Solution Manual for Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry, 4th Edition

Chapter 2 Linear and Quadratic Functions 10. False. The y-intercept is 8. The average rate of change is 2 (the slope). Section 2.1 1. From the equation y  2 x  3 , we see that the yintercept is 3 . Thus, the point  0, 3 is on the 11. a graph. We can obtain a second point by choosing a value for x and finding the corresponding value for y. Let x  1 , then y  2 1  3  1 . Thus, 12. d 13. f  x   2x  3 the point 1, 1 is also on the graph. Plotting a. Slope = 2; y-intercept = 3 the two points and connecting with a line yields the graph below. b. Plot the point (0, 3). Use the slope to find an additional point by moving 1 unit to the right and 2 units up.   2. m  3. y2  y1 3  5 2 2    x2  x1 1  2 3 3 average rate of change = 2 d. increasing 14. g  x   5 x  4 f (2)  3(2) 2  2  10 2 f (4)  3(4)  2  46 y f (4)  f (2) 46  10 36     18 x 42 42 2 a. Slope = 5; y-intercept = 4 b. Plot the point (0, 4) . Use the slope to find an additional point by moving 1 unit to the right and 5 units up. 4. 60 x  900  15 x  2850 75 x  900  2850 75 x  3750 x  50 The solution set is {50}. 5. c. f  2    2   4  4  4  0 2 6. True 7. slope; y-intercept c. average rate of change = 5 d. increasing 8. positive 9. True 149 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 15. h  x   3x  4 a. Slope = 3 ; y-intercept = 4 b. Plot the point (0, 4). Use the slope to find an additional point by moving 1 unit to the right and 3 units down. c. d. c. average rate of change = d. increasing 1 4 2 18. h  x    x  4 3 2 a. Slope =  ; y-intercept = 4 3 b. Plot the point (0, 4). Use the slope to find an additional point by moving 3 units to the right and 2 units down. average rate of change = 3 decreasing 16. p  x    x  6 a. Slope = 1 ; y-intercept = 6 b. Plot the point (0, 6). Use the slope to find an additional point by moving 1 unit to the right and 1 unit down. c. average rate of change =  d. decreasing 2 3 19. F  x   4 a. Slope = 0; y-intercept = 4 b. Plot the point (0, 4) and draw a horizontal line through it. c. d. 17. average rate of change = 1 decreasing 1 x 3 4 1 a. Slope = ; y-intercept = 3 4 b. Plot the point (0, 3) . Use the slope to find an additional point by moving 4 units to the right and 1 unit up. f  x  c. d. average rate of change = 0 constant 150 Copyright © 2019 Pearson Education, Inc. Section 2.1: Properties of Linear Functions and Linear Models 20. G  x   2 23. a. b. Slope = 0; y-intercept = 2 Plot the point (0, 2) and draw a horizontal line through it. c. d. average rate of change = 0 constant zero: 0  5 x  10 : y-intercept = 10 x2 b. Plot the points 1 unit to the right and 5 units down. a. 21. g  x   2 x  8 a. b. f  x   5 x  10 24. zero: 0  2 x  8 : y-intercept = 8 x4 Plot the points (4, 0), (0, 8) . f  x   6 x  12 a. b. 22. g  x   3 x  12 a. zero: 0  3x  12 : y-intercept = 12 x  4 b. Plot the points ( 4, 0), (0,12) . zero: 0  6 x  12 : y-intercept = 12 x2 Plot the points (2, 0), (0,12) . 1 25. H  x    x  4 2 1 a. zero: 0   x  4 : y-intercept = 4 2 x8 b. Plot the points (8, 0), (0, 4) . 151 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 1 x4 3 1 a. zero: 0  x  4 : y-intercept = 4 3 x  12 b. Plot the points (12, 0), (0, 4) . 26. G  x   29. x y 2 8 1 3 0 0 Avg. rate of change = 3   8  1   2  0   3 0   1  5 5 1  3 3 1 y x 1 1 2 0 Since the average rate of change is not constant, this is not a linear function. 30. 27. x y Avg. rate of change = y x y 2 4 1 0 0  (4) 4  4 1  (2) 1 2 4 1 1 1 4 3   3 1   2  1 0 4 40 4  4 0  (1) 1 0 2 2  1 3   3 0   1 1 1 8 84 4  4 1 0 1 1 5 5   2  1 0 8   5   x 2 8 y  31. y Avg. rate of change = x 1 4 1 1 2 0 1 y x 12  8 4  4 2 1 1 Since the average rate of change is constant at 4, this is a linear function with slope = 4. The y-intercept is (0, 4) , so the equation of the line is y  4x  4 . 3  3 1 3  3 2 1 1 Since the average rate of change is constant at 3 , this is a linear function with slope = –3. The y-intercept is (0, 2) , so the equation of the line is y  3 x  2 . 2 28. Avg. rate of change = x  12  14  14 1   1   2  1 4 1  12  12 1   0   1 1 2 2 12 x y 2 26 1 4 0 2 Avg. rate of change = 4   26  1   2  2   4  0   1 y x  22  22 1  6 6 1 1 –2 2 –10 Since the average rate of change is not constant, this is not a linear function. 1 2 2 4 Since the average rate of change is not constant, this is not a linear function. 152 Copyright © 2019 Pearson Education, Inc. Section 2.1: Properties of Linear Functions and Linear Models 32. x y 2 4 Avg. rate of change = y x 1 3.5 0 3 3  (3.5) 0.5   0.5 0  (1) 1 2.5 2.5  (3) 0.5   0.5 1 0 1 a. 33. b. 2 8 1 8 88 0  0 1  (2) 1 0 8 88 0  0 0  (1) 1 1 8 88 0  0 1 0 1 1 4  1 1  The solution set is  x x   or  ,   . 4 4     c. f  x  g  x 4 x  1  2 x  5 6x  6 x 1 d. f  x  g  x 4 x  1  2 x  5 6x  6 x 1 The solution set is  x x  1 or  , 1 . e. 88 0  0 2 1 1 Since the average rate of change is constant at 0, this is a linear function with slope = 0. The yintercept is (0, 8) , so the equation of the line is y  0 x  8 or y  8 . 34. f  x  0 x y Avg. rate of change = x y 1 4 4x 1  0 2 x 2 g  x   2 x  5 f  x  0 x 2  (2.5) 0.5   0.5 2 1 1 Since the average rate of change is constant at 0.5, this is a linear function with slope = 0.5. The y-intercept is (0, 3) , so the equation of the line is y  0.5 x  3 . 2 f  x   4 x  1; 4x 1  0 3.5  (4) 0.5   0.5 1  (2) 1 1 35. 8 Avg. rate of change = x y 2 0 1 1 1 0 1  1 1  (2) 1 0 4 4 1 3  3 0  (1) 1 36. f  x   3 x  5; a. y x g  x   2 x  15 f  x  0 3x  5  0 x b. 5 3 f  x  0 3x  5  0 x 1 9 2 16 Since the average rate of change is not constant, this is not a linear function. 5 3  5 5  The solution set is  x x    or  ,   . 3 3    153 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions c. f  x  g  x 38. a. 3x  5  2 x  15 5 x  10 b. The point (15, 60) is on the graph of y  g ( x) , so the solution to g ( x)  60 is x  15 . x2 d. The point (5, 20) is on the graph of y  g ( x) , so the solution to g ( x)  20 is x  5 . f  x  g  x c. 3x  5  2 x  15 5 x  10 The point (15, 0) is on the graph of y  g ( x) , so the solution to g ( x)  0 is x  15 . d. The y-coordinates of the graph of y  g ( x) are above 20 when the x-coordinates are smaller than 5. Thus, the solution to g ( x)  20 is x2 The solution set is  x x  2 or  2,   . e.  x x  5 or (, 5) . e. The y-coordinates of the graph of y  f ( x) are below 60 when the x-coordinates are larger than 15 . Thus, the solution to g ( x)  60 is  x x  15 or [15, ) . f. 37. a. The point (40, 50) is on the graph of y  f ( x) , so the solution to f ( x)  50 is x  40 .  x 15  x  15 or (15, 15) . b. The point (88, 80) is on the graph of y  f ( x) , so the solution to f ( x)  80 is x  88 . c. 39. a. b. f  x   g  x  when the graph of f is above the graph of g. Thus, the solution is  x x  4 or (, 4) . 40. a. f  x   g  x  when their graphs intersect. Thus, x  2 .  x x  40 or (40, ) . f. f  x   g  x  when their graphs intersect. Thus, x  4 . The point (40, 0) is on the graph of y  f ( x) , so the solution to f ( x)  0 is x  40 . d. The y-coordinates of the graph of y  f ( x) are above 50 when the x-coordinates are larger than 40. Thus, the solution to f ( x)  50 is e. The y-coordinates of the graph of y  f ( x) are between 0 and 60 when the xcoordinates are between 15 and 15. Thus, the solution to 0  f ( x)  60 is b. The y-coordinates of the graph of y  f ( x) are below 80 when the x-coordinates are smaller than 88. Thus, the solution to f ( x)  80 is  x x  88 or (, 88] . f  x   g  x  when the graph of f is below or intersects the graph of g. Thus, the solution is  x x  2 or  , 2 . 41. a. f  x   g  x  when their graphs intersect. Thus, x  6 . The y-coordinates of the graph of y  f ( x) are between 0 and 80 when the x-coordinates are between 40 and 88. Thus, the solution to 0  f ( x)  80 is  x 40  x  88 or b. g  x   f  x   h  x  when the graph of f is above or intersects the graph of g and below the graph of h. Thus, the solution is  x 6  x  5 or  6, 5 . (40, 88) . 154 Copyright © 2019 Pearson Education, Inc. Section 2.1: Properties of Linear Functions and Linear Models 42. a. f  x   g  x  when their graphs intersect. d. Thus, x  7 . b. g  x   f  x   h  x  when the graph of f is above or intersects the graph of g and below the graph of h. Thus, the solution is  x 4  x  7 or  4, 7  . e. 43. C  x   2.5 x  85 a. C  40   2.5  40   85  $185 . b. Solve C  x   2.5 x  85  245 2.5 x  85  245 2.5 x  100 160 x  64 miles 2.5 c. Solve C  x   0.35 x  45  150 2.5 x  85  150 2.5 x  105 65  26 miles x 2.5 d. The number of mile towed cannot be negative, so the implied domain for C is {x | x  0} or [0, ) . The cost of being towed increases $2.50 for each mile, or there is a charge of $2.50 per mile towed in addition to a fixed charge of $85. It costs $85 for towing 0 miles, or there is a fixed charge of $85 for towing in addition to a charge that depends on mileage. e. f. f. 45. S  p    600  50 p; D  p   1200  25 p a. b. Solve D  p   S  p  . 1200  25 p   600  50 p 1800  75 p 1800 p 75 24  p The demand will exceed supply when the price is less than $24 (but still greater than $0). That is, $0  p  $24 . C  50   0.07  50   24.99  $28.49 . c. b. Solve C  x   0.07 x  24.99  31.85 0.07 x  24.99  31.85 0.07 x  6.86 6.86  98 minutes x 0.07 c. Solve S  p   D  p  .  600  50 p  1200  25 p 75 p  1800 1800  24 p 75 S  24   600  50  24  600 Thus, the equilibrium price is $24, and the equilibrium quantity is 600 T-shirts. 44. C  x   0.07 x  24.99 a. The number of minutes cannot be negative, so x  0 . If there are 30 days in the month, then the number of minutes can be at most 30  24  60  43, 200 . Thus, the implied domain for C is {x | 0  x  43, 200} or [0, 43200] . The monthly cost of the plan increases $0.07 for each minute used, or there is a charge of $0.07 per minute to use the phone in addition to a fixed charge of $24.99. It costs $24.99 per month for the plan if 0 minutes are used, or there is a fixed charge of $24.99 per month for the plan in addition to a charge that depends on the number of minutes used. The price will eventually be increased. 46. S  p    2000  3000 p; D  p   10000  1000 p a. Solve C  x   0.07 x  24.99  36 0.07 x  24.99  36 0.07 x  11.01 11.01  157 minutes x 0.07 Solve S  p   D  p  .  2000  3000 p  10000  1000 p 4000 p  12000 12000 3 p 4000 S  3   2000  3000  3  7000 Thus, the equilibrium price is $3, and the equilibrium quantity is 7000 hot dogs. 155 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b. Solve D  p   S  p  . e. 0.15  x  9325   932.50  3673.75 10000  1000 p  2000  3000 p 12000  4000 p c. 47. a. b. c. d. We must solve T  x   3673.75 . 0.15 x  1398.75  932.50  3673.75 12000 p 4000 3 p The demand will be less than the supply when the price is greater than $3. The price will eventually be decreased. 0.15 x  466.25  3673.75 0.15 x  4140 f. We are told that the tax function T is for adjusted gross incomes x between $9,325 and $37,950, inclusive. Thus, the domain is  x 9,325  x  37,950 or 9325, 37950 . 48. a. T  20, 000   0.15  20, 000  9325   932.50  2533.75 If a single filer’s adjusted gross income is $20,000, then his or her tax bill will be $2533.75. The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T. b. x  27600 A single filer with an adjusted gross income of $27,600 will have a tax bill of $3673.75. For each additional dollar of taxable income between $9325 and $37,950, the tax bill of a single person in 2013 increased by $0.15. The independent variable is payroll, p. The payroll tax only applies if the payroll exceeds $189 million. Thus, the domain of T is  p | p  189 or (189, ) . T  243.8   0.5  243.8  189   27.4 The luxury tax for the New York Yankees was $27.4 million. c. Evaluate T at x  9325, 20000, and 37950 . T  9325   0.15  9325  9325   932.50 Evaluate T at p  189 , 243.8, and 300 million. T 189   0.5 189  189   0 million T  243.8   0.5  243.8  189   27.4 million T  300   0.5  300  189   55.5 million  932.50 T  20, 000   0.15  20, 000  9325   932.50 Thus, the points 189 million, 0 million  ,  2533.75 T  37,950   0.15  37950  9325   932.50  243.8 million, 27.4 million  , and  300 million, 55.5 million  are on the graph.  5226.25 Thus, the points  9325,932.50  ,  20000, 2533.75  , and  37950,5226.25 are on the graph. d. We must solve T  p   31.8 . 0.5  p  189   31.8 0.5 p  94.5  31.8 0.5 p  126.3 p  252.6 156 Copyright © 2019 Pearson Education, Inc. Section 2.1: Properties of Linear Functions and Linear Models e. If the luxury tax is $31.8 million, then the payroll of the team is $252.6 million. For each additional million dollars of payroll in excess of $189 million in 2016, the luxury tax of a team increased by $0.5 million. $0 after 3 years. Thus, the implied domain for V is {x | 0  x  3} or [0, 3]. c. The graph of V ( x)  1000 x  3000 d. V (2)  1000(2)  3000  1000 The computer’s book value after 2 years will be $1000. e. Solve V ( x)  2000 1000 x  3000  2000 1000 x  1000 x 1 The computer will have a book value of $2000 after 1 year. 49. R  x   8 x; C  x   4.5 x  17,500 a. Solve R  x   C  x  . 8 x  4.5 x  17,500 3.5 x  17,500 x  5000 The break-even point occurs when the company sells 5000 units. b. Solve R  x   C  x  8 x  4.5 x  17,500 3.5 x  17,500 x  5000 The company makes a profit if it sells more than 5000 units. 50. R ( x)  12 x; C ( x)  10 x  15, 000 a. Solve R ( x)  C ( x) 12 x  10 x  15, 000 2 x  15, 000 x  7500 The break-even point occurs when the company sells 7500 units. 52. a. slope formula yields: y 0  120000 120000 m    12000 x 10  0 10 The y-intercept is  0,120000  , so b. Solve R ( x)  C ( x) 12 x  10 x  15, 000 2 x  15, 000 x  7500 The company makes a profit if it sells more than 7500 units. 51. a. Consider the data points  x, y  , where x = the age in years of the machine and y = the value in dollars of the machine. So we have the points  0,120000  and 10, 0  . The b  120, 000 . Therefore, the linear function is V  x   mx  b  12, 000 x  120, 000 . b. The age of the machine cannot be negative, and the book value of the machine will be $0 after 10 years. Thus, the implied domain for V is {x | 0  x  10} or [0, 10]. Consider the data points ( x, y ) , where x = the age in years of the computer and y = the value in dollars of the computer. So we have the points (0,3000) and (3, 0) . The slope formula yields: y 0  3000 3000 m    1000 x 30 3 The y-intercept is (0,3000) , so b  3000 . Therefore, the linear function is V ( x)  mx  b  1000 x  3000 . c. The graph of V  x   12, 000 x  120, 000 b. The age of the computer cannot be negative, and the book value of the computer will be 157 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions d. V  4   12000  4   120000  72000 c. The graph of C  x   90 x  1805 d. The cost of manufacturing 14 bicycles is given by C 14   90 14   1805  $3065 . e. Solve C  x   90 x  1805  3780 90 x  1805  3780 90 x  1975 x  21.94 So approximately 21 bicycles could be manufactured for $3780. The machine’s value after 4 years is given by $72,000. e. 53. a. Solve V  x   72000 . 12000 x  120000  72000 12000 x  48000 x4 The machine will be worth $72,000 after 4 years. Let x = the number of bicycles manufactured. We can use the cost function C  x   mx  b , with m = 90 and b = 1800. Therefore C  x   90 x  1800 b. The graph of C  x   90 x  1800 55. a. b. c. The cost of manufacturing 14 bicycles is given by C 14   90 14   1800  $3060 . b. C 110    0.89 110   39.95  $137.85 C  230    0.89  230   39.95  $244.65 56. a. d. Solve C  x   90 x  1800  3780 90 x  1800  3780 90 x  1980 x  22 So 22 bicycles could be manufactured for $3780. 54. a. Let x = number of miles driven, and let C = cost in dollars. Total cost = (cost per mile)(number of miles) + fixed cost C  x   0.89 x  39.95 Let x = number of megabytes used, and let C = cost in dollars. Total cost = (cost per megabyte)(number of megabytes over 200) + fixed cost: C ( x)  0.25( x  200)  40  0.25 x  50  40  0.25 x  10, x  200 The new daily fixed cost is 100 1800   $1805 20 b. C  265    0.25  265   10  $56.25 C  300    0.25  300   10  $65 Let x = the number of bicycles manufactured. We can use the cost function C  x   mx  b , with m = 90 and b = 1805. Therefore C  x   90 x  1805 158 Copyright © 2019 Pearson Education, Inc. Section 2.1: Properties of Linear Functions and Linear Models 57. a. e. f. b. Avg. rate of change = n m m n 4 30 16 120 120  30 90 15   16  4 12 2 64 480 480  120 360 15   64  16 48 2 128 960 960  480 480 15   128  64 64 2 58. a. Since each input (memory) corresponds to a single output (recording time), we know that recording time is a function of memory. Also, because the average rate of change is constant at 7.5 minutes per gigabyte, the function is linear. c. If memory increases by 1 GB, then the number of songs increases by 218.75. b. From part (b), we know slope  7.5 . Using (m1 , n1 )  (4, 30) , we get the equation: t  t1  s (m  m1 ) t  30  7.5(m  4) t  30  7.5m  30 t  7.5m Using function notation, we have t (m)  7.5m . d. The memory cannot be negative, so m  0 . Likewise, the time cannot be negative, so, t ( m)  0 . 7.5m  0 m0 Thus, the implied domain for n(m) is {m | m  0} or  0,   . c. Avg. rate of change = h s s h 20 0 15 3 30 3   0.6 15  20 5 10 6 63 3   0.6 10  15 5 5 9 96 3   0.6 5  10 5 Since each input (soda) corresponds to a single output (hot dogs), we know that number of hot dogs purchased is a function of number of sodas purchased. Also, because the average rate of change is constant at 0.6 hot dogs per soda, the function is linear. From part (b), we know m  0.6 . Using ( s1 , h1 )  (20, 0) , we get the equation: h  h1  m( s  s1 ) h  0  0.6( s  20) h  0.6s  12 Using function notation, we have h( s )  0.6s  12 . 159 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions mx  b  mx  b  mxb  mx 0  2mx m0 So, yes, a linear function f  x   mx  b cab be d. The number of sodas cannot be negative, so s  0 . Likewise, the number of hot dogs cannot be negative, so, h( s )  0 . 0.6 s  12  0 0.6 s  12 s  20 Thus, the implied domain for h(s) is {s | 0  s  20} or [0, 20] . even provided m  0 . e. 62. If you solve the linear function f  x   mx  b for 0 you are actually finding the x-intercept. Therefore using x-intercept of the graph of f  x   mx  b would be same x-value as solving mx  b  0 for x. Then the appropriate interval could be determined x 2  4 x  y 2  10 y  7  0 63. ( x 2  4 x  4)  ( y 2  10 y  25)  7  4  25 f. g. If the number of hot dogs purchased increases by $1, then the number of sodas purchased decreases by 0.6. s-intercept: If 0 hot dogs are purchased, then 20 sodas can be purchased. h-intercept: If 0 sodas are purchased, then 12 hot dogs may be purchased. ( x  2) 2  ( y  5) 2  62 Center: (2, -5); Radius = 6 59. The graph shown has a positive slope and a positive y-intercept. Therefore, the function from (d) and (e) might have the graph shown. 60. The graph shown has a negative slope and a positive y-intercept. Therefore, the function from (b) and (e) might have the graph shown. 61. A linear function f  x   mx  b will be odd provided f   x    f  x  . 64. That is, provided m   x   b    mx  b  . mx  b   mx  b b  b 2b  0 b0 So a linear function f  x   mx  b will be odd provided b  0 . 2x  B x3 2(5)  B f (5)  8  53 10  B 8 2 16  10  B B6 f ( x)  A linear function f  x   mx  b will be even provided f   x   f  x  . That is, provided m   x   b  mx  b . 160 Copyright © 2019 Pearson Education, Inc. Section 2.2: Building Linear Models from Data 65. f (3)  f (1) 3 1 12  ( 2)  2 14  2 7 6. Nonlinear relation 7. Linear relation, m  0 8. Linear relation, m  0 9. Nonlinear relation 10. Nonlinear relation 11. a. 66. b.   Answers will vary. We select (4, 6) and (8, 14). The slope of the line containing these points is: 14  6 8 m  2 84 4 The equation of the line is: y  y1  m( x  x1 ) y  6  2( x  4) Section 2.2 1.  y  6  2x  8 y  2x  2 y  c.  d.   Using the LINear REGression program, the line of best fit is: y  2.0357 x  2.3571 e.    x No, the relation is not a function because an input, 1, corresponds to two different outputs, 5 and 12. 2. Let  x1 , y1   1, 4  and  x2 , y2    3, 8  . m y2  y1 8  4 4   2 x2  x1 3  1 2 y  y1  m  x  x1  y  4  2  x  1 y  4  2x  2 y  2x  2 12. a.    3. scatter diagram 4. decrease; 0.008  5. Linear relation, m  0  161 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b. Answers will vary. We select (5, 2) and (11, 9). The slope of the line containing 92 7  these points is: m  11  5 6 The equation of the line is: y  y1  m( x  x1 )  7 ( x  5) 6 7 35 y2  x 6 6 7 23 y  x 6 6  y2  c.  c. e.  Using the LINear REGression program, the line of best fit is: y  2.2 x  1.2     d. e.  b.   13. a.   b.  14. a.  Using the LINear REGression program, the line of best fit is: y  1.1286 x  3.8619   Answers will vary. We select (–2,–4) and (2, 5). The slope of the line containing 5  ( 4) 9 these points is: m   . 2  ( 2) 4 The equation of the line is: y  y1  m( x  x1 ) 9 y  ( 4)  ( x  ( 2)) 4 9 9 y4 x 4 2 9 1 y  x 4 2 c.  Answers will vary. We select (–2, 7) and (2, 0). The slope of the line containing 07 7 7 these points is: m    . 2  (2) 4 4 The equation of the line is: y  y1  m( x  x1 ) 7 y  7   ( x  ( 2)) 4 7 7 y7   x 4 2 7 7 y   x 4 2   d.  Using the LINear REGression program, the line of best fit is: y  1.8 x  3.6 162 Copyright © 2019 Pearson Education, Inc. Section 2.2: Building Linear Models from Data e.  b.  The equation of the line is: y  y1  m( x  x1 )   15. a. b. 1  x  (30)  2 1 y  10  x  15 2 1 y  x  25 2 y  10    Answers will vary. We select (–20,100) and (–10,140). The slope of the line containing these points is: 140  100 40 m  4 10   20  10 c. The equation of the line is: y  y1  m( x  x1 ) d. y  100  4  x  (20)  y  100  4 x  80 e. y  4 x  180 c. Selection of points will vary. We select (–30, 10) and (–14, 18). The slope of the line containing these points is: 18  10 8 1 m   14   30  16 2    Using the LINear REGression program, the line of best fit is: y  0.4421x  23.4559     d. e.   Using the LINear REGression program, the line of best fit is: y  3.8613 x  180.2920  17. a.   16. a. b. Linear. c. Answers will vary. We will use the points (39.52, 210) and (66.45, 280) .    163 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions d. 280  210 70   2.5993316 66.45  39.52 26.93 y  210  2.5993316( x  39.52) y  210  2.5993316 x  102.7255848 y  2.599 x  107.274 m d. e. e. f. f. x  62.3 : y  2.599(62.3)  107.274  269 We predict that a candy bar weighing 62.3 grams will contain 269 calories. If the weight of a candy bar is increased by one gram, then the number of calories will increase by 2.599. 19. a. 18. a. L(450)  0.011(450)  0.3  5.25 We predict that the approximately length of a 450 yard wide tornado is 5.25 miles. For each 1-yard increase in the width of a tornado, the length of the tornado increases by 0.011 mile, on average. The independent variable is the number of hours spent playing video games and cumulative grade-point average is the dependent variable because we are using number of hours playing video games to predict (or explain) cumulative grade-point average. b. b. c. Linear with positive slope. Answers will vary. We will use the points (200, 2.5) and (500, 5.8) . c. Using the LINear REGression program, the line of best fit is: G (h)  0.0942h  3.2763 m d. L  2.5  0.011 w  200 L  2.5  0.011w  2.2 L  0.011w  0.3 e. If the number of hours playing video games in a week increases by 1 hour, the cumulative grade-point average decreases 0.09, on average. G (8)  0.0942(8)  3.2763  2.52 We predict a grade-point average of approximately 2.52 for a student who plays 8 hours of video games each week. 2.40  0.0942(h)  3.2763 2.40  3.2763  0.0942h 0.8763  0.0942h 9.3  h 5.8  2.5 3.3   0.011 500  200 300 L  L1  m  w  w1  f. 164 Copyright © 2019 Pearson Education, Inc. Section 2.2: Building Linear Models from Data A student who has a grade-point average of 2.40 will have played approximately 9.3 hours of video games. 20. a.  d. For each 1-mph increase in the speed off bat, the homerun distance increases by 3.3641 feet, on average. e. d ( s)  3.3641s  51.8233 f. Since the speed off bat must be greater than 0 the domain is  s | s  0 . g. d (103)  3.3641(103)  51.8233  398 ft A hurricane with a wind speed of 85 knots would have a pressure of approximately 967 millibars. a. The relation is a function because none of the invariables are repeated.   b. 22. Using the LINear REGression program, the line of best fit is: w( p)  1.1857 p  1231.8279 c. d. e. 21. a. b. For each 10-millibar increase in the atmospheric pressure, the wind speed of the tropical system decreases by 1.1857 knots, on average. w(990)  1.1857(990)  1231.8279  58 knots To find the pressure, we solve the following equation: 85  1.1857 p  1231.8279 1146.8279  1.1857 p 967  p A hurricane with a wind speed of 85 knots would have a pressure of approximately 967 millibars. c. d. If Internet ad spending increases by 1%, magazine ad spending goes down by about 0.2277%, on average. This relation does not represent a function since the values of the input variable s are repeated. e. m  n   0.2277 n  15.9370 f. Domain: n 0  n  70.0 Note that the m-intercept is roughly 15.9 and that the percent of Internet sales cannot be negative. b. g. c. Using the LINear REGression program, the line of best fit is: m  0.2277n  15.9370 . D(28)  0.2277(26.0)  15.9370  10.0 Percent of magazine sales is about 10.0%. Using the LINear REGression program, the line of best fit is: d  3.3641s  51.8233 165 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions x 2  25  0 x 2  25  x  5 So the domain is:  x | x  5, 5 23. f ( x)  5 x  8 and g ( x)  x 2  3x  4 30. ( g  f )( x)  ( x 2  3 x  4)  (5 x  8)  x 2  3x  4  5 x  8  x 2  8 x  12 31. Since y is shifted to the left 3 units we would use y  ( x  3) 2 . Since y is also shifted down 4 units,we would use y  ( x  3) 2  4 . The data do not follow a linear pattern so it would not make sense to find the line of best fit. Section 2.3 24. Using the ordered pairs (1, 5) and (3, 8) , the line of best fit is y  1.5 x  3.5 . 1. a. b. 2. x 2  5 x  6   x  6  x  1 2 x 2  x  3   2 x  3 x  1 82  4  2  3  64  24  40  4 10  2 10 The correlation coefficient is r  1 . This is reasonable because two points determine a line. 3. x  3  0 or 3x  5  0 25. A correlation coefficient of 0 implies that the data do not have a linear relationship. x3 3x  5 5 3 5   The solution set is  ,3 .  3  x 26. The y-intercept would be the calories of a candy bar with weight 0 which would not be meaningful in this problem. 27. G (0)  0.0942(0)  3.2763  3.2763 . The approximate grade-point average of a student who plays 0 hours of video games per week would be 3.28. 28. m   x  3 3x  5  0 2 1  4. add;   6   9 2  5. If f (4)  10 , then the point (4, 10) is on the graph of f. 3  5 8   2 3  ( 1) 4 6. y  y1  m  x  x1  f  3  (3) 2  4(3)  3  9  12  3  0 3 is a zero of f  x  . y  5  2  x  1 y  5  2 x  2 y  2 x  3 or 2x  y  3 7. repeated; multiplicity 2 8. discriminant; negative 29. The domain would be all real numbers except those that make the denominator zero. 9. A quadratic functions can have either 0, 1 or 2 real zeros. 166 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros 10. x  F  x  0 17. b  b 2  4ac 2a 2 11. False; the equation will have only two real solution but not necessarily negatives of one another. x  x6  0 ( x  3)( x  2)  0 x  3  0 or x  2  0 x  3 x2 12. b The zeros of F  x   x 2  x  6 are 3 and 2. 13. The x-intercepts of the graph of F are 3 and 2. f  x  0 x2  7 x  6  0 x  x  9  0 x  0 or ( x  6)( x  1)  0 x  6  0 or x  1  0 x  6 x  1 x9  0 x9 The zeros of f  x   x 2  9 x are 0 and 9. The x- The zeros of H  x   x 2  7 x  6 are 6 and 1 . intercepts of the graph of f are 0 and 9. 14. The x-intercepts of the graph of H are 6 and 1 . f  x  0 2 2 x  5x  3  0 (2 x  1)( x  3)  0 x  x  4  0 x40 x  4 2x  1  0 x The zeros of f  x   x 2  4 x are 4 and 0. The x-intercepts of the graph of f are 4 and 0. 15. or x  3  0 x3 1 2 1 and 3. 2 1 The x-intercepts of the graph of g are  and 3. 2 The zeros of g  x   2 x 2  5 x  3 are  g  x  0 x 2  25  0 ( x  5)( x  5)  0 x  5  0 or x  5  0 x  5 x5 20. f  x  0 3x 2  5 x  2  0 (3x  2)( x  1)  0 The zeros of g  x   x 2  25 are 5 and 5. The 3x  2  0 x-intercepts of the graph of g are 5 and 5. 16. g  x  0 19. x2  4 x  0 x  0 or H  x  0 18. x2  9 x  0 or x  1  0 2 x  1 3 The zeros of f  x   3 x 2  5 x  2 are 1 and x G  x  0 x2  9  0 ( x  3)( x  3)  0 x  3  0 or x  3  0 x  3 x3 2  . The x-intercepts of the graph of f are 1 3 2 and  . 3 The zeros of G  x   x 2  9 are 3 and 3. The x-intercepts of the graph of G are 3 and 3. 167 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions P  x  0 21. 3 . 2 3 The only x-intercept of the graph of G is . 2 The only zero of G  x   4 x 2  9  12 x is 2 3 x  48  0 3( x 2  16)  0 3( x  4)( x  4)  0 t  4  0 or t  4  0 F  x  0 26. 2 t4 25 x  16  40 x  0 The zeros of P  x   3 x 2  48 are 4 and 4. 25 x 2  40 x  16  0 (5 x  4)(5 x  4)  0 t  4 The x-intercepts of the graph of P are 4 and 4. 5 x  4  0 or 5 x  4  0 4 4 x x 5 5 H  x  0 22. 2 x 2  50  0 2( x 2  25)  0 The only zero of F  x   25 x 2  16  40 x is 2( x  5)( x  5)  0 y  5  0 or y  5=0 y  5 The only x-intercept of the graph of F is y5 The zeros of H  x   2 x 2  50 are 5 and 5. 27. The x-intercepts of the graph of H are 5 and 5. x   8  2 2 The zeros of f  x   x 2  8 are 2 2 and 2 2 . x 2  8 x  12  0 The x-intercepts of the graph of f are 2 2 and 2 2. ( x  6)  x  2   0 x  6 or x  2 The zeros of g  x   x  x  8   12 are 6 and 2 . 28. The x-intercepts of the graph of g are 6 and 2 . g  x  0 x 2  18  0 x 2  18 f  x  0 x  x  4   12  0 x   18  3 3 The zeros of g  x   x 2  18 are 3 3 and x 2  4 x  12  0 ( x  6)  x  2   0 3 3 . The x-intercepts of the graph of g are 3 3 and 3 3 . x  2 or x  6 The zeros of f  x   x  x  4   12 are 2 and 6. 29. The x-intercepts of the graph of f are 2 and 6. g  x  0  x  12  4  0  x  12  4 G  x  0 25. f  x  0 x2  8 x  x  8   12  0 24. 4 . 5 x2  8  0 g  x  0 23. 4 . 5 2 4 x  9  12 x  0 x 1   4 x  1  2 x  1  2 or x  1  2 x3 x  1 4 x 2  12 x  9  0 (2 x  3)(2 x  3)  0 2 x  3  0 or 2 x  3  0 3 3 x x 2 2 The zeros of g  x    x  1  4 are 1 and 3. 2 The x-intercepts of the graph of g are 1 and 3. 168 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros G  x  0 30. f  x  0 33.  x  2  1  0  x  2 2  1 2 2 x  4x  8  0 x2  4 x  8 x2  4 x  4  8  4 x2  1 x  2  1 x  2  1 or x  2  1 x  1 x  3  x  2 2  12 x  2   12 x  2  2 3 The zeros of G  x    x  2   1 are 3 and 1 . 2 x  2  2 3 x  2  2 3 or x  2  2 3 The x-intercepts of the graph of G are 3 and 1 . The zeros of f  x   x 2  4 x  8 are 2  2 3 F  x  0 31.  2 x  32  32  0  2 x  32  32 and 2  2 3 . The x-intercepts of the graph of f are 2  2 3 and 2  2 3 . 2 x  3   32 2 x  3  4 2 x2  6 x  9  0 2 x  3  4 2 x f  x  0 34. x2  6 x  9  9  9  x  32  18 3  4 2 2 x  3   18 The zeros of F  x    2 x  3  32 are 2 x  33 2 3  4 2 3  4 2 and . The x-intercepts of 2 2 3  4 2 3  4 2 and . the graph of F are 2 2 The zeros of f  x   x 2  6 x  9 are 3  3 2 and 3  3 2 . The x-intercepts of the graph of f are 3  3 2 and 3  3 2 . F  x  0 32. 1 3 x 0 2 16 1 3 x2  x  2 16 1 1 3 1 x2  x    2 16 16 16  3x  2   75  0  3x  2 2  75 2 x2  3 x  2   75 3 x  2  5 3 3x  2  5 3 x g  x  0 35. 2 1 1  x 4  4   25 3 3 1 1 1   4 4 2 1 1 x  4 2 3 1 x or x   4 4 1 3 1 3 are  and . The zeros of g  x   x 2  x  2 16 4 4 1 3 The x-intercepts of the graph of g are  and . 4 4 x 25 3 The zeros of G  x    3 x  2   75 are 3 2 and 25 3 . The x-intercepts of the graph of G 3 are 25 3 25 3 and . 3 3 169 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions g  x  0 36. 2 1 x 0 3 3 2 1 2 x  x 3 3 2 1 1 1 x2  x    3 9 3 9 2 2 x  3x  1  0 3 1 x2  x   0 2 2 3 1 2 x  x 2 2 3 9 1 9 2 x  x   2 16 2 16 x2  2 1 4  x 3  9   x 2 3  17   x  4   16   1 4 2   3 9 3 1 2 x  3 3 1 x  or x  1 3 x 4 3  17 . The x-intercepts of the graph of G are 4 3  17 4 F  x  0 4 x2  4 x  2  0 a  1, b   4, c  2 x ( 4)  ( 4) 2  4(1)(2) 4  16  8  2(1) 2 4 8 42 2   2 2 2 2 The zeros of f  x   x 2  4 x  2 are 2  2 and  2 1 7   x  6   36   2  2 . The x-intercepts of the graph of f are 2  2 and 2  2 . 1 7 7   6 36 6 f  x  0 40. 1  7 6 The zeros of F  x   3 x 2  x  and 3  17 . f  x  0 39. 1 0 2 1 1 x2  x   0 3 6 1 1 x2  x  3 6 1 1 1 1 2 x  x   3 36 6 36 3x 2  x  x 3  17 4 The zeros of G  x   2 x 2  3x  1 are 3  17 and 2 1 1 The zeros of g  x   x  x  are 1 and . 3 3 3 1 The x-intercepts of the graph of g are 1 and . 3 x 3 17 17   4 16 4 x 2 37. G  x  0 38. x2  4 x  2  0 a  1, b  4, c  2 1 are 1  7 and 6 2  4  42  4(1)(2)  4  16  8  2(1) 2 1  7 . The x-intercepts of the graph of F are 6 x 1  7 6 4 8 4 2 2   2 2 2 2 The zeros of f  x   x 2  4 x  2 are  2  2 and 1  7 .  6 and  2  2 . The x-intercepts of the graph of f are  2  2 and  2  2 . 170 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros g  x  0 41. x  4x 1  0 a  1, b   4, c  1 x 2 4x  x  2  0 a  4, b   1, c  2 ( 4)  ( 4) 2  4(1)(1) 2(1)  4  16  4 2 x (1)  (1) 2  4(4)(2) 1  1  32  2(4) 8 4  20 4  2 5   2 5 2 2 The zeros of g  x   x 2  4 x  1 are 2  5 and 1  31  not real 8 The function P  x   4 x 2  x  2 has no real 2  5 . The x-intercepts of the graph of g are zeros, and the graph of P has no x-intercepts.   2  5 and 2  5 . 4×2  x  1  0 a  4, b  1, c  1 x  6x  1  0 a  1, b  6, c  1 2 t  6  6  4(1)(1)  6  36  4  2(1) 2 2 x 1  15  not real 8 The function H  x   4 x 2  x  1 has no real zeros, and the graph of H has no x-intercepts. and 3  2 2 . The x-intercepts of the graph of g are 3  2 2 and 3  2 2 . 2 4x 1 2x  0 4 x2  2 x  1  0 a  4, b  2, c  1 2 x  5x  3  0 a  2, b   5, c  3 2 x 2(2) f  x  0 47. F  x  0 ( 5)  ( 5) 2  4(2)(3) 1  12  4(4)(1) 1  1  16  2(4) 8   6  32  6  4 2    3  2 2 2 2 The zeros of g  x   x 2  6 x  1 are 3  2 2 43. H  x  0 46. g  x  0 42. 5  25  24  4 5 1 3   or 1 4 2 x  2  22  4(4)(1)  2  4  16  2(4) 8   2  20  2  2 5 1  5   8 8 4 The zeros of f  x   4 x 2  1  2 x are 1  5 The zeros of F  x   2 x 2  5 x  3 are 1 and 3 . 4 2 3 The x-intercepts of the graph of F are 1 and . 2 44. P  x  0 45. 2 and 1  5 . The x-intercepts of the graph of f 4 are 1  5 and 1  5 . g  x  0 4 2 x2  5x  3  0 a  2, b  5, c  3 4 f  x  0 48. 2 2x 1 2x  0 5  52  4(2)(3) 5  25  24 x  2(2) 4 5  1 3   1 or  4 2 2 x2  2 x  1  0 a  2, b  2, c  1 x  2  22  4(2)(1)  2  4  8  2(2) 4   2  12  2  2 3 1  3   4 4 2 The zeros of g  x   2 x 2  5 x  3 are  3 and 1 . 2 3 The x-intercepts of the graph of g are  and 1 . 2 171 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 2 2 4 x  20 x  25  0 a  4, b  20, c  25 and 1  3 . The x-intercepts of the graph of f 2 are 1  3 2 and 1  3 . x 2 x( x  2)  3  0 2 x2  4 x  3  0 a  2, b  4, c  3 x    4    4   4(2)(3) 2 4  16  24 4  2(2) f  x  g  x x  6x  3  3 x2  6x  0  x  x  6  0 x  0 or x  6  0 x  6 The x-coordinates of the points of intersection are 6 and 0. The y-coordinates are g  6   3 and 2 and 2  10 . The x-intercepts of the graph of G 2 are 2  10 and 2  10 . 2 2 g  0   3 . The graphs of the f and g intersect at F  x  0 the points (6,3) and (0,3) . 3x( x  2)  1  0  3 x  6 x  1  0 a  3, b  6, c  1 2 x   6 2  4(3)(1) 2(3) 2 x  4x  3  3 6  36  12  6 x2  4 x  0 x  x  4  0 6  48 6  4 3 3  2 3  = 6 6 3 x  0 or x  4  0 x4 The x-coordinates of the points of intersection are 0 and 4. The y-coordinates are g  0   3 and 3 and 3  2 3 . The x-intercepts of the graph of G 3 g  4   3 . The graphs of the f and g intersect at are 3  2 3 and 3  2 3 . 3 3 the points (0,3) and (4,3) . p  x  0 55. 2 9x  6x  1  0 a  9, b  6, c  1 x   6    6   4(9)(1) f  x  g  x 2 x 2  1  3 x  2 0  2 x 2  3x  1 2 2(9) f  x  g  x 54. The zeros of F  x   3 x( x  2)  2 are 3  2 3 51. 2 2 4  40 4  2 10 2  10  = 4 4 2   6   20   4(4)(25) 5 5 . The only x-intercept of the graph of F is  . 2 2 53. The zeros of G  x   2 x( x  2)  3 are 2  10 50. 20  20  400  400  2(4) 8 20  0 20 5    8 8 2 The only real zero of q  x   4 x 2  20 x  25 is 2 G  x  0 49. q  x  0 52. The zeros of f  x   2 x 2  1  2 x are 1  3  0   2 x  1 x  1 6  36  36 18 2x 1  0 or x  1  0 1 x  1 x 2 The x-coordinates of the points of intersection 60 1   18 3 The only real zero of p  x   9 x 2  6 x  1 is 1 . 3 1 2 are 1 and  . The y-coordinates are 1 The only x-intercept of the graph of g is . 3 g  1  3  1  2  3  2  1 and 172 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros f  6    6   5  6   3  36  30  3  3 and 3 1  1  1 g     3    2    2  . 2 2 2 2     The graphs of the f and g intersect at the points 2 f  4   42  5  4   3  16  20  3  33 . The graphs of the f and g intersect at the points (6, 3) and  4, 33 .  1 1   (1, 1) and   ,  . 2 2 f  x  g  x 56. P  x  0 59. 4 x  6 x  16  0 3x 2  7  10 x  1  x  2 x  8  0 2 3x 2  10 x  8  0  3x  2  x  4   0 or x  4  0 2 x4 x 3 The x-coordinates of the points of intersection or x 2  8  0 x 2  2 x2  8 x   2  not real x 8 The zeros of P  x   x  6 x 2  16 are 2 2 2 and 4. The y-coordinates are 3 and 2 2 . The x-intercepts of the graph of P are 2 2 and 2 2 . H  x  0 60. The graphs of the f and g intersect at the points 4 x  3×2  4  0  2 17    ,   and  4, 41 .  3 3  x  1 x  4   0 2 f  x  g  x x 2  x  1  2 x 2  3x  14 2 x2  1  0 or x 2  4  0 x 2  1 x2  4 x   1 x 4  not real  2 4 The zeros of H  x   x  3 x 2  4 are 2 and 2. 0  x 2  2 x  15 0   x  3 x  5  x  3  0 or x  5  0 x  3 x5 The x-coordinates of the points of intersection are 3 and 5. The y-coordinates are The x-intercepts of the graph of H are 2 and 2. f  x  0 61. x4  5×2  4  0 f  3   3   3  1  9  3  1  13 and 2  x  4 x  1  0 2 f  5   52  5  1  25  5  1  21 . 2 x 2  4  0 or x 2  1  0 x  2 or x  1 The zeros of f  x   x 4  5 x 2  4 are 2 , 1 , The graphs of the f and g intersect at the points (3, 13) and  5, 21 . 58.  2 2 4 20 17  2  2 g     10     1    1   and 3 3  3  3 g  4   10  4   1  40  1  41 . 57. 2 x2  2  0 3x  2  0 are  2 f  x  g  x 1, and 2. The x-intercepts of the graph of f are 2 , 1 , 1, and 2. x 2  5 x  3  2 x 2  7 x  27 0  x 2  2 x  24 0   x  6  x  4  x  6  0 or x  4  0 x  6 x4 The x-coordinates of the points of intersection are 6 and 4. The y-coordinates are 173 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions f  x  0 The zeros of g  x   x 6  7 x3  8 are 2 and 1. x 4  10 x 2  24  0 The x-intercepts of the graph of g are 2 and 1. 62.  x  4  x  6  0 2 2 2 x 4  0 x 6  0 or 2 x 6  7 x3  8  0  x  8 x  1  0 2 x 4 x  2 x 6 3 x 6 2 6 , 2 and 2 . The x-intercepts of the graph of f are  6 , 6 , 2 and 2 . 1 3 x2  1 The x-intercepts of the graph of G are 1 and 1. The zeros of G  x    x  2   7  x  2   12 are 2 F  x  0 6 and 5 . The x-intercepts of the graph of G are 6 and 5 . 2 2 x  5 x  12  0  2 x  3 x  4  0 2 2x  3  0 x2   f  x  0 68. or  2 x  5   2 x  5  6  0 2 Let u  2 x  5  u 2   2 x  5  2 2 x 4  0 3 2 x2  4 x 4 3 x  2  not real u2  u  6  0  u  3 u  2   0  2 u 3  0 u 3 The zeros of F  x   2 x 4  5 x 2  12 are 2 and 2. 2x  5  3 x  1 The x-intercepts of the graph of F are 2 and 2. g  x  0 65. G  x  0 u  3  0 or u  4  0 u  3 u  4 x  2  3 x  2  4 x  5 x  6  1 The zeros of G  x   3x 4  2 x 2  1 are 1 and 1. 2 x  1 u 2  7u  12  0  u  3 u  4   0 x 1 1 x  3 x  not real 2 x2  x  2 2  7  x  2   12  0 2 Let u  x  2  u 2   x  2  x2  1  0 or 4 x3  1 67.  3x 2  1 x 2  1  0 64. x3  8 The x-intercepts of the graph of g are 1 and 2. 3x 4  2 x 2  1  0 x2   x3  1  0 or The zeros of g  x   x 6  7 x3  8 are 1 and 2. G  x  0 3x 2  1  0 3 x3  8  0 The zeros of f  x   x  10 x  24 are  6 , 4 63. g  x  0 66. 2 or u20 u  2 2 x  5  2 7 x 2 x  7 x3  8  0 The zeros of f  x    2 x  5    2 x  5   6 are 3 7 and 1 . The x-intercepts of the graph of f 2 7 are  and 1 . 2 2 6  x  8 x  1  0 3 x3  8  0 or  x3  1  0 x 3  8 x3  1 x  2 x 1 174 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros f  x  0 69. 3 and 2. The x-intercepts of the graph of P 2 3 are  and 2. 2   3x  4   6  3x  4   9  0 2 Let u  3 x  4  u 2   3x  4  2 u 2  6u  9  0  u  3  0 3 1  x   5 1  x   2  0 2 u 3  0 u3 3x  4  3 x Let u  1  x  u 2  1  x   3u  2  u  1  0 1 3 3u  2  0 or u  1  0 2 u  1 u 3 1  x  1 2 x2 1 x   3 5 x 3 2 1 3 1 3 is  . The x-intercept of the graph of f is  . H  x  0  2  x    2  x   20  0 2 Let u  2  x  u 2   2  x  2 The zeros of H  x   3 1  x   5 1  x   2 are 2 5 and 2. The x-intercepts of the graph of H are 3 5 and 2. 3 u 2  u  20  0  u  5  u  4   0 u 5  0 or u  4  0 u  5 u4 2  x  5 2 x  4 x7 x  2 73. G  x  0 x4 x  0 The zeros of H  x    2  x    2  x   20 are 2 Let u  x  u 2  x u 2  4u  0 2 and 7. The x-intercepts of the graph of H are 2 and 7. u u  4  0 P  x  0 71. 2 3u 2  5u  2  0 The only zero of f  x    3 x  4   6  3 x  4   9 70. H  x  0 72. 2 u0 or u  4  0 u4 2  x  1  5  x  1  3  0 2 Let u  x  1  u 2   x  1 x 0 2 x 4 2 x0 0 2 2u  5u  3  0 x  42  16 Check: G  0  0  4 0  0  2u  1 u  3  0 2u  1  0 or u  3  0 u3 1 u 2 x 1  3 1 x2 x 1   2 3 x 2 G 16   16  4 16  16  16  0 The zeros of G  x   x  4 x are 0 and 16. The x-intercepts of the graph of G are 0 and 16. The zeros of P  x   2  x  1  5  x  1  3 are 2 175 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 74. f  x  0 x 8 x  0 2 x  50  0 2 Let u  x  u  x x 2  50  x   50  5 2 2 u  8u  0 The zeros of f  x   x 2  50 are 5 2 and u u  8  0 u0 5 2 . The x-intercepts of the graph of f are or u  8  0 5 2 and 5 2 . u  8 x 0 78. x  8 x  not real x  02  0 f  x  0 x 2  20  0 x 2  20  x   20  2 5 Check: f  0   0  8 0  0 The zeros of f  x   x 2  6 are 2 5 and The only zero of f  x   x  8 x is 0. The only 2 5 . The x-intercepts of the graph of f are x-intercept of the graph of f is 0. 75. f  x  0 77. 2 5 and 2 5 . g  x  0 79. x  x  20  0 Let u  x  u 2  x u 2  u  20  0 g  x  0 16 x 2  8 x  1  0  4 x  12  0 4x 1  0  x   u  5  u  4   0 u 5  0 or u  4  0 u  5 u4 x  5 x  not real x 4 The only real zero of g  x   16 x 2  8 x  1 is The only x-intercept of the graph of g is x  42  16 F  x  0 80. Check: g 16   16  16  20  16  4  20  0 4 x 2  12 x  9  0 The only zero of g  x   x  x  20 is 16. The  2 x  32  0 only x-intercept of the graph of g is 16. 76. 1 4 2x  3  0  x  f  x  0 1 . 4 3 2 x x 2  0 The only real zero of F  x   4 x 2  12 x  9 is Let u  x  u 2  x u2  u  2  0 The only x-intercept of the graph of F is  u  1 u  2   0 or u  2  0 10 x  19 x  15  0 u 1 u  2  5 x  3 2 x  5   0 x 1 x  2 x  not real 5x  3  0 x 1 1 3 . 2 3 . 2 G  x  0 81. u 1  0 2 1 . 4 2 3 x 5 Check: f 1  1  1  2  1  1  2  0 or 2 x  5  0 5 x 2 The zeros of G  x   10 x 2  19 x  15 are  The only zero of f  x   x  x  2 is 1. The only x-intercept of the graph of f is 1. 176 Copyright © 2019 Pearson Education, Inc. 3 and 5 Section 2.3: Quadratic Functions and Their Zeros 5 3 . The x-intercepts of the graph of G are  2 5 5 and . 2 x 2 2  8  8 2 2  16  4 4 2 2  4  2  2   4 2 1  2 2 The zeros of G  x   x 2  2 x  are 2 2 2 6 x  7 x  20  0  3x  4  2 x  5  0 3x  4  0 or 2 x  5  0 4 3 x x 5 2 5 4 and . 2 3 5 4 The x-intercepts of the graph of f are  and . 2 3 The zeros of f  x   6 x 2  7 x  20 are   3x  2  2 x  1  0 are  2 2  22 and . 2 2 F  x  0 x2  2 2x  2  0 3x  2  0 or 2 x  1  0 2 1 x x 3 2 a  1, b  2 2, c  2 x 1 2 The zeros of P  x   6 x 2  x  2 are  and . 2 3 1 2 The x-intercepts of the graph of P are  and . 2 3 (2 2)  (2 2) 2  4(1)  2  2(1) 2 2  16 2 2  4 22   2 2 1 1 The zeros of F  x   x 2  2 x  1 are 2  2 2 and 2  2 . The x-intercepts of the graph of F  H  x  0 6 x2  x  2  0  3x  2  2 x  1  0 are 3x  2  0 or 2 x  1  0 2 1 x x 3 2 2  2 and 2 2. f  x  0 87. 2 x  x4  0 a  1, b  1, c  4 2 1 and . 3 2 2 1 The x-intercepts of the graph of H are  and . 3 2 The zeros of H  x   6 x 2  x  2 are  85.  22 . The x-intercepts of the graph of G 2 1 2 x  2x 1  0 2 1  2  x 2  2 x  1   0  2  2  6 x2  x  2  0 84. and 86. P  x  0 83. 2(2)  f  x  0 82. (2 2)  (2 2) 2  4(2)  1 x G  x  0  1 0 2 1  2  x 2  2 x     0  2  2  x2  2 x  (1)  (1) 2  4 1 4  2(1) 1  1  16 1  17  2 2 The zeros of f  x   x 2  x  4 are 1  17 and 2 1  17 . The x-intercepts of the graph of f are 2 2 x2  2 2 x  1  0 1  17 1  17 and . 2 2 a  2, b  2 2, c  1 177 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions g  x  0 88. ( x  3) 2  9  0 x  x 1  0 a  1, b  1, c  1 x F  x  0 b. 2 (1)  (1) 2  4 1 1 2(1) x2  6 x  9  9  0  x2  6 x  0 1  5 2 x( x  6)  0  x  0 or x  6 1  5 and The zeros of g  x   x 2  x  1 are 2 91. a. f ( x)  2( x  4) 2  8 1  5 . The x-intercepts of the graph of g are 2 Using the graph of y  x 2 , horizontally shift to the left 4 units, vertically stretch by a factor of 2, and then vertically shift downward 8 units. 1  5 1  5 and . 2 2 89. a. g ( x)  ( x  1) 2  4 Using the graph of y  x 2 , horizontally shift to the right 1 unit, and then vertically shift downward 4 units. f  x  0 b. 2( x  4) 2  8  0 2( x 2  8 x  16)  8  0 2 x 2  16 x  32  8  0 b. 2 x 2  16 x  24  0 2( x  2)( x  6)  0  x  2 or x  6 g  x  0 ( x  1) 2  4  0 92. a. h( x )  3( x  2) 2  12 x2  2 x  1  4  0 Using the graph of y  x 2 , horizontally shift to the right 2 units, vertically stretch by a factor of 3, and then vertically shift downward 12 units. x2  2 x  3  0 ( x  1)( x  3)  0  x  1 or x  3 90. a. F ( x )  ( x  3) 2  9 Using the graph of y  x 2 , horizontally shift to the left 3 units, and then vertically shift downward 9 units. 178 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros h  x  0 b. vertically shift upward 12 units. 3( x  2) 2  12  0 3( x 2  4 x  4)  12  0 3x 2  12 x  12  12  0 3 x 2  12 x  0 3 x( x  4)  0  x  0 or x  4 93. a. H ( x)  3( x  3) 2  6 f  x  0 b. Using the graph of y  x 2 , horizontally shift to the right 3 units, vertically stretch by a factor of 3, reflect about the x-axis, and then vertically shift upward 6 units. 2 2( x  1)  12  0 2( x 2  2 x  1)  12  0 2 x 2  4 x  2  12  0 2 x 2  4 x  10  0 2( x 2  2 x  5)  0 a  1, b  2, c  5 x H  x  0 b.  3( x  3) 2  6  0 3( x 2  6 x  9)  6  0 95. 3x 2  18 x  27  6  0 2  4  20 2 2  24 2  2 6   1  6 2 2 f  x  g  x 5 x 2  5 x  7 x 2  2 3( x 2  6 x  7)  0 a  1, b  6, c  7  2(1)  5 x( x  1)  7 x 2  2 3x 2  18 x  21  0 x (2)  (2) 2  4 1 5  12 x 2  5 x  2  0 2 1 or x   3 4 2  2   2   f    5      1 3  3   3   10  10   1         3 3 9    (3x  2)(4 x  1)  0  x  (6)  (6) 2  4 1 7  6  36  28  2(1) 2 6 8 62 2   3 2 2 2 94. a. f ( x)  2( x  1) 2  12  1  1   1   f     5        1  4  4   4    5  5  25         4  4  16 2 Using the graph of y  x , horizontally shift to the left 1 unit, vertically stretch by a factor of 2, reflect about the x-axis, and then The points of intersection are:  2 10   1 25   ,   and   ,  3 9    4 16  179 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions f  x  g  x 96. 10 x( x  2)  3x  5 5 3x 5   x  2 x  1 x 2  3x  2 5 3x 5   x  2 x  1 ( x  2)( x  1) 3x( x  1)  5( x  2)  5 10 x  20 x  3x  5 2 10 x 2  23x  5  0 5 1 or x  2 5  5  5   5   f     10        2   2  2   2    1  25   25       2 2 (2 x  5)(5 x  1)  0  x   3 x 2  3 x  5 x  10  5 3x 2  2 x  5  0 (3x  5)( x  1)  0 5 x  or x  1 3 5 3  5 5 3   f       3   5   2  5  1     3 3  5 5    11   8       3  3 15 15   11 8 45  88 1  1   1   f    10      2  5    5   5    11  22   2    5 5 The points of intersection are:  5 25   1 22    ,  and  ,   2 2  5 5  f  x  g  x 97. 3( x  4)  3x 2  2 x  4 2 3 x 2  12  3x 2  2 x  4  5 45  The point of intersection is:  ,    3 88  12  2 x  4 16  2 x  x  8 2 f  8   3  8   4     3 64  4  180 f  x  g  x 100. 2x 3 2 x  18   x  3 x  1 x2  2 x  3 2x 3 2 x  18   x  3 x  1 ( x  3)( x  1) 2 x( x  1)  3( x  3)  2 x  18 The point of intersection is:  8,180  98. f  x  g  x 99. f  x  g  x 4( x  1)  4 x 2  3 x  8 2 x 2  2 x  3 x  9  2 x  18 4 x 2  4  4 x 2  3x  8 2 x 2  3x  9  0 4  3x  8 (2 x  3)( x  3)  0 3 x   or x  3 2 2 12  3x  x  4 2 f  4   4  4   1    4 16  1  68 The point of intersection is:  4, 68  180 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros  3 2   3 2  3 f       2    3   3   3  1      2  2  3   x 2  3 x  18    x 2  2 x  3  x 2  3x  18  x 2  2 x  3  5 x  15 5 x  15  0  x  3 3  9  1      2  2 6 2 20  6  6  9 3 3  3 20  The point of intersection is:   ,   2 3      x 2  3 x  18  x 2  2 x  3   x  3 x  6  x  3 x  1 ( f  g)  x  0 0   x  3 x  6  x  3 x  1  x  3 or x  6 or x  1  x 2  5 x  14  x 2  3 x  4 x( x  2)  143 2 x  2 x  143  0 x2  4x  9  0 x (4)  (4) 2  4 1 9  2(1) ( x  13)( x  11)  0  4  16  36 2 x  13 or x  11 Discard the negative solution since width cannot be negative. The width of the rectangular window is 11 feet and the length is 13 feet. 4  52 4  2 13   2  13 2 2 ( f  g)  x  A( x )  306 104.   x  5 x  14    x  3x  4  2 x( x  1)  306 2 2 x  x  306  0 ( x  18)( x  17)  0  x 2  5 x  14  x 2  3x  4  2 x  10 2 x  10  0  x  5 c. A( x )  143 103.  2 x 2  8 x  18 2 2 x  8 x  18  0 b. ( f  g)  x  c. ( f  g)  x  101. a.  ( f  g)  x  b. x  18 or x  17 Discard the negative solution since width cannot be negative. The width of the rectangular window is 17 cm and the length is 18 cm. ( f  g)  x    x 2  5 x  14  x 2  3x  4    x  7  x  2  x  4  x  1 105. V  x  4  x  2 2  4 ( f  g)  x  0 x2   4 x  2  2 0   x  7  x  2  x  4  x  1 x  22 x  4 or x  0 Discard x  0 since that is not a feasible length for the original sheet. Therefore, the original sheet should measure 4 feet on each side.  x  7 or x  2 or x  4 or x  1 102. a. ( f  g)  x   x 2  3x  18  x 2  2 x  3  2 x 2  x  21 2 x 2  x  21  0  2 x  7  x  3  0  x  7 or x  3 2 181 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions V  x  4 106. t  x  2   16 2 x  2   16  x  2  4 2  4.9  20  106 9.8 20  106 9.8 t  0.99 or t  3.09 The object will be 15 meters above the ground after about 0.99 seconds (on the way up) and about 3.09 seconds (on the way down). x  24 x  6 or x  2 Discard x  2 since width cannot be negative. Therefore, the original sheet should measure 6 feet on each side. 107. a. 20  202  4  4.9  15   When the ball strikes the ground, the distance from the ground will be 0. Therefore, we solve s0 b. The object will strike the ground when the distance from the ground is 0. Thus, we solve s0 4.9t 2  20t  0 2 96  80t  16t  0 t  4.9t  20   0 16t 2  80t  96  0 t0 4.9t  20  0 or t 2  5t  6  0 4.9t  20  t  6  t  1  0 t  4.08 The object will strike the ground after about 4.08 seconds. t  6 or t  1 Discard the negative solution since the time of flight must be positive. The ball will strike the ground after 6 seconds. 2 4.9t  20t  100 b. When the ball passes the top of the building, it will be 96 feet from the ground. Therefore, we solve s  96 2 4.9t  20t  100  0 a  4.9, b  20, c  100 t 2 96  80t  16t  96 2 16t  80t  0 20  202  4  4.9  100  2  4.9  20  1560 9.8 There is no real solution. The object never reaches a height of 100 meters.  2 t  5t  0 t t  5  0 t  0 or t  5 The ball is at the top of the building at time t  0 seconds when it is thrown. It will pass the top of the building on the way down after 5 seconds. 108. a. s  100 c. 109. For the sum to be 210, we solve S (n)  210 1 n(n  1)  210 2 n(n  1)  420 To find when the object will be 15 meters above the ground, we solve s  15 n 2  n  420  0 (n  20)(n  21)  0 n  20  0 or n  21  0 4.9t 2  20t  15 4.9t 2  20t  15  0 a  4.9, b  20, c  15 n  20 n  21 Discard the negative solution since the number of consecutive integers must be positive. For a sum of 210, we must add the 20 consecutive integers, starting at 1. 182 Copyright © 2019 Pearson Education, Inc. Section 2.3: Quadratic Functions and Their Zeros 110. To determine the number of sides when a polygon has 65 diagonals, we solve D (n)  65 1 n(n  3)  65 2 n(n  3)  130 so the product of the roots is  b  b 2  4ac  b  b 2  4ac    x1  x2   2a 2a     2 n  3n  130  0  (n  10)(n  13)  0 n  10  0 or n  13  0  b 2   b 2  4ac   2a 2 b 2  b 2  4ac 4ac c  2  a 4a 2 4a 1  4k 2  0 4k 2  1 1 k2  4 k  2 n  3n  160  0 a  1, b  3, c  160 k (3)  (3) 2  4(1)(160) 2(1) 3  649 2 Since the solutions are not integers, a polygon with 80 diagonals is not possible. 1 2 or k   1 2  k 2  4 1 4   0 k 2  16  0  k  4  k  4   0 111. The roots of a quadratic equation are 2 b  b  4ac b  b  4ac and x2  , 2a 2a so the sum of the roots is x1  x2  1 4 114. In order to have one repeated real zero, we need the discriminant to be 0. b 2  4ac  0  x1  b 2   b 2  4ac  4a 2 12  4  k  k   0 To determine the number of sides if a polygon has 80 diagonals, we solve D(n)  80 1 n(n  3)  80 2 n(n  3)  160 2  113. In order to have one repeated real zero, we need the discriminant to be 0. b 2  4ac  0 n  13 n  10 Discard the negative solution since the number of sides must be positive. A polygon with 65 diagonals will have 13 sides. t 2 k  4 or k  4 115. For f  x   ax 2  bx  c  0 : b  b 2  4ac b  b 2  4ac  2a 2a x1  b  b 2  4ac  b  b 2  4ac 2a 2b b   2a a  b  b 2  4ac b  b 2  4ac and x2  2a 2a For f  x   ax 2  bx  c  0 : x1*  112. The roots of a quadratic equation are  b  b 2  4ac b  b 2  4ac x1  and x2  , 2a 2a   b    b 2  4ac 2a  b  b 2  4ac  b  b 2  4ac    x2     2a 2a   183 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions and x2*     b   118. Answers may vary. Methods discussed in this section include factoring, the square root method, completing the square, and the quadratic formula.  b   4ac 2 2a  b  b 2  4ac  b  b  4ac    x1     2a 2a   2 119. Answers will vary. Knowing the discriminant allows us to know how many real solutions the equation will have. 116. For f  x   ax 2  bx  c  0 : 2 120. Answers will vary. One possibility: Two distinct: f  x   x 2  3 x  18 2 b  b  4ac b  b  4ac and x2  2a 2a x1  One repeated: f  x   x 2  14 x  49 For f  x   cx  bx  a  0 : 2 b  b  4  c  a  2 x1*     2c  No real: f  x   x 2  x  4 b  b 2  4ac 2c 121. Answers will vary. b  b 2  4ac b  b 2  4ac  2c b  b 2  4ac  b 2  b 2  4ac   2 2c b  b  4ac 2a 2 b  b  4ac   122. Two quadratic functions can intersect 0, 1, or 2 times. 123. The graph is shifted vertically by 4 units and is reflected about the x-axis. 4ac  2c  b  b  4ac  2 1 x2 and x2*     117. a. b  b 2  4  c  a  2c  b  b 2  4ac 2c 2 b  b  4ac b  b 2  4ac  2c b  b 2  4ac  b 2  b 2  4ac   2c b  b 2  4ac 2a 2 b  b  4ac   124. Domain:  3, 1,1,3 Range:  2, 4 4ac  2c  b  b  4ac  10  2 8   4 2 2 4  ( 1) 3 y  2 2 2 125. x  1 x1 3  So the midpoint is:  4,  .  2 x 2  9 and x  3 are not equivalent because they do not have the same solution set. In the first equation we can also have x  3 . b. x  9 and x  3 are equivalent because 9 3. c.  x  1 x  2    x  12 and x  2  x  1 are 126. If the graph is symmetric with respect to the yaxis then x and –x are on the graph. Thus if  1, 4 is on the graph, then so is 1, 4 . not equivalent because they do not have the same solution set. The first equation has the solution set 1 while the second equation has no solutions. 184 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions 17. G Section 2.4 18. B 1. y  x 2  9 To find the y-intercept, let x  0 : y  02  9  9 . To find the x-intercept(s), let y  0 : 19. H 20. D x2  9  0 21. x2  9 x   9  3 The intercepts are (0, 9), (3, 0), and (3, 0) . 1 2 x 4 Using the graph of y  x 2 , compress vertically f ( x)  by a factor of 1 . 4 2 x2  7 x  4  0 2.  2 x  1 x  4   0 2 x  1  0 or x  4  0 2 x  1 or x  4 1 x or x  4 2 1  The solution set is 4,  . . 2  22. 2 25 1  3.   (5)   4 2  f ( x)  2 x 2  4 Using the graph of y  x 2 , stretch vertically by a factor of 2, then shift up 4 units. 4. right; 4 5. parabola 6. axis (or axis of symmetry) 7.  b 2a 8. True; a  2  0 . 9. True;  b 4  2 2a 2  1 23. f ( x)  ( x  2) 2  2 Using the graph of y  x 2 , shift left 2 units, then shift down 2 units. 10. True 11. a 12. d 13. C 14. E 15. F 16. A 185 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 24. f ( x)  ( x  3) 2  10 27. Using the graph of y  x 2 , shift right 3 units, then shift down 10 units. f ( x)  2 x 2  4 x  1    2 x2  2x  1 2  2( x  2 x  1)  1  2  2( x  1) 2  1 Using the graph of y  x 2 , shift right 1 unit, stretch vertically by a factor of 2, then shift down 1 unit. 25. f ( x)  x 2  4 x  1  ( x 2  4 x  4)  1  4  ( x  2) 2  3 Using the graph of y  x 2 , shift left 2 units, then shift down 3 units. 28. f ( x)  3 x 2  6 x   3 x2  2 x   3( x 2  2 x  1)  3  3( x  1) 2  3 Using the graph of y  x 2 , shift left 1 unit, stretch vertically by a factor of 3, then shift down 3 units. 26. f ( x)  x 2  6 x  1  ( x 2  6 x  9)  1  9  ( x  3) 2  10 Using the graph of y  x 2 , shift right 3 units, then shift down 10 units. 29. f ( x)   x 2  2 x    x2  2x  2  ( x  2 x  1)  1  ( x  1) 2  1 Using the graph of y  x 2 , shift left 1 unit, reflect across the x-axis, then shift up 1 unit. 186 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions down 30. f ( x)  2 x 2  6 x  2    2 x 2  3x  2 32. 9 9   2  x 2  3 x    2  4 2   2 2 4 x  x 1 3 3 2 2  x  2x 1 3 2 2  x2  2 x  1  1  3 3 2 5 2   x  1  3 3 Using the graph of y  x 2 , shift left 1 unit, f ( x)    2 3  13   2  x    2 2  3 Using the graph of y  x , shift right units, 2 reflect about the x-axis, stretch vertically by a 13 factor of 2, then shift up units. 2 2   compress vertically by a factor of down 31. 3 units. 2 1 2 x  x 1 2 1  x2  2 x  1 2 1 1  x2  2 x  1  1  2 2 1 3 2   x  1  2 2 Using the graph of y  x 2 , shift left 1 unit, 2 , then shift 3 5 unit. 3 f ( x)     33. a.  compress vertically by a factor of For f ( x)  x 2  2 x , a  1 , b  2 , c  0. Since a  1  0 , the graph opens up. The x-coordinate of the vertex is x b (2) 2    1 . 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  (1) 2  2(1)  1  2  1.  2a  Thus, the vertex is (1,  1) . The axis of symmetry is the line x  1 . The discriminant is b 2  4ac  (2) 2  4(1)(0)  4  0 , so the graph has two x-intercepts. 1 , then shift 2 187 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The x-intercepts are found by solving: x2  2 x  0 x( x  2)  0 x  0 or x  2 The x-intercepts are –2 and 0 . The y-intercept is f (0)  0 . b. The domain is (, ) . The range is [1, ) . c. Decreasing on (,  1] . Increasing on [1, ) . 34. a. b. The domain is (, ) . The range is [4, ) . c. Decreasing on (, 2] . Increasing on [2, ) . 35. a. For f ( x)  x 2  4 x , a  1 , b  4 , c  0 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b (4) 4 x   2. 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (2)  (2) 2  4(2)  4  8  4.  2a  Thus, the vertex is (2,  4) . The axis of symmetry is the line x  2 . The discriminant is: b 2  4ac  (4) 2  4(1)(0)  16  0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2  4 x  0 x( x  4)  0 x  0 or x  4. The x-intercepts are 0 and 4. The y-intercept is f (0)  0 . For f ( x)   x 2  6 x , a  1 , b  6 , c  0 . Since a  1  0, the graph opens down. The x-coordinate of the vertex is b (6) 6 x    3. 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (3)  (3) 2  6(3)  2a   9  18  9. Thus, the vertex is (3, 9) . The axis of symmetry is the line x  3 . The discriminant is: b 2  4ac  (6) 2  4(1)(0)  36  0 , so the graph has two x-intercepts. The x-intercepts are found by solving:  x2  6 x  0  x( x  6)  0 x  0 or x  6. The x-intercepts are 6 and 0 . The y-intercepts are f (0)  0 . 188 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions b. The domain is (, ) . The range is (, 9] . c. Increasing on (,  3] . Decreasing on [3, ) . 36. a. b 2 2    1 . 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  (1) 2  2(1)  8  2a   1  2  8  9. Thus, the vertex is (1,  9) . The axis of symmetry is the line x  1 . The discriminant is: b 2  4ac  22  4(1)( 8)  4  32  36  0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2  2 x  8  0 ( x  4)( x  2)  0 x  4 or x  2. The x-intercepts are 4 and 2 . The y-intercept is f (0)  8 . x For f ( x)   x 2  4 x, a  1, b  4 , c  0 . Since a  1  0 , the graph opens down. The x-coordinate of the vertex is b 4 4 x    2. 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (2)  2a   (2) 2  4(2)  4. Thus, the vertex is (2, 4) . The axis of symmetry is the line x  2 . The discriminant is: b 2  4ac  42  4( 1)(0)  16  0, so the graph has two x-intercepts. The x-intercepts are found by solving:  x2  4 x  0  x( x  4)  0 x  0 or x  4. The x-intercepts are 0 and 4. The y-intercept is f (0)  0 . b. The domain is (, ) . The range is [9, ) . c. 38. a. b. The domain is (, ) . The range is (, 4] . c. Increasing on (, 2] . Decreasing on [2, ) . 37. a. Decreasing on (,  1] . Increasing on [1, ) . For f ( x)  x 2  2 x  3, a  1, b  2, c  3. Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b (2) 2 x    1. 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  12  2(1)  3  4.  2a  Thus, the vertex is (1,  4) . The axis of symmetry is the line x  1 . The discriminant is: For f ( x)  x 2  2 x  8 , a  1 , b  2 , c  8 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b 2  4ac  ( 2) 2  4(1)( 3)  4  12  16  0 , 189 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The y-intercept is f (0)  1 . so the graph has two x-intercepts. The x-intercepts are found by solving: x2  2 x  3  0 ( x  1)( x  3)  0 x  1 or x  3. The x-intercepts are 1 and 3 . The y-intercept is f (0)  3 . b. The domain is (, ) . The range is [0, ) . c. Decreasing on (,  1] . Increasing on [1, ) . 40. a. b. The domain is (, ) . The range is [4, ) . c. Decreasing on (, 1] . Increasing on [1, ) . 39. a. For f ( x)  x 2  2 x  1 , a  1 , b  2 , c  1 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b 2 2 x    1 . 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  2a  For f ( x)  x 2  6 x  9 , a  1 , b  6 , c  9 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b 6 6 x    3 . 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (3)  2a   (3) 2  6(3)  9  9  18  9  0. Thus, the vertex is (3, 0) . The axis of symmetry is the line x  3 . The discriminant is: b 2  4ac  62  4(1)(9)  36  36  0 , so the graph has one x-intercept. The x-intercept is found by solving: x2  6 x  9  0  (1) 2  2(1)  1  1  2  1  0. Thus, the vertex is (1, 0) . The axis of symmetry is the line x  1 . The discriminant is: b 2  4ac  22  4(1)(1)  4  4  0 , so the graph has one x-intercept. The x-intercept is found by solving: x2  2 x  1  0 ( x  3) 2  0 x  3. The x-intercept is 3 . The y-intercept is f (0)  9 . ( x  1) 2  0 x  1. The x-intercept is 1 . b. The domain is (, ) . The range is [0, ) . 190 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions c. 41. a.   Decreasing on (,  3] . Increasing on [3, ) . Thus, the vertex is 1 , 3 . 4 4 The axis of symmetry is the line x  1 . 4 The discriminant is: b 2  4ac  (2) 2  4(4)(1)  4  16  12 , so the graph has no x-intercepts. The y-intercept is f (0)  1 . For f ( x)  2 x 2  x  2 , a  2 , b  1 , c  2 . Since a  2  0 , the graph opens up. The x-coordinate of the vertex is b (1) 1 x   . 2a 2(2) 4 The y-coordinate of the vertex is 2  b  1 1 1 f    f    2    2  2a  4 4 4 1 1 15   2 . 8 4 8  1 15  Thus, the vertex is  ,  . 4 8  The axis of symmetry is the line x  1 . 4 The discriminant is: b 2  4ac  (1) 2  4(2)(2)  1  16  15 , so the graph has no x-intercepts. The y-intercept is f (0)  2 . b. The domain is (, ) . The range is  c. 1 1 Decreasing on  ,  . Increasing on  ,   . 4  4  42. a.  The range is  3 ,  . 4 c. 43. a. b. The domain is (, ) . 15  ,  . 8   Decreasing on , 1  . 4 1 Increasing on  ,  . 4  For f ( x)  2 x 2  2 x  3 , a  2 , b  2 , c  3 . Since a  2  0 , the graph opens down. The x-coordinate of the vertex is b (2) 2 1 x    . 2a 2(2) 4 2 The y-coordinate of the vertex is 2  b  1 1 1 f    f    2    2    3 2 2 2 a       2 1 5   1 3   . 2 2 1 5 Thus, the vertex is  ,   . 2 2 1 The axis of symmetry is the line x  . 2 The discriminant is: b 2  4ac  22  4(2)(3)  4  24  20 , so the graph has no x-intercepts. For f ( x)  4 x 2  2 x  1 , a  4 , b  2 , c  1 . Since a  4  0 , the graph opens up. The x-coordinate of the vertex is b (2) 2 1 x    . 2a 2(4) 8 4 The y-coordinate of the vertex is 2  b  1 1 1 f    f    4   2  1 2 4 4 a       4 1 1 3   1  . 4 2 4 191 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The y-intercept is f (0)  3 . b. The domain is (, ) . 5  The range is  ,   . 4  c. 45. a. b. The domain is (, ) . 5  The range is  ,   . 2  c. 44. 1  Increasing on  ,  . 2  1  Decreasing on  ,   . 2  a. For f ( x)  3x 2  3 x  2 , a  3 , b  3 , c  2 . Since a  3  0 , the graph opens down. The x-coordinate of the vertex is b 3 3 1    . x 2a 2(3) 6 2 The y-coordinate of the vertex is 1  Increasing on  ,  . 2  1  Decreasing on  ,   . 2  For f ( x)  3 x 2  6 x  2 , a  3 , b  6 , c  2 . Since a  3  0 , the graph opens up. The x-coordinate of the vertex is b 6 6    1 . x 2a 2(3) 6 The y-coordinate of the vertex is  b  f    f (1)  3(1) 2  6(1)  2  2a   3  6  2  1. Thus, the vertex is (1,  1) . The axis of symmetry is the line x  1 . The discriminant is: b 2  4ac  62  4(3)(2)  36  24  12 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2  6 x  2  0 x 2  b  1 1 1 f    f    3    3    2  2a  2 2 2 3 3 5    2   . 4 2 4 1 5  Thus, the vertex is  ,   . 2 4 1 The axis of symmetry is the line x  . 2 The discriminant is: b 2  4ac  32  4(3)(2)  9  24  15 , so the graph has no x-intercepts. The y-intercept is f (0)  2 . b  b 2  4ac 2a 6  12 6  2 3 3  3   6 6 3 3 3 The x-intercepts are 1  and 1  . 3 3 The y-intercept is f (0)  2 .  b. The domain is (, ) . The range is  1,   . c. Decreasing on  ,  1 . Increasing on  1,   . 192 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions 46. a. c. 2 For f ( x)  2 x  5 x  3 , a  2 , b  5 , c  3 . Since a  2  0 , the graph opens up. The x-coordinate of the vertex is b 5 5   . x 2a 2(2) 4 The y-coordinate of the vertex is  b   5 f    f    2a   4 47. a. 2  5  5  2    5    3  4  4 25 25   3 8 4 1  . 8  5 1 Thus, the vertex is   ,   .  4 8 The axis of symmetry is the line x   For f ( x)  4 x 2  6 x  2 , a  4 , b  6 , c  2 . Since a  4  0 , the graph opens down. The x-coordinate of the vertex is b (6) 6 3    . x 2a 2(4) 8 4 The y-coordinate of the vertex is 2  b   3  3  3 f    f     4     6     2  2a   4  4  4 9 9 17   2 . 4 2 4  3 17  Thus, the vertex is   ,  .  4 4 3 The axis of symmetry is the line x   . 4 The discriminant is: b 2  4ac  (6) 2  4(4)(2)  36  32  68 , so the graph has two x-intercepts. The x-intercepts are found by solving: 4 x 2  6 x  2  0 5 . 4 The discriminant is: b 2  4ac  52  4(2)(3)  25  24  1 , so the graph has two x-intercepts. The x-intercepts are found by solving: 2 x2  5x  3  0 (2 x  3)( x  1)  0 3 x   or x  1. 2 3 The x-intercepts are  and  1 . 2 The y-intercept is f (0)  3 . b. 5  Decreasing on  ,   . 4   5  Increasing on   ,   .  4  x b  b 2  4ac (6)  68  2a 2(4) 6  68 6  2 17 3  17   8 8 4 3  17 3  17 and . The x-intercepts are 4 4 The y-intercept is f (0)  2 .  The domain is (, ) .  1  The range is   ,   .  8  b. The domain is (, ) . 17   The range is  ,  . 4  193 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions c. 48. a.  3  Decreasing on   ,   .  4  3  Increasing on  ,   . 4  c. 4  Decreasing on  ,  . 3  4  Increasing on  ,   . 3  49. Consider the form y  a  x  h   k . From the 2 For f ( x)  3 x 2  8 x  2, a  3, b  8, c  2 . Since a  3  0 , the graph opens up. The x-coordinate of the vertex is b (8) 8 4    . x 2a 2(3) 6 3 The y-coordinate of the vertex is graph we know that the vertex is  1, 2  so we have h  1 and k  2 . The graph also passes through the point  x, y    0, 1 . Substituting these values for x, y, h, and k, we can solve for a: 1  a  0   1    2  2 2  b  4 4 4 f    f    3   8    2  2a  3 3 3 16 32 10   2  . 3 3 3  4 10  Thus, the vertex is  ,   . 3 3 1  a 1  2 2 1  a  2 1 a The quadratic function is f  x    x  1  2  x 2  2 x  1 . 2 4 The axis of symmetry is the line x  . 3 The discriminant is: b 2  4ac  (8) 2  4(3)(2)  64  24  40 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2  8 x  2  0 50. Consider the form y  a  x  h   k . From the 2 graph we know that the vertex is  2,1 so we have h  2 and k  1 . The graph also passes through the point  x, y    0,5  . Substituting these values for x, y, h, and k, we can solve for a: b  b 2  4ac (8)  40  x 2a 2(3) 5  a  0  2  1 2 5  a  2   1 2 8  40 8  2 10 4  10   6 6 3 4  10 4  10 and . The x-intercepts are 3 3 The y-intercept is f (0)  2 .  5  4a  1 4  4a 1 a The quadratic function is f  x    x  2  1  x2  4 x  5 . 2 51. Consider the form y  a  x  h   k . From the 2 graph we know that the vertex is  3,5  so we have h  3 and k  5 . The graph also passes through the point  x, y    0, 4  . Substituting these values for x, y, h, and k, we can solve for a: 4  a  0  (3)   5 2 4  a  3  5 2 b. The domain is (, ) . 4  9a  5  10  The range is   ,   .  3  9  9a 1  a 194 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions The quadratic function is 2  a  4   b(4)  c  2  16a  4b  c 2 f  x     x  3  5   x  6 x  4 . 2 2 and 4  a  1  b(1)  c  4  a  b  c 2 and 52. Consider the form y  a  x  h   k . From the 2 2  a  0   b(0)  c  2  c Since 2  c , we have the following equations: 2  16a  4b  2, 4  a  b  2,  2  c Solving the first two simultaneously we have 2  16a  4b  2   4  a b2  2 graph we know that the vertex is  2,3 so we have h  2 and k  3 . The graph also passes through the point  x, y    0, 1 . Substituting these values for x, y, h, and k, we can solve for a: 1  a  0  2   3 2 0  16a  4b   6  a b  1  a  2   3 2 1  4a  3 a  2, b  8 The quadratic function is f  x   2 x 2  8 x  2 . 4  4a 1  a The quadratic function is 55. For f ( x)  2 x 2  12 x, a  2, b  12, c  0 . Since a  2  0, the graph opens up, so the vertex is a minimum point. The minimum b 12 12 occurs at x     3. 2a 2(2) 4 The minimum value is f (3)  2(3) 2  12(3)  18  36  18 . f  x     x  2  3   x2  4 x  1 . 2 53. Consider the form y  ax 2  bx  c . Substituting the three points from the graph into the general form we have the following three equations. 5  a  1  b(1)  c  5  a  b  c 2 and 56. For f ( x)  2 x 2  12 x, a  2, b  12, c  0, . Since a  2  0, the graph opens down, so the vertex is a maximum point. The maximum b 12 12    3. occurs at x  2a 2(2) 4 The maximum value is f (3)  2(3) 2  12(3)  18  36  18 . 5  a  3  b(3)  c  5  9a  3b  c 2 and 1  a  0   b(0)  c  1  c Since 1  c , we have the following equations: 5  a  b  1, 5  9a  3b  1,  1  c Solving the first two simultaneously we have 5  a  b 1   5  9a  3b  1 2 57. For f ( x)  2 x 2  12 x  3, a  2, b  12, c  3. Since a  2  0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b 12 12    3. The minimum value is x 2a 2(2) 4 6  a b    a  2, b  4 6  9a  3b  The quadratic function is f  x   2 x 2  4 x  1 . f (3)  2(3) 2  12(3)  3  18  36  3  21 . 54. Consider the form y  ax 2  bx  c . Substituting the three points from the graph into the general form we have the following three equations. 58. For f ( x)  4 x 2  8 x  3, a  4, b  8, c  3. Since a  4  0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b (8) 8    1. The minimum value is x 2a 2(4) 8 f (1)  4(1) 2  8(1)  3  4  8  3  1 . 195 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The x-intercepts are found by solving: x 2  2 x  15  0 ( x  3)( x  5)  0 x  3 or x  5 The x-intercepts are 3 and 5 . The y-intercept is f (0)  15 . 59. For f ( x)   x 2  10 x  4 , a  1, b  10 , c   4 . Since a  1  0, the graph opens down, so the vertex is a maximum point. The maximum occurs b 10 10    5 . The maximum at x  2a 2(1)  2 value is f (5)  (5) 2  10(5)  4   25  50  4  21 . 60. For f ( x)   2 x 2  8 x  3 , a  2, b  8, c  3. Since a   2  0, the graph opens down, so the vertex is a maximum point. The maximum 8 8 b occurs at x     2 . The 2a 2( 2)  4 maximum value is f (2)   2(2) 2  8(2)  3   8  16  3  11 . 61. For f ( x )  3x 2  12 x  1 , a  3, b  12, c  1. Since a  3  0, the graph opens down, so the vertex is a maximum point. The maximum occurs b 12 12    2 . The maximum value at x  2a 2(3)  6 The domain is (, ) . The range is [16, ) . c. Decreasing on (, 1] . Increasing on [1, ) . 64. a. is f (2)  3(2) 2  12(2)  1  12  24  1  13 . 62. For f ( x)  4 x 2  4 x , a  4, b  4, c  0. Since a  4  0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b ( 4) 4 1    . The minimum value is x 2a 2(4) 8 2 2 1 1 1 f    4    4    1  2  1 . 2 2 2 63. a. b. For f ( x)  x 2  2 x  15 , a  1 , b  2 , c  15 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b (2) 2   1. x 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  (1) 2  2(1)  15  2a   1  2  15  16. Thus, the vertex is (1, 16) . The discriminant is: b 2  4ac  (2) 2  4(1)(15)  4  60  64  0 , so the graph has two x-intercepts. For f ( x)  x 2  2 x  8 , a  1 , b  2 , c  8 . Since a  1  0 , the graph opens up. The x-coordinate of the vertex is b (2) 2   1. x 2a 2(1) 2 The y-coordinate of the vertex is  b  f    f (1)  (1) 2  2(1)  8  1  2  8  9.  2a  Thus, the vertex is (1, 9) . The discriminant is: b 2  4ac  (2) 2  4(1)( 8)  4  32  36  0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2  2 x  8  0 ( x  2)( x  4)  0 x  2 or x  4 The x-intercepts are 2 and 4 . The y-intercept is f (0)  8 . 196 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions b. The domain is (, ) . The range is [9, ) . c. Decreasing on (, 1] . Increasing on [1, ) . 65. a. F ( x)  2 x  5 is a linear function. The x-intercept is found by solving: 2x  5  0 b. The domain is (, ) . The range is (, ) . c. Increasing on (, ) . 67. a. Using the graph of y  x 2 , shift right 3 units, reflect about the x-axis, stretch vertically by a factor of 2, then shift up 2 units. 2x  5 x g ( x)  2( x  3) 2  2 5 2 5 . 2 The y-intercept is F (0)  5 . The x-intercept is b. The domain is (, ) . The range is (, ) . c. Increasing on (, ) . 66. a. b. The domain is (, ) . The range is (, 2] . c. Increasing on (, 3] . Decreasing on [3, ) . 68. a. h( x)  3( x  1) 2  4 Using the graph of y  x 2 , shift left 1 unit, reflect about the x-axis, stretch vertically by a factor of 3, then shift up 4 units. 3 x  2 is a linear function. 2 The x-intercept is found by solving: 3 x2  0 2 3 x2 2 2 4 x  2  3 3 4 The x-intercept is . 3 The y-intercept is f (0)  2 . f ( x)  197 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b. The domain is (, ) . The range is (, 4] . c. Increasing on (, 1] . Decreasing on [1, ) . 69. a. 2  b   1  1  1 G    G     3    2     5  2a   3  3  3 1 2 14   5  . 3 3 3  1 14  Thus, the vertex is   ,  .  3 3 The discriminant is: b 2  4ac  22  4(3)(5)  4  60  56 , so the graph has no x-intercepts. The y-intercept is G (0)  5 . For f ( x)  2 x 2  x  1 , a  2 , b  1 , c  1 . Since a  2  0 , the graph opens up. The x-coordinate of the vertex is b 1 1 1 x    . 2a 2(2) 4 4 The y-coordinate of the vertex is 2  b   1  1  1 f    f     2       1  2a   4  4  4 1 1 7   1  . 8 4 8  1 7 Thus, the vertex is   ,  .  4 8 The discriminant is: b 2  4ac  12  4(2)(1)  1  8  7 , so the graph has no x-intercepts. The y-intercept is f (0)  1 . b. 14  The range is  ,   . 3  c. 71. a. b. The domain is (, ) . 7  The range is  ,   . 8  c. 70. a. The domain is (, ) . 1  Decreasing on  ,   . 4   1  Increasing on   ,   .  4  1  Decreasing on  ,   . 3   1  Increasing on   ,   .  3  2 h( x)   x  4 is a linear function. 5 The x-intercept is found by solving: 2  x40 5 2  x  4 5  5 x  4     10  2 The x-intercept is 10. The y-intercept is h(0)  4 . For G ( x)  3 x 2  2 x  5 , a  3 , b  2 , c  5 . Since a  3  0 , the graph opens up. The x-coordinate of the vertex is b 2 2 1    . x 2a 2(3) 6 3 The y-coordinate of the vertex is 198 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions  1  Thus, the vertex is   , 0  .  2  The discriminant is: b 2  4ac  (4) 2  4(4)(1)  16  16  0 , so the graph has one x-intercept. The x-intercept is found by solving: 4 x 2  4 x  1  0 4×2  4 x  1  0 b. The domain is (, ) . The range is (, ) . (2 x  1) 2  0 c. Decreasing on (, ) . x 72. a. 2x 1  0 1 2 1 The x-intercept is  . 2 The y-intercept is H (0)  1 . f ( x)  3 x  2 is a linear function. The x-intercept is found by solving: 3 x  2  0 3x  2 x 2 2  3 3 2 . 3 The y-intercept is f (0)  2 . The x-intercept is b. The domain is (, ) . The range is  , 0 . c. b. c. 73. a. The domain is (, ) . The range is (, ) . 74. a. Decreasing on (, ) . For H ( x)  4 x 2  4 x  1 , a  4 , b  4 , c  1 . Since a  4  0 , the graph opens down. The x-coordinate of the vertex is b (4) 4 1    . x 2a 2(4) 8 2 The y-coordinate of the vertex is 1  Increasing on  ,   . 2   1  Decreasing on   ,   .  2  For F ( x)  4 x 2  20 x  25 , a  4 , b  20 , c  25 . Since a  4  0 , the graph opens down. The x-coordinate of the vertex is b 20 20 5 x    . 2a 2(4) 8 2 The y-coordinate of the vertex is 2  b  5 5 5 F    F    4    20    25  2a  2 2 2  25  50  25  0 5  Thus, the vertex is  , 0  . 2  The discriminant is: 2  b   1  1  1 H    H     4     4     1 2 2 2 a        2  1  2  1  0 199 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b 2  4ac  (20) 2  4(4)(25) f (1)  8 .  400  400  0, so the graph has one x-intercept. The x-intercept is found by solving: 4 x 2  20 x  25  0 8  a (1  1) 2  4 8  a (2) 2  4 8  4a  4 12  4a 4 x 2  20 x  25  0 3  a f ( x)  3( x  1) 2  4 (2 x  5) 2  0 2x  5  0  3( x 2  2 x  1)  4 5 2 5 The x-intercept is . 2 The y-intercept is F (0)  25 .  3 x 2  6 x  3  4 x  3 x 2  6 x  1 a  3, b  6, c  1 77. a and d. b. The domain is (, ) . The range is  , 0 . c. 5  Increasing on  ,  . 2  5  Decreasing on  ,   . 2  b. f ( x)  g ( x) 2x 1  x2  4 0  x2  2x  3 0  ( x  1)( x  3) 75. Use the form f ( x)  a ( x  h) 2  k . The vertex is (0, 2) , so h = 0 and k = 2. x  1  0 or x  3  0 x  1 x3 f ( x)  a( x  0) 2  2  ax 2  2 . The solution set is {1, 3}. Since the graph passes through (1, 8) , f (1)  8 . c. f ( x)  ax  2 2 8  a (1) 2  2 8 a2 6a f  x   6 x2  2 . f (1)  2(1)  1  2  1  3 g (1)  (1) 2  4  1  4  3 f (3)  2(3)  1  6  1  5 g (3)  (3) 2  4  9  4  5 Thus, the graphs of f and g intersect at the points (1, 3) and (3, 5) . a  6, b  0, c  2 76. Use the form f ( x)  a ( x  h) 2  k . The vertex is (1, 4) , so h  1 and k  4 . f ( x)  a ( x  1) 2  4 . Since the graph passes through (1,  8) , 200 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions 78. a and d. b. f  x  g  x  x 2  4  2 x  1 0  x2  2x  3 0   x  1 x  3 x  1  0 or x  3  0 x  1 x3 The solution set is {1, 3}. c. f 1    1  4  1  4  3 2 g 1  2  1  1  2  1  3 f  3    3  4  9  4  5 2 g  3  2  3  1  6  1  5 Thus, the graphs of f and g intersect at the points  1, 3 and  3, 5  . b. f ( x)  g ( x) 2 x  1  x 2  9 0  x2  2 x  8 0  ( x  4)( x  2) 80. a and d. x  4  0 or x  2  0 x  4 x2 The solution set is {4, 2}. c. f (4)  2(4)  1  8  1  7 g (4)  (4) 2  9  16  9  7 f (2)  2(2)  1  4  1  5 g (2)  (2) 2  9  4  9  5 Thus, the graphs of f and g intersect at the points  4, 7  and  2, 5  . 79. a and d. b. f  x  g  x  x2  9  2 x  1 0  x2  2x  8 0   x  4  x  2  x  4  0 or x  2  0 x  4 x2 The solution set is {4, 2}. c. f  4     4   9  16  9  7 2 g  4   2  4   1  8  1  7 f  2     2   9  4  9  5 2 g  2  2  2  1  4  1  5 201 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions Thus, the graphs of f and g intersect at the points  4, 7  and  2, 5  . f  x  g  x b.  x  7 x  6  x2  x  6 2 0  2 x2  6 x 81. a and d. 0  2 x  x  3 2 x  0 or x3  0 x0 x3 The solution set is {0, 3}. c. f  0     0   7  0   6  6 2 g  0   02  0  6  6 f  3    3  7  3  6  9  21  6  6 2 g  3  32  3  6  9  3  6  6 f  x  g  x b. Thus, the graphs of f and g intersect at the points  0, 6  and  3, 6  .  x 2  5 x  x 2  3x  4 0  2 x2  2 x  4 83. a. 0  x2  x  2 0   x  1 x  2   1( x  (3))( x  1) x  1  0 or x  2  0 x  1 x2 The solution set is {1, 2}. c. For a  1: f ( x)  a ( x  r1 )( x  r2 )  ( x  3)( x  1)  x 2  2 x  3 For a  2 : f ( x)  2( x  (3))( x  1)  2( x  3)( x  1) f  1    1  5  1  1  5  6 2  2( x 2  2 x  3)  2 x 2  4 x  6 For a  2 : f ( x)  2( x  (3))( x  1) g  1   1  3  1  4  1  3  4  6 2 f  2     2   5  2   4  10  6 2 g  2   22  3  2   4  4  6  4  6  2( x  3)( x  1)  2( x 2  2 x  3)  2 x 2  4 x  6 For a  5 : f ( x)  5( x  (3))( x  1) Thus, the graphs of f and g intersect at the points  1, 6  and  2, 6  . 82. a and d.  5( x  3)( x  1)  5( x 2  2 x  3)  5 x 2  10 x  15 b. The x-intercepts are not affected by the value of a. The y-intercept is multiplied by the value of a . c. The axis of symmetry is unaffected by the value of a . For this problem, the axis of symmetry is x  1 for all values of a. d. The x-coordinate of the vertex is not affected by the value of a. The y-coordinate of the vertex is multiplied by the value of a . e. The x-coordinate of the vertex is the mean of the x-intercepts. 202 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions 84. a. For a  1: f ( x)  1( x  (5))( x  3) x  4 x  21  21 2  ( x  5)( x  3)  x 2  2 x  15 For a  2 : f ( x)  2( x  (5))( x  3) x2  4x  0 x  x  4  0 x  0 or  2( x  5)( x  3) x40 x  4  2( x 2  2 x  15)  2 x 2  4 x  30 For a  2 : f ( x)  2( x  (5))( x  3) The solutions f  x   21 are 4 and 0. Thus, the points  4, 21 and  0, 21 are  2( x  5)( x  3) 2 f  x   21 c. on the graph of f. 2 d.  2( x  2 x  15)  2 x  4 x  30 For a  5 : f ( x)  5( x  (5))( x  3)  5( x  5)( x  3)   5( x 2  2 x  15)  5 x 2  10 x  75 b. The x-intercepts are not affected by the value of a. The y-intercept is multiplied by the value of a . c. The axis of symmetry is unaffected by the value of a . For this problem, the axis of symmetry is x  1 for all values of a.  d. The x-coordinate of the vertex is not affected by the value of a. The y-coordinate of the vertex is multiplied by the value of a . e. 85. a. 86. a. The x-coordinate of the vertex is the mean of the x-intercepts. x y  f  1   1  2  1  8  9 2 The vertex is  1, 9  . 4 b x   2 2a 2 1 f  x  0 b. x  2x  8  0 2 y  f  2    2   4  2   21  25 2  x  4  x  2   0 The vertex is  2, 25  . x40 f  x  0 b. x  7 x2 The x-intercepts of f are (4, 0) and (2, 0).  x  7  x  3  0 or x2  0 or x  4 x 2  4 x  21  0 x7  0 b 2   1 2a 2 1 f  x   8 c. x3  0 x  2 x  8  8 2 x3 x2  2x  0 The x-intercepts of f are (7, 0) and (3, 0). x  x  2  0 x  0 or x20 x  2 The solutions f  x   8 are 2 and 0. Thus, 203 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions the points  2, 8  and  0, 8  are on the coordinate of the minimum point of  d ( x)  will graph of f. also give us the x-coordinate of the minimum of b (8) 8 d ( x) : x     2 . So, 2 is the x2a 2(2) 4 coordinate of the point on the line y = x + 1 that is closest to the point (4, 1). The y-coordinate is y = 2 + 1 = 3. Thus, the point is (2, 3) is the point on the line y = x + 1 that is closest to (4, 1). 2 d.   89. R ( p )  4 p 2  4000 p , a   4, b  4000, c  0. Since a  4  0 the graph is a parabola that opens down, so the vertex is a maximum point. The b  4000   500 . maximum occurs at p  2a 2( 4) Thus, the unit price should be $500 for maximum revenue. The maximum revenue is R (500)   4(500) 2  4000(500)  1000000  2000000  $1, 000, 000 87. Let (x, y) represent a point on the line y = x. Then the distance from (x, y) to the point (3, 1) is d  x  3   y  1 . Since y = x, we can 2 2 replace the y variable with x so that we have the distance expressed as a function of x: d ( x)   x  32   x  12  x2  6x  9  x2  2 x  1 1 2 1 p  1900 p , a   , b  1900, c  0. 2 2 1 Since a    0, the graph is a parabola that 2 opens down, so the vertex is a maximum point. The maximum occurs at b 1900 1900 p    1900 . Thus, the 2a 2  1/ 2  1 90. R ( p )    2 x 2  8 x  10 Squaring both sides of this function, we obtain  d ( x)2  2 x 2  8 x  10 . Now, the expression on the right is quadratic. Since a = 2 > 0, it has a minimum. Finding the xcoordinate of the minimum point of  d ( x)  will 2 also give us the x-coordinate of the minimum of b (8) 8    2 . So, 2 is the xd ( x) : x  2a 2(2) 4 coordinate of the point on the line y = x that is closest to the point (3, 1). Since y = x, the ycoordinate is also 2. Thus, the point is (2, 2) is the point on the line y = x that is closest to (3, 1). unit price should be $1900 for maximum revenue. The maximum revenue is 1 2 R 1900    1900   1900 1900  2  1805000  3610000  $1,805, 000 88. Let (x, y) represent a point on the line y = x + 1. Then the distance from (x, y) to the point (4, 1) is d 91. a.  x  4    y  1 . Replacing the y variable 2 2 with x + 1, we find the distance expressed as a function of x: d ( x)   x  4 2   ( x  1)  12  x 2  8 x  16  x 2 C ( x)  x 2  140 x  7400 , a  1, b  140, c  7400. Since a  1  0, the graph opens up, so the vertex is a minimum point. The minimum marginal cost b (140) 140    70 , occurs at x  2a 2(1) 2 70,000 digital music players produced. b. The minimum marginal cost is 2  b  f    f  70    70   140  70   7400 2 a    4900  9800  7400  $2500  2 x 2  8 x  16 Squaring both sides of this function, we obtain  d ( x)2  2 x 2  8 x  16 . Now, the expression on the right is quadratic. Since a = 2 > 0, it has a minimum. Finding the x204 Copyright © 2019 Pearson Education, Inc. Section 2.4: Properties of Quadratic Functions 92. a. C ( x)  5 x 2  200 x  4000 , a  5, b  200, c  4000. Since a  5  0, the graph opens up, so the vertex is a minimum point. The minimum marginal cost b   200  200    20 , occurs at x  2a 2(5) 10 20,000 thousand smartphones manufactured. x  118.75  119 boxes of candy The maximum revenue is: R (119)  9.5 119   0.04 119   $564.06 2 b.  0.04 x 2  8.25 x  250 c. R ( x)  75 x  0.2 x 2 a  0.2, b  75, c  0 The maximum revenue occurs when b 75 75 x    187.5 2a 2  0.2  0.4 P (103)  0.04 103  8.25 103  250 2  $175.39 d. Answers will vary. R (187)  75 187   0.2 187   $7031.20 2 95. a. R (188)  75 188   0.2 188   $7031.20 P( x)  R  x   C  x   75 x  0.2 x 2   32 x  1750  b. x   1.1  1.1  4  0.06  200  P( x)  0.2 x 2  43 x  1750 2 2  0.06  a  0.2, b  43, c  1750 b 43 43    107.5 2a 2  0.2  0.4  The maximum profit occurs when x  107 or x  108 watches. The maximum profit is:  1.1  49.21 0.12 1.1  7.015 0.12 v  49 or v  68 Disregard the negative value since we are talking about speed. So the maximum speed you can be traveling would be approximately 49 mph. P (107)  0.2 107   43 107   1750 2  $561.20 P (108)  0.2 108   43 108   1750 2  $561.20 c. d. Answers will vary. 94. a. 200  1.1v  0.06v 2 0  200  1.1v  0.06v 2  0.2 x 2  43 x  1750 x d (v)  1.1v  0.06v 2 d (45)  1.1(45)  0.06(45) 2  49.5  121.5  171 ft. 2 c. P( x)  0.04 x 2  8.25 x  250 a  0.04, b  8.25, c  250 The maximum profit occurs when b 8.25 8.25 x   2a 2  0.04  0.08  103.125  103 boxes of candy The maximum profit is: The maximum revenue occurs when x  187 or x  188 watches. The maximum revenue is: b. P( x)  R  x   C  x   9.5 x  0.04 x 2  1.25 x  250  b. The minimum marginal cost is 2  b  f    f  20   5  20   200  20   4000 2 a    2000  4000  4000  $2000 93. a. b 9.5 9.5   2a 2  0.04  0.08 R( x)  9.5 x  0.04 x 2 a  0.04, b  9.5, c  0 The maximum revenue occurs when 96. a. The 1.1v term might represent the reaction time. a b 19.09 19.09    28.1 years old 2a 2  0.34  0.68 205 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b. B (28.1)  0.34(28.1) 2  19.09(28.1)  203.98  63.98 births per 1000 unmarried women c. B (40)  0.34(40) 2  19.09(40)  203.98  15.62 births/1000 unmarried women over 40 97. If x is even, then ax 2 and bx are even. When two even numbers are added to an odd number the result is odd. Thus, f ( x) is odd. If x is odd, then ax 2 and bx are odd. The sum of three odd numbers is an odd number. Thus, f ( x) is odd. Each member of this family will be a parabola with the following characteristics: (i) opens upwards since a > 0 (ii) y-intercept occurs at (0, 1). 98. Answers will vary. 99. y  x 2  2 x  3 ; y  x 2  2 x  1 ; y  x 2  2 x 101. The graph of the quadratic function f  x   ax 2  bx  c will not have any x-intercepts whenever b 2  4ac  0 . 102. By completing the square on the quadratic function f  x   ax 2  bx  c we obtain the 2 b  b2  . We can then equation y  a  x    c  2a  4a  draw the graph by applying transformations to the graph of the basic parabola y  x 2 , which opens up. When a  0 , the basic parabola will either be stretched or compressed vertically. When a  0 , the basic parabola will either be stretched or compressed vertically as well as reflected across the x-axis. Therefore, when a  0 , the graph of f ( x)  ax 2  bx  c will open up, and when a  0 , the graph of f  x   ax 2  bx  c will open down. Each member of this family will be a parabola with the following characteristics: (i) opens upwards since a > 0; b 2   1 ; (ii) vertex occurs at x   2a 2(1) (iii) There is at least one x-intercept since b 2  4ac  0 . 103. No. We know that the graph of a quadratic function f  x   ax 2  bx  c is a parabola with 100. y  x 2  4 x  1 ; y  x 2  1 ; y  x 2  4 x  1   vertex  2ba , f  2ba   . If a > 0, then the vertex is a minimum point, so the range is  f  b ,  . If a 0. Since the maximum height is 25 feet, when x  0, y  k  25 . Since the point (60, 0) is on the parabola, we can find the constant a : Since 0   a (60) 2  25 then 2 x 2  12 x  16  0 x2  6 x  8  0 ( x  4)( x  2)  0 x  4, x  2 25 . The equation of the parabola is: 602 25 h( x)   2 x 2  25 . 60 a The graph of A  2 x 2  12 x is above the graph of A  16 where the depth is between 2 and 4 inches. (0,25) (–60,0) (0,0) 10 20 40 16. Let x  width of the window and y  height of the rectangular part of the window. The x perimeter of the window is: x  2 y   20. 2 40  2 x  x . Solving for y : y  4 The area of the window is: (60,0) 2  40  2 x  x  1  x  A( x)  x    2  2  4     x 2 x 2 x 2  10 x    2 4 8  1  2      x  10 x.  2 8 This equation is a parabola opening down; thus, it has a maximum when At x  10 : 25 25 (10) 2  25    25  24.3 ft. 36 602 At x  20 : 25 25 h(20)   2 (20) 2  25    25  22.2 ft. 9 60 At x  40 : 25 100 h(40)   2 (40) 2  25    25  13.9 ft. 9 60 h(10)   230 Copyright © 2019 Pearson Education, Inc. Section 2.6: Building Quadratic Models from Verbal Descriptions and From Data 10 b 10    5.6 feet 2a 1     2     1    2 8  4 40  2(5.60)  (5.60)  2.8 feet y 4 The width of the window is about 5.6 feet and the height of the rectangular part is approximately 2.8 feet. The radius of the semicircle is roughly 2.8 feet, so the total height is about 5.6 feet. has a maximum when x x 8 b  2a 2  3   2 3 4 8   3  3 2  16  3.75 ft. 6  3 The window is approximately 3.75 feet wide.  16  48 16  3   16  6  3   6  3  8  24  y  2 2 6  3 The height of the equilateral triangle is 3  16  8 3  feet, so the total height is 2  6  3  6  3 17. Let x  the width of the rectangle or the diameter of the semicircle and let y  the length of the x rectangle. The perimeter of each semicircle is . 2 The perimeter of the track is given x x by:   y  y  1500 . 2 2 Solving for x :  x  2 y  1500 x  1500  2 y 1500  2 y x  8 8 3 24   5.62 feet. 6  3 6  3 19. We are given: V ( x)  kx(a  x)  kx 2  akx . The reaction rate is a maximum when: ak a b ak x    . 2a 2(k ) 2k 2 20. We have: a( h) 2  b(h)  c  ah 2  bh  c  y0 The area of the rectangle is:  2 2 1500  1500  2 y  A  xy   y y  y.       This equation is a parabola opening down; thus, it has a maximum when 1500 b 1500  y    375. 2a 4 2    2     1500  2(375) 750 Thus, x    238.73   The dimensions for the rectangle with maximum 750  238.73 meters by 375 meters. area are  a(0) 2  b(0)  c  c  y1 a(h) 2  b(h)  c  ah 2  bh  c  y2 Equating the two equations for the area, we have: y0  4 y1  y2  ah 2  bh  c  4c  ah 2  bh  c  2ah 2  6c. Therefore, h h Area  2ah 2 6c   y0  4 y1  y2  sq. units. 3 3  21.  f ( x)  5 x 2  8, h  1    h 1 2ah 2  6c  2(5)(1) 2  6(8) 3 3 1 38  (10  48)  sq. units 3 3 Area  18. Let x = width of the window and y = height of the rectangular part of the window. The perimeter of the window is: 3x  2 y  16 16  3 x y 2 The area of the window is 3 2 3 3 2  16  3 x  A( x)  x  x  8x  x2  x  2 4  2  4 22.  f ( x)  2 x 2  8, h  2   h 2 (2ah 2  6c)  2(2)(2) 2  6(8) 3 3 2 2 128  16  48   (64)  sq. units 3 3 3 Area   3 3 2     x  8x  2 4  This equation is a parabola opening down; thus, it 231 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 23. 26. a. f ( x)  x 2  3 x  5, h  4    h 4 2ah 2  6c  2(1)(4) 2  6(5) 3 3 4 248  (32  30)  sq. units 3 3 Area  24.    From the graph, the data appear to follow a quadratic relation with a  0 . f ( x)   x 2  x  4, h  1   h 1 (2ah 2  6c)  2(1)(1) 2  6(4) 3 3 1 1 22    2  24   (22)  sq. units 3 3 3 Area   b. Using the QUADratic REGression program 25. a. h( x)  0.0037 x 2  1.0318 x  5.6667 c. b 1.0318   139.4 2a 2(0.0037) The ball will travel about 139.4 feet before it reaches its maximum height. x d. The maximum height will be: h(139.4)  From the graph, the data appear to follow a quadratic relation with a  0 . 0.0037(139.4) 2  1.0318(139.4)  5.6667  77.6 feet b. Using the QUADratic REGression program e.    27. a. I ( x)  58.56 x 2  5301.617 x  46, 236.523 c. b 5301.617   45.3 2a 2(58.56) An individual will earn the most income at about 45.3 years of age. x d. The maximum income will be: I(48.0) = 58.56(45.3) 2  5301.617(45.3)  46, 236.523 From the graph, the data appear to be linearly related with m  0 .  $73, 756 b. Using the LINear REGression program e. R ( x)  1.321x  920.161 232 Copyright © 2019 Pearson Education, Inc. Section 2.6: Building Quadratic Models from Verbal Descriptions and From Data c. 28. a. R (875)  1.321(875)  920.161  2076 The rent for an 875 square-foot apartment in San Diego will be about $2076 per month.  30. a.    From the graph, the data appear to be linearly related with m  0 . b. Using the LINear REGression program   From the graph, the data appear to follow a quadratic relation with a  0 . b. Using the QUADratic REGression program C ( x)  0.233x  2.037 C (80)  0.233(80)  2.037  16.6 When the temperature is 80F , there will be about 16.6 chirps per second. c. M ( s )  0.017 s 2  1.935s  25.341 c. 31. Answers will vary. One possibility follows: If the price is $140, no one will buy the calculators, thus making the revenue $0. M (63)  0.017(63) 2  1.935(63)  25.341  29.1 A Camry traveling 63 miles per hour will get about 29.1 miles per gallon. 32. m  2  ( 2) 4 2   5  1 6 3 y  y1  m( x  x1 ) 29. a. 2 y  ( 2)   ( x  1) 3 2 2 y2  x 3 3 2 4 y   x 3 3 or 3 y  2 x  4 2 x  3 y  4 From the graph, the data appear to follow a quadratic relation with a  0 . b. Using the QUADratic REGression program 33. d  ( x2  x1 ) 2  ( y2  y1 ) 2  (( 1)  4) 2  (5  ( 7))2  ( 5) 2  (12) 2  25  144  169  13 2 B (a)  0.563a  32.520a  368.118 c. 34. B (35)  0.563(35) 2  32.520(35)  368.118  80.4 The birthrate of 35-year-old women is about 80.4 per 1000. ( x  h) 2  ( y  k ) 2  r 2 ( x  ( 6)) 2  ( y  0) 2  ( 7) 2 ( x  6) 2  y 2  7 233 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 35. 3(0) 2  4 y  48 4 y  48 y  12 The y intercept is (0, 12) 10. f  x  0 x2  9  0 x2  9 x   9  3 The zeros are 3 and 3. 3x 2  4(0)  48 3x 2  48 x 2  16 x  4 The x intercepts are: (4, 0), ( 4, 0) Section 2.7 1. Integers: 3, 0  Rationals: 3, 0, 6 5 11.  f  x  0 2 x  16  0 x 2  16 2. True; the set of real numbers consists of all rational and irrational numbers. x   16  4 The zeros are 4 and 4. 3. 10  5i 4. 2  5i 5. True 6. 9i 7. 2  3i 8. True 9. f  x  0 x2  4  0 x 2  4 x   4  2i The zero are 2i and 2i . 234 Copyright © 2019 Pearson Education, Inc. Section 2.7: Complex Zeros of a Quadratic Function f  x  0 12. 2 x  25  0 x 2   25 x    25  5i The zeros are 5i and 5i . f  x  0 15. 2 x  6 x  10  0 a  1, b   6, c  10 b 2  4ac  ( 6) 2  4(1)(10)  36  40   4  ( 6)   4 6  2i   3i 2(1) 2 The zeros are 3  i and 3  i . x f  x  0 13. 2 x  6 x  13  0 a  1, b   6, c  13, b 2  4ac  ( 6) 2  4(1)(13)  36  52  16  ( 6)  16 6  4i   3  2i 2(1) 2 The zeros are 3  2i and 3  2i . x f  x  0 16. 2 x  2x  5  0 a  1, b   2, c  5 b 2  4ac  ( 2) 2  4(1)(5)  4  20  16  ( 2)   16 2  4i   1  2i 2(1) 2 The zeros are 1  2i and 1  2i . x f  x  0 14. x2  4 x  8  0 a  1, b  4, c  8 b 2  4ac  42  4(1)(8)  16  32  16  4  16  4  4i    2  2i 2(1) 2 The zeros are  2  2i and  2  2i . x 235 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions f  x  0 17. 2 x  4x 1  0 a  1, b   4, c  1 b 2  4ac  ( 4) 2  4(1)(1)  16  4  12 x  ( 4)  12 4  2 3   2 3 2(1) 2 The zeros are 2  3 and 2  3 , or approximately 0.27 and 3.73. f  x  0 20. 3x 2  6 x  4  0 a  3, b  6, c  4 b 2  4ac   6   4(3)(4)  36  48  12 2 x  6  12 6  2 3i 3   1  i 2(3) 6 3 The zeros are 1  f  x  0 18. 3 3 i and 1  i. 3 3 x2  6 x  1  0 a  1, b  6, c  1 b 2  4ac  62  4(1)(1)  36  4  32 x  6  32  6  4 2   3  2 2 2(1) 2 The zeros are 3  2 2 and 3  2 2 , or approximately 5.83 and 0.17 . f  x  0 21. x2  x  1  0 a  1, b  1, c  1, b 2  4ac  12  4(1)(1)  1  4  3 x 1 3 1 3 i and   i. The zeros are   2 2 2 2 f  x  0 19. 1  3 1  3 i 1 3    i 2(1) 2 2 2 2 2x  2x  1  0 a  2, b  2, c  1 b 2  4ac   2   4(2)(1)  4  8  4 2  2  4  2  2i 1 1    i 2(2) 4 2 2 1 1 1 1 The zeros are   i and   i . 2 2 2 2 x 236 Copyright © 2019 Pearson Education, Inc. Section 2.7: Complex Zeros of a Quadratic Function f  x  0 22. 2 x  x 1  0 a  1, b  1, c  1 b 2  4ac  (1) 2  4(1)(1)  1  4  3 x  (1)  3 1  3 i 1 3    i 2(1) 2 2 2 The zeros are 1 3 1 3  i and  i. 2 2 2 2 25. 3x 2  3 x  4  0 a  3, b   3, c  4 b 2  4ac  ( 3) 2  4(3)(4)  9  48  39 The equation has two complex solutions that are conjugates of each other. 26. 2 x 2  4 x  1  0 a  2, b   4, c  1 b 2  4ac  ( 4) 2  4(2)(1)  16  8  8 The equation has two unequal real number solutions. f  x  0 23. 2 2 x  8 x  1  0 a  2, b  8, c  1 2 27. 2 x 2  3x  4  0 a  2, b  3, c   4 2 b  4ac  8  4(2)(1)  64  8  72 x 8  72 8  6 2 4  3 2 3 2    2 2(2) 4 2 2 b 2  4ac  32  4(2)( 4)  9  32  41 The equation has two unequal real solutions. 43 2 43 2 and , or 2 2 approximately 0.12 and 4.12. The zeros are 28. x 2  2 x  6  0 a  1, b  2, c  6 b 2  4ac   2   4(1)(6)  4  24   20 The equation has two complex solutions that are conjugates of each other. 2 29. 9 x 2  12 x  4  0 a  9, b  12, c  4 b 2  4ac  (12) 2  4(9)(4)  144  144  0 The equation has a repeated real solution. 24. 30. 4 x 2  12 x  9  0 a  4, b  12, c  9 f  x  0 3x 2  6 x  1  0 a  3, b  6, c  1 b 2  4ac  122  4(4)(9)  144  144  0 The equation has a repeated real solution. b 2  4ac  62  4(3)(1)  36  12  48 31. t 4  16  0 6  48 6  4 3 3  2 3 2 3 x    1 2(3) 6 3 3 (t 2  4)(t 2  4)  0 t 2  4 t 2  4 t  2 t  2i 3 2 3 3 2 3 and , or 3 3 approximately 0.15 and 2.15. The zeros are 237 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 32. y 4  81  0 2 3 3 3 1  3 The solution set is   i,  i, 3, 1 2 2 2 2   2 ( y  9)( y  9)  0 y 2  9 y 2  9 y  3 y  3i 35. 33. F ( x)  x 6  9 x3  8  0 ( x3  8)( x3  1)  0 ( x  2)( x 2  2 x  4)( x  1)( x 2  x  1)  0  2 x  2 x  4  0  a  1, b  2, c  4 x x 1 g ( x)  x 2  3x  2 x2  x( x  1) x( x  1) x 2  3x  2  x2 x( x  1) 3x  2  x( x  1) 2  22  4  4 2  12 2  2i 3   x 2(1) 2 2   1  3i Domain:  x | x  1, x  0 x 2  x  1  0  a  1, b  1, c  1 x x2 x x2 x ( g  f )( x)   x x 1 x( x) ( x  2)( x  1)   x( x  1) x( x  1) f ( x)  1  12  4 1 1  3 1  i 3   2(1) 2 2 36. a. Domain:  3,3 Range:  2, 2 b. Intercepts:  3, 0 ,  0, 0 ,  3, 0 1 3 i   2 2 c. Symmetric with respect to the orgin. d. The relation is a function. It passes the vertical line test.   1 3 , 2,1 The solution set is 1  i 3,   i 2 2   37. 34. P ( z )  z 6  28 z 3  27  0 ( z 3  27)( z 3  1)  0 ( z  3)( z 2  3 z  9)( z  1)( z 2  z  1)  0 z 2  3z  9  0 a  1, b  3, c  9 z    3   32  4  9  2(1) Local maximum: (0,0) Local Minima: (-2.12,-20.25), (2.12,-20.25) Increasing: (-2.12,0), (2.12,4) Decreasing: (-4, -2.12), (0,2.12) 3  27  2 3  3i 3 3 3 3   i 2 2 2 k x2 k k 24  2  25 5 k  600 38. y  z 2  z  1  0  a  1, b  1, c  1 z    1   1 2  4 1 2(1)  1  3 1  i 3  2 2 y 1 3  i 2 2 600 x2 238 Copyright © 2019 Pearson Education, Inc. Section 2.8: Equations and Inequalities Involving the Absolute Value Function Section 2.8 13. a. Since the graphs of f and g intersect at the points (9, 6) and (3, 6) , the solution set of f ( x)  g ( x) is {9, 3} . b. Since the graph of f is below the graph of g when x is between 9 and 3 , the solution set of f ( x)  g ( x) is {x | 9  x  3} or , using interval notation, [9, 3] . c. Since the graph of f is above the graph of g to the left of x  9 and to the right of x  3 , the solution set of f ( x)  g ( x) is {x | x  9 or x  3} or , using interval notation, (, 9)  (3, ) . 14. a. Since the graphs of f and g intersect at the points (0, 2) and (4, 2) , the solution set of f ( x)  g ( x) is {0, 4} . b. Since the graph of f is below the graph of g when x is between 0 and 4, the solution set of f ( x)  g ( x) is {x | 0  x  4} or , using interval notation, [0, 4] . c. Since the graph of f is above the graph of g to the left of x  0 and to the right of x  4 , the solution set of f ( x)  g ( x) is {x | x  0 or x  4} or , using interval notation, (, 0)  (4, ) . 15. a. Since the graphs of f and g intersect at the points (2,5) and (3,5) , the solution set of f ( x)  g ( x) is {2, 3} . 1. x  2  2. The distance on a number line from the origin to a is a for any real number a . 3. 4 x  3  9 4 x  12 x3 The solution set is {3}. 4. 3x  2  7 3x  9 x3 The solution set is {x | x > 3} or, using interval notation,  3,   . 5. 1  2 x  5  13 6  2 x  8 3  x  4 The solution set is  x | 3  x  4 or, using interval notation,  3, 4  . 6. To graph f  x   x  3 , shift the graph of y  x to the right 3 units. y f ( x)  x  3  b. Since the graph of f is above the graph of g to the left of x  2 and to the right of x  3 , the solution set of f ( x)  g ( x) is {x | x  2 or x  3} or , using interval notation, (, 2]  [3, ) .  x   c. Since the graph of f is below the graph of g when x is between 2 and 3, the solution set of f ( x)  g ( x) is {x | 2  x  3} or , using interval notation, (2, 3) . 16. a. Since the graphs of f and g intersect at the points (4, 7) and (3, 7) , the solution set of f ( x)  g ( x) is {4, 3} . 7. a ; a 8. a  u  a 9.  10. True 11. False. Any real number will be a solution of x  2 since the absolute value of any real b. Since the graph of f is above the graph of g to the left of x  4 and to the right of x  3 , the solution set of f ( x)  g ( x) is {x | x  4 or x  3} or , using interval notation, (, 4]  [3, ) . number is positive. 12. False. u  a is equivalent to u  a or u  a . 239 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions c. Since the graph of f is below the graph of g when x is between 4 and 3, the solution set of f ( x)  g ( x) is {x | 4  x  3} or , using interval notation, (4, 3) . 17. x 6 x  6 or x   6 The solution set is {–6, 6}. 18. x  12 x  12 or x  12 23.  2x  8 or  2 x   8 x   4 or x4 The solution set is {–4, 4}. 24.  2 x  1 2x  1 2 x  1 or 2 x  1 1 1 x or x   2 2 2x  3  5 2 x  3  5 or 2 x  3   5 2 x  2 or 2x   8 x  1 or x  4 The solution set is {–4, 1}. 20. 3x  1  2 3x  1  2 or 3x  1   2 3x  3 or 3x  1 1 x  1 or x 3 1 The solution set is  , 1 . 3 26. 5  1 x 3 2  1 x  2 2 1 x 2 2 1 1 x  2 or x  2 2 2 x  4 or x  4 The solution set is 4, 4 . 1  4t  8  13 1  4t  5 1  4t  5 or 1  4t  5 4t  4 or  4t  6 3 t  1 or t 2 3 The solution set is 1, . 2 27.   22.   1 1 The solution set is  , . 2 2   21.  x 1  x  1 or  x  1 The solution set is {–1, 1}. 25. 4  2 x  3 The solution set is 12, 12 . 19.  2x  8 2 x 9 3 27 x  2 27 27 x or x   2 2 27 27 . The solution set is  , 2 2  1 2z  6  9 1 2z  3 28. 1  2 z  3 or 1  2 z  3 2 z  2 or  2 z  4 z  1 or z2  3 x 9 4 x  12 x  12 or x   12 The solution set is {–12, 12}. The solution set is 1, 2 . 240 Copyright © 2019 Pearson Education, Inc. Section 2.8: Equations and Inequalities Involving the Absolute Value Function 29. x 2  2 3 5 35. x 2  2 x  3 or x 2  2 x  3 x 2 x 2   2 or   2 3 5 3 5 5 x  6  30 or 5 x  6   30 5 x  24 or 24 x or 5 x 2  2 x  3  0 or x 2  2 x  3  0 ( x  3)( x  1)  0 or x 2  2 x  3  0 5 x  36   x  3 or x  1 3x  8 36. x 1   1 2 3 or 3 x  2   6 or or 8 x or 3 32. ( x  3)( x  4)  0 or x 2  x  3  0 1  1  48 2 1 47 1  47   i 2 2 2 x   x  3 or x  4 The solution set is  1 47 1 47  i,   i . 4,3,   2 2 2 2   1 2 No solution, since absolute value always yields a non-negative number. u2   2  v  1 37. x2  x  1  1 x2  x  1  1 or x 2  x  1  1 x2  x  2  0 x2  9  0 or x 2  x  0  x  1 x  2   0 or x  x  1  0 x2  9  0 x  1, x  2 or x  0, x  1 x2  9 The solution set is 2,  1, 0,1 . x  3 The solution set is 3, 3 . 34. x 2  x  12 x 2  x  12  0 or x 2  x  12  0 3x   4 No solution, since absolute value always yields a non-negative number. 33.  x 2  x  12 or x 2  x  12 4 x 3 4 8 The solution set is  , . 3 3 31.  The solution set is 1,3,1  2i,1  2i . x 1  1 2 3 x 1  1 2 3 3x  2  6 2  4  12 2 2  8   1  2i 2 x 36 x 5 36 24 The solution set is  , . 5 5 30. x2  2 x  3 38. x 2  3x  2  2 x2  3x  2  2 x 2  16  0 x2  3x  4 x 2  16  0 or x 2  3 x  2  2 or x 2  3x  0 x 2  3 x  4  0 or x  x  3  0 x 2  16 x  4 The solution set is 4, 4 .  x  4  x  1  0 or x  0, x  3 x  4, x  1 The solution set is 4, 3, 0,1 . 241 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 39. x 6 6  x  6 46. 2 x  6 or 2 x  6  x  6  x  6 or   6, 6  x   3 or x  3   40. 2x  6  x x  3 or x  3 or  , 3   3,     x 9 9  x  9 47.   x2 23  x  9  x  9 or  9, 9  x 2 1 1  x  2  1  41.   1 x 3  x 1  x  3 or 1,3 x 4 x  4 or x  4   x x   4 or x  4 or  ,  4    4,     42. x4 3 5 x4  2 2  x  4  2 x  1 or x  1 6  x  2  x x  1 or x  1 or  , 1  1,    x  6  x  2 or  6,  2     49. 2x  8  8  2x  8 4  x  4  2  3t  6   2 t 2 3  2   2  t   t  2  or   , 2  3  3      3 x  15 15  3 x  15 2 3 5  x  5  x  5  x  5 or  5,5  45.  50.   7  2u  5  7  12  2u  2 6  u 1 3x  12 or 3 x  12 x   4 or x  4 u  6  u  1 or  6, 1  x x  4 or x  4 or  , 4    4,     2u  5  7 3 x  12    3t  2  4  4  3t  2  4  x  4  x  4 or  4,4  44.  x 1  43. 48.     242 Copyright © 2019 Pearson Education, Inc.   Section 2.8: Equations and Inequalities Involving the Absolute Value Function 51.  x x  1 or x  2 or  , 1   2,   x3  2 x  3  2 or x  3  2  x  1 or x  5  x x  1 or x  5 or  ,1  5,    52. 56. 3x   3 or  3x  1 x 1 x  6 or x  2  x x  6 or x  2 or  , 6   2,     1  4 x  7  2 57.  2 x  1  1  or 1  x  x  1  x  32 or  1, 32    3 2 58. 3x  4  0 All real numbers since absolute value is always non-negative.  x x is any real number or  ,   3 2  1  2 x  4  1 59. 1  2x  3 1  3x  2   7  2 1 2 1 1   3x  9  2 2 17 19  3x  2 2 17 19 x 6 6 3  1  2 x  3 3x  9  4  2 x  2 4 2 x 2 2 2  x  1 or  1  x  2  x  1  x  2 or  1,2     x | 176  x  196  or  176 , 196  1  2 x  3 1 2x  3 17 6 1  2 x  3 or 1  2 x  3 2 x   4 or  2 x  2 x2 1 3 No solution since absolute value is always nonnegative. 6  4 x  4 6 4 x 4 4 3  x  1 2 1 3   5  1  4 x  5 55. x or  1  1   x x  or x  1 or  ,   1,   3 3    1 4x  5 54. 2  3 x  1 2  3x  1 or 2  3 x  1 x4  2 x  4  2 or x  4  2 53.  2  3x  1    or x  1 243 Copyright © 2019 Pearson Education, Inc. 19 6 Chapter 2: Linear and Quadratic Functions 60. f ( x)  g ( x) b. 1  4 x  1  11  4 3 5 x  2  9 1 4 x  12  4 1 1   4 x  12  4 4 47 49  4x  4 4 47 49 x 16 16 5x  2  3 3  5 x  2  3 1  5 x  5 1   x 1 5 x |  15  x  1 or   15 ,1 x | 1647  x  1649 or  1647 , 1649  f ( x)  g ( x) c. 3 5 x  2  9 5x  2  3 49 16 47 16 5 x  2  3 or 5 x  2  3 5 x  5 or 61. 5  x  1  2  x  1  3  x 1  3 3  x  1  3 2  x  4  x | 2  x  4 or  64. a.  2, 4   f ( x)  g ( x) 2x  3  6   1 5 1  x | x   15 or x  1 or  ,    1,   5  2 x  3  6 or 2 x  3  6  x  3  4 2 x  9 or 2 x  3 9 or 2 x x x3  4 4  x  3  4 7  x  1  x | 7  x  1 or  7,1  b. f ( x)  g ( x) 2x  3  6 6  2 x  3  6 3  2 x  9  f ( x)  g ( x)  3 5 x  2  9 5 x  2  3 or 5 x  2  3 c. 5 x  1 x 3 9 x 2 2 x |  32  x  92 or  32 , 92  5x  2  3 x  1 or 3 2 2 2 x  3  12  5 x  5 or x 2 2 x  3  12 62. 6  x  3  2 63. a. 5 x  1 x  1 or f ( x)  g ( x) 2 2 x  3  12 1 5 2x  3  6 244 Copyright © 2019 Pearson Education, Inc. Section 2.8: Equations and Inequalities Involving the Absolute Value Function 2 x  3  6 or 2 x  3  6 2 x  9 or 2 x  3 9 or 2 x x Look at the graph of f ( x) and g ( x) and see where the graph of f ( x)  g ( x) . We see that this occurs where x  15 or x  53 . 3 2 65. a.   , 15    53 ,   f ( x)  g ( x) c. 3 x  2  x  10  3x  2    x  10   3x  2  x  10 or 4 x  8 3x  2   x  10 or x  2 2 x  12 or x6 Look at the graph of f ( x) and g ( x) and see where the graph of f ( x)  g ( x) . We see that this occurs where x is between 15 and 53 . So the solution set is: x | 15  x  53 or  15 , 53  . b. 67. x  10  2 2  x  10  2 8  x  12 Solution set:  x | 8  x  12 or (8, 12) Look at the graph of f ( x ) and g ( x) and see where the graph of f ( x)  g ( x) . We see that this occurs where x  2 or x  6 . So the solution set is:  x | x  2 or x  6 68. 66. a. x   6   3 x6 3 3  x  6  3 9  x  3 Solution set:  x | 9  x  3 or (9, 3) or  , 2   6,   . c.  So the solution set is: x | x  15 or x  53 or x | x   32 or x  92 or  ,  32    92 ,   Look at the graph of f ( x ) and g ( x) and see where the graph of f ( x)  g ( x) . We see that this occurs where x is between -2 and 6. So the solution set is:  x | 2  x  6 or  2, 6  . 69. 2 x   1  5 2x 1  5 2 x  1  5 or 2 x  1  5 2 x  6 or 2x  4 x  3 or x2 f ( x)  g ( x) 4x  3  x  2 Solution set:  x | x  3 or x  2 or 4x  3    x  2 4x  3  x  2 or 4 x  3   x  2 3x  5 or 5 5x  1 x or 3 1 x 5  , 3   2,   70. 2x  3  1 2 x  3  1 or 2 x  3  1 2 x  2 or 2x  4 x  1 or x2 b. Solution set:  x | x  1 or x  2 or  ,1   2,   245 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 71. the equation must always be zero or larger. Thus, the only solution to the inequality above will be when the absolute value expression equals 0: 2x 1  0 x  5.7  0.0005 0.0005  x  5.7  0.0005 5.6995  x  5.7005 The acceptable lengths of the rod is from 5.6995 inches to 5.7005 inches. 72. 2x 1  0 2x  1 1 x 2 x  6.125  0.0005 0.0005  x  6.125  0.0005 6.1245  x  6.1255 The acceptable lengths of the rod is from 6.1245 inches to 6.1255 inches. 73. 74. 1  Thus, the solution set is   . 2 x  100  1.96 15 x  100 x  100  1.96 or  1.96 15 15 x  100  29.4 or x  100  29.4 x  70.6 or x  129.4 Since IQ scores are whole numbers, any IQ less than 71 or greater than 129 would be considered unusual. 78. f ( 4)  2( 4)  7  8  7  15  15 79. 2( x  4)  x  4( x  2) 2x  8  x  4x  8 3x  8  4 x  8 x  0 x0 x  266  1.96 16 x  266 x  266  1.96 or  1.96 16 16 x  266  31.36 or x  266  31.36 x  234.64 or x  297.36 Pregnancies less than 235 days long or greater than 297 days long would be considered unusual. 80. c. Increasing: 3,5 :Decreasing:  2,1 Constant: 1, 3 5 x  1  2 No matter what real number is substituted for x, the absolute value expression on the left side of the equation must always be zero or larger. Thus, it can never equal 2 . d. Neither 2 x  5  3  1  2 x  5  2 Chapter 2 Review Exercises No matter what real number is substituted for x, the absolute value expression on the left side of the equation must always be zero or larger. Thus, it will always be larger than 2 . Thus, the solution is the set of all real numbers. 77. (5  i )(3  2i )  15  10i  3i  2i 2  15  7i  2  17  7i 81. a. Intercepts: (0,0), (4,0) b. Domain:  2,5 , Range:  2, 4 75. 5 x  1  7  5 76. f ( x)  2 x  7 1. f  x  2x  5 a. Slope = 2; y-intercept = 5 b. Plot the point (0, 5) . Use the slope to find an additional point by moving 1 unit to the right and 2 units up. 2x 1  0 No matter what real number is substituted for x, the absolute value expression on the left side of 246 Copyright © 2019 Pearson Education, Inc. Chapter 2 Review Exercises c. Domain and Range:  ,   Domain:  ,   c. Range:  y | y  4 d. Average rate of change = slope = 2 e. Increasing 2. h( x)  a. d. e. 4 x6 5 4. 4 Slope = ; y-intercept = 6 5 Domain and Range:  ,   d. Average rate of change = slope = e. Increasing 5. 4 5 3. G  x   4 a. f  x   2 x  14 zero: f  x   2 x  14  0 2 x  14 x  7 y-intercept = 14 b. Plot the point (0, 6) . Use the slope to find an additional point by moving 5 units to the right and 4 units up. c. Average rate of change = slope = 0 Constant Slope = 0; y-intercept = 4 b. Plot the point (0, 4) and draw a horizontal line through it. x y  f  x –2 –7 0 3 1 8 3 18 6 33 247 Copyright © 2019 Pearson Education, Inc. Avg. rate of change = 3   7  0   2   10 5 2 83 5  5 1 0 1 18  8 10  5 3 1 2 33  18 15  5 63 3 y x Chapter 2: Linear and Quadratic Functions This is a linear function with slope = 5, since the average rate of change is constant at 5. To find the equation of the line, we use the point-slope formula and one of the points.  x  32  4  0  x  32  4 y  y1  m  x  x1  x3   4 x  3  2 x  3 2 x  3  2  1 or x  3  2  5 y  3  5  x  0 y  5x  3 6. x y  f  x –1 –3 0 4 1 7 2 3 6 1 Avg. rate of change = 4   3 0   1  y x The zeros of g  x    x  3  4 are 1 and 5. The 2 x-intercepts of the graph of g are 1 and 5. 7 7 1 2 9x  6x 1  0 74 3  3 1 0 1 (3x  1)(3 x  1)  0 3x  1  0 x  x  72  0  x  9  x  8  0 2 2x  4x 1  0 1 x2  2 x   0 2 1 2 x  2x  2 1 2 x  2x  1   1 2 3 2  x  1  2 3 3 2 6    x 1   2 2 2 2 x 8 The zeros of f  x   x 2  x  72 are 9 and 8. The x-intercepts of the graph of f are 9 and 8. P t   0 8. 2 6t  13t  5  0 (3t  1)(2t  5)  0 3t  1  0 t or 2t  5  0 1 3 t 1 3 G  x  0 11. x 8  0 x  9 x 1 The only zero of h  x   9 x 2  6 x  1 is  . 3 1 The only x-intercept of the graph of h is  . 3 f  x  0 or or 3 x  1  0 1 x 3 2 x9  0 h  x  0 10. This is not a linear function, since the average rate of change is not constant. 7. g  x  0 9. 5 2 x  1 1 5 and . 3 2 1 5 The t-intercepts of the graph of P are  and . 3 2 The zeros of P  t   6t 2  13t  5 are  6 2 6  2 2 The zeros of G  x   2 x 2  4 x  1 are 2 6 2 2 6 . The x-intercepts of the graph of G 2 2 6 2 6 are and . 2 2 and 248 Copyright © 2019 Pearson Education, Inc. Chapter 2 Review Exercises f  x  0 g  2   4  2   1  8  1  7 . The graphs of the f 2 x  x  1  0 and g intersect at the points (2, 9) and  2, 7  . 12. 2 2 2x  x 1  0 (2 x  1)( x  1)  0 2x 1  0 or x  1  0 1 x 1 x 2 1 and 1. 2 1 The x-intercepts of the graph of f are  and 1. 2 The zeros of f  x   2 x 2  x  1 are  13. f  x  g  x f  x  0 15.  x  32  16 x4  5×2  4  0  x  4  x  1  0 2 x  3   16  4 x  3 4 x  3  4  1 or x  3  4  7 The solution set is {1, 7} . The x-coordinates of the points of intersection are 1 and 7. The y-coordinates are g  1  16 and 2 x 2  4  0 or x 2  1  0 x  2 or x  1 The zeros of f  x   x 4  5 x 2  4 are 2 , 1 , 1, and 2. The x-intercepts of the graph of f are 2 , 1 , 1, and 2. g  7   16 . The graphs of the f and g intersect at F  x  0 16. the points (1, 16) and (7, 16) .  x  3  2  x  3  48  0 2 Let u  x  3  u 2   x  3 2 u 2  2u  48  0  u  6  u  8   0 u  6  0 or u  8  0 u  6 u 8 14. x  3  6 x3  8 x  3 x  11 The zeros of F  x    x  3  2  x  3  48 are 2 f  x  g  x 3 and 11. The x-intercepts of the graph of F are 3 and 11. x2  4 x  5  4x  1 x2  4  0  x  2  x  2   0 17. x  2  0 or x  2  0 x  2 x2 The solution set is {2, 2} . The x-coordinates of the points of intersection are 2 and 2. The y-coordinates are g  2   4  2   1  8  1  9 and h  x  0 3x  13 x  10  0 Let u  x  u 2  x 249 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions then shift up 2 units. 3u 2  13u  10  0  3u  2  u  5   0 3u  2  0 or u  5  0 u 5 2 3 2 x  3 x  not real u x 5 x  52  25 Check: h  25   3  25   13 25  10 20.  3  25   13  5   10 f ( x)   ( x  4) 2 Using the graph of y  x 2 , shift the graph 4 units right, then reflect about the x-axis.  75  65  10  0 The only zero of h  x   3 x  13 x  10 is 25. The only x-intercept of the graph of h is 25. f  x  0 18. 2 1 1  x   4  x   12  0     Let u  1 1  u2    x x 2 u 2  4u  12  0  u  2  u  6   0 21. u20 or u  6  0 u  2 u6 1 1  2 6 x x 1 1 x x 2 6 f ( x)  2( x  1) 2  4 Using the graph of y  x 2 , stretch vertically by a factor of 2, then shift 1 unit left, then shift 4 units up. 2 1 1 1 The zeros of f  x      4    12 are  2 x x     1 . The x-intercepts of the graph of f are 6 1 1  and . 2 6 and 19. 22. a. f ( x)   x  2   2 2 f  x    x  2  2 2  x2  4 x  4  2 Using the graph of y  x 2 , shift right 2 units,  x2  4 x  6 a  1, b  4, c  6. Since a  1  0, the graph opens up. The x-coordinate of the b 4 4 vertex is x      2. 2a 2(1) 2 The y-coordinate of the vertex is  b  f     f (2)  (2) 2  4  2   6  2 .  2a  250 Copyright © 2019 Pearson Education, Inc. Chapter 2 Review Exercises The y-intercept is f (0)  16 . Thus, the vertex is (2, 2). The axis of symmetry is the line x  2 . The discriminant is: b 2  4ac  (4) 2  4 1 (6)  8  0 , so the graph has no x-intercepts. The y-intercept is f (0)  6 . b. Domain: (, ) . Range: [16, ) . c. Decreasing on  , 0 ; increasing on  0,   . b. Domain: (, ) . Range: [2, ) . c. Decreasing on  , 2 ; increasing on 24. a. a   4, b  4, c  0. Since a   4  0, the graph opens down. The x-coordinate of the b 4 4 1 vertex is x      . 8 2 2a 2( 4) The y-coordinate of the vertex is  2,   . 23. a. f ( x)  f ( x)   4 x 2  4 x 1 2 x  16 4 2  b  1 1 1 f     f     4   4   2a  2 2 2  1  2  1 1  Thus, the vertex is  , 1 . 2  1 The axis of symmetry is the line x  . 2 The discriminant is: b 2  4ac  42  4( 4)(0)  16  0 , so the graph has two x-intercepts. The x-intercepts are found by solving:  4×2  4 x  0 1 1 , b  0, c  16. Since a   0, the 4 4 graph opens up. The x-coordinate of the 0 b 0     0. vertex is x   1 2a 1 2  2 4   The y-coordinate of the vertex is 1  b  f     f (0)  (0) 2  16  16 . 4  2a  Thus, the vertex is (0, –16). The axis of symmetry is the line x  0 . The discriminant is: 1 b 2  4ac  (0) 2  4   (16)  16  0 , so 4 the graph has two x-intercepts. The x-intercepts are found by solving: 1 2 x  16  0 4 x 2  64  0 a  4 x( x  1)  0 x  0 or x  1 The x-intercepts are 0 and 1. The y-intercept is f (0)   4(0) 2  4(0)  0 . x 2  64 x  8 or x   8 The x-intercepts are –8 and 8. 251 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions b. c. 25. a. Domain: (, ) . Range:  , 1 . b. 1  Increasing on  ,  ; decreasing on 2  1   2 ,   . f ( x)  c. 9 2 x  3x  1 2 26. a. 9 9 , b  3, c  1. Since a   0, the 2 2 graph opens up. The x-coordinate of the 3 3 b 1    . vertex is x   2a 9 3 9 2  2 The y-coordinate of the vertex is 1  Domain: (, ) . Range:  ,   . 2  1  Decreasing on  ,   ; increasing on 3   1    3 ,   . f ( x)  3x 2  4 x  1 a  3, b  4, c  1. Since a  3  0, the graph opens up. The x-coordinate of the b 4 4 2 vertex is x      . 2a 2(3) 6 3 The y-coordinate of the vertex is a 2  b   2  2  2 f     f     3    4    1 2 a 3 3        3 4 8 7   1   3 3 3  2 7 Thus, the vertex is   ,   .  3 3 2 The axis of symmetry is the line x   . 3 The discriminant is: b 2  4ac  (4) 2  4(3)(1)  28  0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2  4 x  1  0 . 2  b   1 9 1  1 f     f         3   1 a 2 3 2 3        3 1 1  1 1  2 2  1 1 Thus, the vertex is   ,  .  3 2 1 The axis of symmetry is the line x   . 3 The discriminant is: 9 b 2  4ac  32  4   (1)  9  18   9  0 , 2 so the graph has no x-intercepts. The y9 2 intercept is f  0    0   3  0   1  1 . 2 x  b  b 2  4ac  4  28  2a 2(3) 4  2 7 2  7  6 3 The x-intercepts are 252 Copyright © 2019 Pearson Education, Inc. 2  7  1.55 and 3 Chapter 2 Review Exercises 12 12 b   2. 2a 2(3) 6 The maximum value is 2  b  f     f  2   3  2   12  2   4  2a   12  24  4  16 2  7  0.22 . 3 x The y-intercept is f (0)  3(0) 2  4(0)  1  1 . 30. Consider the form y  a  x  h   k . The vertex 2 is  2, 4 so we have h  2 and k  4 . The function also contains the point  0, 16 . Substituting these values for x, y, h, and k, we can solve for a: 16  a  0   2    4 2 b. c. 27.  7  Domain: (, ) . Range:   ,   .  3  16  a  2  4 2 16  4a  4 2  Decreasing on  ,   ; increasing on 3   2    3 ,   . 12  4a a  3 The quadratic function is f  x   3  x  2  4  3 x 2  12 x  16 . 2 f ( x)  3 x 2  6 x  4 31. Use the form f ( x)  a ( x  h) 2  k . The vertex is ( 1, 2) , so h  1 and k  2 . a  3, b   6, c  4. Since a  3  0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b 6 6 x   1. 2a 2(3) 6 The minimum value is 2  b  f     f 1  3 1  6 1  4 2 a    36 4 1 28. f ( x)  a( x  1) 2  2 . Since the graph passes through (1, 6) , f (1)  6 . 6  a (1  1) 2  2 6  a (2) 2  2 6  4a  2 4  4a 1 a f ( x)  ( x  1) 2  2 f ( x)   x 2  8 x  4 a  1, b  8, c   4. Since a  1  0, the graph opens down, so the vertex is a maximum point. The maximum occurs at b 8 8 x    4. 2a 2(1) 2 The maximum value is 2  b  f     f  4    4  8  4  4  2a   16  32  4  12 29.  ( x 2  2 x  1)  2  x2  2 x  3 32. x 2  6 x  16  0 f ( x)  x 2  6 x  16 x 2  6 x  16  0 ( x  8)( x  2)  0 x   8, x  2 are the zeros of f . Interval ( ,  8)  8, 2   2,   Test Number 9 0 3 Value of f 11 16 11 Conclusion Positive Negative Positive f ( x)  3x 2  12 x  4 a  3, b  12, c  4. Since a  3  0, the graph opens down, so the vertex is a maximum point. The maximum occurs at 253 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The solution set is  x | 8  x  2 or, using interval notation,  8, 2  . 3x 2  14 x  5 33. 3x 2  14 x  5  0 f ( x)  3 x 2  14 x  5 3x 2  14 x  5  0 (3 x  1)( x  5)  0 1 x   , x  5 are the zeros of f . 3 2 2 x  4 x  3  0 a  2, b  4, c  3 1  1    5,    ,     , 5  3  3   Test Number 1 0 2 Value of f 12 5 19 Conclusion Positive Negative Positive Interval b 2  4ac  42  4(2)(3)  16  24  8 x 1   The solution set is  x x   or x  5 or, 3   1  using interval notation,  ,    5,   . 3  34. p  x  0 36. 4  8 4  2 2 i 2   1 i 4 2(2) 2 The zeros are 1  2 2 i and 1  i. 2 2  f  x  0 x2  8  0 x 2  8 x   8  2 2 i The zero are 2 2 i and 2 2 i . f  x  0 37. 4 x2  4 x  3  0 a  4, b  4, c  3 b 2  4ac  42  4(4)(3)  16  48  32 x 4  32 4  4 2 i 1 2    i 2(4) 8 2 2 g  x  0 35. 2 x  2x  4  0 a  1, b  2, c  4 b 2  4ac  22  4(1)(4)  4  16  20 x  2  20  2  2 5   1  5 2(1) 2 The zeros are 1  5 and 1  5 . 254 Copyright © 2019 Pearson Education, Inc. Chapter 2 Review Exercises  x x   2 or x  7 or  ,  2  7,   1 2 1 2 The zeros are   i and   i. 2 2 2 2 42. 2  2  3x  4 2  3x  2 2  2  3 x  2 38. 39. 4  3 x  0 4 x0 3  4  4  x 0  x   or 0,  3  3   2x  3  7 2 x  3  7 or 2x  3  7 2 x  4 or 2 x   10 x  2 or x  5 The solution set is 5, 2 . 43. 1  2  3 x  4  2  3 x  5 2  3x  2  9 2  3x  5 2  3x  7 2  3x  7 3x  5 x 2  3x  5 or 2  3x  5 7  3x or  3  3x 7  x or  1  x 3 7 x  1 or x  3  7 7   x x   1 or x   or  ,  1   ,   3 3    or 2  3x  7 or  3 x   9 5 or 3 x3   5 The solution set is  , 3 . 3 40. 3x  4  1 2 1 1  3x  4  2 2 9 7   3x   2 2 3 7   x  2 6  3 7  3 7  x   x    or   ,   2 6  2 6    41. 44. a. 2x  5  9 2 x  5   9 or 2 x  5  9 2 x   4 or 2x  14 x   2 or x7 S ( x)  0.01x  25, 000 b. S (1, 000, 000)  0.01(1, 000, 000)  25, 000  10, 000  25, 000  35, 000 Bill’s salary would be $35,000. c. 0.01x  25, 000  100, 000 0.01x  75, 000 x  7,500, 000 Bill’s sales would have to be $7,500,000 in order to earn $100,000. d. 0.01x  25, 000  150, 000 0.01x  125, 000 x  12,500, 000 Bill’s sales would have to be more than 255 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions Disregard the negative answer because the width of a rectangle must be positive. Thus, the width is 6 inches, and the length is 8 inches $12,500,000 in order for his salary to exceed $150,000. 45. a. If x  1500  10 p, then p  1500  x . 10 47. C ( x)  4.9 x 2  617.4 x  19, 600 ; a  4.9, b  617.4, c  19, 600. Since a  4.9  0, the graph opens up, so the vertex is a minimum point. a. The minimum marginal cost occurs at  617.40 617.40 b x    63 . 2a 2(4.9) 9.8 Thus, 63 golf clubs should be manufactured in order to minimize the marginal cost. R ( p )  px  p (1500  10 p )  10 p 2  1500 p b. Domain:  p 0  p  150 c. p b 1500 1500    $75 2a 2  10  20 d. The maximum revenue is R(75)  10(75) 2  1500(75)  56250  112500  $56, 250 e. x  1500  10(75)  1500  750  750 f. Graph R  10 p 2  1500 p and R  56000 . b. The minimum marginal cost is  b  C     C  63  2a   4.9  63    617.40  63   19600 2  $151.90 48. Since there are 200 feet of border, we know that 2 x  2 y  200 . The area is to be maximized, so A  x  y . Solving the perimeter formula for y : 2 x  2 y  200 2 y  200  2 x y  100  x The area function is: A( x)  x(100  x)   x 2  100 x The maximum value occurs at the vertex: b 100 100 x    50 2a 2(1) 2 The pond should be 50 feet by 50 feet for maximum area. 49. The area function is: A( x)  x(10  x)   x 2  10 x The maximum value occurs at the vertex: b 10 10 x   5 2a 2(1) 2 The maximum area is: Find where the graphs intersect by solving 56000  10 p 2  1500 p . 10 p 2  1500 p  56000  0 p 2  150 p  5600  0 ( p  70)( p  80)  0 p  70, p  80 The company should charge between $70 and $80. 46. Let w = the width. Then w + 2 = the length. w  in. w By the Pythagorean Theorem we have: w2   w  2   10  2 2 w2  w2  4w  4  100 2w2  4 w  96  0 w2  2w  48  0  w  8 w  6   0 w  8 or w  6 256 Copyright © 2019 Pearson Education, Inc. Chapter 2 Review Exercises Since each input (price) corresponds to a single output (quantity demanded), we know that the quantity demanded is a function of price. Also, because the average rate of change is constant at $0.8 per LCD monitor, the function is linear. A(5)  (5)2  10(5)   25  50  25 square units 10 (x,10-x) (0,10-x) c. (x,0) 10 50. Locate the origin at the point directly under the highest point of the arch. Then the equation is in the form: y  ax 2  k , where a  0 . Since the maximum height is 10 feet, when x  0, y  k  10 . Since the point (10, 0) is on the parabola, we can find the constant: 0   a(10) 2  10 10 1  0.10 a 2  10 10 The equation of the parabola is: 1 y   x 2  10 10 At x  8 : 1 y   (8) 2  10   6.4  10  3.6 feet 10 From part (b), we know m  0.8 . Using ( p1 , q1 )  (75, 100) , we get the equation: q  q1  m( p  p1 ) q  100  0.8( p  75) q  100  0.8 p  60 q  0.8 p  160 Using function notation, we have q( p )  0.8 p  160 . d. The price cannot be negative, so p  0 . Likewise, the quantity cannot be negative, so, q( p)  0 . 0.8 p  160  0 0.8 p  160 p  200 Thus, the implied domain for q(p) is { p | 0  p  200} or [0, 200] . e. 51. a. f. b. Avg. rate of change = q p p q 75 100 100 80 80  100 20   0.8 100  75 25 125 60 60  80 20   0.8 125  100 25 150 40 40  60 20   0.8 150  125 25 g. If the price increases by $1, then the quantity demanded of LCD monitors decreases by 0.8 monitor. p-intercept: If the price is $0, then 160 LCD monitors will be demanded. q-intercept: There will be 0 LCD monitors demanded when the price is $200. 257 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions The maximum revenue occurs at b   411.88  A  2a 2(7.76) 411.88   $26.5 thousand 15.52 52. a. c. The maximum revenue is  b  R   R  26.53866   2a   7.76  26.5    411.88  26.5   942.72 2  $6408 thousand b. Yes, the two variables appear to have a linear relationship. c. Using the LINear REGression program, the line of best fit is: y  1.390171918 x  1.113952697 d. d. y  1.390171918  26.5   1.113952697  38.0 mm Chapter 2 Test 53. a. 1. The data appear to be quadratic f  x   4 x  3 a. The slope f is 4 . b. The slope is negative, so the graph is decreasing. c. Plot the point (0, 3) . Use the slope to find an additional point by moving 1 unit to the right and 4 units down. with a < 0. b. Using the QUADratic REGression program, the quadratic function of best fit is: y  7.76 x 2  411.88 x  942.72 . 258 Copyright © 2019 Pearson Education, Inc. Chapter 2 Test 2. x y 2 12 Avg. rate of change = y x f  x  g  x 5. 2 x  3x  5 x  3 2 1 7 7  12 5   5 1  (2) 1 0 2 5 27   5 0  (1) 1 1 3 3  2 5   5 1 0 1 2 8 8  (3) 5   5 2 1 1 x  2x  3  0 ( x  1)( x  3)  0 x  1  0 or x  3  0 x  1 x3 The solution set is 1, 3 . Since the average rate of change is constant at 5 , this is a linear function with slope = 5 . To find the equation of the line, we use the point-slope formula and one of the points. y  y1  m  x  x1  y  2  5  x  0  y  5 x  2 3.  x  1  5  x  1  4  0 2 Let u  x  1  u 2   x  1 2 f  x  0 3x 2  2 x  8  0 (3x  4)( x  2)  0 u 2  5u  4  0  u  4  u  1  0 3x  4  0 u  4  0 or u  1  0 u  4 u  1 x  1  4 x  1  1 x  3 x0 The zeros of G are 3 and 0. or x  2  0 4 x2 x 3 The zeros of f are  4. f  x  0 6. 4 and 2. 3 G  x  0 7. 2 x 2  4 x  1  0 a  2, b  4, c  1 f ( x)   x  3  2 2 Using the graph of y  x 2 , shift right 3 units, then shift down 2 units. y 2 b  b 2  4ac 4  4  4  2 1 x  2a 2  2   4  24 4  2 6 2  6    2 4 4 2 6 2 6 The zeros of G are and . 2 2      259 Copyright © 2019 Pearson Education, Inc.    x Chapter 2: Linear and Quadratic Functions b. The x-coordinate of the vertex is b 4 4 x    1. 2a 2(2) 4 The y-coordinate of the vertex is 2  b  g     g 1  2 1  4 1  5 2 a    2  4  5  3 Thus, the vertex is 1, 3 . f ( x)  3 x 2  12 x  4 a  3, b  12, c  4. Since a  3  0, the graph opens up. b. The x-coordinate of the vertex is b  12 12 x   2. 2a 2  3 6 8. a. The y-coordinate of the vertex is 2  b  f     f  2   3  2   12  2   4 2 a    12  24  4  8 Thus, the vertex is  2, 8  . c. c. The axis of symmetry is the line x  1 . d. The discriminant is: b 2  4ac   4   4  2  5   24  0 , so the 2 graph has no x-intercepts. The y-intercept is g (0)  2(0) 2  4(0)  5  5 . The axis of symmetry is the line x  2 . d. The discriminant is: b 2  4ac   12   4  3 4   96  0 , so the 2 e. graph has two x-intercepts. The x-intercepts are found by solving: 3x 2  12 x  4  0 . x  b  b 2  4ac (12)  96  2a 2(3) 12  4 6 6  2 6  6 3 The x-intercepts are 62 6  0.37 and 3 62 6  3.63 . The y-intercept is 3 f (0)  3(0) 2  12(0)  4  4 . f. The domain is (, ) . The range is (, 3] . g. Increasing on  , 1 . Decreasing on 1,   . e. 10. Consider the form y  a  x  h   k . From the 2 graph we know that the vertex is 1, 32  so we have h  1 and k  32 . The graph also passes through the point  x, y    0, 30  . Substituting these values for x, y, h, and k, we can solve for a: 30  a (0  1) 2  (32) The quadratic function is f. The domain is (, ) . The range is [8, ) . g. Decreasing on  , 2 . 30  a (1) 2  32 30  a  32 2a f ( x)  2( x  1) 2  32  2 x 2  4 x  30 . Increasing on  2,   . 9. a. g ( x)  2 x 2  4 x  5 a  2, b  4, c  5. Since a  2  0, the graph opens down. 260 Copyright © 2019 Pearson Education, Inc. Chapter 2 Test 11. f ( x)  2 x 2  12 x  3 15. a  2, b  12, c  3. Since a  2  0, the graph opens down, so the vertex is a maximum point. The maximum occurs at b 12 12 x   3. 2a 2(2) 4 The maximum value is x3 2 4 x3 2  2 4 8  x  3  8 11   x  11  x  5 or  11, 5 f  3  2  3  12  3  3  18  36  3  21 . 2 12. x 2  10 x  24  0 f ( x)  x 2  10 x  24 16. x 2  10 x  24  0 2x  3  4  3 2x  3  7 ( x  4)( x  6)  0 x  4, x  6 are the zeros of f. 2 x  3  7 Interval  , 4   4, 6   6,   Test Number 0 5 7 Value of f 24 1 3 Conclusion Positive Negative Positive interval notation,  , 4   6,   . 17. a. b. f  x  0 x 2 b  b 2  4ac 4  4  4  2  5   2a 2  2  6 4  24 4  2 6 i i   1  4 4 2 The complex zeros of f are 1  14. 2x  4 x   5 or x2 C  m   0.15m  129.50 C  860   0.15  860   129.50 C  m   213.80 c. 0.15m  129.50  213.80 0.15m  84.30 m  562 The rental cost is $213.80 if 562 miles were driven. 6 i and 2 6 i. 2 18. a. 3x  1  8 b. 3x  1  8 2 x  10 or  129  129.50  258.50 If 860 miles are driven, the rental cost is $258.50. 2 x2  4 x  5  0 a  2, b  4, c  5 1  or 2 x  3  7  x x  5 or x  2 or  , 5   2,   The solution set is  x x  4 or x  6 or, using 13. x 5 1  1  R ( x )  x   x  1000    x 2  1000 x 10 10   1 (400) 2  1000(400) 10  16, 000  400, 000 R (400)   or 3x  1  8 3x  7 or 3x  9 7 x 3 or x  3  $384, 000 c.   The solution set is 3, 7 . 3 x b 1000 1000    5000 2a 2   1    1  10 5 The maximum revenue is 261 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions Chapter 2 Cumulative Review 1 (5000) 2  1000(5000) 10  250, 000  5, 000, 000 R (5000)   1. P   1,3 ; Q   4, 2   $2,500, 000 Thus, 5000 units maximizes revenue at $2,500,000. d. Distance between P and Q: d  P, Q    1 p   (5000)  1000 10  500  1000 Set A:  5   5 2 2 2  50  5 2 Midpoint between P and Q:  1  4 3  2   3 1  ,     ,   1.5, 0.5  2  2 2  2  2. y  x 3  3x  1  a.  2, 1 : 1   2   3  2   1 3 1  8  6  1 1  1 Yes,  2, 1 is on the graph.  The data appear to be linear with a negative slope. Set B: 2  25  25  $500 19. a.  4   1    2  3 b.  2,3 : 3   2   3  2   1 3 3  8  6 1  33 Yes,  2,3 is on the graph. c.  3,1 : 1   3  3  3  1 3 1  27  9  1   The data appear to be quadratic and opens up. 1  35 No,  3,1 is not on the graph. 3. 5 x  3  0 5 x  3 3 x 5 b. Using the LINear REGression program, the linear function of best fit is: y  4.234 x  2.362 .  3  3  The solution set is  x x    or   ,   . 5 5     c. Using the QUADratic REGression program, the quadratic function of best fit is: y  1.993 x 2  0.289 x  2.503 . 262 Copyright © 2019 Pearson Education, Inc. Chapter 2 Cumulative Review 4. (–1,4) and (2,–2) are points on the line. 2  4 6 Slope    2 2   1 3 y  y1  m  x  x1  y  4  2  x   1  y  4  2  x  1 y  4  2 x  2 y  2 x  2 7. Yes, this is a function since each x-value is paired with exactly one y-value. 8. f ( x)  x 2  4 x  1 a. f (2)  22  4  2   1  4  8  1  3 b. f ( x)  f  2   x 2  4 x  1   3  x2  4x  2 5. Perpendicular to y  2 x  1 ; Containing (3,5) 1 Slope of perpendicular =  2 y  y1  m( x  x1 ) c. f ( x)    x   4   x   1  x 2  4 x  1 d.  f ( x)    x 2  4 x  1   x 2  4 x  1 e. f ( x  2)   x  2   4  x  2   1 2 2  x2  4 x  4  4 x  8  1  x2  3 1  x  3 2 1 3 y 5   x 2 2 1 13 y   x 2 2 y 5   f. f ( x  h)  f  x  h  x  h   4  x  h   1   x 2  4 x  1 2  h x  2 xh  h  4 x  4h  1  x 2  4 x  1  h 2 2 xh  h  4h  h h  2x  h  4   2x  h  4 h 2 2 3z  1 6z  7 The denominator cannot be zero: 6z  7  0 6z  7 9. h( z )  6. x 2  y 2  4 x  8 y  5  0 x2  4x  y2  8 y  5 ( x 2  4 x  4)  ( y 2  8 y  16)  5  4  16 7 6  7 Domain:  z z   6  z ( x  2)  ( y  4)  25 2 2 ( x  2) 2  ( y  4) 2  52 Center: (2,–4) Radius = 5 10. Yes, the graph represents a function since it passes the Vertical Line Test. 263 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions 11. f ( x)  a. b. c. 12. x x4 15. a. Range:  y | 1  y  3 or  1, 3 1 1 1  1 f (1)    , so  1,  is not on 1 4 5 4  4 the graph of f. b. Intercepts:  1, 0  ,  0, 1 , 1, 0  2 2   1, so  2,  1 is a 2  4 2 point on the graph of f. c. x-intercepts: 1, 1 y-intercept: 1 f (2)  The graph is symmetric with respect to the y-axis. d. When x  2 , the function takes on a value of 1. Therefore, f  2   1 . Solve for x: x 2 x4 2x  8  x x  8 So, (8, 2) is a point on the graph of f. e. The function takes on the value 3 at x  4 and x  4 . f. f  x   0 means that the graph lies below the x-axis. This happens for x values between 1 and 1. Thus, the solution set is  x | 1  x  1 or  1, 1 . x2 f ( x)  2x 1 x x2   f  x  or  f  x  2   x   1 2 x  1 2 f ( x)  g. The graph of y  f  x   2 is the graph of y  f  x  but shifted up 2 units. Therefore, f is neither even nor odd. 13. Domain:  x | 4  x  4 or  4, 4 y  f  x   x  5 x  4 on the interval  4, 4  3 5  Use MAXIMUM and MINIMUM on the graph of y1  x3  5 x  4 .      5 5 x 5 h. The graph of y  f   x  is the graph of y  f  x  but reflected about the y-axis. Local maximum is 5.30 and occurs at x  1.29 ; Local minimum is –3.30 and occurs at x  1.29 ; f is increasing on  4, 1.29 or 1.29, 4 ; y 5  f is decreasing on  1.29,1.29 . 14. f  x   3 x  5; a.  5 g  x  2x 1 f  x  g  x  5 x 5 3x  5  2 x  1 3x  5  2 x  1 b.     i. The graph of y  2 f  x  is the graph of x  4 y  f  x  but stretched vertically by a f  x  g  x factor of 2. That is, the coordinate of each point is multiplied by 2. 3x  5  2 x  1 3x  5  2 x  1 x  4 The solution set is  x x  4 or  4,   . 264 Copyright © 2019 Pearson Education, Inc. Chapter 2 Projects y 10       5  5 x  j. Since the graph is symmetric about the yaxis, the function is even. k. The function is increasing on the open interval  0, 4  . Chapter 2 Projects Project I – Internet-based Project Answers will vary. Project II a. b. The data would be best fit by a quadratic function. 1000 m/sec   y  0.085 x 2  14.46 x  1069.52 kg 1000 m/sec   kg These results seem reasonable since the function fits the data well. 265 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions c. s0 = 0m Type Weight kg Velocity m/sec MG 17 10.2 905 2 v0 t  s0 2 s (t )  4.9t 2  639.93t Best. (It goes the highest) MG 131 19.7 710 s (t )  4.9t 2  502.05t MG 151 41.5 850 s (t )  4.9t 2  601.04t MG 151/20 42.3 695 s (t )  4.9t 2  491.44t MG/FF 35.7 575 s (t )  4.9t 2  406.59t MK 103 145 860 s (t )  4.9t 2  608.11t MK 108 58 520 s (t )  4.9t 2  367.70t WGr 21 111 315 s (t )  4.9t 2  222.74t Type Weight kg Velocity m/sec MG 17 10.2 905 2 v0 t  s0 2 s (t )  4.9t 2  639.93t  200 Best. (It goes the highest) MG 131 19.7 710 s (t )  4.9t 2  502.05t  200 MG 151 41.5 850 s (t )  4.9t 2  601.04t  200 MG 151/20 42.3 695 s (t )  4.9t 2  491.44t  200 MG/FF 35.7 575 s (t )  4.9t 2  406.59t  200 MK 103 145 860 s (t )  4.9t 2  608.11t  200 MK 108 58 520 s (t )  4.9t 2  367.70t  200 WGr 21 111 315 s (t )  4.9t 2  222.74t  200 Type Weight kg Velocity m/sec MG 17 10.2 905 2 v0 t  s0 2 s (t )  4.9t 2  639.93t  30 Best. (It goes the highest) MG 131 19.7 710 s (t )  4.9t 2  502.05t  30 MG 151 41.5 850 s (t )  4.9t 2  601.04t  30 MG 151/20 42.3 695 s (t )  4.9t 2  491.44t  30 MG/FF 35.7 575 s (t )  4.9t 2  406.59t  30 MK 103 145 860 s (t )  4.9t 2  608.11t  30 MK 108 58 520 s (t )  4.9t 2  367.70t  30 WGr 21 111 315 s (t )  4.9t 2  222.74t  30 Equation in the form: s (t )  4.9t 2  s0 = 200m Equation in the form: s (t )  4.9t 2  s0 = 30m Equation in the form: s (t )  4.9t 2  Notice that the gun is what makes the difference, not how high it is mounted necessarily. The only way to change the true maximum height that the projectile can go is to change the angle at which it fires. 266 Copyright © 2019 Pearson Education, Inc. Chapter 2 Projects Project III a. x 1 2 3 4 5 y  2 x  5 3 1 1 3 5 b. y y2  y1 1  3    2 1 x x2  x1 y y2  y1 1  1    2 1 x x2  x1 y y2  y1 3  (1)    2 1 x x2  x1 y y2  y1 5  (3)    2 1 x x2  x1 All of the values of y are the same. x  c. Median Income ($)  d. e.  Age Class Midpoint I 30633  9548   2108.50 x 10 I 37088  30633   645.50 x 10 I 41072  37088   398.40 x 10 I 34414  41072   665.80 10 x I 19167  34414   1524.70 x 10 I values are not all equal. The data are not linearly related. These x x 2 1 0 1 y 23 9 3 5 15 33 59 y x 2 3 4 14 6 2 10 18 26 As x increases, y increases. This makes sense because the parabola is increasing (going up) steeply as x x increases. 267 Copyright © 2019 Pearson Education, Inc. Chapter 2: Linear and Quadratic Functions f. x 2 1 0 1 y 23 2 y x2 9 2 3 4 3 5 15 33 59 8 8 8 8 8 The second differences are all the same. g. The paragraph should mention at least two observations: 1. The first differences for a linear function are all the same. 2. The second differences for a quadratic function are the same. Project IV a. – i. Answers will vary , depending on where the CBL is located above the bouncing ball. j. The ratio of the heights between bounces will be the same. 268 Copyright © 2019 Pearson Education, Inc.