the results in a report to corporate headquarters to demonstrate the level of customer satisfaction at her store. Answer: The manager is using a voluntar response sample (or convenience sample ), which would not represent all customers ver well. A sample like this tends to overrepresent those with the strongest opinions. d. The student council is tring to determine how much space the will need for the prom. Carmine, the Junior Class President, walks around during lunchtime at her school with a clipboard, asking, Are ou planning on going to the best prom ever? Answer: Carmine is not taking a random sample. She is more likel to talk to students she knows, and the, as a group, are likel to be biased about the prom. There ma also be a strong desire to please the interviewer in this situation. The wording of the question itself is biased, using best prom ever to describe the dance. Parent Guide with Etra Practice 105

116 Problems Determine whether each situation describes a sample or census. If it is a sample, discuss the sampling technique and potential sources of bias. 1. A math teacher wants to determine whether plaing classical music during testing benefits high school math students. He plas classical music to half of his classes while the test, and compares their scores to the other half of his classes, who tested without the music. 2. A sports news reporter wants to know the win-loss record of the local high school girl s lacrosse team, so he looks at the league website and sees that the have 8 wins and 5 losses. 3. A batter manufacturer wants to monitor the durabilit and peak voltage of its products. A machine is programmed to select ever 1000 th batter from the production line and subject it to a series of tests. 4. Scott is doing a report on the risk factors for cancer. He asks all of the members of his PE class: Do ou have a famil histor of cancer? Do ou eat 5 servings of fruits and vegetables each da? Answers 1. Because the math teacher is using his data to make a statement about all high school math students (the population), his classes represent a sample of those students. His students would have man things in common, like where the live and how the were taught, which other high school math students do not have. The would be a poor representation of the population. The teacher ma also be biased in grading the eams to obtain the results he favors. 2. This is a census. The reporter has access to ever game outcome for the team he is researching and is not using the information to draw conclusions about other teams or the sport of lacrosse in general. 3. The batteries tested represent a sample of those being made. The might be a reasonable representation of the population of batteries. However, if the 1000 th batter is usuall made at the same time ever da, or corresponds to some other variable like periodic machine maintenance, it might be subject to bias. 4. Scott s PE class represents a sample of people or students. However this is a convenience sample and cannot be considered a reliable representation of the population of people or students. The question order ma cause bias in the results as people being reminded of a disease like cancer would likel want to appear concerned enough to be eating a health diet with lots of fruits and vegetables. Members of a PE class ma be more inclined to be thinking about health issues such as diet. 106 Core Connections Algebra 2

117 Chapter 9 TESTING CAUSE AND EFFECT WITH EXPERIMENTS AND CONCLUSIONS FROM OBSERVATIONAL STUDIES and Two was to gather information for statistical studies are observational studies and eperiments. Observational studies seek to collect data about subjects without changing them, while eperiments impose treatments (variables of interest) upon the subjects. Observational studies are generall easier to perform but cannot demonstrate cause-and-effect relationships because it is not possible to separate the man effects of all the lurking variables (those outside of what is being observed directl). For additional information, see the Math Notes bo in Lesson Eamples For each question below: Does a census or a sample make more sense in this situation? Should an observational stud or an eperiment be carried out to answer the question? If an observational stud is suitable and asks for an association between variables, discuss the effects of possible lurking variables. If an eperiment would be suitable, outline an eperimental design. a. Is there a relationship between drinking green tea and longevit? Answer: Because it would be impossible to find ever green tea consumer this stud would need to rel on sampling. Green tea would not be a difficult treatment to administer in an eperiment but because life-spans are being measured, an eperiment would be impractical. An observational stud ma find a relationship between these two variables but would be unable to establish cause and effect because of the man possible lurking variables. Perhaps people who choose to drink green tea also choose to be more health conscious in other was such as smoke less than the general population. The ma also tend to be of higher income and have better access to health care. Hence their longevit causes could reall be smoking less and having better health care, not from drinking green tea. b. What is the mean verbal SAT score in the state of Colorado for a given ear? Answer: This information is easil found b a census. c. What percentage of razorback clams show signs of stress from acidic ocean water? Answer: One could not find ever clam for stud and the stress testing of the clams ma harm them, so a sample would be appropriate. It would be important to gather razorback clams from a variet of locations as environmental conditions ma var considerabl. d. Would brighter classroom lighting increase eam scores? Answer: This would require using a sample of students in a controlled eperiment. One could take student volunteers and give them the same instruction on a topic, then randoml assign them to classrooms with different lighting levels for the same eam. The mean test scores from each group could be compared. Parent Guide with Etra Practice 107

118 Problems For each question: Does a census or a sample make more sense in this situation? Should an observational stud or an eperiment be carried out to answer the question? If an observational stud is suitable and asks for an association between variables, discuss the effects of possible lurking variables. If an eperiment would be suitable, outline an eperimental design. 1. Is there a relationship between the temperature of water and its electrical resistance? 2. What is the mean distance students commute to school in our count? 3. What percentage of car crashes causing injur occurred when at least one driver was impaired b drugs or alcohol? 4. Is there an association between baldness and being colorblind? Answers 1. One would have to work with samples of water in this situation. This question would best be answered as an eperiment. Begin with water from the same source so its other variables like impurities remain constant and divide it into several like containers. Randoml assign different temperatures to the water containers and measure the electrical resistance between two wires inserted an equal distance apart in each container, then compare results. 2. For most counties, surveing ever student about commute distances is not possible, so a random sample of students could be used to represent the population. This is clearl an observational stud because the question seeks to determine how students commute without imposing an changes upon them. 3. This kind of information is routinel collected b law enforcement, so census data is likel available. 4. This stud would require samples from the population. An observational stud is the onl reasonable alternative because an eperiment would require imposing baldness or colorblindness upon subjects. A lurking variable here is gender. Men are more often colorblind and more often bald than women, so if the stud included both genders then it would likel find an association. 108 Core Connections Algebra 2

119 Chapter 9 RELATIVE FREQUENCY HISTOGRAMS When the frequencies in a frequenc table are divided b the total number of data points, the frequencies become ratios, and the frequenc table becomes a relative frequenc table. Histograms show the information from frequenc tables and relative frequenc histograms displa information from relative frequenc tables. In this section histograms or their associated tables are used to translate ranges of data into percentages. Note: Man of these lessons require the use of a graphing calculator. Instructions for using a TI-83/84+ calculator can be found at cpm.org under Technolog Support. Eample 1 Fort applicants interviewed for an administrative assistant position. Each applicant was given a tping test and earned a score in words per minute (wpm). Their results are shown in the table a. Create a frequenc table, histogram, relative frequenc table, and relative frequenc histogram. b. What percentage of applicants tped 40 to 49 wpm? c. What percentage of applicants tped at least 60 wpm? d. If 50 wpm is the minimum job qualification, what percentage of applicants did not qualif? Answers: a. wpm frequenc total relative frequenc 1/40 = /40 = /40 = /40 = /40 = /40 = Histogram frequenc Tping score (wpm) b. 15% (from the relative frequenc table or histogram) c = 0.55 = 55% d = = 17.5% relative e Relative Frequenc frequenc Histogram Tping score (wpm) Parent Guide with Etra Practice 109

120 Eample 2 A random sample of 35 cars was taken from a car rental agenc, and the highwa mileage of each car was measured in miles per gallon (mpg) checksum 1264 a. Use a calculator and find the mean and standard deviation for this sample of cars. b. Create a relative frequenc table and relative frequenc histogram for the data. Begin our table at 24 mpg and use intervals of 3 mpg. c. The rental agenc advertises that its fleet of compact cars gets at least 30 mpg. What percentage of the sample does not meet this standard? d. What interval would contain the median car? e. The State awards a gold star mileage rating to cars that get a minimum of 39 mpg. What percentage of cars in the sample attained this rating? Answers: a. mean mpg, sample standard deviation mpg b. Highwa Mileage (mpg) frequenc relative frequenc 24 up to up to up to up to up to up to up to = = = = = = = total Relative Frequenc Histogram relative frequenc Highwa Mileage (mpg) c = = 14.28% d. Counting the cars from either end of the mileage distribution, ou would find that the 18 th car would be in the middle. The 18 th car would also be in the 36 up to 39 mpg group. e = = 31.42% 110 Core Connections Algebra 2

121 Chapter 9 THE NORMAL PROBABILITY DENSITY FUNCTION AND PERCENTILES and There are man situations when a relative frequenc histogram will closel resemble a bellshaped curve. Sometimes an equation called the normal probabilit densit function is fitted to the bell-shaped relative frequenc histogram. This function is commonl called the normal distribution and can be used to find percentages and probabilities associated with what is being measured. The normal probabilit densit function is also used to find percentiles. Percentiles identif the measurement below which a specified percentage of data points are found. For eample, if the 90 th percentile for men s heights were 72 inches, 90% of men would be shorter than 72 inches. For additional information, see the Math Notes bo in Lesson Eample 1 Half liter soda bottles are filled with cola at a bottling plant. The equipment used to fill the bottles is not capable of putting eactl 500 ml in each bottle, so it is adjusted such that the amount of cola placed in the bottles follows a normal distribution with a mean of 502 ml and a standard deviation of 1.8 ml. a. What proportion of bottles will have more than 502 ml of cola in them? b. What proportion of bottles will have between 500 and 502 ml in them? c. What proportion of bottles are under-filled (hold less than 500 ml)? d. What percentile corresponds to 503 ml? e. What percentile corresponds to 505 ml? Answers: a. normalcdf(502, 10^99, 502, 1.8) = One ma recognize that 502 ml is the mean and half (0.500) of the probabilit in a normal distribution is alwas above and below the mean. b. normalcdf(500, 502, 502, 1.8) = c. normalcdf( 10^99, 500, 502, 1.8) = d. normalcdf( 10^99, 503, 502, 1.8) = or the 71 st percentile e. normalcdf( 10^99, 505, 502, 1.8) = or the 95 th percentile Parent Guide with Etra Practice 111

122 Eample 2 Commercial cherr trees produce fruit according to a normal distribution with an average of 90 kg of cherries per tree and a standard deviation of 11 kg per tree. a. What is the probabilit that a tree produces at least 95 kg of cherries? b. What is the probabilit that a tree ields between 85 and 105 kg of cherries? c. What percentile would a harvest of 80 kg be for a single tree? Answers: a. normalcdf(95, 10^99, 90, 11) = b. normalcdf(85, 105, 90, 11) = c. normalcdf( 10^99, 80, 90, 11) = or the 18 th percentile Problems 1. Concrete is loaded into trucks for a large construction project according to a normal distribution. The mean amount loaded is 230 cubic feet with a standard deviation of 7 cubic feet. Concrete trucks are rated b the number of cubic ards the can carr, and 9 cubic ards is a common maimum load. Hence the epression The whole nine ards. a. What proportion of the trucks will be loaded with between 220 and 240 cubic feet? b. What proportion of trucks are loaded with more concrete than the 9 cubic ard maimum. (3 feet = 1 ard) c. 235 cubic feet is what percentile load? d. If a particular job requires at least 220 cubic ards of concrete, what is the probabilit there will not be enough concrete in a single truck? 2. Assume the annual snowfall in the village of Thalhammer follows a normal distribution with a mean of 185 cm and standard deviation of 36 cm. a. Less than 100 cm of snow in a ear is considered drought conditions in Thalhammer. What is the chance that a particular ear will have drought conditions? b. The Rivera Valle lies below Thalhammer. Residents of Rivera Valle are placed on a flood-warning from snowmelt in the spring if annual snowfall in Thalhammer is greater than 250 cm. What is the probabilit the Rivera Valle residents will be placed on a flood-warning this spring? c. What percentile corresponds to 160 cm of annual snowfall in Thalhammer? 112 Core Connections Algebra 2

123 Chapter 9 Answers 1. a. normalcdf(220, 240, 230, 7) = b. (3 ft)(3 ft)(3 ft) = 27 ft 3, so 27cubic feet per cubic ard. (9)(27) = 243 cubic feet normalcdf(243, 10^99, 230, 7) = c. normalcdf( 10^99, 235, 230, 7) = or the 76 th percentile d. normalcdf( 10^99, 220, 230, 7) = a. normalcdf( 10^99, 100, 185, 36) = b. normalcdf(250, 10^99, 185, 36) = c. normalcdf( 10^99, 160, 185, 36) = or the 24 th percentile Parent Guide with Etra Practice 113

124 SAT PREP 1. In the rectangle ABCD at right, the area of the shaded region is given b πlw. If the area of the shaded region is 7π, what is 6 the total area, to the nearest whole number, of the unshaded regions of the rectangle ABCD? D A l C w B a. 14 b. 15 c. 20 d. 22 e Consider the following equations: a = p b = p c = (p 0.61) 2 If p is a negative integer, what is the ordering of a, b, and c from least to greatest? a. c < a < b b. a < c < b c. b < a < c d. a < b < c e. c < b < a 3. The figure at right represents si offices that will be assigned randoml to si different emploees, one emploee per office. If Maranne and Ginger are two of the si emploees, what is the probabilit that the will be assigned an office indicated with an *? * * a. 1 6 b. 1 8 c d e Raul needed wire pieces 7 inches long. He cut as man as he possibl could from a wire 6 feet long. What is the total length of the wire that is left over? a. 2 inches b. 3 inches c. 4 inches d. 5 inches e. 8 inches 5. The n th term of a sequence is defined to be 5n + 2. The 35 th term is how much greater than the 30 th term? a. 5 b. 18 c. 25 d. 36 e Matilda remembers onl the first four digits of a seven-digit phone number. She is certain that none of the last three digits is zero. If she dials the first four digits, then dials the last three digits randoml from the non-zero digits, what is the probabilit that she will dial the correct number? 7. Let a b be defined as 1 + b where a 0. What is the value of 6 7? a 114 Core Connections Algebra 2

125 Chapter 9 8. If = 1, what is the value of? 9. Eight consecutive integers are arranged in order from smallest to the largest. If the sum of the first four integers is 206, what is the sum of the last four integers? 10. If the points A(4, 1), B(4, 8), and C( 3, 8) form the vertices of a triangle, what is the area of the triangle? Answers 1. C 2. D 3. C 4. A 5. C Parent Guide with Etra Practice 115

126 SERIES This chapter revisits sequences arithmetic then geometric to see how these ideas can be etended, and how the occur in other contets. A sequence is a list of ordered numbers, whereas a series is the sum of those numbers. Students develop methods for finding those sums and learn a compact wa to write the sums, known as summation notation. These ideas are etended with eplorations into Pascal s Triangle, the Binomial theorem, and natural logarithms. For more information, see the Math Notes boes in Lessons , , and Eample 1 Find the sum of each of the following series. a b. Twelve terms with t(1) = 3 and t(12) = 69. c. t(n) = 3n + 10, for integer values starting at 1 and ending at 15. Each of these problems represents an arithmetic series because there is a constant difference between each consecutive term. The are series because the are the sums of the terms. For part (a), we could simpl add each term, filling in the terms represented b the, but that would take some time and there is a good chance for an arithmetic error. Instead, we will use one of the methods developed in the chapter. We will need to know the formula for the n th term of this sequence, as well as which term the number 59 is. The formula for the n th term is t(n) = 5(n 1) + 4 (verif this!) and b setting this equal to 59, we find that 59 is the 12 th term. To find the sum, we write out the sum labeling it S. We repeat this b writing it in reverse order. Adding these two equations gives us a new equation that makes it easier to solve for S. The sum of these twelve terms is 378. S = S = S = S = 12(63) S = of these In part (b) we determine the formula for the n th term b using the given information to write an equation and solve. This gives us the t(1) = 3 So: t(12) = d(12 1) + 3 formula t(n) = 6(n 1) + 3. Knowing the t(12) = = 11d + 3 first and last terms is enough information to use the method of the previous eample to t(n) = d(n 1) + t(1) 66 = 11d find the sum. The method generalizes as d = 6 follows: add the first and last terms, multipl the result b the number of terms in the S = 12(3+69) = 432 series, then divide b two Core Connections Algebra 2

127 Chapter 10 In part (c) we have the formula for the n th term and we know there will be 15 terms. We will calculate the first and the last term, and use them in the same procedure as above, using S(n) = nt(1)+t(n) ( ) 2. t(n) = 3n +10 t(1) = 3(1) +10 = 7 t(15) = 3(15) +10 = = 35 S = 15(7 35) 2 S = 210 Eample 2 15 Epand the series (5k 7) and find the sum. k=1 The epression above is known as summation notation, and it is a shorthand wa to write out a series. The S is the Greek letter sigma and stands for sum. We let the variable k (called the inde when we are using summation notation) start at 1, and equal each integer up to 15 in the epression 5k 7. This looks like: 15 k=1 (5k 7) = [ 5(1) 7]+ [ 5(2) 7]+ [ 5(3) 7]+ [ 5(4) 7]+...+ [ 5(15) 7] = To find the sum, use the same method as we used in Eample 1 with t (1) = 2 and t (15) = 68 to find S = 495. Eample 3 Write the series in summation notation. First we need the formula for the n th term. Here, t(1) = 5 and the common difference is 2. We can write t (n) = 2(n 1) + 5 = 2n + 7. If we set the right side equal to 29, we can determine which term number 29 is. 29 = 2n = 2n n = 18 Since the last term is the 18 th term, we can now put this all together in summation notation. Parent Guide with Etra Practice 117

128 Problems Write the following series in summation notation and find each sum Sum the positive even integers less than or equal to 100. Epand the following series and find each sum (3k + 1) 9. (11 j 8) 10. (50 4k) k= j=1 11. (4i + 9) 12. (1 6k) 13. j 2 i=1 9 k=1 22 k=1 4 j=1 Answers (2k + 3), S = (10k 9), S = 1377 k= ( 6k + 1), S = k, S = 5050 k= (3k 1), S = (6 k), S = 0 k= (2k), S = = 375 k= = 25, = = = = k=1 100 k=1 11 k=1 118 Core Connections Algebra 2

129 Chapter 10 GEOMETRIC SERIES and Just as arithmetic sequences lead us to arithmetic series (the sum of the terms of the arithmetic sequence), geometric sequences lead us to geometric series as well. In these sections students develop a formula to find the sum of geometric series, write geometric series with summation notation, and eplore infinite geometric series. See the Math Notes bo in Lesson for more information. Eample 1 Write the series in summation notation and find the sum. Since each term of this series is one-half the preceding term, this is a ( ) n 1. geometric series. The formula for the n th term of this series is To write it in summation notation, we just need to determine how man terms are in this series. We could write out all the terms and count them, but instead we will use the formula for the n th term, and solve for n = = = 2 n 1 ( ) n 1 ( ) n = 2 n 1 9 = n 1 n = Therefore, we can write: k=1 ( ) k 1 To find the sum, we will use the formula(s) that students developed in class for the sum of the first n terms of a geometric series. Note: The two given formulas are equivalent. S(n) = r t(n) t(1) r 1 or S(n) = a(1 rn ) 1 r Here we can write: S(10) = = 1 ( )( ) = = = or S(10) = 32 ( 1 ( 1 2 ) 10 ) ( ) = Parent Guide with Etra Practice 119 = = ( ) = =

130 Eample 2 Epand and find the sum of the geometric series: 12 k=1 3(4) k 1 Letting the inde k run through its values gives the series: ,582,912. We will use the formula from the previous eample to find the sum. S(12) = 4( ) = = 16, 777, 215 Eample 3 Find the sum of each infinite geometric series. a b k=1 ( ) k 1 If the common ratio in an infinite geometric series is between 1 and 1, the series does in fact equal a number, even though the series goes on forever. The sum is given b the formula: In part (a) the common ratio is 1 3, therefore: S = t(1) 1 r = a 1 r. S = = = = For part (b), the first term, t (1) = 25, and the common ratio, r, is 1 5. Therefore: S = = = = Core Connections Algebra 2

131 Chapter 10 Problems Epand and find the sum of each geometric series k k k= k 4. 2 k=1 8 k=1 k=1 ( ) k 5 Write each series using summation notation and find the sum ,531, , Answers k 1 = 24, 414, k 1 = 88, 573 k= = = 176 k=1 ( ) k k=1 k=1 ( ) k 1 2 Parent Guide with Etra Practice 121

132 PASCAL S TRIANGLE and After eploring patterns in Pascal s Triangle, students see how the numbers in the triangle relate to the counting principles the have learned earlier. The connection to the Binomial Theorem allows students to quickl epand binomials raised to different powers as well as determine probabilities in situations involving two outcomes. See the Math Notes bo in Lesson for more information. Eample 1 Use Pascal s Triangle to epand (a + b) 4, then appl the same pattern to epand (2 3) 4. Students wrote out several lines of Pascal s Triangle when the were looking for patterns. The fourth line of Pascal s Triangle gives us the coefficients of the terms in the epansion of this binomial since its degree is 4. The fourth line is Therefore the epansion is: (a + b) 4 = 1a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4. Notice that the powers of a decrease while the powers of b increase, but the sum of the two eponents in each term is alwas four. We appl this pattern to the second binomial, replacing each a with (2) and each b with ( 3): (2 3) 4 = 1(2) 4 + 4(2) 3 ( 3) + 6(2) 2 ( 3) 2 + 4(2)( 3) 3 + ( 3) 4 = ( 3) (9) 2 + 8( 27) = Eample 2 Use the Binomial Theorem to find the eighth term in the epansion of ( p + 3q) 14. The Binomial Theorem ties Pascal s Triangle to combinations, which the students learned in Core Connections Geometr. The Math Notes bo in Lesson provides the formula for using combinations to generate the numbers in Pascal s Triangle. The Math Notes bo in Lesson shows how to use the Binomial Theorem in combinations form to find the epansion of an binomial raised to an positive integer value. In particular, the k th term of the epansion (a + b) n is given b n C n (k 1) a n (k 1) b k 1. In this case the term is: 14 C 14 (8 1) a14 (8 1) b 8 1 = 14 C 7 a 7 b 7 = 14! 7!7! a7 b 7 = 3432a 7 b Core Connections Algebra 2

133 Chapter 10 Problems Epand each of the following binomials. 1. (5 6) 6 2. (8 + 3q) 5 3. Find the fifth term in the epansion of ( + ) Find the seventh term in the epansion of ( + ) Find the 15 th term in the epansion of (3p 2q) Eight coins are tossed. What is the probabilit that eactl five are heads? Answers q q q q q C 18 (3p) 18 ( 2q) 14 = ( )p 18 (16384)q C 5 ( ) 5 1 ( 2 ) 3 = Parent Guide with Etra Practice 123

134 SAT PREP 1. If is directl proportional to and if = 24 when = 4, what is the value of when = 7? 6 a. 7 b c. 27 d. 28 e For all positive integers a and b, let the operation {} be defined b a{}b = a b+2 6a b For how man positive integers a is a{}2 equal to 9? a. 0 b. 1 c. 2 d. 3 e In the figure at right, the line l has a slope of 2. What is the -intercept of l? a. 5 b. 8 c. 11 d. 13 e. 15 (0, 5) l O (3, 0) 4. Which of the following numbers is divisible b 7 and 11 but not divisible b 10? a. 49 b. 66 c. 70 d. 308 e In the figure below right, the intersection of ra AC with ra BA is: a. Segment AC b. Segment AB A B C c. Ra AC d. Ra BA e. Line AC 6. What is the greatest integer value of for which 4 28 < 0? 7. The mean of a list of 87 consecutive integers is 33. What is the greatest integer in the list? 124 Core Connections Algebra 2

135 Chapter When a positive integer p is divided b 5 the remainder is 4. What is the remainder when 12p is divided b 10? 9. When a certain number is divided b What is the number? and the quotient is then multiplied b 9 the result is 10. The diameter of a circle is 5. If the circle is cut in half, what is the total perimeter of the two pieces? Answers 1. E 2. B 3. C 4. D 5. B π + 10 Parent Guide with Etra Practice 125

136 SIMULATION When theoretical probabilit calculations become too comple, statisticians often use simulations instead. Simulations can also be used to check statistical computations, or the can be used in place of a stud which is too epensive, time-consuming, or unethical. Simulations require random numbers or outcomes. In this chapter the simulations are simple enough that students can use coins, dice, tables of random digits, or their calculators to generate required random numbers or outcomes. Eample 1 Dana is looking for clear wood (without knots) to make furniture. The chair he has designed requires 4 three-foot boards of clear wood. If there is a 20% chance of a knot being in a particular foot of wood, on average, how man boards will Dana have to sort through to get enough clear wood for one chair? Solution: To simulate each 3-foot board, ou could use a calculator or random number table to produce random digits in groups of three from 0 to 9. Consider each digit to be a foot of wood within the board with zeros and ones to represent knots. Simulated boards would look like {6 0 3}, {9 9 5}, {1 4 7}, etc. with each foot having a 20% chance of having a knot. Continue generating our simulated boards counting the number created in order to get 4 sets without a 0 or 1. Repeat this process enough times to get a meaningful average. 8 boards, 7 boards, 4 boards, 9 boards, 10 boards, 11 boards The theoretical mean is No Clear Clear Clear No Clear 126 Core Connections Algebra 2

137 Chapter 11 Eample 2 Would it be easier for Dana to find a single 12-foot board of clear wood rather than man 3-foot pieces? Use a simulation to estimate the average number of 12-foot boards he would have to consider. Solution: The mean will be about fifteen boards (theoretical mean 14.55). This is considerabl more than the mean boards eamined when four 3-foot boards were used, so finding four 3-foot clear boards would be easier. Problems 1. Jack is planning flights to Fort Lauderdale to catch a cruise ship. The least epensive route will require three separate flights to arrive in Miami and then a shuttle bus connection to Fort Lauderdale. Jack believes that if more than one of the legs is delaed he will miss his ship. Jack researches and finds the airline and bus compan both have an 80% on-time rating. If Jack bus the least epensive ticket, what is the probabilit that Jack misses his ship? 2. Jack looks for an alternative and finds a more epensive flight to Fort Lauderdale with another airline, eliminating the need for the shuttle bus. To afford this ticket he will need to avoid airport parking fees and have his friend Clara drive him to the airport. This airline has an 85% on-time rating but he estimates Clara s on-time percentage at around 65%. a. Use simulation to estimate the probabilit that more than one leg of this journe is delaed. b. Which travel plan would ou recommend to Jack, the one with the shuttle bus or the one involving Clara? Wh? Parent Guide with Etra Practice 127

138 Answers 1. To simulate each trip ou could use a calculator or random number table to produce random digits in groups of four from 0 to 9. randint(0, 9, 4) Consider each digit to be a leg of his journe with zeros and ones to represent delas. Simulated journes would look like { }, { }, { }, etc. with each leg having a 20% chance of a dela. Continue generating simulated trips (about 30 to 50) until ou feel ou have an accurate proportion of missed cruise ships out of total trip attempts. ok ok miss ok ok ok ok ok ok ok ok ok ok miss ok ok ok miss miss miss ok ok ok ok ok ok ok ok In this eample, 5 missed cruise ships out of 28 trips is 28 5 probabilit is = The theoretical 2. a. To simulate each trip ou could use a calculator or random number table to produce random numbers in groups of 3 from 0 to 99. randint(0, 99, 3) Consider each number to be a leg of Jack s journe with 0 34 to represent delas in the first leg with Clara and 0 14 to represent delas in the flights. Simulated journes would now look like { }, { }, { }, etc. Continue generating simulated trips (about 30 50) until ou feel ou have an accurate proportion of missed cruise ships out of total trip attempts. ok ok ok ok ok ok ok ok ok ok ok miss ok ok ok ok ok ok miss ok miss ok ok ok ok ok ok ok In this eample, 3 missed ships out of 28 trips is 28 3 probabilit is = The theoretical b. Option 2 looks better with a lower chance of missing the ship. The advantages of eliminating one leg of the trip and choosing a more reliable airline make up for Clara s lower on-time rating. 128 Core Connections Algebra 2