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Solution Manual for Intermediate Algebra Functions and Authentic Applications, 6th Edition

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    S OLUTIONS M ANUAL DIACRI T ECH I NTERMEDIATE A LGEBRA : F UNCTIONS AND A UTHENTIC A PPLICATIONS S IXTH E DITION Jay Lehmann College of San Mateo

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    i Contents Chapter 1 Linear Equations and Linear Functions 1.1 Using Qualitative Graphs to Describe Situations 1 1.2 Graphing Linear Equations 3 1.3 Slope of a Line 8 1.4 Meaning of Slope for Equations, Graphs, and Tables 12 1.5 Finding Linear Equations 17 1.6 Functions 23 Chapter 1 Review Exercises 25 Chapter 1 Test 29 Chapter 2 Modeling with Linear Functions 2.1 Using Lines to Model Data 33 2.2 Finding Equations of Linear Models 35 2.3 Function Notation and Making Predictions 39 2.4 Slope Is a Rate of Change 46 Chapter 2 Review Exercises 51 Chapter 2 Test 55 Chapter 3 Systems of Linear Equations and Systems of Linear Inequalities 3.1 Using Graphs and Tables to Solve Systems 58 3.2 Using Substitution and Elimination to Solve Systems 63 3.3 Using Systems to Model Data 70 3.4 Value, Interest, and Mixture Problems 75 3.5 Using Linear Inequalities in One Variable to Make Predictions 80 3.6 Linear Inequalities in Two Variables; Systems of Linear Inequalities 85 Chapter 3 Review Exercises 90 Chapter 3 Test 98 Cumulative Review of Chapters 1–3 103 Chapter 4 Exponential Functions 4.1 Properties of Exponents 110 4.2 Rational Exponents 114 4.3 Graphing Exponential Functions 117 4.4 Finding Equations of Exponential Functions 122 4.5 Using Exponential Functions to Model Data 128 Chapter 4 Review Exercises 135 Chapter 4 Test 139

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    ii Chapter 5 Logarithmic Functions 5.1 Composite Functions 142 5.2 Inverse Functions 147 5.3 Logarithmic Functions 153 5.4 Properties of Logarithms 156 5.5 Using the Power Property with Exponential Models to Make Predictions 161 5.6 More Properties of Logarithms 168 5.7 Natural Logarithm 172 Chapter 5 Review Exercises 177 Chapter 5 Test 184 Cumulative Review of Chapters 1–5 187 Chapter 6 Polynomial Functions 6.1 Adding and Subtracting Polynomial Expressions and Functions 194 6.2 Multiplying Polynomial Expressions and Functions 196 6.3 Dividing Polynomials: Long Division and Synthetic Division 201 6.4 Factoring Trinomials of the Form x 2 + bx + c ; Factoring Out the GCF 206 6.5 Factoring Polynomials 208 6.6 Factoring Special Binomials; A Factoring Strategy 210 6.7 Using Factoring to Solve Polynomial Equations 212 Chapter 6 Review Exercises 217 Chapter 6 Test 220 Chapter 7 Quadratic Functions 7.1 Graphing Quadratic Functions in Vertex Form 223 7.2 Graphing Quadratic Functions in Standard Form 228 7.3 Using the Square Root Property to Solve Quadratic Equations 237 7.4 Solving Quadratic Equations by Completing the Square 243 7.5 Using the Quadratic Formula to Solve Quadratic Equations 249 7.6 Solving Systems of Linear Equations in Three Variables; Finding Quadratic Functions 259 7.7 Finding Quadratic Models 266 7.8 Modeling with Quadratic Functions 270 Chapter 7 Review Exercises 275 Chapter 7 Test 282 Cumulative Review of Chapters 1–7 287 Chapter 8 Rational Functions 8.1 Finding the Domains of Rational Functions and Simplifying Rational Expressions 297 8.2 Multiplying and Dividing Rational Expressions; Converting Units 301 8.3 Adding and Subtracting Rational Expressions 307 8.4 Simplifying Complex Rational Expressions 312 8.5 Solving Rational Equations 317 8.6 Modeling with Rational Functions 326

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    iii 8.7 Variation 331 Chapter 8 Review Exercises 335 Chapter 8 Test 343 Chapter 9 Radical Functions 9.1 Simplifying Radical Expressions 348 9.2 Adding, Subtracting, and Multiplying Radical Expressions 351 9.3 Rationalizing Denominators and Simplifying Quotients of Radical Expressions 355 9.4 Graphing and Combining Square Root Functions 360 9.5 Solving Radical Equations 365 9.6 Modeling with Square Root Functions 372 Chapter 9 Review Exercises 378 Chapter 9 Test 384 Chapter 10 Sequences and Series 10.1 Arithmetic Sequences 388 10.2 Geometric Sequences 390 10.3 Arithmetic Series 395 10.4 Geometric Series 398 Chapter 10 Review Exercises 401 Chapter 10 Test 403 Cumulative Review of Chapters 1–10 405 Chapter 11 Additional Topics 11.1 Absolute Value: Equations and Inequalities 417 11.2 Performing Operations with Complex Numbers 421 11.3 Pythagorean Theorem, Distance Formula, and Circles 425 11.4 Ellipses and Hyperbolas 431 11.5 Solving Nonlinear Systems of Equations 440 Appendix A Reviewing Prerequisite Material A.1 Plotting Points 448 A.2 Identifying Types of Numbers 448 A.3 Absolute Value 448 A.4 Performing Operations with Real Numbers 448 A.5 Exponents 448 A.6 Order of Operations 449 A.7 Constants, Variables, Expressions, and Equations 449 A.8 Distributive Law 449 A.9 Combining Like Terms 450 A.10 Solving Linear Equations in One Variable 450 A.11 Solving Equations in Two or More Variables 451 A.12 Equivalent Expressions and Equivalent Equations 452

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    Chapter 1 Linear Equations and Linear Functions Homework 1.1 2. a. (c) b. (b) c. (a) d. (d) 4. The more pencils there are to buy, the more the total cost. Total cost c is the response variable and the number of pencils n is the explanatory variable. 6. The greater the rate water is added to a pool, the less time it takes to fill the pool. The number of hours to fill a pool t is the response variable and the rate water is added r is the explanatory variable. 8. The older the car in years, the greater the annual cost of repairs. The number of annual cost of repairs c is the response variable and the age of a car a is the explanatory variable. 10. The hotter the temperature, the more people go to the beach. The number of people at the beach n is the response variable and the temperature at the beach F is the explanatory variable. 12. The greater your annual income, the more federal taxes you pay. The federal taxes T is the response variable and the person’s annual income I is the explanatory variable. 14. The response variable is s , the runner’s speed, and the explanatory variable is t , time after she began her run. The graph shows her speed increases to a certain point and then decreases to a stop when s = 0. The rest of her time is spent walking at a moderate but slightly increasing speed. 16. Height h is the response variable and time, t is the explanatory variable. The height of the tennis ball decreases as it drops. The height then increases after the ball bounces from h = 0. This pattern is repeated with the maximum height decreasing after each bounce until the ball stops at h = 0. 18. P , the percentage of smokers, is the response variable and t , time in years since 2010, is the explanatory variable. This graph shows that the percentage of smokers approximately steadily decreases as time increases from 2010. 20. The foreign born percentage , p, is the response variable and time since 1910, t, is the explanatory variable. The graph shows that the percentage of the U.S. population that is foreign born decreased from 1910 to 1970 and then increased over time. 22. Volume V is the response variable and time t is the explanatory variable. The graph shows the volume of air in a person’s lungs alternately increases and decreases as air is taken in and out over time.

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    2 ISM: Intermediate Algebra 24. Angle A is the response variable and minutes t is the explanatory variable. The graph shows that as the number of minutes past midnight increases, the angle (in degrees) between the hour hand and the minute hand of a clock increases. 26. The response variable is n (species in existence) and the explanatory variable is d (amount of deforestation). The graph shows that the number of species in existence decreases steadily due to increased deforestation. 28. Speed, s , is the response variable and current in miles per hour, c , is the explanatory variable. a. The graph shows that a speedboat travels faster downstream as the river current increases. b. The graph shows that a speedboat travels slower upstream as the “opposing” river current increases. 30. The percentage of times a person hits the bull’s-eye p is the response variable and the distance in feet from the dartboard d is the explanatory variable. The graph shows that as the distance in feet increases, the percentage of bull’s-eye hits decreases. 32. The number of hours to paint a house T is the response variable and the number of people on the painting crew N is the explanatory variable. The graph shows that as the number of people on the crew increases, the number of hours to paint decreases. 34. The driving time in hours T is the response variable and the speed in miles per hour the person drives S is the explanatory variable. The graph shows that as the speed in miles per hour increases, the driving time decreases. 36. Pressure P is the response variable and volume V is the explanatory variable. The graph shows that pressure in the balloon is greater when the volume is lower. 38. An increasing curve can have at most two intercepts where it crosses each axis.

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    Chapter 1: Linear Equations and Linear Functions 3 40. First, determine what the response variable is and place it on the vertical axis. Then place the explanatory variable on the horizontal axis. There is no need for scaling on these axes. Determine from the given situation how to describe the relationship between the two variables. Use this relationship to see if an increasing or decreasing line or curve is necessary. Note where a point of intersection is needed on one or both axes. Homework 1.2 2. 4. 6. 8. 10. 12. 14. 16. 0 4 20 20 4 4 4 5 y x x y x y = − = =

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    4 ISM: Intermediate Algebra 18. 10 5 20 5 10 20 5 5 2 4 x y y x y x − = − − + = − − = − 20. 30 6 12 0 6 30 12 6 6 5 2 x y y x y x + − = − + = = − + 22. 3 3 2 7 3 3 9 3 3 3 y x y x y x + − = − + = = − + 24. 6 4 1 7 2 4 2 3 2 3 1 1 2 3 y x y x y x y x y x − − = − − − = − − − = − − = − + 26. ( ) ( ) 2 3 4 1 2 6 4 4 2 4 10 2 2 2 5 y x y x y x y x − = + − = + + = = + 28. 5 2(3 1) 2 3( 2) 5 6 2 2 3 6 6 2 3 5 4 4 8 4 4 4 2 1 x y y x x y y x y y x x y x y x − − = − − − − + = − − + − + = − − + − − + = − − = −

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    Chapter 1: Linear Equations and Linear Functions 5 30. 32. 34. 36. a. i. ii. iii. b. An equation of the form x a = is a vertical line passing through ( a , 0). 38. 40. 42. 44. ( ) 2 3 6 3 2 6 2 6 2 2 3 3 x y y x x y x − = − = − + − + = = − −

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    6 ISM: Intermediate Algebra (0, − 2) is a point on the line. ( ) ( ) 2 3 6 2 0 3 2 6 0 6 6 6 6 true x y − = − − +    46. a. b. Answers may vary. Example: 0 1.890 0.2 1.404 0.4 0.918 0.6 0.432 0.8 0.054 − x y 48. 3 5 2 4 9 7 5 1 3 9 2 8 2 2 4 x x x x x x x − − = + − − + = − + − = − − = − 50. 4(2 1) 5 7 8 4 5 7 13 11 13 13 11 13 t t t t t t − = − + − = − + = = 52. 7 2(5 4 ) 3 (8 3 ) 7 10 8 3 8 3 3 8 6 8 2 5 2 2 5 2 x x x x x x x x x x − − = − − − + = − + − + = − − = = − 54. 4( 6) 5( 3) 3( 1) 4 24 5 15 3 3 39 3 3 2 36 2 2 18 a a a a a a a a a a − + + − = − − − + − = − − = − − = − − = − 56. 1 1 2 3 4 3 1 1 2 12 12 3 4 3 4 3 8 4 11 4 4 11 4 x x x x x − =     − =         − = = = 58. 3 5 3 1 4 8 2 4 3 5 3 1 8 8 4 8 2 4 6 5 12 2 18 7 18 18 7 18 w w w w w w w w − − = +     − − = +         − − = + − = − − = − 60. 6.54 87.35 4.66 99.03 1.88 186.38 1.88 1.88 99.14 x x x x − + = − − − − = − − ≈ 62. ax by c by c ax b b c ax y b + = − = − = 64. ( ) 1 1 x y a a x y a a a a x y a y a x + =   + =     + = = − 66. 3 12 y x = − − To find the x -intercept, let y = 0 and solve for x . 0 3 12 12 3 4 x x x = − − = − − = The x -intercept is ( − 4, 0). To find the y -intercept, let x = 0 and solve for y . ( ) 3 0 12 0 12 12 y = − − = − = − The y -intercept is (0, − 12). 68. 5 4 20 x y − = To find the x -intercept, let y = 0 and solve for x .

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    Chapter 1: Linear Equations and Linear Functions 7 ( ) 5 4 0 20 5 0 20 5 20 4 x x x x − = − = = = The x -intercept is (4, 0). To find the y -intercept, let x = 0 and solve for y . ( ) 5 0 4 20 0 4 20 4 20 5 y y y y − = − = − = = − The y -intercept is (0, − 5). 70. 2 y x = − To find the x -intercept, let y = 0 and solve for x . 0 2 0 x x = − = The x -intercept is (0, 0). This is also the y -intercept. 72. 2 x = − Since x = − 2 is a vertical line, it never intersects the y -axis. Therefore, there is no y -intercept. Since the graph passes through ( − 2, 0), this is the x -intercept. 74. ax by c + = To find the x -intercept, let y = 0 and solve for x . (0) ax b c ax c c x a + = = = The x -intercept is , 0 c a       . To find the y -intercept, let x = 0 and solve for y . (0) a by c by c c y b + = = = The y -intercept is 0, c b       . 76. ( ) a x by c − = To find the x -intercept, let y = 0 and solve for x . ( ) 0 a x b c ax c c x a − ⋅ = = = The x -intercept is , 0 c a       . To find the y -intercept, let x = 0 and solve for y . ( ) 0 a by c aby c c y ab − = − = = − The y -intercept is 0, c ab   −     . 78. ( ) ay b c dx = + To find the x -intercept, let y = 0 and solve for x . ( ) ( ) 0 0 a b c dx bc bdx bdx bc c x d = + = + − = = − The x -intercept is , 0 c d   −     . To find the y -intercept, let x = 0 and solve for y . ( ) 0 ay b c d ay bc bc y a = + ⋅ = = The y -intercept is 0, bc a       . 80. y b x m − = To find the x -intercept, let y = 0 and solve for x . (0) b x m b x m − = − = The x -intercept is , 0 b m −       . To find the y -intercept, let x = 0 and solve for y . 0 0 y b m y b y b − = − = = The y -intercept is (0, b ). 82. Answers may vary. Example: 3 3 0 1 1.5 0 3 1 6 3 − − − x y 84. 2 x =

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    8 ISM: Intermediate Algebra 86. 1 y = 88. 1 2 y ≈ − 90. 8 x = 92. 6 x = − 94. 1 x ≈ − 96. a. b. To find the y -intercept, let x = 0 and solve for y . ( ) 2 0 10 0 10 10 y = − + = + = The y -intercept is (0, 10). This means that there are 10 gallons of gas in the tank at the time of refueling. c. To find the x -intercept, let y = 0 and solve for x . 0 2 10 2 10 5 x x x = − + = = The x -intercept is (5, 0). This means that when 5 hours have passed since filling the tank, the tank will be empty. 98. The graph of 2 2 5 1 y x x x + = + + is a line. When combining like terms, note that the equation becomes 5 1 y x = + . The 2 x term cancels out. It is now an equation in the form y mx b = + . 100. Since (2, 11) is a point on the graph of the equation 3 y mx = + , (2, 11) satisfies the equation. Substitute x = 2 and y = 11 into 3 y mx = + and solve for m . ( ) 11 2 3 8 2 4 m m m = + = = Therefore, m = 4 in 3 y mx = + . 102. a. Points A and F satisfy the equation y ax b = + since these points lie on the line for this equation. b. Points D and F satisfy the equation y cx d = + since these points lie on the line for this equation. c. Point F satisfies both equations since it lies on both lines. It lies at the point where the lines intersect. d. Points B , C , and E do not satisfy either equation since they do not lie on either line. 104. The y -coordinate of the x -intercept is 0 because the x -intercept is a point on the x -axis. All points on the x -axis have a y -coordinate equal to 0. The x -coordinate of the y -intercept is 0 because the y -intercept is a point on the y -axis. All points on the y -axis have an x -coordinate equal to 0. 106. No, every line does not have an x -intercept. Answers may vary. Example: Most horizontal lines do not have x -intercepts. For example, the line 1 y = . does not have an x -intercept. Homework 1.3 2. 2500 0.3125 8000 A m = = 3100 0.326 9500 B m = ≈ Airplane B is making a steeper climb than airplane A. 4. The slope of the ski run from the top of the mountain to the chairlift is 100 1 400 4 − = . If the vertical distance from the top of the mountain to the chairlift is 100 yards and the vertical distance for the entire run is 415 yards, then the vertical distance from the chairlift to the restaurant is 415 − 100 = 315 yards. Also, since the horizontal distance from the top of the mountain to the chairlift is 400 yards and the horizontal distance for the entire run is 1300 yards, the horizontal distance from the chairlift to the restaurant is 1300 − 400 = 900 yards. Therefore, the slope of the ski run from the chairlift to the restaurant is 315 315 7 or 900 900 20 − = .

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    Chapter 1: Linear Equations and Linear Functions 9 6. 2 1 2 1 8 4 4 1 1 5 4 y y m x x − − = = = = − − − − Since m is negative 1, the line is decreasing. 8. 2 1 2 1 6 ( 2) 8 4 2 8 6 3 y y m x x − − − = = = = − − − − Since m is negative 4 3 , the line is decreasing. 10. 2 1 2 1 6 ( 10) 16 4 3 1 4 y y m x x − − − = = = = − − − − − Since m is negative 4, the line is decreasing. 12. 2 1 2 1 3 ( 9) 6 3 2 6 4 2 y y m x x − − − − = = = = − − − − Since m is negative 3 2 , the line is decreasing. 14. 2 1 2 1 12 ( 2) 10 5 1 7 8 4 y y m x x − − − − − = = = = − − − − Since m is positive 5 4 , the line is increasing. 16. ( ) ( ) 2 1 2 1 6 5 6 5 1 1 1 2 1 2 1 y y m x x − − − − − + − = = = = = − − − − − − + Since m is negative 1, the line is decreasing. 18. 2 1 2 1 100 0 100 1 100 0 100 y y m x x − − = = = = − − Since m is positive 1, the line is increasing. 20. 2 1 2 1 1 ( 1) 0 0 3 5 8 y y m x x − − − − = = = = − − − − Since m is zero, the line is horizontal. 22. 2 1 2 1 1 8 9 undefined 4 4 0 y y m x x − − − − = = = = − − Since m is undefined, the line is vertical. 24. ( ) 2 1 2 1 0 3 3 1 6 0 6 2 y y m x x − − − = = = = − − − − − Since m is negative 1 2 , the line is decreasing. 26. 2 1 2 1 7.6 2.2 9.8 8.17 5.1 ( 3.9) 1.2 y y m x x − − − − = = = = − − − − − Since m is positive 8.17, the line is increasing. 28. 2 1 2 1 82.78 ( 66.66) 149.44 25.41 ( 11.26) 14.15 10.56 y y m x x − − − = = = − − − − − = − Since m is negative 10.56, the line is decreasing. 30. The points (0, 4) and ( − 2, 1) lie on the line. ( ) 2 1 2 1 4 1 3 0 2 2 y y m x x − − = = = − − − The slope of the line is 3 2 . 32. Since 2 1 m m = , the lines are parallel. 34. Since 2 1 1 m m = − , the lines are perpendicular. 36. Since 2 1 1 m m = − , the lines are perpendicular. 38. Since 2 1 m m = , the lines are parallel. 40. Since 1 m and 2 m are undefined, 1 l and 2 l are vertical lines. Vertical lines are parallel. 42. The two lines are not parallel, since their slopes, 1 3 and 3 10 , are not equal. 44. Answers may vary. Example: 46. Answers may vary. Example:

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    10 ISM: Intermediate Algebra 48. Answers may vary. Example: 50. Answers may vary. Example: 52. -4 -2 2 4 -4 -2 2 4 x y Answers may vary. Example: Since 2 1 6 3 − = − and 1 1 3 3 = − − , the slopes are the same. Both lines have the same slope and the same y -intercept, so they must be the same line. 54. Answers may vary. Example: The lines have the same steepness. 56. Answers may vary. Example: The points (0, 0) and (10, − 1) are on the line and can be used to find the slope. 2 1 2 1 1 0 1 1 10 0 10 10 y y m x x − − − − = = = = − − − 58. Answers may vary. Example: The points (0, 0) and (1, 10) are on the line and can be used to find the slope. 2 1 2 1 10 0 10 10 1 0 1 y y m x x − − = = = = − − 60. Answers may vary. Example: The line must be increasing and almost horizontal.

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    Chapter 1: Linear Equations and Linear Functions 11 62. Answers may vary. Example: The line must be decreasing and almost vertical. 64. The student exchanged 1 y with 2 y in calculating the slope. The correct slope is negative instead of positive since 2 1 2 1 1 9 8 8 7 2 5 5 y y m x x − − − = = = = − − − . 66. Answers may vary. Example: Three points that lie on the line are (3, 11), (0, 6), and ( − 9, − 9) since the line equation is 5 6 3 y x = + . 68. a. 2 1 2 1 5 3 2 7 2 5 y y m x x − − = = = − − b. 1 2 1 2 3 5 2 2 2 7 5 5 y y m x x − − − = = = = − − − c. The results are the same. It does not matter which point is taken first. d. 2 1 1 2 2 1 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 y y y y x x x x y y y y x x x x y y y y x x y y y y y y x x y y − − = − − − − − = ⋅ − − − − − = ⋅ − − − − = − − e. It does not matter which point is used first and which point is used second. The slope will be the same regardless of which point is considered first. 70. a. There are two possibilities for the other two vertices, since the figure is a square and sides must have equal length. The coordinates for the vertices are ( − 3, 1) and ( − 3, 7). Another possibility is (9, 1) and (9, 7). b. One possibility is the point (0, − 3). A second possibility is the point (6, 5). Another possibility is ( − 14, − 1). 72. Answers may vary. Example: 74. Answers may vary. 5 ft Road with 5% grade 100 ft

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