Chapter 3 Transformations of Graphs and Data
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Chapter 3 Transformations of Graphs and Data Chapter Review
(pp. 216–219)
1. n(m(6)) = n(3) = 27
10. D; Multiplying times x stretches horizontally and dividing y shrinks vertically.
2. (n m)(2) = n(m(2)) = n(1) = 1
11. B; Adding to x moves to the right and subtracting from y moves down.
3. a. f(g(-2)) = f(6) = 33 b. g(f(-2)) = g(-23) = 552
12. Use the Graph Translation Theorem. y 7 = (x + 2)2 so y = x 2 + 4x +11
4. f(f(t)) = f(7t – 9) = 7(7t – 9) – 9 = 49t – 72 5. a. ( f g)(-8) = f (g(-8)) = f (-16) = - 14 b. (g g)(-8) = g(g(-8)) = g(-16) = -24 6. (g f )(x) = g( f (x)) = g ( 4x ) = 4x 8 7. a. The range of the function is the nonnegative real numbers so the domain of the inverse will be the nonnegative real numbers. To find the equation of the inverse we see that the original function can be written as ±y = x . Switching x and y and solving for y we find the inverse to be y = ±x . b. This is not a function because each x is mapped to y = x and y = -x. 8. a. Set l(x) = y. Switch x and y and solve for y to find the inverse. y = 4x + 3 b. The inverse is a function. This is the function y = 1x shrunk by a factor of 4 in the y direction and translated up three, which is still a function. 9. a. The inverse is the set of points where T(x, y) = (y, x). {(4, 3), (12, 4), (9, 5), (0, 6), (11, 7)} b. This is a function because each input has exactly one output.
13. Use the Graph Scale-Change Theorem. y 3
= (2x)2 so y = 12x 2
14. Use the Graph Scale-Change Theorem. x 4y = 5x so y = 20 15. Use the Graph Translation Theorem. y = x +1 16. Use the Graph Scale-Change theorem. The graph will be shrunk by a factor of 5 in the horizontal direction so the transformation is S(x, y) = ( 5x , y) . 17. Use the Graph Translation Theorem. The graph will be translated up by 4 units so the transformation is T (x, y) = (x, y + 4) . 18. By the Graph Translation Theorem we know the image is translated 90 units to the right. 19. By the Graph Translation Theorem we know the image is translated 15.3 units down. 20. By the Graph Scale-Change Theorem we know the image is shrunk by a factor of 3 in the horizontal direction. 21. By the Graph Scale-Change Theorem we know the image is stretched by a factor of 8 in the vertical direction. 22. a. 20; Graph Translation Theorem
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b. 22; Graph Scale-Change Theorem c. 21; Graph Scale-Change Theorem d. 19; Graph Translation Theorem 23. True. Under a translation everything is moved proportionally, including asymptotes. 24. We are trying to map y to -y so the scale change is S(x, y) = (x, -y) . 25. neither; k(-t) = -3t + 4 3t + 4 = k(t) and k(-t) = -3t + 4 - 3t + 4 = -k(t) 26. odd; j(-m) = 10(-m)3 = -10m 3 = - j(m) 27. even; f (-x) = 3 -x +1 = 3 x +1 = f (x) 28. even; s(-y) = 7(-y)2 3(-y)4 = 7y 2 3y 4 = s(y)
29. Asymptotes will be at places where x or y cannot reach. x cannot reach 12 so
x=
1 2
is an asymptote and y cannot
reach 0 so y = 0 is an asymptote. 30. The graph is probably centered around the intersection of the asymptotes, which is the point (-2, 3). 31. f(x) can only input non-negative numbers so x 2 0 . The domain is { x x 2} . 32. The translation is where x is mapped to y and y is mapped to x so T (x, y) = (y, x) . 33. false; The inverse of a function is the function such that f -1 ( f ( x)) = x . 34. false;
f g(x) = f ( x + 2) = ( x + 2) = 2
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x + 4 x + 4 but
( )
g f (x) = g x 2 = x 2 + 2 = x + 2 .
35. true; This is the definition of an inverse function. 36. false; See question 35 answer. 37. false; Taking a number and adding 2 and then multiplying by 3 gives x+2 but 3 switching the order gives
x +2, 3
which
are not equal. 38. 1; definition of a standardized data set 39. 0; definition of a standardized data set 40. This means that the person’s raw score is 0.04 standard deviations above the mean. 41. This means that the person’s raw score is 1.3 standard deviations below the mean. 42. a. The mode is the score with the highest frequency so mode = 8; there are 15 scores so the median is the 8th from the bottom, or 8; to find the mean use the frequencies as weights so mean = 7. b. T(x) = 4x c. The mode is the score with the highest frequency so mode = 32; there are 15 scores so the median is the 8th from the bottom, or 32; to find the mean use the frequencies as weights so mean = 28. d. This is the Central Tendency of Scale Changes of Data Theorem. 43. a. 9 – 4 = 5 b. 36 – 16 = 20 c. This is the Spread of Scale Changes of Data Theorem. 44. Subtract 100 from each score. The mean of the translated scores is 6 so the mean of the original scores is 106.
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45. a. The mean will be the transformation of the original mean so mean = 46.2 mm. b. The range will be the transformation of the original range so range = 68.8 mm. c. The variance will be the square of the transformed standard deviation, so variance 136.52 mm2. 46. Transform each score to its z-score by subtracting the mean and dividing by the standard deviation. The z-score of the FST test is 1.463 and the z-score of the English test is 1.698. His z-score is higher on his English test so he did better compared to his classmates on his English test. 47. Standardize each weight by subtracting the mean and dividing by the standard deviation. The standard score of the porpoise is 1.517. The standard score of the tuna is 1.412. The porpoise has a higher standard score so the porpoise is heaver relative to its group. 48. The function appears to have an inverse because each input has one output and each output has one input. 49. The function does not appear to have an inverse because some outputs have more than one input, for example y = 0 has two inputs.
57. odd; The function appears to have origin symmetry. 58.
59. a.
b. Looking at the graph the asymptotes are at y = 12 and x = -4. c. Set y = 0 and solve the equation for x. 2 0 = 12 x+4 , so 12(x + 4) = 2 and
x = - 23 , 0) . . The intercept is at (- 23 6 6
60. Replace each x with -x and multiply each y-coordinate by 4.
50. E 51. B 52. D 53. A 54. C 55. F
61. All points are transformed the same way so choose one point on the preimage and find the translation needed to map it to the same point on the image. The point (1, 3) went to (4, 4) so the translation is T(x, y) = (x + 3, y + 2).
56. even; The function appears to be symmetric to the y-axis. 93
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62. a.
c. The inverse has exactly one output for each input so it is a function. 67. a. The inverse can be found by switching y and x and solving for y. x = 2y so y = 2x is the inverse. b.
b. false; g is the image of f under the transformation S(x, y) = (x, 2y) . 63. The scale is the same (as determined by the three points given) so it is only a translation. The vertex moved from (0, 0) to (-2, 5) so the equation for the translation is y 5 = (x + 2)2 or
c. The inverse is the same as the function so it is a function.
y = x 2 + 4x + 9 .
64. The point (0, 0) was mapped to itself so there was no translation. The scale change can be determined by testing a point. 2 = a(2) , so we find that a is 2. This checks with the other points so the equation of the scale change is y = 2x . 65. This is a translation because the asymptotes have moved. The asymptotes are translated the same amount as the rest of the function, so the equation of the translation is y2 = 1 2 . (x+4)
66. a. The inverse can be found by switching y and x and solving for y. x = -2y 3 so y = 3
-x 2
is the inverse.
b.
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68. a. The inverse can be found by switching y and x and solving for y. x = -y 2 + 4 so y = ± 4 x is the inverse. b.
c. The inverse has more than one output for several inputs, such as x = 0, so it is not a function. 69. The graphs are inverses of each other because they are symmetric over the line y = x. This implies that f(g(x)) = g(f(x)) = x.
Chapter 3, Chapter Review
Chapter 3 Transformations of Graphs and Data Chapter Review
(pp. 216–219)
1. n(m(6)) = n(3) = 27
10. D; Multiplying times x stretches horizontally and dividing y shrinks vertically.
2. (n m)(2) = n(m(2)) = n(1) = 1
11. B; Adding to x moves to the right and subtracting from y moves down.
3. a. f(g(-2)) = f(6) = 33 b. g(f(-2)) = g(-23) = 552
12. Use the Graph Translation Theorem. y 7 = (x + 2)2 so y = x 2 + 4x +11
4. f(f(t)) = f(7t – 9) = 7(7t – 9) – 9 = 49t – 72 5. a. ( f g)(-8) = f (g(-8)) = f (-16) = - 14 b. (g g)(-8) = g(g(-8)) = g(-16) = -24 6. (g f )(x) = g( f (x)) = g ( 4x ) = 4x 8 7. a. The range of the function is the nonnegative real numbers so the domain of the inverse will be the nonnegative real numbers. To find the equation of the inverse we see that the original function can be written as ±y = x . Switching x and y and solving for y we find the inverse to be y = ±x . b. This is not a function because each x is mapped to y = x and y = -x. 8. a. Set l(x) = y. Switch x and y and solve for y to find the inverse. y = 4x + 3 b. The inverse is a function. This is the function y = 1x shrunk by a factor of 4 in the y direction and translated up three, which is still a function. 9. a. The inverse is the set of points where T(x, y) = (y, x). {(4, 3), (12, 4), (9, 5), (0, 6), (11, 7)} b. This is a function because each input has exactly one output.
13. Use the Graph Scale-Change Theorem. y 3
= (2x)2 so y = 12x 2
14. Use the Graph Scale-Change Theorem. x 4y = 5x so y = 20 15. Use the Graph Translation Theorem. y = x +1 16. Use the Graph Scale-Change theorem. The graph will be shrunk by a factor of 5 in the horizontal direction so the transformation is S(x, y) = ( 5x , y) . 17. Use the Graph Translation Theorem. The graph will be translated up by 4 units so the transformation is T (x, y) = (x, y + 4) . 18. By the Graph Translation Theorem we know the image is translated 90 units to the right. 19. By the Graph Translation Theorem we know the image is translated 15.3 units down. 20. By the Graph Scale-Change Theorem we know the image is shrunk by a factor of 3 in the horizontal direction. 21. By the Graph Scale-Change Theorem we know the image is stretched by a factor of 8 in the vertical direction. 22. a. 20; Graph Translation Theorem
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b. 22; Graph Scale-Change Theorem c. 21; Graph Scale-Change Theorem d. 19; Graph Translation Theorem 23. True. Under a translation everything is moved proportionally, including asymptotes. 24. We are trying to map y to -y so the scale change is S(x, y) = (x, -y) . 25. neither; k(-t) = -3t + 4 3t + 4 = k(t) and k(-t) = -3t + 4 - 3t + 4 = -k(t) 26. odd; j(-m) = 10(-m)3 = -10m 3 = - j(m) 27. even; f (-x) = 3 -x +1 = 3 x +1 = f (x) 28. even; s(-y) = 7(-y)2 3(-y)4 = 7y 2 3y 4 = s(y)
29. Asymptotes will be at places where x or y cannot reach. x cannot reach 12 so
x=
1 2
is an asymptote and y cannot
reach 0 so y = 0 is an asymptote. 30. The graph is probably centered around the intersection of the asymptotes, which is the point (-2, 3). 31. f(x) can only input non-negative numbers so x 2 0 . The domain is { x x 2} . 32. The translation is where x is mapped to y and y is mapped to x so T (x, y) = (y, x) . 33. false; The inverse of a function is the function such that f -1 ( f ( x)) = x . 34. false;
f g(x) = f ( x + 2) = ( x + 2) = 2
92
Functions, Statistics and Trigonometry Solution Manual
x + 4 x + 4 but
( )
g f (x) = g x 2 = x 2 + 2 = x + 2 .
35. true; This is the definition of an inverse function. 36. false; See question 35 answer. 37. false; Taking a number and adding 2 and then multiplying by 3 gives x+2 but 3 switching the order gives
x +2, 3
which
are not equal. 38. 1; definition of a standardized data set 39. 0; definition of a standardized data set 40. This means that the person’s raw score is 0.04 standard deviations above the mean. 41. This means that the person’s raw score is 1.3 standard deviations below the mean. 42. a. The mode is the score with the highest frequency so mode = 8; there are 15 scores so the median is the 8th from the bottom, or 8; to find the mean use the frequencies as weights so mean = 7. b. T(x) = 4x c. The mode is the score with the highest frequency so mode = 32; there are 15 scores so the median is the 8th from the bottom, or 32; to find the mean use the frequencies as weights so mean = 28. d. This is the Central Tendency of Scale Changes of Data Theorem. 43. a. 9 – 4 = 5 b. 36 – 16 = 20 c. This is the Spread of Scale Changes of Data Theorem. 44. Subtract 100 from each score. The mean of the translated scores is 6 so the mean of the original scores is 106.
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45. a. The mean will be the transformation of the original mean so mean = 46.2 mm. b. The range will be the transformation of the original range so range = 68.8 mm. c. The variance will be the square of the transformed standard deviation, so variance 136.52 mm2. 46. Transform each score to its z-score by subtracting the mean and dividing by the standard deviation. The z-score of the FST test is 1.463 and the z-score of the English test is 1.698. His z-score is higher on his English test so he did better compared to his classmates on his English test. 47. Standardize each weight by subtracting the mean and dividing by the standard deviation. The standard score of the porpoise is 1.517. The standard score of the tuna is 1.412. The porpoise has a higher standard score so the porpoise is heaver relative to its group. 48. The function appears to have an inverse because each input has one output and each output has one input. 49. The function does not appear to have an inverse because some outputs have more than one input, for example y = 0 has two inputs.
57. odd; The function appears to have origin symmetry. 58.
59. a.
b. Looking at the graph the asymptotes are at y = 12 and x = -4. c. Set y = 0 and solve the equation for x. 2 0 = 12 x+4 , so 12(x + 4) = 2 and
x = - 23 , 0) . . The intercept is at (- 23 6 6
60. Replace each x with -x and multiply each y-coordinate by 4.
50. E 51. B 52. D 53. A 54. C 55. F
61. All points are transformed the same way so choose one point on the preimage and find the translation needed to map it to the same point on the image. The point (1, 3) went to (4, 4) so the translation is T(x, y) = (x + 3, y + 2).
56. even; The function appears to be symmetric to the y-axis. 93
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62. a.
c. The inverse has exactly one output for each input so it is a function. 67. a. The inverse can be found by switching y and x and solving for y. x = 2y so y = 2x is the inverse. b.
b. false; g is the image of f under the transformation S(x, y) = (x, 2y) . 63. The scale is the same (as determined by the three points given) so it is only a translation. The vertex moved from (0, 0) to (-2, 5) so the equation for the translation is y 5 = (x + 2)2 or
c. The inverse is the same as the function so it is a function.
y = x 2 + 4x + 9 .
64. The point (0, 0) was mapped to itself so there was no translation. The scale change can be determined by testing a point. 2 = a(2) , so we find that a is 2. This checks with the other points so the equation of the scale change is y = 2x . 65. This is a translation because the asymptotes have moved. The asymptotes are translated the same amount as the rest of the function, so the equation of the translation is y2 = 1 2 . (x+4)
66. a. The inverse can be found by switching y and x and solving for y. x = -2y 3 so y = 3
-x 2
is the inverse.
b.
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68. a. The inverse can be found by switching y and x and solving for y. x = -y 2 + 4 so y = ± 4 x is the inverse. b.
c. The inverse has more than one output for several inputs, such as x = 0, so it is not a function. 69. The graphs are inverses of each other because they are symmetric over the line y = x. This implies that f(g(x)) = g(f(x)) = x.
Chapter 3, Chapter Review
