Solution Manual for Precalculus, 6th Edition

Chapter 2 GRAPHS AND FUNCTIONS Section 2.1 Rectangular Coordinates and Graphs 8. True 1. The point (–1, 3) lies in quadrant II in the rectangular coordinate system. 2. The point (4, 6) lies on the graph of the equation y = 3x – 6. Find the y-value by letting x = 4 and solving for y. y  3  4  6  12  6  6 3. Any point that lies on the x-axis has y-coordinate equal to 0. 5. The x-intercept of the graph of 2x + 5y = 10 is (5, 0). Find the x-intercept by letting y = 0 and solving for x. 2 x  5 0  10  2 x  10  x  5 6. The distance from the origin to the point (–3, 4) is 5. Using the distance formula, we have d ( P, Q)  (3  0) 2  (4  0) 2 7. True 32  42  9  16  10. False. The distance between the point (0, 0) and (4, 4) is d ( P, Q)  (4  0) 2  (4  0) 2  42  42  16  16  32  4 2 4. The y-intercept of the graph of y = –2x + 6 is (0, 6).  9. False. The midpoint of the segment joining (0, 0) and (4, 4) is 40 40 4 4 ,     ,    2, 2 . 2 2  2 2 25  5 11. Any three of the following: 2, 5 , 1, 7  , 3, 9 , 5, 17  , 6, 21 12. Any three of the following: 3, 3 , 5, 21 , 8,18 , 4, 6 , 0, 6 13. Any three of the following: (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20), (2009, 20) 14. Any three of the following: 2002,86.8 , 2004, 89.8 , 2006, 90.7 , 2008, 97.4 , 2010, 106.5 , 2012,111.4 , 2014, 111.5 15. P(–5, –6), Q(7, –1) (a) d ( P, Q)  [7 – (–5)]2  [1 – (–6)]2  122  52  169  13 (b) The midpoint M of the segment joining points P and Q has coordinates  –5  7 6  (1)   2 7  ,     ,   2 2 2 2 7   1,   .  2 Copyright © 2017 Pearson Education, Inc. 170 Section 2.1 Rectangular Coordinates and Graphs 16. P(–4, 3), Q(2, –5)   2, – 5  21. P 3 2, 4 5 , Q 2 (a) d ( P, Q)  [2 – (– 4)]  (–5 – 3) 2  62  (–8) 2  100  10 (b) The midpoint M of the segment joining points P and Q has coordinates  – 4  2 3  (–5)   2 2  ,   ,  2 2   2 2    1, 1 . 17. P (8, 2), Q(3, 5) (a) d ( P, Q)  (3 – 8) 2  (5 – 2) 2   171 52  32  25  9  34 (b) The midpoint M of the segment joining points P and Q has coordinates  8  3 2  5   11 7  ,  , .  2 2   2 2 18. P (−8, 4), Q (3, −5) 2 (a) d ( P, Q)  3 –  8  5  4 2  112  (–9) 2  121  81  202 (b) The midpoint M of the segment joining points P and Q has coordinates  –8  3 4  (–5)   5 1  ,    ,  .  2 2   2 2 19. P(–6, –5), Q(6, 10) (a) d ( P, Q)  [6 – (– 6)]2  [10 – (–5)]2  122  152  144  225  369  3 41 (b) The midpoint M of the segment joining points P and Q has coordinates  – 6  6 –5  10   0 5   5  ,    ,    0,  .  2 2  2 2  2 20. P(6, –2), Q(4, 6) 2  (–2)  8  2 – 3 2   – 5 – 4 5    –2 2    –5 5  2  4  64  68  2 17 2  2 2 2  8  125  133 (b) The midpoint M of the segment joining points P and Q has coordinates  3 2  2 4 5  (– 5)  ,   2 2   4 2 3 5   3 5 , .    2 2,  2   2   2    22. P – 7, 8 3 , Q 5 7, – 3  (a) d ( P, Q)  [5 7 – (– 7 )]2  (– 3 – 8 3) 2  (6 7 ) 2  (–9 3) 2  252  243  495  3 55 (b) The midpoint M of the segment joining points P and Q has coordinates  – 7  5 7 8 3  (– 3)  ,   2 2   4 7 7 3  7 3 , .   2 7,  2   2   2 23. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the length of each side of the triangle. d ( A, B )  [0 – (– 6)]2  [–2 – (– 4)]2  62  22  36  4  40 d ( B, C )  (–10 – 0)2  [8 – (–2)]2  (10) 2  102  100  100  200 d ( A, C )  [–10 – (– 6)]2  [8 – (– 4)]2  (– 4) 2  122  16  144  160 2 (a) d ( P, Q)  (4 – 6)  [6 – (–2)] 2 (a) d ( P, Q) Because  40    160    200  , 2 2 triangle ABC is a right triangle. (b) The midpoint M of the segment joining points P and Q has coordinates  6  4 2  6   10 4  ,     ,   5, 2 2 2   2 2 Copyright © 2017 Pearson Education, Inc. 2 172 Chapter 2 Graphs and Functions 24. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle. 2 d ( B, C )   (– 3) 2  (–11) 2 2 d ( A, B )  [0 – (–2)]  [– 4 – (–8)] 2  9  121  130 2  2  4  4  16  20 2 2 d ( B, C )  (– 4 – 0)  [–7 – (– 4)] 2 2 d ( A, C )   –1 –  –4   6  3 2  (– 4)  (–3)  16  9 Because  25  5 d ( A, C )  [– 4 – (–2)]2  [–7 – (–8)]2  (–2) 2  12  4  1  5 Because ( 5) 2  ( 20 ) 2  5  20  25  52 , triangle ABC is a right triangle. 2 2 2  5  3  25  9  34 2 d ( B, C )  (– 6 – 1)  (–1 – 4) 2 d ( A, C )  [– 6 – (– 4)]  (–1 – 1) 2 26. Label the points A(–2, –5), B(1, 7), and C(3, 15). d ( A, B )  [1  (2)]2  [7  (5)]2  32  122  9  144  153 2 d ( B, C )  (3  1)  (15  7) 2  22  82  4  64  68 2 2 d ( A, C )  [3  (2)]  [15  (5)]  425  because 2 68  153  221  425 , triangle ABC is not a right triangle. 27. Label the points A(–4, 3), B(2, 5), and C(–1, –6). 2 d ( A, B )   2 –  –4  5  3 2 2  132   6 2 2  169  36  205 d ( B, C )   0  62   –15 –  –2 2  – 62   –132  36  169  205 2 d ( A, C )  0 –  –7    15  4 2  7 2   19  49  361  410 2 Because  205    205    410  , 2 2 2 triangle ABC is a right triangle. 29. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points. 2 d ( A, B )  3  02  5 –  –7   –32  122  9  144  153  3 17 2 d ( B, C )   2 –  3   –15 – 5 2  52   –20  25  400 2  52  202  25  400  425 Because ( 68) 2  ( 153) 2  2 d ( A, B )  6 –  –7    2  4  (–2) 2  (–2) 2  4  4  8 Because ( 8) 2  ( 34 ) 2  ( 74 )2 because 8  34  42  74, triangle ABC is not a right triangle. 2 28. Label the points A(–7, 4), B(6, –2), and C(0, –15). 2  (–7) 2  (–5) 2  49  25  74  40    90    130  , triangle ABC is a right triangle. 25. Label the points A(–4, 1), B(1, 4), and C(–6, –1). d ( A, B )  [1 – (– 4)]  (4 – 1) 2  32   9  9  81  90 2 2 1  22  6  52 2  425  5 17 d ( A, C )  2  02   15 – 7  2  22   –8  68  2 17 2 Because d ( A, B )  d ( A, C )  d ( B, C ) or 3 17  2 17  5 17, the points are collinear.  62  22  36  4  40 Copyright © 2017 Pearson Education, Inc. Section 2.1 Rectangular Coordinates and Graphs 30. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of points. 2 d ( A, B )   –2 –  –1   –1 – 4  2 2 2 d ( B, C )  1 –  –2  14 –  –1 2  32  152  234  3 26 2 d ( A, C )  1 –  –1  14 – 4 2  22  102  104  2 26 Because 26  2 26  3 26 , the points are collinear. 31. Label the points A(0, 9), B(–3, –7), and C(2, 19). 2 d ( A, B )  (–3 – 0)  (–7 – 9) 2  (–3) 2  (–16) 2  9  256  265  16.279 2 d ( B, C )   2 –  –3  19 –  –7  2 32. Label the points A(–1, –3), B(–5, 12), and C(1, –11). 2 d ( A, B )   –5 –  –1  12 –  –3  – 42  152  16  225  241  15.5242 d ( B, C )  1 –  –5   –11 – 12 2  6   –23  36  529  565  23.7697 2  132   6  169  36 2  205  14.3178 d ( B, C )  1  62  1   –2  72  32  49  9 2 2 2 265  104  701 16.279  10.198  26.476, 26.477  26.476, the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.) 2 2 d ( A, B)  6 –  –7    2  4  62   –3  36  9 or 2 33. Label the points A(–7, 4), B(6,–2), and C(–1,1). 2 2 – 02  19 – 92 2 2 d ( A, C )   1 –  –7   1  4  22  102  4  100  104  10.198 Because d ( A, B )  d ( A, C )  d ( B, C )   22   –8  4  64  58  7.6158  52  262  25  676  701  26.476 d ( A, C )  2  68  8.2462 Because d(A, B) + d(A, C)  d(B, C) or 241  68  565 15.5242  8.2462  23.7697 23.7704  23.7697, the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)  –1   –5  26 2 2 d ( A, C )  1 –  –1   –11 –  –3 173 2  45  6.7082 Because d(B, C) + d(A, C)  d(A, B) or 58  45  205 7.6158  6.7082  14.3178 14.3240  14.3178, the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.) 34. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points. 2 d ( A, B)   2 –  –4  5  3  62  22  36  4  40  2 10 2 d ( B, C )  (1  2) 2  (4  5) 2  32  (–1)2  9  1  10 2 d ( A, C )   1 –  4   4  3 2  32  12  9  1  10 Because d ( B, C )  d ( A, C )  d ( A, B ) or 10  10  2 10, the points are collinear. Copyright © 2017 Pearson Education, Inc. 174 Chapter 2 Graphs and Functions 35. Midpoint (5, 8), endpoint (13, 10) 13  x 10  y  5 and 8 2 2 13  x  10 and 10  y  16 x  –3 and y  6. The other endpoint has coordinates (–3, 6). 36. Midpoint (–7, 6), endpoint (–9, 9) –9  x 9 y  –7 and 6 2 2 –9  x  –14 and 9  y  12 x  –5 and y  3. The other endpoint has coordinates (–5, 3). 37. Midpoint (12, 6), endpoint (19, 16) 19  x 16  y  12 and 6 2 2 19  x  24 and 16  y  12 x  5 and y  – 4. The other endpoint has coordinates (5, –4). 38. Midpoint (–9, 8), endpoint (–16, 9) –16  x 9 y  –9 and 8 2 2 –16  x  –18 and 9  y  16 x  –2 and y7 The other endpoint has coordinates (–2, 7). 39. Midpoint (a, b), endpoint (p, q) px q y and a b 2 2 and q  y  2b p  x  2a x  2a  p and y  2b  q The other endpoint has coordinates (2a  p, 2b  q ) . 40. Midpoint 6a, 6b  , endpoint 3a, 5b  3a  x 5b  y  6a and  6b 2 2 3a  x  12a and 5b  y  12b x  9a and y  7b The other endpoint has coordinates (9a, 7b). 41. The endpoints of the segment are (1990, 21.3) and (2012, 30.1).  1990  2012 21.3  30.9  M  ,   2 2   2001, 26.1 The estimate is 26.1%. This is very close to the actual figure of 26.2%. 42. The endpoints are (2006, 7505) and (2012, 3335)  2006  2012 7505  3335  M  ,   2 2   2009, 5420 According to the model, the average national advertising revenue in 2009 was $5420 million. This is higher than the actual value of $4424 million. 43. The points to use are (2011, 23021) and (2013, 23834). Their midpoint is  2011  2013 23, 021  23,834  ,   2 2  (2012, 23427.5). In 2012, the poverty level cutoff was approximately $23,428. 44. (a) To estimate the enrollment for 2003, use the points (2000, 11,753) and (2006, 13,180)  2000  2006 11, 753  13,180  M  ,   2 2   2003, 12466.5 The enrollment for 2003 was about 12,466.5 thousand. (b) To estimate the enrollment for 2009, use the points (2006, 13,180) and (2012, 14,880)  2006  2012 13,180  14,880  M  ,   2 2   2009, 14030 The enrollment for 2009 was about 14,030 thousand. 45. The midpoint M has coordinates  x1  x2 y1  y2   2 , 2  . d ( P, M ) 2 x x   y  y2  – y1    1 2 – x1    1  2   2  2 2 2 x   y  y2 2 y1  x x   1 2 – 1  1 –  2 2   2 2  2  x  x   y  y1    2 1  2  2   2   2 2  x2  x1 2   y2  y1 2 4 4  x2  x1    y2  y1 2 2   12 4  x2  x1    y2  y1 2 2 (continued on next page) Copyright © 2017 Pearson Education, Inc. Section 2.1 Rectangular Coordinates and Graphs (continued) (b) d ( M , Q) 2 x x   y  y2     x2  1 2    y2  1  2   2  2 2 x  x   2y y  y2   2x   2  1 2  2  1  2 2   2 2  2  x  x   y  y1    2 1  2  2   2   x2  x1    y2  y1  2  4 2 2 48. (a) 2 x 0 y 2 4 0 y-intercept: x0 1 y   0   2  2 2 x-intercept: y0 1 0 x2 2 1 2   x  x  4 2 2 1 additional point x 0 y 5 3 y-intercept: x0 2 0   3 y  5  3 y  5  y  53 5 2 0 x-intercept: y0 2 x  3 0   5  2 x  5  x  52 4 −1 additional point 4  x2  x1    y2  y1 2 2   12 4  x2  x1    y2  y1 2 2 d ( P, Q )   x2  x1 2   y2  y1 2 Because 12  x2  x1 2   y2  y1 2 2 2  12  x2  x1    y2  y1  2 2   x2  x1    y2  y1  , (b) this shows d ( P, M )  d ( M , Q )  d ( P, Q ) and d ( P, M )  d ( M , Q ). 46. The distance formula, d  ( x2 – x1 ) 2  ( y2 – y1 ) 2 , can be written as d  [( x2 – x1 ) 2  ( y2 – y1 ) 2 ]1/ 2 . In exercises 47−58, other ordered pairs are possible. 47. (a) x 0 y −2 4 0 49. (a) y-intercept: x0 y  12 0  2  2 x-intercept: y0 0  12 x  2  2  12 x  4  x 2 −1 additional point (b) Copyright © 2017 Pearson Education, Inc. 175 176 Chapter 2 Graphs and Functions 50. (a) x 0 y −3 2 0 4 3 (b) y-intercept: x0 3 0  2 y  6  2 y  6  y  3 x-intercept: y0 3 x  2 0   6  3x  6  x  2 53. (a) additional point x 3 y 0 4 1 (b) x-intercept: y0 0 x3  0 x33 x additional point 7 2 additional point no y-intercept: x  0  y  0  3  y  3 (b) 51. (a) x 0 y 0 1 1 additional point −2 4 additional point x- and y-intercept: 0  02 (b) 54. (a) 52. (a) x 0 y 2 −1 3 y-intercept: x0 y  02  2  y  02 y  2 additional point x 0 y −3 4 −1 9 0 (b) 2 6 additional point no x-intercept: y  0  0  x2  2  2  x 2   2  x Copyright © 2017 Pearson Education, Inc. y-intercept: x0 y  0 3 y  0  3  y  3 additional point x-intercept: y0 0 x 3 3 x 9 x Section 2.1 Rectangular Coordinates and Graphs 55. (a) x 0 y 2 2 0 x-intercept: y0 0 x2  0 x22 x −2 4 additional point 4 2 additional point y-intercept: x0 y  02  y  2  y  2 177 (b) 58. (a) (b) x 0 y 0 1 2 −1 −8 x- and y-intercept: 0  03 additional point additional point (b) 56. (a) x −2 y −2 −4 0 0 −4 additional point x-intercept: y0 0 x4  0 x4  0  x  4  4  x y-intercept: x0 y   04  y   4  y  4 (b) 57. (a) x 0 y 0 −1 2 −1 8 59. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0). 60. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3). 61. Because (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinateand will lie in quadrant I. (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. x- and y-intercept: 0  03 additional point additional point Copyright © 2017 Pearson Education, Inc. 178 Chapter 2 Graphs and Functions 62. Label the points A(2, 2), B (13,10), C (21, 5), and D(6, 13). To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points. 3  52  4  22 2   2  22  4  4  8 d ( B, C )  1  32  3  42 2 2   4   1 d (C , D )   16  1  17 2 d ( D, A)  1   1  1  3 2  22   2  4  4  8 2 Use the distance formula to find the length of each side. 2 d ( A, B)  13   2  10  2 2  152  82  225  64  289  17 21  132  5  102 2  82   15  64  225 d ( B, C )   289  17 d (C , D)   6  212   13  5 2 152  82  225  64  289  17 d ( D, A)   2  62   2  13 2 Because d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Because d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus. 64. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8). Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11). We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D). d ( A, B)  82  152  22  32  4  9  13  64  225  289  17 Because all sides have equal length, the four points form a rhombus. d ( B, C )  63. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points. d (C , D )  d ( A, B )  8  62  11  82  22  32  4  9  13 10  82  14  112  22  32  4  9  13 d ( A, D)  Use the distance formula to find the length of each side. 6  42  8  52 10  42  14  52  62  92  36  81  117  9(13)  3 13 d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) Because 3 13  13  13  13. 5  12  2  12  42  12  16  1  17 Copyright © 2017 Pearson Education, Inc. Section 2.2 Circles Section 2.2 Circles (b) 1. The circle with equation x 2  y 2  49 has center with coordinates (0, 0) and radius equal to 7. 2. The circle with center (3, 6) and radius 4 has equation  x  3   y  6  16. 2 3. The graph of  x  4   y  7   9 has center with coordinates (4, –7). 2 2 13. (a) Center (2, 0), radius 6  x  2 2   y  0 2  6  x  2 2   y  0 2  6 2 4. The graph of x   y  5  9 has center with coordinates (0, 5). 2 2 5. This circle has center (3, 2) and radius 5. This is graph B. ( x – 2) 2  y 2  36 (b) 6. This circle has center (3, –2) and radius 5. This is graph C. 7. This circle has center (–3, 2) and radius 5. This is graph D. 8. This circle has center (–3, –2) and radius 5. This is graph A. 9. The graph of x 2  y 2  0 has center (0, 0) and radius 0. This is the point (0, 0). Therefore, there is one point on the graph. 10. 14. (a) Center (3, 0), radius 3  x  32   y  02  3  x  32  y 2  9 100 is not a real number, so there are no points on the graph of x 2  y 2  100. (b) 11. (a) Center (0, 0), radius 6  x  02   y  02  6  x  02   y  02  62  x 2  y 2  36 (b) 15. (a) Center (0, 4), radius 4  x  0 2   y  4 2  4 2 x 2   y  4  16 (b) 12. (a) Center (0, 0), radius 9  x  02   y  02  9  x  02   y  02  92  x 2  y 2  81 Copyright © 2017 Pearson Education, Inc. 179 180 Chapter 2 Graphs and Functions 16. (a) Center (0, –3), radius 7 (b)  x  0   y  3  7 2  x  02   y  3  7 2 2 2 x 2  ( y  3)2  49 (b) 20. (a) Center (–3, –2), radius 6 2 2 2 2  x   3   y   2  6  x   3   y   2  62 ( x  3) 2  ( y  2) 2  36 17. (a) Center (–2, 5), radius 4  x   2   y  5  4 [ x – (–2)]2  ( y – 5) 2  42 ( x  2) 2  ( y – 5) 2  16 2 2 (b) (b)  2, 2  , radius 2 x  2    y  2   2 x  2    y  2   2 21. (a) Center 18. (a) Center (4, 3), radius 5  x  42   y  32  5  x  42   y  32  52  x  42   y  32  25 2 2 2 2 (b) (b) 3    x    3    y    3    3      x    3    y    3     3         x 3 y 3     3 22. (a) Center  3,  3 , radius 2 2 2 2 2 19. (a) Center (5, –4), radius 7  x  52   y  4  7 2 ( x – 5) 2  [ y – (– 4)]2  7 2 ( x – 5) 2  ( y  4)2  49 Copyright © 2017 Pearson Education, Inc. 2 2 Section 2.2 Circles (b) 23. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the midpoint formula, we have 1 5 1 1 C ,   3,1 . The radius is  2 2  one-half the length of the diameter: 1 r 5  12  1  12  2 2 The equation of the circle is  x  32   y  12  4 (b) Expand  x  3   y  1  4 to find the equation of the circle in general form: 2 2  x  32   y  12  4 x2  6 x  9  y 2  2 y  1  4 x2  y 2  6x  2 y  6  0 24. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−1, 1) and (−1, −5). Using the midpoint formula, we have  1  (1) 1  (5)  C ,    1, 2 .  2 2  The radius is one-half the length of the diameter: 2 1 2  1   1   5  1  3 r 2  The equation of the circle is  x  12   y  22  9 (b) Expand  x  1   y  2  9 to find the equation of the circle in general form: 2 2  x  1   y  2  9 x2  2 x  1  y 2  4 y  4  9 x2  y 2  2 x  4 y  4  0 2 2 181 25. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have  2  (2) 4  0  C ,    2, 2 .  2 2  The radius is one-half the length of the diameter: 2 1 2  2   2   4  0  2 r 2 The equation of the circle is  x  2 2   y  2 2  4 (b) Expand  x  2   y  2  4 to find the equation of the circle in general form: 2 2  x  22   y  22  4 x2  4 x  4  y 2  4 y  4  4 x2  y 2  4 x  4 y  4  0 26. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have  0  6 3  (3)  C ,   3, 3 .  2 2 The radius is one-half the length of the diameter: 2 1 r 6  02   3  3  3 2 The equation of the circle is  x  32   y  32  9 (b) Expand  x  3   y  3  9 to find the equation of the circle in general form: 2 2  x  32   y  32  9 x2  6 x  9  y 2  6 y  9  9 x2  y 2  6 x  6 y  9  0 27. x 2  y 2  6 x  8 y  9  0 Complete the square on x and y separately.  x  6 x   y  8 y   –9  x  6 x  9   y  8 y  16  –9  9  16 2 2 2 2  x  32   y  42  16 Yes, it is a circle. The circle has its center at (–3, –4) and radius 4. Copyright © 2017 Pearson Education, Inc. 182 Chapter 2 Graphs and Functions 28. x 2  y 2  8 x – 6 y  16  0 Complete the square on x and y separately.  x  8x   y – 6 y   16  x  8x  16   y – 6 y  9  –16  16  9 2 2 2 2  x  42   y – 32  9 Yes, it is a circle. The circle has its center at (–4, 3) and radius 3. 29. x 2  y 2  4 x  12 y  4 Complete the square on x and y separately.  x – 4 x   y  12 y   – 4  x – 4 x  4   y  12 y  36  – 4  4  36 2 2 2 2  x – 22   y  62  36 Yes, it is a circle. The circle has its center at (2, –6) and radius 6. 2 2 30. x  y – 12 x  10 y  –25 Complete the square on x and y separately.  x – 12 x   y  10 y   –25  x – 12 x  36   y  10 y  25  2 2 – 25  36  25  x – 62   y  52  36 Yes, it is a circle. The circle has its center at (6, –5) and radius 6. 31. 4 x 2  4 y 2  4 x – 16 y – 19  0 Complete the square on x and y separately.     4  x  x    4  y – 4 y  4   1 4 2 2 x  12  4 2   19  4 14  4 4  y – 2  36 2  x  12    y – 22  9   radius 3. 32. 9 x 2  9 y 2  12 x – 18 y – 23  0 Complete the square on x and y separately.     9  x  x    9  y – 2 y  1  9 x 2  43 x  9 y 2 – 2 y  23 4 3 radius 2. 33. x 2  y 2  2 x – 6 y  14  0 Complete the square on x and y separately.  x  2 x   y – 6 y   –14  x  2 x  1   y – 6 y  9  –14  1  9 2 2 2 2  x  12   y – 32  – 4 The graph is nonexistent. 34. x 2  y 2  4 x – 8 y  32  0 Complete the square on x and y separately.  x  4 x   y – 8 y   –32  x  4 x  4   y – 8 y  16  2 4 9 2  2 2 2 – 32  4  16  x  2   y – 42  –12 2 The graph is nonexistent. 35. x 2  y 2  6 x  6 y  18  0 Complete the square on x and y separately.  x  6 x   y  6 y   18  x  6 x  9   y  6 y  9  18  9  9 2 2 2 2  x  32   y  32  0 36. x 2  y 2  4 x  4 y  8  0 Complete the square on x and y separately.  x  4 x   y  4 y   8  x  4 x  4   y  4 y  4  8  4  4 2 2 2 Yes, it is a circle with center  12 , 2 and 2 2 The graph is the point (3, 3). 4 x 2  x  4 y 2 – 4 y  19 4   9  y – 12  36 2  x  23    y – 12  4 Yes, it is a circle with center  23 , 1 and 2 2 2  9 x  23 23  9 94  9 1 2  x  22   y  22  0 The graph is the point (−2, −2). 37. 9 x 2  9 y 2  6 x  6 y  23  0 Complete the square on x and y separately. 9 x  6 x  9 y  6 y   23 9  x  x   9  y  y   23 x  x     y  y      2 2 2 2 3 2 2 2 3 1 9 2 2 3 2 3 1 9 2 23 9 1 9 1 9 5 2  x  13    y  13   259   3  Yes, it is a circle with center  13 ,  13  and 2 radius 53 . Copyright © 2017 Pearson Education, Inc. 2 Section 2.2 Circles 183 38. 4 x 2  4 y 2  4 x  4 y  7  0 Complete the square on x and y separately.  4 x       4y – y    4 x2  x  4 y2  y  7 2  x  14 2 1 4 7  4 14 1 2    4  14  2 4  x  12   4  y – 2   9 2 2  x  12    y – 12   94 Yes, it is a circle with center   12 , 12  and radius 32 . 39. The equations of the three circles are ( x  7) 2  ( y  4) 2  25 , ( x  9) 2  ( y  4) 2  169 , and ( x  3) 2  ( y  9) 2  100 . From the graph of the three circles, it appears that the epicenter is located at (3, 1). Check algebraically: ( x  3) 2  ( y  1) 2  5 (5  3) 2  (2  1) 2  5 22  12  5  5  5 ( x  5) 2  ( y  4) 2  36 (5  5) 2  (2  4) 2  36 62  36  36  36 ( x  1) 2  ( y  4) 2  40 (5  1) 2  (2  4) 2  40 62  (2) 2  40  40  40 (5, 2) satisfies all three equations, so the epicenter is at (5, 2). 41. From the graph of the three circles, it appears that the epicenter is located at (−2, −2). Check algebraically: ( x  7) 2  ( y  4) 2  25 (3  7)2  (1  4) 2  25 42  32  25  25  25 ( x  9) 2  ( y  4) 2  169 (3  9) 2  (1  4) 2  169 122  52  169  169  169 ( x  3) 2  ( y  9) 2  100 (3  3) 2  (1  9) 2  100 62  (8) 2  100  100  100 (3, 1) satisfies all three equations, so the epicenter is at (3, 1). 40. The three equations are ( x  3) 2  ( y  1) 2  5 , ( x  5) 2  ( y  4) 2  36 , and ( x  1)2  ( y  4) 2  40 . From the graph of the three circles, it appears that the epicenter is located at (5, 2). Check algebraically: ( x  2) 2  ( y  1) 2  25 (2  2) 2  (2  1) 2  25 (4) 2  (3) 2  25 25  25 2 ( x  2)  ( y  2) 2  16 (2  2) 2  (2  2) 2  16 02  (4) 2  16 16  16 ( x  1) 2  ( y  2) 2  9 (2  1) 2  (2  2) 2  9 (3) 2  02  9 99 (−2, −2) satisfies all three equations, so the epicenter is at (−2, −2). Copyright © 2017 Pearson Education, Inc. 184 Chapter 2 Graphs and Functions 42. From the graph of the three circles, it appears that the epicenter is located at (5, 0). 1  x 2  3  x 2  16 1  2 x  x 2  9  6 x  x 2  16 2 x 2  8 x  10  16 2 x2  8x  6  0 x2  4 x  3  0 To solve this equation, we can use the quadratic formula with a = 1, b = –4, and c = –3. x Check algebraically: ( x  2) 2  ( y  4) 2  25 (5  2) 2  (0  4) 2  25 32  (4) 2  25 25  25 2 ( x  1)  ( y  3) 2  25 (5  1) 2  (0  3) 2  25 42  32  25 25  25 ( x  3) 2  ( y  6) 2  100 (5  3) 2  (0  6) 2  100 82  62  100 100  100 (5, 0) satisfies all three equations, so the epicenter is at (5, 0). 43. The radius of this circle is the distance from the center C(3, 2) to the x-axis. This distance is 2, so r = 2. ( x – 3) 2  ( y – 2) 2  22  ( x – 3) 2  ( y – 2) 2  4 44. The radius is the distance from the center C(–4, 3) to the point P(5, 8). r  [5 – (– 4)]2  (8 – 3) 2  92  52  106 The equation of the circle is [ x – (– 4)]2  ( y – 3) 2  ( 106) 2  ( x  4) 2  ( y – 3)3  106 45. Label the points P(x, y) and Q(1, 3). If d ( P, Q)  4 , 1  x 2  3  y 2  4  1  x 2  3  y 2  16. If x = y, then we can either substitute x for y or y for x. Substituting x for y we solve the following:   4  42  4 13 2 1 4  16  12 4  28  2 2 42 7   2 7 2 Because x = y, the points are  2  7 , 2  7  and 2 – 7 , 2  7 . 46. Let P(–2, 3) be a point which is 8 units from Q(x, y). We have d ( P, Q )  2  x 2  3  y 2  8  2  x 2  3  y 2  64. Because x + y = 0, x = –y. We can either substitute  x for y or  y for x. Substituting  x for y we solve the following: 2 2  x 2  3   x   64 2  x 2  3  x 2  64 4  4 x  x 2  9  6 x  x 2  64 2 x 2  10 x  13  64 2 x 2  10 x  51  0 To solve this equation, use the quadratic formula with a = 2, b = 10, and c = –51. x 10  102  4  2 51 2  2 10  100  408 4 10  508 10  4 127    4 4 10  2 127 5  127   4 2 Because y   x the points are   5  127 5  127  ,   and 2 2    –5  127 5  127  ,  . 2 2   Copyright © 2017 Pearson Education, Inc. Section 2.2 Circles 47. Let P(x, y) be a point whose distance from A(1, 0) is 10 and whose distance from 49. Label the points A(3, y) and B(–2, 9). If d(A, B) = 12, then 2  32  9  y 2  12 52  9  y 2  12 52  9  y 2  122 B(5, 4) is 10 . d(P, A) = 10 , so (1  x) 2  (0  y ) 2  10  (1  x) 2  y 2  10. d ( P, B )  10, so (5  x) 2  (4  y ) 2  10  (5  x) 2  (4  y ) 2  10. Thus, (1  x) 2  y 2  (5  x) 2  (4  y ) 2 1  2x  x2  y2  25  10 x  x 2  16  8 y  y 2 1  2 x  41  10 x  8 y 8 y  40  8 x y  5 x Substitute 5 – x for y in the equation (1  x) 2  y 2  10 and solve for x. (1  x) 2  (5  x) 2  10  1  2 x  x 2  25  10 x  x 2  10 2 x 2  12 x  26  10  2 x 2  12 x  16  0 x 2  6 x  8  0  ( x  2)( x  4)  0  x  2  0 or x  4  0 x  2 or x4 To find the corresponding values of y use the equation y = 5 – x. If x = 2, then y = 5 – 2 = 3. If x = 4, then y = 5 – 4 = 1. The points satisfying the conditions are (2, 3) and (4, 1). 48. The circle of smallest radius that contains the points A(1, 4) and B(–3, 2) within or on its boundary will be the circle having points A and B as endpoints of a diameter. The center will be M, the midpoint:  1   3 4  2   2 6   2 , 2    2 , 2   (−1, 3).   The radius will be the distance from M to either A or B: d ( M , A)  [1  (1)]2  (4  3) 2 185 25  81  18 y  y 2  144 y 2  18 y  38  0 Solve this equation by using the quadratic formula with a = 1, b = –18, and c = –38: y    18  182  4 138 2 1 18  324  152 18  476  2 1 2 18  4 119 18  2 119   9  119 2 2 The values of y are 9  119 and 9  119 .  50. Because the center is in the third quadrant, the radius is 2 , and the circle is tangent to both axes, the center must be at ( 2,  2 ). Using the center-radius of the equation of a circle, we have      x  2    y  2   2. 2 2 x   2    y   2       2  2  2 2 51. Let P(x, y) be the point on the circle whose distance from the origin is the shortest. Complete the square on x and y separately to write the equation in center-radius form: x 2  16 x  y 2  14 y  88  0 x 2  16 x  64  y 2  14 y  49   88  64  49 ( x  8) 2  ( y  7) 2  25 So, the center is (8, 7) and the radius is 5.  22  12  4  1  5 The equation of the circle is 2  x   1   y  3  2  x  12   y  32  5.  5  2 d (C , O )  82  7 2  113 . Because the length of the radius is 5, d ( P, O )  113  5 . Copyright © 2017 Pearson Education, Inc. 186 Chapter 2 Graphs and Functions 52. Using compasses, draw circles centered at Wickenburg, Kingman, Phoenix, and Las Vegas with scaled radii of 50, 75, 105, and 180 miles respectively. The four circles should intersect at the location of Nothing. 58. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates  3  (–7) –5  3   4 –2  ,     ,   (–2, – 1). 2 2   2 2  The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5) 2 d (C , P)  3 –  –2   –5 –  –1 2  52   –4  25  16  41 2 We could also use points C(–2, –1).and Q(–7, 3). 2 d (C , Q )   7 –  –2  3 –  –1 53. The midpoint M has coordinates  –1+5 3   –9   4 6   ,   (2, – 3).  2 , 2   2 2     –1 – 22  3 –  –3   –32  62  9  36 2 d ( P, Q )    45  3 5 2  32   –6  9  36 2  45  3 5 The radius is 3 5. 56. Use the points P(–1, 3) and Q(5, –9). 2 Because d ( P, Q )  5 –  –1   –9 – 3  62   –12  36  144  180 2  6 5 ,the radius is r 1 d ( P, Q ). Thus 2   1 6 5  3 5. 2 57. The center-radius form for this circle is ( x – 2) 2  ( y  3) 2  (3 5) 2  ( x – 2) 2  ( y  3) 2  45. 2 102  82  100  64  55. Use points C(2, –3) and Q(5, –9). 5 – 22   –9 –  –3 7 – 32  3 –  –5  164  2 41 1 1 d ( P, Q )  2 41  41 2 2 The center-radius form of the equation of the circle is [ x – (–2)]2  [ y – (–1)]2  ( 41) 2 ( x  2) 2  ( y  1) 2  41 The radius is 3 5. d (C , Q )  25  16  41 We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the diameter. 54. Use points C(2, –3) and P(–1, 3). d (C , P )  52  42  2 2  59. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates  1  11 2  7   9  ,     5,  . 2 2   2   can use points C 5, 92  and P(−1, 2). 2 2 d (C , P)  5 –  –1   92  2 2  62   52   169  132 4 We could also use points C 5, 92  and The center is C 5, 92 . To find the radius, we Q(11, 7). d (C , Q )  5  112   92 – 7   62   52   2 2 169  132 4 (continued on next page) Copyright © 2017 Pearson Education, Inc. Section 2.3 Functions (continued) 187 The length of the diameter PQ is Using the points P and Q to find the length of the diameter, we have d  P, Q    1  112  2  7 2 12  5 2 The center-radius form of the equation of the circle is 2 2 2 5  12  4  12   The center is C 1, 52 . The length of the diameter PQ is 2 3  52  10  5  82  152  289  17. The length of the radius is 12 17   172 . The center-radius form of the equation of the circle is 2 d (C , Q)  1   3  1   –2 2 62. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates  3  5 10  (5)  5 ,    1, 2 . 2 2  x  12   y  52    172  2  x  12   y  52   289 4  42  32  25  5 We could also use points C(1, 1) and Q(−3, −2). 2 2   60. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the segment joining P and Q has coordinates  5  (3) 4  (2)  ,    1, 1 . 2 2 The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4). d (C , P )   x  32   y  52    52  2  x  32   y  52   254 2  169  13 1 1 13 d  P, Q   13  2 2 2 The center-radius form of the equation of the circle is  x  52   y  92    132  2  x  52   y  92   1694 1  52  4  12  42  32  25  5. The length of the radius is 12 5  52 . 2 Section 2.3 2 2 Functions  4  3  25  5 Using the points P and Q to find the length of the diameter, we have 1. The domain of the relation 3, 5 , 4, 9 , 10,13 is 3, 4,10. d  P, Q   5   3   4   2 2. The range of the relation in Exercise 1 is 5, 9,13. 2 2  82  62  100  10 1 1 d ( P, Q )  10  5 2 2 The center-radius form of the equation of the circle is  x  1   y  1  5  x  12   y  12  25 2 2 2  3. The equation y = 4x – 6 defines a function with independent variable x and dependent variable y. 4. The function in Exercise 3 includes the ordered pair (6, 18). 5. For the function f  x   4 x  2, 61. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates 1 5 4 1 5 ,    3, 2 . 2 2    The center is C 3, 52  .  f 2  4  2  2  8  2  10. 6. For the function g  x   x , g 9  9  3. 7. The function in Exercise 6, g  x   x , has domain  0,  . Copyright © 2017 Pearson Education, Inc. 188 Chapter 2 Graphs and Functions 8. The function in Exercise 6, g  x   x , has range  0,  . For exercises 9 and 10, use this graph. 17. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 18. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 9. The largest open interval over which the function graphed here increases is , 3. 10. The largest open interval over which the function graphed here decreases is 3,  . 11. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 12. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 13. Two ordered pairs, namely (2, 4) and (2, 6), have the same x-value paired with different y-values, so the relation is not a function. 14. Two ordered pairs, namely (9, −2) and (9, 1), have the same x-value paired with different y-values, so the relation is not a function. 15. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 16. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 19. Two sets of ordered pairs, namely (1, 1) and (1, −1) as well as (2, 4) and (2, −4), have the same x-value paired with different y-values, so the relation is not a function. domain: {0, 1, 2}; range: {−4, −1, 0, 1, 4} 20. The relation is not a function because the x-value 3 corresponds to two y-values, 7 and 9. This correspondence can be shown as follows. domain: {2, 3, 5}; range: {5, 7, 9, 11} 21. The relation is a function because for each different x-value there is exactly one y-value. domain: {2, 3, 5, 11, 17}; range: {1, 7, 20} 22. The relation is a function because for each different x-value there is exactly one y-value. domain: {1, 2, 3, 5}; range: {10, 15, 19, 27} 23. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. Domain: {0, −1, −2}; range: {0, 1, 2} Copyright © 2017 Pearson Education, Inc. Section 2.3 Functions 24. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows. 189 33. y  x 2 represents a function because y is always found by squaring x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the square of any real number is not negative, the range would be zero or greater. Domain: {0, 1, 2}; range: {0, −1, −2} 25. The relation is a function because for each different year, there is exactly one number for visitors. domain: {2010, 2011, 2012, 2013} range: {64.9, 63.0, 65.1, 63.5} 26. The relation is a function because for each basketball season, there is only one number for attendance. domain: {2011, 2012, 2013, 2014} range: {11,159,999, 11,210,832, 11,339,285, 11,181,735} 27. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: ,   ; range:  ,   domain:  ,   ; range:  0,   34. y  x3 represents a function because y is always found by cubing x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the cube of any real number could be negative, positive, or zero, the range would be any real number. 28. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: ,   ; range:  , 4 29. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: 3,   ; range:  ,   30. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: [−4, 4]; range: [−3, 3] 31. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: ,   ; range:  ,   domain:  ,   ; range:  ,   35. The ordered pairs (1, 1) and (1, −1) both satisfy x  y 6 . This equation does not represent a function. Because x is equal to the sixth power of y, the values of x are nonnegative. Any real number can be raised to the sixth power, so the range of the relation is all real numbers. 32. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: [−2, 2]; range: [0, 4] domain:  0,   range:  ,   Copyright © 2017 Pearson Education, Inc. 190 Chapter 2 Graphs and Functions 36. The ordered pairs (1, 1) and (1, −1) both satisfy x  y 4 . This equation does not represent a function. Because x is equal to the fourth power of y, the values of x are nonnegative. Any real number can be raised to the fourth power, so the range of the relation is all real numbers. 39. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x + y < 3 corresponds to many values of y. The ordered pairs (1, –2), (1, 1), (1, 0), (1, −1), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers. domain:  0,   range:  ,   37. y  2 x  5 represents a function because y is found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers. domain:  ,   ; range:  ,   38. y  6 x  4 represents a function because y is found by multiplying x by −6 and adding 4. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is −6 times x, plus 4, y also may be any real number, and so the range is also all real numbers. domain:  ,   ; range:  ,   40. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x − y 2. We can write the function as 1 if x  –1 f ( x)  –1 if x  2.  domain: (– , – 1]  (2, ); range: {–1, 1} 39. For x  0 , that piece of the graph goes through the points (−1, −1) and (0, 0). The slope is 1, so the equation of this piece is y = x. For x > 0, that piece of the graph is a horizontal line passing through (2, 2), so its equation is y = 2. We can write the function as x if x  0 . f ( x)  2 if x  0  221 41. For x < 1, that piece of the graph is a curve passes through (−8, −2), (−1, −1) and (1, 1), so the equation of this piece is y  3 x . The right piece of the graph passes through (1, 2) and 23 (2, 3). m   1 , and the equation of the 1 2 line is y  2  x  1  y  x  1 . We can write 3 the function as f ( x)   x if x  1  x  1 if x  1 domain: (– , ) range: (,1)  [2, ) 42. For all values except x = 2, the graph is a line. It passes through (0, −3) and (1, −1). The slope is 2, so the equation is y = 2x −3. At x = 2, the graph is the point (2, 3). We can write 3 if x  2 the function as f ( x)  . 2 x  3 if x  2  domain: (– , ) range: (,1)  (1, ) 43. f(x) =   x  Plot points. x –x f(x) =   x  –2 –1.5 2 1.5 2 1 –1 –0.5 0 0.5 1 0.5 0 –0.5 1 0 0 –1 1 –1 –1 1.5 –1.5 –2 2 –2 –2 More generally, to get y = 0, we need 0  – x  1  0  x  1  1  x  0. To get y = 1, we need 1  – x  2  1  x  2  –2  x  1. Follow this pattern to graph the step function. domain: (– , ) range: (, 0]  {2} 40. For x < 0, that piece of the graph is a horizontal line passing though (−3, −3), so the equation of this piece is y = −3. For x  0 , the curve passes through (1, 1) and (4, 2), so the equation of this piece is y  x . We can  3 if x  0 write the function as f ( x)   .  x if x  0 domain: (– , ) range: {3}  [0, ) domain:  ,   ; range: {…,–2,–1,0,1,2,…} Copyright © 2017 Pearson Education, Inc. 222 Chapter 2 Graphs and Functions 44. f(x) =   x  Plot points. x  x f(x) =   x  –2 −2 2 –1.5 –1 –0.5 0 −2 −1 −1 0 2 1 1 0 0.5 1 1.5 2 0 1 1 2 0 −1 –1 –2 domain:  ,   ; range: {…, 2,–1,0,1,2,…} Follow this pattern to graph the step function. 47. The cost of mailing a letter that weighs more than 1 ounce and less than 2 ounces is the same as the cost of a 2-ounce letter, and the cost of mailing a letter that weighs more than 2 ounces and less than 3 ounces is the same as the cost of a 3-ounce letter, etc. domain:  ,   ; range: {…,–2,–1,0,1,2,…} 45. f(x) =  2x  To get y = 0, we need 0  2 x  1  0  x  12 . To get y = 1, we need 1  2 x  2  12  x  1. To get y = 2, we need 2  2 x  3  1  x  32 . 48. The cost is the same for all cars parking between 12 hour and 1-hour, between 1 hour and 1 12 hours, etc. Follow this pattern to graph the step function. domain:  ,   ; range: {…,–2,–1,0,1,2,…} 49. 46. g(x) =  2 x  1 To get y = 0, we need 0  2 x  1  1  1  2 x  2  12  x  1. To get y = 1, we need 1  2 x – 1  2  2  2 x  3  1  x  32 . Follow this pattern to graph the step function. Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 50. 56. (a) 49.8  34.2  1.95 , 80 so y  1.95 x  34.2. For 8  x  13 , 51. (a) For 0  x  8, m  223 if 0  x  4 6.5 x  f ( x)  5.5 x  48 if 4  x  6  –30 x  195 if 6  x  6.5 Draw a graph of y = 6.5x between 0 and 4, including the endpoints. Draw the graph of y = –5.5x + 48 between 4 and 6, including the endpoint at 6 but not the one at 4. Draw the graph of y = –30x + 195, including the endpoint at 6.5 but not the one at 6. Notice that the endpoints of the three pieces coincide. 52.2  49.8  0.48 , so the equation 13  8 is y  52.2  0.48( x  13)  y  0.48 x  45.96 m (b) f ( x)   1.95 x  34.2 if 0  x  8 0.48 x  45.96 if 8  x  13 52. When 0  x  3 , the slope is 5, which means that the inlet pipe is open, and the outlet pipe is closed. When 3  x  5 , the slope is 2, which means that both pipes are open. When 5  x  8 , the slope is 0, which means that both pipes are closed. When 8  x  10 , the slope is −3, which means that the inlet pipe is closed, and the outlet pipe is open. 53. (a) The initial amount is 50,000 gallons. The final amount is 30,000 gallons. (b) The amount of water in the pool remained constant during the first and fourth days. (c) f (2)  45, 000; f (4)  40, 000 (d) The slope of the segment between (1, 50000) and (3, 40000) is −5000, so the water was being drained at 5000 gallons per day. 54. (a) There were 20 gallons of gas in the tank at x = 3. (b) From the graph, observe that the snow depth, y, reaches its deepest level (26 in.) when x = 4, x = 4 represents 4 months after the beginning of October, which is the beginning of February. (c) From the graph, the snow depth y is nonzero when x is between 0 and 6.5. Snow begins at the beginning of October and ends 6.5 months later, in the middle of April. Section 2.7 Graphing Techniques 1. To graph the function f  x   x 2  3, shift the graph of y  x 2 down 3 units. 2. To graph the function f  x   x 2  5, shift the graph of y  x 2 up 5 units. 3. The graph of f  x    x  4 is obtained by 2 shifting the graph of y  x 2 to the left 4 units. (b) The slope is steepest between t = 1 and t ≈ 2.9, so that is when the car burned gasoline at the fastest rate. 4. The graph of f  x    x  7  is obtained by 55. (a) There is no charge for additional length, so we use the greatest integer function. The cost is based on multiples of two feet, so f ( x)  0.8  2x  if 6  x  18 . 5. The graph of f  x    x is a reflection of (b)   0.8(4)  $3.20 f (8.5)  0.8  8.5  2    0.8(7)  $5.60 f (15.2)  0.8  15.2  2  2 shifting the graph of y  x 2 to the right 7 units. the graph of f  x   x across the x-axis. 6. The graph of f  x    x is a reflection of the graph of f  x   x across the y-axis. Copyright © 2017 Pearson Education, Inc. 224 Chapter 2 Graphs and Functions 7. To obtain the graph of f  x    x  2  3, 3 shift the graph of y  x 3 2 units to the left and 3 units down. 8. To obtain the graph of f  x    x  3  6, 3 shift the graph of y  x 3 3 units to the right and 6 units up. 9. The graph of f  x    x is the same as the graph of y  x because reflecting it across the y-axis yields the same ordered pairs. 10. The graph of x  y 2 is the same as the graph of x    y  because reflecting it across the x-axis yields the same ordered pairs. 2 2 2 11. (a) B; y  ( x  7) is a shift of y  x , 7 units to the right. (b) D; y  x 2  7 is a shift of y  x 2 , 7 units downward. y  x 2 , by a factor of 7. 2 (d) A; y  ( x  7) is a shift of y  x , 7 units to the left. (e) C; y  x 2  7 is a shift of y  x 2 , 7 units upward. (b) C; y   x is a reflection of y  x , over the x-axis. 3 (c) D; y  3  x is a reflection of y  3 x , over the y-axis. (d) A; y  x  4 is a shift of y  x , 4 units to the right. 3 (e) B; y  3 x  4 is a shift of y  3 x , 4 units down. 2 2 13. (a) B; y  x  2 is a shift of y  x , 2 units upward. (b) A; y  x 2  2 is a shift of y  x 2 , 2 units downward. (f) D; y   x 2 is a reflection of y  x 2 , across the x-axis. (g) H; y  ( x  2) 2  1 is a shift of y  x 2 , 2 units to the right and 1 unit upward. (h) E; y  ( x  2) 2  1 is a shift of y  x 2 , 2 units to the left and 1 unit upward. (i) I; y  ( x  2) 2  1 is a shift of y  x 2 , 2 units to the left and 1 unit down. 14. (a) G; y  x  3 is a shift of y  x , 3 units to the left. (c) E; y  x  3 is a shift of y  x , 3 units upward. (d) B; y  3 x is a vertical stretch of (e) C; y   x is a reflection of y  x across the x-axis. of y  3 x , by a factor of 4. 3 (e) F; y  2 x 2 is a vertical stretch of y  x 2 , by a factor of 2. y  x , by a factor of 3. 12. (a) E; y  4 3 x is a vertical stretch 3 (d) C; y  ( x  2) 2 is a shift of y  x 2 , 2 units to the right. (b) D; y  x  3 is a shift of y  x , 3 units downward. (c) E; y  7 x 2 is a vertical stretch of 2 (c) G; y  ( x  2) 2 is a shift of y  x 2 , 2 units to the left. (f) A; y  x  3 is a shift of y  x , 3 units to the right. (g) H; y  x  3  2 is a shift of y  x , 3 units to the right and 2 units upward. (h) F; y  x  3  2 is a shift of y  x , 3 units to the left and 2 units upward. (i) I; y  x  3  2 is a shift of y  x , 3 units to the right and 2 units downward. 15. (a) F; y  x  2 is a shift of y  x 2 units to the right. (b) C; y  x  2 is a shift of y  x 2 units downward. (c) H; y  x  2 is a shift of y  x 2 units upward. Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 225 (d) D; y  2 x is a vertical stretch of y  x by a factor of 2. (e) G; y   x is a reflection of y  x across the x-axis. (f) A; y   x is a reflection of y  x across the y-axis. (g) E; y  2 x is a reflection of y  2 x 19. f  x   23 x across the x-axis. y  2 x is a vertical x h  x  x f  x   23 x stretch of y  x by a factor of 2. −3 3 2 (h) I; y  x  2  2 is a shift of y  x 2 units to the right and 2 units upward. −2 2 4 3 −1 1 2 3 0 0 0 16. The graph of f  x   2  x  1  6 is the graph 1 1 2 3 of f  x   x 3 stretched vertically by a factor of 2, shifted left 1 unit and down 6 units. 2 2 4 3 3 3 2 x h  x  x f  x   34 x −4 4 3 −3 3 9 4 −2 2 3 2 −1 1 3 4 0 0 0 1 1 3 4 2 2 3 2 3 3 9 4 4 4 (i) B; y  x  2  2 is a shift of y  x 2 units to the left and 2 units downward. 3 17. f  x  3 x x h  x  x f  x  3 x −2 2 6 −1 1 3 0 0 0 1 1 3 2 2 6 20. 18. f  x  4 x x h  x  x f  x  4 x −2 2 8 −1 1 4 0 0 0 1 1 4 2 2 8 f  x   34 x Copyright © 2017 Pearson Education, Inc. 3 (continued on next page) 226 Chapter 2 Graphs and Functions (continued) 21. 23. x h  x  x2 −2 4 2 −1 1 1 2 0 0 0 1 1 1 2 2 4 2 f  x  2 x2 x h  x  x2 f  x  2 x2 −2 4 8 −1 1 2 0 0 0 1 1 2 2 4 8 24. 22. f  x   12 x 2 f  x   3x 2 f  x   12 x 2 f  x   13 x 2 f  x   13 x 2 x h  x  x2 −3 9 3 −2 4 4 3 −1 1 1 3 0 0 0 x h  x  x2 f  x   3x 2 1 1 1 3 −2 4 12 −1 1 3 2 4 4 3 0 0 0 3 9 3 1 1 3 2 4 12 Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 25. f  x    12 x 2 27. x h  x  x 9  92 −2 2 −6 −2 4 2 −1 1 −3 0 0 1  12 0 −1 1 1 −3 0 0 0 2 2 −6 1 1  12 2 4 2 3 9  92 h  x  x2 −3 28. 26. f  x   3 x f  x    12 x 2 x f  x    13 x 2 f  x x h  x  x −3 9 −3 −2 4  43 −1 1  13 0 0 0 1 1  13 2 4  43 3 9 −3 2 227 f  x   3 x f  x   2 x x h  x  x −2 2 −4 −1 1 −2 0 0 0 1 1 −2 2 2 −4   13 x 2 f  x   2 x 29. h  x    12 x h  x    12 x x f  x  x −4 4 2 −3 3 3 2 −2 2 1 −1 1 1 2 0 0 0   12 x  12 x (continued on next page) Copyright © 2017 Pearson Education, Inc. 228 Chapter 2 Graphs and Functions (continued) f  x  x x f  x  x x h  x    12 x   12 x  12 x 1 1 1 2 2 2 1 3 3 3 2 4 4 2 h  x  4 x  2 x 3 3 2 3 4 2 4 32. h  x   9 x f  x  x x 30. h  x    13 x h  x    13 x h  x  9x  3 x 0 0 0 1 1 3 2 2 3 2 3 3 3 3 4 2 6 x f  x −3 3 1 −2 2 2 3 −1 1 1 3 0 0 0 1 1 1 3 2 2 2 3 x 3 3 1 −4 2 −2 −3 3  3 −2 2  2 −1 1 −1 0 0 0   13 x   13 x  13 x 33. f  x    x h  x   x 31. h  x   4 x f  x  x x h  x  4x  2 x 0 0 0 1 1 2 2 2 2 2 Copyright © 2017 Pearson Education, Inc. f  x    x Section 2.7 Graphing Techniques 34. f  x    x x h  x   x f  x    x −3 3 −3 −2 2 −2 −1 1 −1 0 0 0 1 1 −1 2 2 −2 3 3 −3 35. (a) y  f  x  4 is a horizontal translation of f, 4 units to the left. The point that corresponds to (8, 12) on this translated function would be 8  4,12   4,12 . (b) y  f  x   4 is a vertical translation of f, 4 units up. The point that corresponds to (8, 12) on this translated function would be 8,12  4  8,16 . 36. (a) 229 38. (a) The point that corresponds to (8, 12) when reflected across the x-axis would be (8, −12). (b) The point that corresponds to (8, 12) when reflected across the y-axis would be (−8, 12). 39. (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3). (b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3). (c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3). 40. (a) The point that is symmetric to (–6, 1) with respect to the x-axis is (–6, –1). (b) The point that is symmetric to (–6, 1) with respect to the y-axis is (6, 1). (c) The point that is symmetric to (–6, 1) with respect to the origin is (6, –1). y  14 f  x  is a vertical shrinking of f, by a factor of 14 . The point that corresponds to (8, 12) on this translated function   would be 8, 14  12  8, 3 . (b) y  4 f  x  is a vertical stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function would be 8, 4  12  8, 48 . 41. (a) The point that is symmetric to (–4, –2) with respect to the x-axis is (–4, 2). y  f (4 x ) is a horizontal shrinking of f, by a factor of 4. The point that corresponds to (8, 12) on this translated (c) The point that is symmetric to (–4, –2) with respect to the origin is (4, 2). 37. (a)  (b) The point that is symmetric to (–4, –2) with respect to the y-axis is (4, –2).  function is 8  14 , 12   2, 12 . (b) y  f  14 x is a horizontal stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function is 8  4, 12  32, 12 . Copyright © 2017 Pearson Education, Inc. 230 Chapter 2 Graphs and Functions 42. (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0) because this point lies on the x-axis. (b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0). (c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0). 47. x 2  y 2  12 Replace x with –x to obtain ( x) 2  y 2  12  x 2  y 2  12 . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain x 2  ( y )2  12  x 2  y 2  12 The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. 48. y 2  x 2  6 Replace x with –x to obtain y 2   x   6  y 2  x 2  6 The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain ( y )2  x 2  6  y 2  x 2  6 The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin. 2 43. The graph of y = |x − 2| is symmetric with respect to the line x = 2. 44. The graph of y = −|x + 1| is symmetric with respect to the line x = −1. 45. y  x 2  5 Replace x with –x to obtain y  ( x) 2  5  x 2  5 . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  ( x) 2  2   y  x 2  2  y   x 2  2. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only. 46. y  2 x 4  3 Replace x with –x to obtain y  2( x) 4  3  2 x 4  3 The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain – y  2( x) 4  3   y  2 x 4  3  y  2 x 4  3 . The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only. 49. y  4 x3  x Replace x with –x to obtain y  4( x)3  ( x)  y  4( x3 )  x  y  4 x 3  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain  y  4 x 3  x  y  4 x3  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  4( x)3  ( x)   y  4( x3 )  x   y  4 x3  x  y  4 x3  x. The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only. Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 50. y  x3  x Replace x with –x to obtain y  ( x)3  ( x)  y   x3  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain  y  x3  x  y   x3  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  ( x)3  ( x)   y   x3  x  54. y  x3  x. The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only. 56. 51. y  x 2  x  8 Replace x with –x to obtain y  ( x) 2  ( x)  8  y  x 2  x  8. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  ( x) 2  ( x)  8  2 52. y = x + 15 Replace x with –x to obtain y  ( x)  15  y   x  15. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  ( x)  15  y  x  15. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries. 53. f  x    x3  2 x f  x     x   2  x  f  x   x5  2 x3 f  x    x   2  x  5   The function is odd. 55. f  x   0.5 x 4  2 x 2  6 f   x   0.5   x   2   x   6 4 2  0.5 x 4  2 x 2  6  f  x  The function is even. f  x   0.75 x 2  x  4 f  x   0.75   x    x  4 2  0.75 x 2  x  4  f  x  The function is even. 57. f  x   x3  x  9 f  x    x    x   9 3     x3  x  9   x3  x  9   f  x  The function is neither. 58. f  x  x4  5x  8 f  x    x   5  x   8 4  x4  5x  8  f  x The function is neither. 59. f  x  x2  1 This graph may be obtained by translating the graph of y  x 2 1 unit downward. 60. f  x  x2  2 This graph may be obtained by translating the graph of y  x 2 2 units downward. 3  3   x5  2 x3   x5  2 x3   f  x  2  y  x  x  8  y   x  x  8. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries. 231   x3  2 x    x3  2 x   f  x  The function is odd. Copyright © 2017 Pearson Education, Inc. 232 Chapter 2 Graphs and Functions 61. f  x  x2  2 This graph may be obtained by translating the graph of y  x 2 2 units upward. 65. g  x    x  2 This graph may be obtained by translating the graph of y  x 2 2 units to the left. 62. f  x  x2  3 This graph may be obtained by translating the graph of y  x 2 3 units upward. 66. g  x    x  3 This graph may be obtained by translating the graph of y  x 2 3 units to the left. 63. g  x    x  4 This graph may be obtained by translating the graph of y  x 2 4 units to the right. 67. g  x   x  1 The graph is obtained by translating the graph of y  x 1 unit downward. 64. g  x    x  2 This graph may be obtained by translating the graph of y  x 2 2 units to the right. 68. g  x   x  3  2 This graph may be obtained by translating the graph of y  x 3 units to the left and 2 units upward. 2 2 2 2 Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 69. h  x   ( x  1)3 This graph may be obtained by translating the graph of y  x3 1 unit to the left. It is then reflected across the x-axis. 70. h  x   ( x  1)3 This graph can be obtained by translating the graph of y  x3 1 unit to the right. It is then reflected across the x-axis. (We may also reflect the graph about the x-axis first and then translate it 1 unit to the right.) 233 72. h  x   3 x 2  2 This graph may be obtained by stretching the graph of y  x 2 vertically by a factor of 3, then shifting the resulting graph down 2 units. 73. f  x   2( x  2) 2  4 This graph may be obtained by translating the graph of y  x 2 2 units to the right and 4 units down. It is then stretched vertically by a factor of 2. 74. f  x   3( x  2) 2  1 This graph may be obtained by translating the graph of y  x 2 2 units to the right and 1 unit up. It is then stretched vertically by a factor of 3 and reflected over the x-axis. 75. f  x  x  2 This graph may be obtained by translating the graph of y  x two units to the left. 71. h  x   2 x 2  1 This graph may be obtained by translating the graph of y  x 2 1 unit down. It is then stretched vertically by a factor of 2. Copyright © 2017 Pearson Education, Inc. 234 Chapter 2 Graphs and Functions 76. f  x  x  3 This graph may be obtained by translating the graph of y  x three units to the right. 77. f  x   x This graph may be obtained by reflecting the graph of y  x across the x-axis. 78. f  x  x  2 This graph may be obtained by translating the graph of y  x two units down. 80. f  x  3 x  2 This graph may be obtained by stretching the graph of y  x vertically by a factor of three and then translating the resulting graph two units down. 81. g  x   12 x3  4 This graph may be obtained by stretching the graph of y  x3 vertically by a factor of 12 , then shifting the resulting graph down four units. 82. g  x   12 x3  2 This graph may be obtained by stretching the graph of y  x3 vertically by a factor of 12 , then shifting the resulting graph up two units. 79. f  x  2 x  1 This graph may be obtained by stretching the graph of y  x vertically by a factor of two and then translating the resulting graph one unit up. Copyright © 2017 Pearson Education, Inc. Section 2.7 Graphing Techniques 83. g  x    x  3 This graph may be obtained by shifting the graph of y  x3 three units left. 3 84. f  x    x  2 This graph may be obtained by shifting the graph of y  x3 two units right. 3 235 87. (a) y = g(–x) The graph of g(x) is reflected across the y-axis. (b) y = g(x – 2) The graph of g(x) is translated to the right 2 units. (c) y = –g(x) The graph of g(x) is reflected across the x-axis. 85. f  x   23  x  2 2 This graph may be obtained by translating the graph of y  x 2 two units to the right, then stretching the resulting graph vertically by a factor of 23 . (d) y = –g(x) + 2 The graph of g(x) is reflected across the x-axis and translated 2 units up. 86. Because g ( x)   x  x  f ( x), the graphs are the same. Copyright © 2017 Pearson Education, Inc. 236 Chapter 2 Graphs and Functions 88. (a) y   f  x  The graph of f(x) is reflected across the x-axis. 90. It is the graph of g  x   x translated 4 units to the left, reflected across the x-axis, and translated two units up. The equation is y   x  4  2. 91. It is the graph of f  x   x translated one unit right and then three units down. The equation is y  x  1  3. (b) y  2 f  x  The graph of f(x) is stretched vertically by a factor of 2. 92. It is the graph of f  x   x translated 2 units to the right, shrunken vertically by a factor of 1 , and translated one unit down. The 2 equation is y  12 x  2  1. 93. It is the graph of g  x   x translated 4 units to the left, stretched vertically by a factor of 2, and translated four units down. The equation is y  2 x  4  4. (c) y  f  x  The graph of f(x) is reflected across the y-axis. 94. It is the graph of f  x   x reflected across the x-axis and then shifted two units down. The equation is y   x  2 . 95. Because f(3) = 6, the point (3, 6) is on the graph. Because the graph is symmetric with respect to the origin, the point (–3, –6) is on the graph. Therefore, f(–3) = –6. 96. Because f(3) = 6, (3, 6) is a point on the graph. The graph is symmetric with respect to the y-axis, so (–3, 6) is on the graph. Therefore, f(–3) = 6. (d)  x The graph of f(x) is compressed vertically by a factor of 12 . y  12 f 97. Because f(3) = 6, the point (3, 6) is on the graph. The graph is symmetric with respect to the line x = 6 and the point (3, 6) is 3 units to the left of the line x = 6, so the image point of (3, 6), 3 units to the right of the line x = 6 is (9, 6). Therefore, f(9) = 6. 98. Because f(3) = 6 and f(–x) = f(x), f(–3) = f(3). Therefore, f(–3) = 6. 99. Because (3, 6) is on the graph, (–3, –6) must also be on the graph. Therefore, f(–3) = –6. 89. It is the graph of f  x   x translated 1 unit to the left, reflected across the x-axis, and translated 3 units up. The equation is y   x  1  3. 100. If f is an odd function, f(–x) = –f(x). Because f(3) = 6 and f(–x) = –f(x), f(–3) = –f(3). Therefore, f(–3) = –6. Copyright © 2017 Pearson Education, Inc. Chapter 2 Quiz (Sections 2.5−2.7) 101. f(x) = 2x + 5 Translate the graph of f ( x ) up 2 units to obtain the graph of t ( x )  (2 x  5)  2  2 x  7. Now translate the graph of t(x) = 2x + 7 left 3 units to obtain the graph of g ( x)  2( x  3)  7  2 x  6  7  2 x  13. (Note that if the original graph is first translated to the left 3 units and then up 2 units, the final result will be the same.) 102. f(x) = 3 – x Translate the graph of f ( x ) down 2 units to obtain the graph of t ( x )  (3  x)  2   x  1. 237 (b) f(x) is even. An even function has a graph symmetric with respect to the y-axis. Reflect the left half of the graph in the y-axis. Now translate the graph of t  x    x  1 right 3 units to obtain the graph of g ( x)   ( x  3)  1   x  3  1   x  4. (Note that if the original graph is first translated to the right 3 units and then down 2 units, the final result will be the same.) 103. (a) Because f(–x) = f(x), the graph is symmetric with respect to the y-axis. Chapter 2 Quiz (Sections 2.5−2.7) 95 2 1  (3) Choose either point, say, (−3, 5), to find the equation of the line: y  5  2( x  (3))  y  2( x  3)  5  y  2 x  11 . 1. (a) First, find the slope: m  (b) Because f(–x) = –f(x), the graph is symmetric with respect to the origin. (b) To find the x-intercept, let y = 0 and solve for x: 0  2 x  11  x   11 . The 2   x-intercept is  11 ,0 . 2 2. Write 3x − 2y = 6 in slope-intercept form to find its slope: 3x  2 y  6  y  32 x  3. Then, the slope of the line perpendicular to this graph is  23 . y  4   23 ( x  (6))  104. (a) f(x) is odd. An odd function has a graph symmetric with respect to the origin. Reflect the left half of the graph in the origin. y   23 ( x  6))  4  y   23 x 3. (a) x  8 (b) y  5 4. (a) Cubing function; domain: (, ) ; range: (, ) ; increasing over (, ) . (b) Absolute value function; domain: (, ) ; range: [0, ) ; decreasing over (, 0) ; increasing over (0, ) Copyright © 2017 Pearson Education, Inc. 238 Chapter 2 Graphs and Functions (c) Cube root function: domain: (, ) ; range: (, ) ; increasing over (, ) . 5. f  x   0.40  x   0.75 f 5.5  0.40 5.5  0.75  0.40 5  0.75  2.75 A 5.5-minute call costs $2.75. 9. This is the graph of g ( x)  x , translated four units to the left, reflected across the x-axis, and then translated two units down. The equation is y   x  4  2 . 10. (a) f  x  x2  7 Replace x with –x to obtain f   x   ( x) 2  7  f  x   x 2  7  f  x  The result is the same as the original function, so the function is even.  if x  0 6. f ( x)   x  x 2 3 if x  0  For values of x < 0, the graph is the line y = 2x + 3. Do not include the right endpoint (0, 3). Graph the line y  x for values of x ≥ 0, including the left endpoint (0, 0). (b) f  x   x3  x  1 Replace x with –x to obtain f  x    x    x   1 3   x3  x  1  f  x  The result is not the same as the original equation, so the function is not even. Because f   x    f  x  , the function is not odd. Therefore, the function is neither even nor odd. (c) 7. f ( x)   x3  1 f  x   x101  x99 Replace x with –x to obtain f  x    x  101 Reflect the graph of f ( x)  x 3 across the x-axis, and then translate the resulting graph one unit up.   x  99    x101   x99  101  x x 99     f  x Because f(−x) = −f(x), the function is odd. Section 2.8 Function Operations and Composition 8. In exercises 1–10, f  x   x  1 and g  x   x 2 . f ( x)  2 x  1  3 Shift the graph of f ( x)  x one unit right, stretch the resulting graph vertically by a factor of 2, then shift this graph three units up. 1.  f  g 2  f 2  g 2   2  1  2 2  7 2.  f  g 2  f 2  g 2   2  1  2 2  1 3.  fg 2  f 2  g 2   2  1  2 2  12 f  2 2  1 3 f   2  4.    2  g  2 4 g 2 5.  f  g 2  f  g 2  f 2 2   2 2  1  5 Copyright © 2017 Pearson Education, Inc. Section 2.8 Function Operations and Composition 6.  g  f 2  g  f 2  g 2  1  2  12  9 ( f  g )( x)  f ( x)  g ( x)  (3x  4)  (2 x  5)  x  9 ( fg )( x)  f ( x)  g ( x)  (3 x  4)(2 x  5)  6 x 2  15 x  8 x  20  6 x 2  7 x  20 f  f ( x) 3 x  4  g  ( x)  g ( x)  2 x  5 The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, 7. f is defined for all real numbers, so its domain is ,   . 8. g is defined for all real numbers, so its domain is ,   . 9. f + g is defined for all real numbers, so its domain is ,   . 10. and fg are all  ,   . The domain of g is f f is defined for all real numbers except those g the set of all real numbers for which g  x   0. This is the set of all real numbers values that make g  x   0, so its domain is except 52 , which is written in interval notation , 0  0,   . In Exercises 11–18, f ( x)  x 2  3 and g ( x)  2 x  6 . 11. ( f  g )(3)  f (3)  g (3)  (3) 2  3   2(3)  6  12  0  12 12. ( f  g )(5)  f (5)  g (5)  [(5) 2  3]  [2(5)  6]  28  16  44 13. ( f  g )(1)  f (1)  g (1)  [(1) 2  3]  [ 2( 1)  6]  4  8  4 14. ( f  g )(4)  f (4)  g (4)  [(4) 2  3]  [2(4)  6]  19  (2)  21     as  , 52  52 ,  . 20. f(x) = 6 – 3x, g(x) = – 4x + 1 ( f  g )( x)  f ( x)  g ( x)  (6  3x)  ( 4 x  1)  7 x  7 ( f  g )( x)  f ( x)  g ( x)  (6  3 x)  ( 4 x  1)  x  5 ( fg )( x)  f ( x)  g ( x)  (6  3x)( 4 x  1)  24 x  6  12 x 2  3 x  12 x 2  27 x  6 f  f ( x) 6  3x  g  ( x)  g ( x)   4 x  1 The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all  ,   . The domain of f g is the set of all real numbers for which g  x   0. This is the set of all real numbers 15. ( fg )(4)  f (4)  g (4)  [42  3]  [2(4)  6]  19  (2)  38 except 14 , which is written in interval notation 16. ( fg )(3)  f (3)  g ( 3)  [(3) 2  3]  [2(3)  6]  12  12  144 2 f  (1)  3 4 1 f (1) 17.   (1)     g (1) 2(1)  6 8 2 g f  (5) 2  3 28 f (5) 18.   (5)     7 g (5) 2(5)  6 4 g 19. 239 f ( x)  3 x  4, g ( x)  2 x  5     as – , 14  14 ,  . 21. f ( x )  2 x 2  3 x, g ( x )  x 2  x  3 ( f  g )( x)  f ( x)  g ( x)  (2 x 2  3 x)  ( x 2  x  3)  3x 2  4 x  3 ( f  g )( x)  f ( x)  g ( x)  (2 x 2  3 x)  ( x 2  x  3)  2 x 2  3x  x 2  x  3  x2  2x  3 ( f  g )( x)  f ( x)  g ( x)  (3 x  4)  (2 x  5)  5 x  1 Copyright © 2017 Pearson Education, Inc. (continued on next page) 240 Chapter 2 Graphs and Functions (continued) ( fg )( x)  f ( x)  g ( x)  (2 x 2  3x)( x 2  x  3)  2 x 4  2 x3  6 x 2  3 x3  3 x 2  9 x  2 x 4  5 x3  9 x 2  9 x f  f ( x) 2 x 2  3x  g  ( x)  g ( x)  x 2  x  3 The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all  ,   . The domain of f g 23. 1 x 1 ( f  g )( x)  f ( x)  g ( x)  4 x  1  x ( fg )( x)  f ( x)  g ( x) 4x  1 1  4x  1    x x f  f ( x)  g  ( x)  g ( x)  4x  1 1 x  x 4x  1 Because 4 x  1  0  4 x  1  x  14 , the g  x   0. If x 2  x  3  0 , then by the  domain of f is  14 ,  . The domain of g is quadratic formula x  1 i2 11 . The equation has no real solutions. There are no real numbers which make the denominator zero. Thus, the domain of 1 x ( f  g )( x)  f ( x)  g ( x)  4 x  1  is the set of all real numbers for which f g f ( x)  4 x  1, g ( x)  , 0  0,   . Considering the intersection of the domains of f and g, the domains of f + g, f – g, and fg are all  14 ,  . Because 1x  0  is also  ,   . f 22. for any value of x, the domain of g is also f ( x )  4 x 2  2 x, g ( x )  x 2  3 x  2 ( f  g )( x)  f ( x)  g ( x)  (4 x 2  2 x)  ( x 2  3x  2)  5×2  x  2 ( f  g )( x)  f ( x)  g ( x)  (4 x 2  2 x)  ( x 2  3x  2)  4 x 2  2 x  x 2  3x  2  3x 2  5 x  2 ( fg )( x)  f ( x)  g ( x)  (4 x 2  2 x)( x 2  3x  2)  4 x 4  12 x3  8 x 2  2 x3  6 x 2  4 x  4 x 4  10 x3  2 x 2  4 x f  f ( x) 4 x2  2 x ( x )    g  g ( x) x 2  3 x  2 The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all  ,   . The domain of g is the f set of all real numbers x such that x 2  3 x  2  0 . Because x 2  3 x  2  ( x  1)( x  2) , the numbers which give this denominator a value of 0 are f x = 1 and x = 2. Therefore, the domain of g is the set of all real numbers except 1 and 2, which is written in interval notation as (– , 1)  (1, 2)  (2, ) .   14 ,  .  24. f ( x)  5 x  4, g ( x)   1 x ( f  g )( x)  f ( x)  g ( x) 1  1  5x  4      5x  4   x x ( f  g )( x)  f ( x)  g ( x) 1  1  5x  4      5x  4   x x ( fg )( x)  f ( x)  g ( x) 5x  4  1  5x  4       x x  f  f ( x)  g  ( x)  g ( x)   5x  4   x 5x  4  1x Because 5 x  4  0  5 x  4  x  54 , the  domain of f is  54 ,  . The domain of g is , 0  0,   . Considering the intersection of the domains of f and g, the domains of f + g, f – g, and fg are all  54 ,  .  1x  0 for any  f value of x, so the domain of g is also  4 ,  . 5 Copyright © 2017 Pearson Education, Inc. Section 2.8 Function Operations and Composition 25. M  2008  280 and F  2008  470, thus T  2008  M  2008  F  2008  280  470  750 (thousand). 26. M  2012  390 and F  2012  630, thus T  2012  M  2012  F  2012  390  630  1020 (thousand). 27. Looking at the graphs of the functions, the slopes of the line segments for the period 2008−2012 are much steeper than the slopes of the corresponding line segments for the period 2004−2008. Thus, the number of associate’s degrees increased more rapidly during the period 2008−2012. 28. If 2004  k  2012, T (k )  r , and F(k) = s, then M(k) = r − s. 29. T  S 2000  T 2000  S 2000  19  13  6 It represents the dollars in billions spent for general science in 2000. 30. T  G 2010  T 2010  G 2010  29  11  18 It represents the dollars in billions spent on space and other technologies in 2010. 31. Spending for space and other technologies spending decreased in the years 1995−2000 and 2010–2015. 32. Total spending increased the most during the years 2005−2010. 33. (a)  f  g 2  f 2  g 2  4   2  2 (b) ( f  g )(1)  f (1)  g (1)  1  (3)  4 241 34. (a) ( f  g )(0)  f (0)  g (0)  0  2  2 (b) ( f  g )(1)  f (1)  g (1)  2  1  3 (c) ( fg )(1)  f (1)  g (1)  2  1  2 f  f (2) 4 (d)   (2)    2 g (2) 2 g 35. (a) ( f  g )(1)  f (1)  g (1)  0  3  3 (b) ( f  g )(2)  f (2)  g (2)  1  4  5 (c) ( fg )(0)  f (0)  g (0)  1  2  2 f  f (2) 3 (d)   (2)    undefined g (2) 0 g 36. (a) ( f  g )(1)  f (1)  g (1)  3  1  2 (b) ( f  g )(0)  f (0)  g (0)  2  0  2 (c) ( fg )(1)  f (1)  g (1)  3( 1)  3 f  f (1) 3 (d)   (1)    3 g (1) 1 g 37. (a)  f  g 2  f 2  g 2  7  2  5 (b) ( f  g )(4)  f (4)  g (4)  10  5  5 (c) ( fg )(2)  f (2)  g (2)  0  6  0 f  f (0) 5 (d)   (0)    undefined g (0) 0 g 38. (a)  f  g 2  f 2  g 2  5  4  9 (c) ( fg )(0)  f (0)  g (0)  0(4)  0 (b) ( f  g )(4)  f (4)  g (4)  0  0  0 f  f (1) 1 1 (d)   (1)    g (1) 3 3 g (c) ( fg )(2)  f (2)  g (2)  4  2  8 f  f (0) 8 (d)   (0)    8 g (0) 1 g Copyright © 2017 Pearson Education, Inc. 242 Chapter 2 Graphs and Functions 39. x f  x g  x  f  g  x   f  g  x   fg  x  f   g   x  −2 0 6 06 6 0  6  6 06  0 0 0 6 0 5 0 50  5 50  5 50  0 5  undefined 0 2 7 −2 7   2  5 7   2  9 7  2  14 7  3.5 2 4 10 5 10  5  15 10  5  5 10  5  50 10 2 5 x f  x g  x  f  g  x   f  g  x   fg  x  f   g   x  −2 −4 2 4  2  2 4  2  6 4  2  8 4  2 2 0 8 −1 8   1  7 8   1  9 8  1  8 8  8 1 2 5 4 5 4  9 5 4 1 5  4  20 5  1.25 4 4 0 0 00 0 00 0 00  0 0  undefined 0 40. 41. Answers may vary. Sample answer: Both the slope formula and the difference quotient represent the ratio of the vertical change to the horizontal change. The slope formula is stated for a line while the difference quotient is stated for a function f. 45. 42. Answers may vary. Sample answer: As h approaches 0, the slope of the secant line PQ approaches the slope of the line tangent of the curve at P. 43. f ( x  h)  6( x  h)  2  6 x  6h  2 (b) f ( x  h)  f ( x )  (6 x  6h  2)  (6 x  2)  6 x  6h  2  6 x  2  6h f ( x  h)  f ( x ) 6h  6 h h f  x   4 x  11 (a) f ( x  h)  2  ( x  h)  2  x  h (a) f ( x  h)  4( x  h)  11  4 x  4h  11 (b) f ( x  h)  f ( x)  (2  x  h)  (2  x)  2  x  h  2  x  h (b) f ( x  h)  f ( x )  (4 x  4h  11)  (4 x  11)  4 x  4h  11  4 x  11  4h (c) 44. (a) (c) 46. f  x  2  x f  x  6 x  2 f ( x  h)  f ( x )  h   1 h h f  x  1  x (a) f ( x  h )  1  ( x  h)  1  x  h (b) f ( x  h)  f ( x)  (1  x  h)  (1  x)  1  x  h  1  x  h (c) (c) 47. f ( x  h )  f ( x ) 4h  4 h h f  x   2 x  5 (a) f ( x  h)  2( x  h)  5  2 x  2h  5 (b) f ( x  h)  f ( x )  (2 x  2h  5)  (2 x  5)  2 x  2h  5  2 x  5  2h f ( x  h)  f ( x )  h   1 h h (c) f ( x  h)  f ( x) 2h   2 h h Copyright © 2017 Pearson Education, Inc. Section 2.8 Function Operations and Composition 48. f ( x )  4 x  2 (a) f ( x  h)  4( x  h)  2  4 x  4h  2 (b) f ( x  h)  f ( x )  4 x  4h  2   4 x  2  4 x  4h  2  4 x  2  4h (c) 49. 51. (a) (b) (c) (a) f ( x  h)  ( x  h) 2  x 2  2 xh  h 2 (b) f ( x  h)  f ( x)  x 2  2 xh  h 2  x 2  2 xh  h 2 (c) 1 x 52. f ( x  h)  h x  x  h f ( x  h)  f ( x )  h h  (b)  (c) h hx  x  h  1 x  x  h   x  h 2 1  x  h  2  x 2   x  h 1  2 x2 x2  x  h x 2  x 2  2 xh  h 2 x  x  h 2   2 xh  h 2   hx 2  x  h  2 x  h x2  x  h 2 2 54. f ( x  h)  f ( x) 2 xh  h 2  h h  h(2 x  h)  h  2 x  h (b) f ( x  h)  f ( x )  (1  x 2  2 xh  h 2 )  (1  x 2 )  1  x 2  2 xh  h 2  1  x 2  2 xh  h 2 (c) 2 2 f ( x  h )  1  ( x  h) 2  1  ( x 2  2 xh  h 2 )  1  x 2  2 xh  h 2 2 f ( x  h)  f ( x ) x2  x  h2 2 xh  h 2   2 h h hx 2  x  h  h 2 x  h    f ( x  h)  f ( x)   x 2  2 xh  h 2   x 2 (a) 2 xh  h 2 (c)  f ( x)  1  x 2 2 x  x  h 2 2   x  2 xh  h  x 2  2 xh  h 2 1 f ( x  h)  f ( x ) 2 2 53. (b)    x  2 xh  h 1 f ( x)  2 x f ( x  h)  f ( x  h )   ( x  h) 2   x 2  2 xh  h 2 f ( x  h)  f ( x ) 1 1 x   x  h    xh x x  x  h h  x  x  h (a) f ( x  h)  f ( x ) 2 xh  h 2  h h h(2 x  h)  h  2x  h f ( x)   x 2 (a) 1 xh  50. f ( x)  x 2 f ( x  h)  f ( x ) 4h   4 h h f ( x)  243 f ( x  h)  f ( x) 2 xh  h 2  h h h(2 x  h)  h  2 x  h f ( x)  1  2 x 2 (a) f ( x  h)  1  2( x  h)2  1  2( x 2  2 xh  h 2 )  1  2 x 2  4 xh  2h 2 Copyright © 2017 Pearson Education, Inc. 244 Chapter 2 Graphs and Functions (b) f ( x  h)  f ( x )  2 2 2 2  1  2 x  4 xh  2h   1  2 x  2  1  2 x  4 xh  2h  1  2 x 2  4 xh  2h 2 (c) 55. 58. g  x    x  3  g  2  2  3  1  f  g 2  f  g 2  f 1  2 1  3  2  3  1 59. g  x    x  3  g  2    2  3  5 f ( x  h)  f ( x) 4 xh  2h 2  h h h(4 x  2h)  h  4 x  2h  f  g 2  f  g 2  f 5  2 5  3  10  3  7 60.  g  f 3  g  f 3  g 3  3  3  0 f ( x)  x 2  3x  1 (a) f ( x  h)   x  h   3  x  h   1 61. 2 2 f ( x  h)  f ( x )    x 2  2 xh  h 2  3x  3h  1  2  62. 2 2  x  2 xh  h  3 x  3h  1  x  3 x  1  2 xh  h 2  3h (c) 56. f ( x  h)  f ( x) 2 xh  h 2  3h  h h h(2 x  h  3)  h  2x  h  3  g  g 2  g  g 2  g 5  5  3  2 65. ( f  g )(2)  f [ g (2)]  f (3)  1 f ( x  h)   x  h   4  x  h   2 2 f ( x  h)  f ( x )   x 2  2 xh  h 2  4 x  4h  2    x2  4 x  2 66. ( f  g )(7)  f [ g (7)]  f (6)  9 2 f ( x  h)  f ( x ) 2 xh  h  4h  h h h(2 x  h  4)  h  2x  h  4 57. g  x    x  3  g  4  4  3  1  f  g 4  f  g 4  f 1  2  1  3  2  3  5 67. ( g  f )(3)  g[ f (3)]  g (1)  9 68. ( g  f )(6)  g[ f (6)]  g (9)  12   x 2  2 xh  h 2  4 x  4h  2  x 2  4 x  2  2 xh  h 2  4h (c) f  x   2 x  3  f 2  2 2  3  4  3  1 64. g  x    x  3  g  2    2  3  5  x 2  2 xh  h 2  4 x  4h  2 (b) 63.  f  f 2  f  f 2  f 1  2 1  3  1 f ( x)  x 2  4 x  2 (a) f  x   2 x  3  f  2  2  2  3  7  g  f 2  g  f 2  g 7    7   3  7  3  10  x  3x  1 2 f  x   2 x  3  f 0  2 0  3  0  3  3  g  f 0  g  f 0  g 3    3  3  3  3  6 2  x  2 xh  h  3x  3h  1 (b) f  x   2 x  3  f 3  2 3  3  6  3  3 69. ( f  f )(4)  f [ f (4)]  f (3)  1 70. ( g  g )(1)  g[ g (1)]  g (9)  12 71. ( f  g )(1)  f [ g (1)]  f (9) However, f(9) cannot be determined from the table given. 72. ( g  ( f  g ))(7)  g ( f ( g (7)))  g ( f (6))  g (9)  12 73. (a) ( f  g )( x)  f ( g ( x))  f (5 x  7)  6(5 x  7)  9  30 x  42  9  30 x  33 The domain and range of both f and g are (, ) , so the domain of f  g is (, ) . Copyright © 2017 Pearson Education, Inc. Section 2.8 Function Operations and Composition (b) ( g  f )( x)  g ( f ( x))  g (6 x  9)  5(6 x  9)  7  30 x  45  7  30 x  52 The domain of g  f is (, ) . 74. (a) ( f  g )( x)  f ( g ( x))  f (3x  1)  8(3x  1)  12  24 x  8  12  24 x  4 The domain and range of both f and g are (, ) , so the domain of f  g is (, ) . (b) ( g  f )( x)  g ( f ( x))  g (8 x  12)  3(8 x  12)  1  24 x  36  1  24 x  35 The domain of g  f is (, ) . 75. (a) ( f  g )( x)  f ( g ( x ))  f ( x  3)  x  3 The domain and range of g are (, ) , however, the domain and range of f are [0, ) . So, x  3  0  x  3 . Therefore, the domain of f  g is [3, ) . (b) ( g  f )( x)  g ( f ( x))  g   x  x 3 The domain and range of g are (, ) , however, the domain and range of f are [0, ) .Therefore, the domain of g  f is [0, ) . 76. (a) ( f  g )( x)  f ( g ( x))  f ( x  1)  x  1 The domain and range of g are (, ) , however, the domain and range of f are [0, ) . So, x  1  0  x  1 . Therefore, the domain of f  g is [1, ) . (b) ( g  f )( x)  g ( f ( x))  g  x   x 1 The domain and range of g are (, ), however, the domain and range of f are [0, ). Therefore, the domain of g  f is [0, ). 77. (a) ( f  g )( x)  f ( g ( x))  f ( x 2  3 x  1)  ( x 2  3x  1)3 The domain and range of f and g are (, ), so the domain of f  g is (, ). 245 (b) ( g  f )( x)  g ( f ( x))  g ( x3 )    3x   1  x3 2 3  x 6  3 x3  1 The domain and range of f and g are (, ), so the domain of g  f is (, ). 78. (a) ( f  g )( x)  f ( g ( x))  f ( x 4  x 2  4)  x4  x2  4  2  x4  x2  2 The domain of f and g is (, ), while the range of f is (, ) and the range of g is  4,   , so the domain of f  g is (, ). (b) ( g  f )( x)  g ( f ( x))  g ( x  2)  ( x  2)4  ( x  2) 2  4 The domain of f and g is (, ), while the range of f is (, ) and the range of g is  4,   , so the domain of g  f is (, ). 79. (a) ( f  g )( x)  f ( g ( x ))  f (3x)  3x  1 The domain and range of g are (, ) , however, the domain of f is [1, ), while the range of f is [0, ). So, 3x  1  0  x  13 . Therefore, the  domain of f  g is  13 ,  . (b) ( g  f )( x)  g ( f ( x))  g  x  1  3 x 1 The domain and range of g are (, ) , however, the range of f is [0, ) . So x  1  0  x  1 . Therefore, the domain of g  f is [1, ) . 80. (a) ( f  g )( x)  f ( g ( x ))  f (2 x)  2 x  2 The domain and range of g are (, ) , however, the domain of f is [2, ) . So, 2 x  2  0  x  1 . Therefore, the domain of f  g is 1,   . Copyright © 2017 Pearson Education, Inc. 246 Chapter 2 Graphs and Functions (b) ( g  f )( x)  g ( f ( x))  g  x  2 2 x2 The domain and range of g are (, ) , however, the range of f is [0, ) . So x  2  0  x  2 . Therefore, the domain of g  f is [2, ) . 81. (a) ( f  g )( x)  f ( g ( x))  f ( x  1)  x21 The domain and range of g are (, ) , however, the domain of f is (, 0)  (0, ) . So, x  1  0  x  1 . Therefore, the domain of f  g is (, 1)  (1, ) .  (b) ( g  f )( x)  g ( f ( x ))  g 2x  2x  1 The domain and range of f is (, 0)  (0, ) , however, the domain and range of g are (, ) . So x  0 . Therefore, the domain of g  f is (, 0)  (0, ) . 82. (a) ( f  g )( x)  f ( g ( x))  f ( x  4)  x 4 4 The domain and range of g are (, ) , however, the domain of f is (, 0)  (0, ) . So, x  4  0  x  4 . Therefore, the domain of f  g is (, 4)  (4, ) .  (b) ( g  f )( x)  g ( f ( x))  g 4x  4x  4 The domain and range of f is (, 0)  (0, ) , however, the domain and range of g are (, ) . So x  0 . Therefore, the domain of g  f is (, 0)  (0, ) .   83. (a) ( f  g )( x)  f ( g ( x))  f  1x   1x  2 The domain and range of g are (, 0)  (0, ) , however, the domain of f is [2, ) . So,  1x  2  0  x  0 or x  12 (using test intervals). Therefore, the domain of f  g is , 0   12 ,   . (b) ( g  f )( x)  g ( f ( x))  g  x  2   1 x2 The domain of f is [2, ) and its range is [0, ). The domain and range of g are (, 0)  (0, ) . So x  2  0  x  2 . Therefore, the domain of g  f is (2, ) .   84. (a) ( f  g )( x)  f ( g ( x))  f  2x   2x  4 The domain and range of g are (, 0)  (0, ) , however, the domain of f is [4, ) . So,  2x  4  0  x  0 or x  12 (using test intervals). Therefore, the domain of f  g is , 0   12 ,   . (b) ( g  f )( x)  g ( f ( x))  g  x  4   2 x4 The domain of f is [4, ) and its range is [0, ). The domain and range of g are (, 0)  (0, ) . So x  4  0  x  4 . Therefore, the domain of g  f is (4, ) . 85. (a) ( f  g )( x)  f ( g ( x))  f  x 1 5   x 1 5 The domain of g is (, 5)  (5, ) , and the range of g is (, 0)  (0, ) . The domain of f is [0, ) . Therefore, the domain of f  g is 5,   . (b) ( g  f )( x)  g ( f ( x))  g  x  1 x 5 The domain and range of f is [0, ). The domain of g is (, 5)  (5, ). Therefore, the domain of g  f is [0, ). 86. (a) ( f  g )( x)  f ( g ( x))  f  x 3 6   x 3 6 The domain of g is (, 6)  (6, ) , and the range of g is (, 0)  (0, ) . The domain of f is [0, ) . Therefore, the domain of f  g is 6,   . (b) ( g  f )( x)  g ( f ( x))  g  x  3 x 6 The domain and range of f is [0, ) . The domain of g is (, 6)  (6, ) . Therefore, the domain of g  f is [0, ). Copyright © 2017 Pearson Education, Inc. Section 2.8 Function Operations and Composition 87. (a) ( f  g )( x)  f ( g ( x))  f  1x   1 x1 2  1x2 x 90. The domain and range of g are (, 0)  (0, ) . The domain of f is (,  2)  (2, ), and the range of f is f (2)   f (2)  1. Thus, ( f  g )(2)  f  g (2)   f (0)  0 and 0  x  12 or x  12 (using test intervals). ( f  g )(1)  f  g (1)   f (2)  1 and Thus, x  0 and x  12 . Therefore, the ( f  g )(2)  f  g (2)   f (0)  0. domain of f  g is   (b) ( g  f )( x)  g ( f ( x))  g x 1 2  1 ( x1 2)  x2 The domain and range of g are (, 0)  (0, ). The domain of f is (, 2)  (2, ), and the range of f is (, 0)  (0, ). Therefore, the domain of g  f is (, 2)  (2, ).   88. (a) ( f  g )( x)  f ( g ( x))  f  1x  1 1x  4  1x 4 x The domain and range of g are (, 0)  (0, ). The domain of f is (, 4)  (4, ), and the range of f is (, 0)  (0, ). So, 1x 4 x  0  x  0 f ( x) is odd, so f (1)   f (1)  (2)  2. Because g ( x) is even, g (1)  g (1)  2 and g (2)  g (2)  0. ( f  g )(1)  1, so f  g (1)   1 and f (2)  1. f ( x) is odd, so (, 0)  (0, ) . So, 1x2 x  0  x  0 or , 0  0, 12    12 ,   . x –2 –1 0 1 2 f  x –1 2 0 –2 1 g  x 0 2 1 2 0  f  g  x  0 1 –2 1 0 91. Answers will vary. In general, composition of functions is not commutative. Sample answer:  f  g  x   f 2 x  3  3 2 x  3  2  6 x  9  2  6 x  11  g f x     g 3x  2  2 3x  2  3  6x  4  3  6x  7 Thus,  f  g  x  is not equivalent to  g  f  x  . 92.  f  g  x   f  g  x   f  3 x  7    3 x7 or 0  x  14 or 1  4 x  0  x  14 (using test intervals). Thus, x ≠ 0 and x ≠ 14 . Therefore, the domain of f  g is  7 3   x  7  7  x  g  f  x   g  f  x   g  x 3  7 , 0  0, 14    14 ,   .    3 x3  7  7  3 x3  x   (b) ( g  f )( x)  g ( f ( x))  g x 1 4   1 ( x1 4)  x  4 The domain and range of g are (, 0)  (0, ) . The domain of f is (, 4)  (4, ) , and the range of f is (, 0)  (0, ) . Therefore, the domain of g  f is (, 4)  (4, ). 89. g  f (2)   g (1)  2 and g  f (3)   g (2)  5 Since g  f (1)  7 and f (1)  3, g (3)  7. x f  x g  x g  f ( x)  1 3 2 7 2 1 5 2 3 2 7 5 247 93. 94.  f  g  x   f  g  x   4  14  x  2  2   4  14   x  2  2   x  2  2  x  2  2  x  g  f  x   g  f  x   14 4 x  2  2  14  4 x  2  2  14  4 x   x  f  g  x   f  g  x   3  13 x      3  13  x  x  g  f  x   g  f  x    13 3x     13  3 x  x Copyright © 2017 Pearson Education, Inc. 248 95. Chapter 2 Graphs and Functions  f  g  x   f  g  x   3 5  15 x3  54   4  3 x3  4  4  3 x3  x  g  f  x   g  f  x   15  3 5 x  4   54  15 5 x  4  54  55x  54  54 3  55x  x 96.  f  g  x   f  g  x   3  x  1  1 3  3 x3  1  1  3 x3  x  g  f  x   g  f  x    3 x  1   1 104. f  x   3 x, g  x   1760 x  f  g  x   f  g  x   f 1760 x   3 1760 x   5280 x  f  g  x  compute the number of feet in x miles. 105.  ( x)  3 2 x 4 (a)  (2 x)  3  x 11  x In Exercises 97−102, we give only one of many possible answers. 97. h( x)  (6 x  2) 2 2 Let g(x) = 6x – 2 and f ( x)  x . ( f  g )( x)  f (6 x  2)  (6 x  2) 2  h( x) 98. h( x)  (11x 2  12 x) 2 Let g ( x)  11x 2  12 x and f ( x)  x 2 . ( f  g )( x)  f (11x 2  12 x)  (11x 2  12 x) 2  h( x) 99. h( x)  x 2  1 Let g ( x)  x 2  1 and f ( x)  x . 3 3 (2 x) 2  (4 x 2 )  3 x 2 4 4 (b)  (16)  A(2  8)  3(8) 2  64 3 square units 106. (a) x  4 s  4x  s  s  4x (b) y  s 2  (c) y  4x   16x 2 2 62 36   2.25 square units 16 16 107. (a) r (t )  4t and  (r )   r 2 (  r )(t )  [r (t )]   (4t )   (4t ) 2  16 t 2 (b) (  r )(t ) defines the area of the leak in terms of the time t, in minutes. (c)  (3)  16 (3) 2  144 ft 2 ( f  g )( x)  f ( x 2  1)  x 2  1  h( x). 100. h( x)  (2 x  3)3 Let g ( x)  2 x  3 and f ( x)  x3 . 3 ( f  g )( x)  f (2 x  3)  (2 x  3)  h( x) 101. h( x)  6 x  12 Let g ( x)  6 x and f ( x)  x  12. ( f  g )( x)  f (6 x)  6 x  12  h( x) 102. h( x)  3 2 x  3  4 Let g ( x)  2 x  3 and f ( x)  3 x  4. ( f  g )( x)  f (2 x  3)  3 2 x  3  4  h( x) 103. f(x) = 12x, g(x) = 5280x ( f  g )( x)  f [ g ( x)]  f (5280 x)  12(5280 x)  63, 360 x The function f  g computes the number of inches in x miles. 108. (a) (  r )(t )  [r (t )]   (2t )   (2t ) 2  4 t 2 (b) It defines the area of the circular layer in terms of the time t, in hours. (c) (  r )(4)  4 (4) 2  64 mi 2 109. Let x = the number of people less than 100 people that attend. (a) x people fewer than 100 attend, so 100 – x people do attend N(x) = 100 – x (b) The cost per person starts at $20 and increases by $5 for each of the x people that do not attend. The total increase is $5x, and the cost per person increases to $20 + $5x. Thus, G(x) = 20 + 5x. (c) C ( x)  N ( x)  G ( x)  (100  x)(20  5 x) Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises (d) If 80 people attend, x = 100 – 80 = 20. C  20  100  20  20  5 20  8020  100  80120  $9600 110. (a) y1  0.02 x (d) ( y1  y2 )(250)  y1 (250)  y2 (250)  0.02(250)  0.015(250  500)  5  0.015(750)  15  11.25  $16.25 3  5  2   9  7  2 2   2  22  4  4  8 d ( A, B )  d ( A, C )  (6  5) 2  (8  7) 2  12  12  1  1  2 6  32  8  92 2  32   1  9  1  10 d ( B, C )  1 x 2 Because (b) f  x  x  1 (c)  f  g  x   f  g  x   f  (d)  f  g 60  60  1  $31 1  1 x  x 1 2  2  8    2    10  , triangle 2 2 2 ABC is a right triangle with right angle at (5, 7). 1 2 112. If the area of a square is x 2 square inches, each side must have a length of x inches. If 3 inches is added to one dimension and 1 inch is subtracted from the other, the new dimensions will be x + 3 and x – 1. Thus, the area of the resulting rectangle is (x) = (x + 3)(x – 1). Chapter 2 d ( A, B )  [ 6  ( 6)]2  (8  3) 2 4. Label the points A(5, 7), B(3, 9), and C(6, 8). y1  y2 represents the total annual interest. 111. (a) g  x   3. A(– 6, 3), B(– 6, 8)  0  52  25  5 Midpoint: 11    6  (6) 3  8   12 11   , ,     6,    2 2   2 2  2 (b) y2  0.015( x  500) (c) 249 Review Exercises 1. P(3, –1), Q(– 4, 5) d ( P, Q)  ( 4  3) 2  [5  (1)]2  (7) 2  62  49  36  85 Midpoint:  3  ( 4) 1  5   1 4   1  ,     ,    , 2 2 2   2 2  2  5. Let B have coordinates (x, y). Using the midpoint formula, we have 6  x 10  y  , 8, 2    2 2  10  y 6  x 8 2 2 2 6  x  16 10  y  4 x  22 y  6 The coordinates of B are (22, −6). 6. P(–2, –5), Q(1, 7), R(3, 15) d ( P, Q)  (1  (2)) 2  (7  (5)) 2  (3) 2  (12) 2  9  144 2. M(–8, 2), N(3, –7) d ( M , N )  [3  (8)]2  (7  2) 2  112  (9) 2  121  81  202 5  8  3 2  (7)   5 , Midpoint:      ,    2 2 2 2  153  3 17 d (Q, R )  (3  1) 2  (15  7) 2  22  82  4  64  68  2 17 d ( P, R)  (3  (2))2  (15  (5)) 2  (5) 2  (20) 2  25  400  5 17 (continued on next page) Copyright © 2017 Pearson Education, Inc. 250 Chapter 2 Graphs and Functions (continued) d ( P, Q)  d (Q, R )  3 17  2 17  5 17  d ( P, R ), so these three points are collinear. 14. The center of the circle is (5, 6). Use the distance formula to find the radius: r 2  (4  5) 2  (9  6) 2  1  9  10 The equation is ( x  5) 2  ( y  6) 2  10 . 15. x 2  4 x  y 2  6 y  12  0 Complete the square on x and y to put the equation in center-radius form. 7. Center (–2, 3), radius 15 ( x  h) 2  ( y  k ) 2  r 2 [ x  (2)]2  ( y  3) 2  152 ( x  2) 2  ( y  3) 2  225  x  4 x   y  6 y   12  x  4 x  4   y  6 y  9  12  4  9 2 2 8. Center ( 5,  7 ), radius 3  x  5    y   7    3  x  5    y  7   3 2 2 2  x  22   y  32  1 ( x  h) 2  ( y  k ) 2  r 2 2 2 The circle has center (2, –3) and radius 1. 2 2 9. Center (–8, 1), passing through (0, 16) The radius is the distance from the center to any point on the circle. The distance between (–8, 1) and (0, 16) is r  (0  (8)) 2  (16  1) 2  82  152 16. x 2  6 x  y 2  10 y  30  0 Complete the square on x and y to put the equation in center-radius form. ( x 2  6 x  9)  ( y 2  10 y  25)  30  9  25 ( x  3) 2  ( y  5) 2  4 The circle has center (3, 5) and radius 2. 2 x 2  14 x  2 y 2  6 y  2  0 x2  7 x  y 2  3 y  1  0 17.  64  225  289  17. The equation of the circle is [ x  (8)]2  ( y  1) 2  17 2 ( x  8) 2  ( y  1) 2  289  x  7 x    y  3 y   1  x  7 x     y  3 y    1  2 2 2 2 49 4 9 4 3 2  x  72    y  2    44  494  94 2 2  x  72    y  23   544 The circle has center  72 ,  23  and radius 2 10. Center (3, –6), tangent to the x-axis The point (3, –6) is 6 units directly below the x-axis. Any segment joining a circle’s center to a point on the circle must be a radius, so in this case the length of the radius is 6 units. ( x  h) 2  ( y  k ) 2  r 2 ( x  3) 2  [ y  ( 6)]2  62 ( x  3) 2  ( y  6) 2  36 11. The center of the circle is (0, 0). Use the distance formula to find the radius: r 2  (3  0) 2  (5  0) 2  9  25  34 The equation is x 2  y 2  34 . 12. The center of the circle is (0, 0). Use the distance formula to find the radius: r 2  (2  0) 2  (3  0) 2  4  9  13 The equation is x 2  y 2  13 . 13. The center of the circle is (0, 3). Use the distance formula to find the radius: r 2  (2  0) 2  (6  3) 2  4  9  13 The equation is x 2  ( y  3) 2  13 . 49  94 4 54  4 54  4 96  3 26 . 4 3 x 2  33 x  3 y 2  15 y  0 x 2  11x  y 2  5 y  0 18.  x  11x   y  5 y   0  x  11x     y  5 y    0  2 2 121 4 2 2 25 4 5 2 121  25 4 4  x  112    y  2   1464 The circle has center  11 , 5 and radius 2 2 2 146 . 2 19. This is not the graph of a function because a vertical line can intersect it in two points. domain: [−6, 6]; range: [−6, 6] 20. This is not the graph of a function because a vertical line can intersect it in two points. domain: ,   ; range:  0,   Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises 21. This is not the graph of a function because a vertical line can intersect it in two points. domain: ,   ; range: ( ,  1]  [1, ) 32. (a) As x is getting larger on the interval 2,   , the value of y is increasing. (b) As x is getting larger on the interval , 2 , the value of y is decreasing. 22. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: ,   ; range:  0,   23. This is not the graph of a function because a vertical line can intersect it in two points. domain:  0,   ; range:  ,   (c) f(x) is constant on (−2, 2). In exercises 33–36, f  x   2 x 2  3 x  6. 33. 24. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: ,   ; range:  ,   f 3  2  32  3  3  6  2  9  3  3  6  18  9  6  15 34. 25. y  6  x 2 Each value of x corresponds to exactly one value of y, so this equation defines a function. f  0.5  2  0.5  3  0.5  6  2 0.25  3  0.5  6  0.5  1.5  6  8 35. f 0  2 0  3 0  6  6 26. The equation x  13 y 2 does not define y as a 36. f  k   2k 2  3k  6 function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 3) and (3, –3) satisfy the relation. Thus, the relation would not be a function. 27. The equation y   x  2 does not define y as a function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 1) and (3, −1) satisfy the relation. 251 2 2 37. 2 x  5 y  5  5 y  2 x  5  y  52 x  1 The graph is the line with slope 52 and y-intercept (0, –)1. It may also be graphed using intercepts. To do this, locate the x-intercept: y  0 2 x  5 0  5  2 x  5  x  52 4 defines y as a function x of x because for every x in the domain, which is (– , 0)  (0, ) , there will be exactly one value of y. 28. The equation y   29. In the function f ( x)  4  x , we may use any real number for x. The domain is  ,   . 30. 31. 8 x 8 x x can be any real number except 8 because this will give a denominator of zero. Thus, the domain is (– , 8)  (8, ). f ( x)  38. 3x  7 y  14  7 y  3 x  14  y   73 x  2 The graph is the line with slope of  73 and y-intercept (0, 2). It may also be graphed using intercepts. To do this, locate the x-intercept by setting y = 0: 3x  7 0  14  3 x  14  x  143 f  x   6  3x In the function y  6  3 x , we must have 6  3x  0 . 6  3x  0  6  3x  2  x  x  2 Thus, the domain is  , 2 . Copyright © 2017 Pearson Education, Inc. 252 Chapter 2 Graphs and Functions 39. 2 x  5 y  20  5 y  2 x  20  y   52 x  4 The graph is the line with slope of  52 and y-intercept (0, 4). It may also be graphed using intercepts. To do this, locate the x-intercept: x-intercept: y  0 2 x  5 0  20  2 x  20  x  10 43. x = –5 The graph is the vertical line through (–5, 0). 40. 3 y  x  y  13 x The graph is the line with slope 13 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (6, 2) to complete the graph. 41. f(x) = x The graph is the line with slope 1 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (1, 1) to complete the graph. 44. f(x) = 3 The graph is the horizontal line through (0, 3). 45. y  2  0  y  2 The graph is the horizontal line through (0, −2). 46. The equation of the line that lies along the x-axis is y = 0. 47. Line through (0, 5), m   23 42. x  4 y  8 4 y   x  8 y  14 x  2 Note that m   23  32 . The graph is the line with slope 14 and y-intercept (0, –2). It may also be graphed using intercepts. To do this, locate the x-intercept: y  0  x  4 0   8  x  8 Begin by locating the point (0, 5). Because the slope is 32 , a change of 3 units horizontally (3 units to the right) produces a change of –2 units vertically (2 units down). This gives a second point, (3, 3), which can be used to complete the graph. (continued on next page) Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises (continued) 253 54. 9x – 4y = 2. Solve for y to put the equation in slopeintercept form.  4 y  9 x  2  y  94 x  12 Thus, the slope is 94 . 55. x  2  0  x  2 The graph is a vertical line, through (2, 0). The slope is undefined. 48. Line through (2, –4), m  34 First locate the point (2, –4). Because the slope is 34 , a change of 4 units horizontally (4 units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (6, –1), which can be used to complete the graph. 56. x – 5y = 0. Solve for y to put the equation in slopeintercept form. 5y   x  y  15 x 49. through (2, –2) and (3, –4) y  y1 4   2 2 m 2    2 x2  x1 3 2 1 58. (a) This is the graph of a function because no vertical line intersects the graph in more than one point. 50. through (8, 7) and  1 , 2 2  y2  y1 2  7 9  1  15 x2  x1 8  2 2 2 18 6    9       15  15 5 m 57. Initially, the car is at home. After traveling for 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrived home. (b) The lowest point on the graph occurs in December, so the most jobs lost occurred in December. The highest point on the graph occurs in January, so the most jobs gained occurred in January. (c) The number of jobs lost in December is approximately 6000. The number of jobs gained in January is approximately 2000. 51. through (0, –7) and (3, –7) 7   7  0 m  0 3 0 3 (d) It shows a slight downward trend. 52. through (5, 6) and (5, –2) y  y1 2  6 8 m 2   55 0 x2  x1 The slope is undefined. 53. 11x  2 y  3 Solve for y to put the equation in slopeintercept form. 2 y  11x  3  y   11 x  23 2 . Thus, the slope is  11 2 Thus, the slope is 15 . 59. (a) We need to first find the slope of a line that passes between points (0, 30.7) and (12, 82.9) y  y1 82.9  30.7 52.2 m 2    4.35 12  0 12 x2  x1 Now use the point-intercept form with b = 30.7 and m = 4.35. y = 4.35x + 30.7 The slope, 4.35, indicates that the number of e-filing taxpayers increased by 4.35% each year from 2001 to 2013. (b) For 2009, we evaluate the function for x = 8. y = 4.35(8) + 30.7 = 65.5 65.5% of the tax returns are predicted to have been filed electronically. Copyright © 2017 Pearson Education, Inc. 254 Chapter 2 Graphs and Functions 60. We need to find the slope of a line that passes between points (1980, 21000) and (2013, 63800) y  y1 63,800  21, 000 m 2  2013  1980 x2  x1 42,800   $1297 per year 33 The average rate of change was about $1297 per year. 61. (a) through (3, –5) with slope –2 Use the point-slope form. y  y1  m( x  x1 ) y  (5)  2( x  3) y  5  2( x  3) y  5  2 x  6 y  2 x  1 (b) Standard form: y  2 x  1  2 x  y  1 62. (a) through (–2, 4) and (1, 3) First find the slope. 1 3 4 m  1  (2) 3 Now use the point-slope form with ( x1 , y1 )  (1, 3) and m   13 . y  y1  m( x  x1 ) y  3   13 ( x  1) 3( y  3)  1( x  1) 3y  9  x  1 3 y   x  10  y   13 x  103 (b) Standard form: y   13 x  103  3 y   x  10  x  3 y  10 63. (a) through (2, –1) parallel to 3x – y = 1 Find the slope of 3x – y = 1. 3x  y  1   y  3 x  1  y  3x  1 The slope of this line is 3. Because parallel lines have the same slope, 3 is also the slope of the line whose equation is to be found. Now use the point-slope form with ( x1 , y1 )  (2,  1) and m  3. y  y1  m( x  x1 ) y  (1)  3( x  2) y  1  3x  6  y  3x  7 (b) Standard form: y  3 x  7  3 x  y  7  3x  y  7 64. (a) x-intercept (–3, 0), y-intercept (0, 5) Two points of the line are (–3, 0) and (0, 5). First, find the slope. 50 5 m  03 3 The slope is 53 and the y-intercept is (0, 5). Write the equation in slopeintercept form: y  53 x  5 (b) Standard form: y  53 x  5  3 y  5 x  15  5 x  3 y  15  5 x  3 y  15 65. (a) through (2, –10), perpendicular to a line with an undefined slope A line with an undefined slope is a vertical line. Any line perpendicular to a vertical line is a horizontal line, with an equation of the form y = b. The line passes through (2, –10), so the equation of the line is y = –10. (b) Standard form: y = –10 66. (a) through (0, 5), perpendicular to 8x + 5y = 3 Find the slope of 8x + 5y = 3. 8 x  5 y  3  5 y  8 x  3  y   85 x  53 The slope of this line is  85 . The slope of any line perpendicular to this line is    1. 5 , because  85 85 8 The equation in slope-intercept form with slope 85 and y-intercept (0, 5) is y  85 x  5. (b) Standard form: y  85 x  5  8 y  5 x  40  5 x  8 y  40  5 x  8 y  40 67. (a) through (–7, 4), perpendicular to y = 8 The line y = 8 is a horizontal line, so any line perpendicular to it will be a vertical line. Because x has the same value at all points on the line, the equation is x = –7. It is not possible to write this in slopeintercept form. (b) Standard form: x = –7 68. (a) through (3, –5), parallel to y = 4 This will be a horizontal line through (3, –5). Because y has the same value at all points on the line, the equation is y = –5. Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises 255 (b) Standard form: y = –5 69. f ( x)  x  3 The graph is the same as that of y  x , except that it is translated 3 units downward. 70. 73. f ( x)   x  3 To get y = 0, we need 0  x  3  1  3  x  4. To get y = 1, we need 1  x  3  2  4  x  5. Follow this pattern to graph the step function. 74. f ( x)  2 3 x  1  2 f ( x)   x The graph of f ( x)   x is the reflection of the graph of y  x about the x-axis. The graph of f ( x)  2 3 x  1  2 is a 71. translation of the graph of y  3 x to the left 1 unit, stretched vertically by a factor of 2, and translated down 2 units. f ( x)    x  1  3 2 The graph of f ( x)    x  1  3 is a 2 translation of the graph of y  x 2 to the left 1 unit, reflected over the x-axis and translated up 3 units. 75. 72. f ( x)   x  2 The graph of f ( x)   x  2 is the reflection   4 x  2 if x  1 3x  5 if x  1 Draw the graph of y = –4x + 2 to the left of x = 1, including the endpoint at x = 1. Draw the graph of y = 3x – 5 to the right of x = 1, but do not include the endpoint at x = 1. Observe that the endpoints of the two pieces coincide. f ( x)  of the graph of y  x about the x-axis, translated down 2 units. Copyright © 2017 Pearson Education, Inc. 256 76. 77. Chapter 2 Graphs and Functions  2 f ( x)   x  3 if x  2  x  4 if x  2 84. False. For example, f ( x)  x3 is odd, and (2, 8) is on the graph but (–2, 8) is not. Graph the curve y  x 2  3 to the left of x = 2, and graph the line y = –x + 4 to the right of x = 2. The graph has an open circle at (2, 7) and a closed circle at (2, 2). 85. x  y 2  10 x if x  3 f ( x)   6  x if x  3 Draw the graph of y  x to the left of x = 3, but do not include the endpoint. Draw the graph of y = 6 – x to the right of x = 3, including the endpoint. Observe that the endpoints of the two pieces coincide. 78. Because x represents an integer,  x   x. Therefore,  x   x  x  x  2 x. 79. True. The graph of an even function is symmetric with respect to the y-axis. 80. True. The graph of a nonzero function cannot be symmetric with respect to the x-axis. Such a graph would fail the vertical line test 81. False. For example, f ( x)  x 2 is even and (2, 4) is on the graph but (2, –4) is not. 82. True. The graph of an odd function is symmetric with respect to the origin. 83. True. The constant function f  x   0 is both even and odd. Because f   x   0  f  x  , the function is even. Also f  x   0  0   f  x  , so the function is odd. Replace x with –x to obtain ( x)  y 2  10. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x  ( y ) 2  10  x  y 2  10. The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( x)  ( y ) 2  10  ( x)  y 2  10. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the x-axis only. 86. 5 y 2  5 x 2  30 Replace x with –x to obtain 5 y 2  5( x) 2  30  5 y 2  5 x 2  30. The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain 5( y ) 2  5 x 2  30  5 y 2  5 x 2  30. The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the y-axis and x-axis, so it must also be symmetric with respect to the origin. Note that this equation is the same as y 2  x 2  6 , which is a circle centered at the origin. 87. x 2  y 3 Replace x with –x to obtain ( x) 2  y 3  x 2  y 3 . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain x 2  ( y )3  x 2   y 3 . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( x) 2  ( y )3  x 2   y 3 . The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only. Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises 88. y 3  x  4 92. Replace x with –x to obtain y 3   x  4 . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain (  y )3  x  4   y 3  x  4  y 3   x  4 The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( y )3  ( x)  4   y 3   x  4  y 3  x  4. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries. 89. 6 x  y  4 Replace x with –x to obtain 6( x)  y  4  6 x  y  4 . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 6 x  ( y )  4  6 x  y  4 . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 6( x)  ( y )  4  6 x  y  4 . This equation is not equivalent to the original one, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries. 90. y  x Replace x with –x to obtain y  ( x)  y  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain  y   x  y   x. The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain  y  ( x)  y  x. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis only. 91. y = 1 This is the graph of a horizontal line through (0, 1). It is symmetric with respect to the y-axis, but not symmetric with respect to the x-axis and the origin. 257 x  y Replace x with –x to obtain x  y  x  y . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain x   y  x  y . The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the xaxis and with respect to the y-axis, it must also by symmetric with respect to the origin. 93. x 2  y 2  0 Replace x with −x to obtain  x 2  y 2  0  x 2  y 2  0. The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain x 2    y   0  x 2  y 2  0. The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin. 2 94. x 2   y  2  4 Replace x with −x to obtain 2  x 2   y  22  4  x 2   y  22  4. The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain x 2    y  2  4. The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 2  x 2   y  22  4  x 2   y  22  4, which is not equivalent to the original equation. Therefore, the graph is not symmetric with respect to the origin. 95. To obtain the graph of g ( x)   x , reflect the graph of f ( x)  x across the x-axis. 96. To obtain the graph of h( x )  x  2 , translate the graph of f ( x)  x down 2 units. Copyright © 2017 Pearson Education, Inc. 258 Chapter 2 Graphs and Functions 97. To obtain the graph of k ( x)  2 x  4 , translate the graph of f ( x)  x to the right 4 units and stretch vertically by a factor of 2. (d) To graph y  f ( x) , keep the graph of y = f(x) as it is where y  0 and reflect the graph about the x-axis where y < 0. 98. If the graph of f ( x)  3x  4 is reflected about the x-axis, we obtain a graph whose equation is y  (3 x  4)  3 x  4. 99. If the graph of f ( x)  3x  4 is reflected about the y-axis, we obtain a graph whose equation is y  f ( x)  3( x)  4  3 x  4. 100. If the graph of f ( x)  3x  4 is reflected about the origin, every point (x, y) will be replaced by the point (–x, –y). The equation for the graph will change from y  3 x  4 to  y  3( x)  4   y  3 x  4  y  3x  4. 101. (a) To graph y  f ( x)  3, translate the graph of y = f(x), 3 units up. 102. No. For example suppose f  x   x  2 and g  x   2 x. Then ( f  g )( x)  f ( g ( x ))  f (2 x)  2 x  2 The domain and range of g are (, ) , however, the domain of f is [2, ) . So, 2 x  2  0  x  1 . Therefore, the domain of f  g is 1,   . The domain of g, (, ), is not a subset of the domain of f  g , 1,   . For Exercises 103–110, f ( x) = 3x 2 − 4 and g ( x) = x 2 − 3 x − 4. (b) To graph y  f ( x  2), translate the graph of y = f(x), 2 units to the right. (c) To graph y  f ( x  3)  2, translate the graph of y = f(x), 3 units to the left and 2 units down. 103. ( fg )( x)  f ( x)  g ( x)  (3 x 2  4)( x 2  3x  4)  3x 4  9 x3  12 x 2  4 x 2  12 x  16  3x 4  9 x3  16 x 2  12 x  16 104. ( f  g )(4)  f (4)  g (4)  (3  42  4)  (42  3  4  4)  (3  16  4)  (16  3  4  4)  (48  4)  (16  12  4)  44  0  44 105. ( f  g )(4)  f (4)  g (4)  [3(4) 2  4]  [(4) 2  3(4)  4]  [3(16)  4]  [16  3(4)  4]  [48  4]  [16  12  4]  44  24  68 106. ( f  g )(2k )  f (2k )  g (2k )  [3(2k ) 2  4]  [(2k ) 2  3(2k )  4]  [3(4)k 2  4]  [4k 2  3(2k )  4]  (12k 2  4)  (4k 2  6k  4)  16k 2  6k  8 Copyright © 2017 Pearson Education, Inc. Chapter 2 Review Exercises f  f (3) 3  32  4 39  4 107.   (3)   2  g (3) 3  3  3  4 9  3  3  4 g 27  4 23 23    9  9  4 4 4 3  1  4 3 1  4 f   108.   (1)  2 g 1  3 1  4 1  3 1  4 3 4 1    undefined 1 3  4 0 For Exercises 113–118, f ( x)  x  2 and g ( x)  x 2 . 113. ( g  f )( x)  g[ f ( x)]  g  2 109. The domain of (fg)(x) is the intersection of the domain of f(x) and the domain of g(x). Both have domain ,   , so the domain of (fg)(x) is  ,   . ,   , we are concerned about values of x that make g  x   0. Thus, the expression is undefined if (x + 1)(x – 4) = 0, that is, if x = –1 or x = 4. Thus, the domain is the set of all real numbers except x = –1 and x = 4, or (– , – 1)  (–1, 4)  (4, ). 111. f  x  2x  9 f ( x  h)  2( x  h)  9  2 x  2h  9 f ( x  h)  f ( x)  (2 x  2h  9)  (2 x  9)  2 x  2h  9  2 x  9  2h f ( x  h)  f ( x ) 2h   2. Thus, h h 112. f ( x)  x 2  5 x  3 f ( x  h)  ( x  h) 2  5( x  h)  3  x 2  2 xh  h 2  5 x  5h  3 f ( x  h)  f ( x )  ( x 2  2 xh  h 2  5 x  5h  3)  ( x 2  5 x  3)  x 2  2 xh  h 2  5 x  5h  3  x 2  5 x  3  2 xh  h 2  5h f ( x  h)  f ( x) 2 xh  h 2  5h  h h h(2 x  h  5)   2x  h  5 h  x  2  x  2  x  2 2 114. ( f  g )( x)  f [ g ( x)]  f ( x 2 )  x 2  2 115. f  x   x  2, so f 3  3  2  1  1. Therefore,  g  f 3  g  f 3  g 1  12  1. 116. g  x   x 2 , so g  6   6  36. 2 Therefore,  f  g  6  f  g  6  f 36 f  3x 2  4 3×2  4 110.   ( x)  2  g x  3x  4 ( x  1)( x  4) Because both f  x  and g  x  have domain 259  36  2  34 . 117.  g  f 1  g  f 1  g  1  2   g  3  Because 3 is not a real number,  g  f  1 is not defined. 118. To find the domain of f  g , we must consider the domain of g as well as the composed function, f  g. Because  f  g  x   f  g  x   x 2  2 we need to determine when x 2  2  0. Step 1: Find the values of x that satisfy x 2  2  0. x2  2  x   2 Step 2: The two numbers divide a number line into three regions. Step 3 Choose a test value to see if it satisfies the inequality, x 2  2  0. Interval Test Value Is x 2  2  0 true or false? ,  2  2 22  2  0 ? 2  0 True  2, 2  0  2,  0 20 ? 2  0 False 2 22  2  0 ? 2  0 True The domain of f  g is ,  2    2, . Copyright © 2017 Pearson Education, Inc. 2 260 Chapter 2 Graphs and Functions 119.  f  g 1  f 1  g 1  7  1  8 (b) The range of f  x   x  3 is all real numbers greater than or equal to 0. In interval notation, this correlates to the interval in D,  0,   . 120. ( f  g )(3)  f (3)  g (3)  9  9  0 121. ( fg )(1)  f (1)  g ( 1)  3  2  6 (c) The domain of f  x   x 2  3 is all real numbers. In interval notation, this correlates to the interval in C,  ,   . f  f (0) 5 122.   (0)    undefined g (0) 0 g 123. ( g  f )(2)  g[ f (2)]  g (1)  2 (d) The range of f  x   x 2  3 is all real numbers greater than or equal to 3. In interval notation, this correlates to the interval in B, 3,   . 124. ( f  g )(3)  f [ g (3)]  f (2)  1 125. ( f  g )(2)  f [ g (2)]  f (2)  1 126. ( g  f )(3)  g[ f (3)]  g (4)  8 127. Let x = number of yards. f(x) = 36x, where f( x) is the number of inches. g(x) = 1760x, where g(x) is the number of yards. Then ( g  f )( x)  g  f ( x)   1760(36 x)  63, 360 x. There are 63,360x inches in x miles 128. Use the definition for the perimeter of a rectangle. P = length + width + length + width P( x)  2 x  x  2 x  x  6 x This is a linear function. by 3 inches, then the amount of volume gained is given by Vg (r )  V (r  3)  V (r )  43  (r  3)3  43  r 3 . 130. (a) V   r 2 h If d is the diameter of its top, then h = d and r  d2 . So,  d2  (d )    d4 2  (d )   4d . 2 3 (b) S  2 r 2  2 rh   d2   2  d2  (d )   2d   d 2 2 S (d )  2 2 2 2   2d  22d  32d Chapter 2 (f) The range of f  x   3 x  3 is all real numbers. In interval notation, this correlates to the interval in C,  ,   . (g) The domain of f  x   x  3 is all real numbers. In interval notation, this correlates to the interval in C,  ,   . 129. If V (r )  43  r 3 and if the radius is increased V (d )   (e) The domain of f  x   3 x  3 is all real numbers. In interval notation, this correlates to the interval in C,  ,   . 2 (h) The range of f  x   x  3 is all real numbers greater than or equal to 0. In interval notation, this correlates to the interval in D,  0,   . (i) The domain of x  y 2 is x  0 because when you square any value of y, the outcome will be nonnegative. In interval notation, this correlates to the interval in D,  0,   . (j) The range of x  y 2 is all real numbers. In interval notation, this correlates to the interval in C,  ,   . 2. Consider the points  2,1 and 3, 4 . Test m 1. (a) The domain of f  x   x  3 occurs when x  0. In interval notation, this correlates to the interval in D,  0,   . 4 1 3  3  (2) 5 3. We label the points A  2,1 and B 3, 4 . d ( A, B)  [3  (2)]2  (4  1)2  52  32  25  9  34 Copyright © 2017 Pearson Education, Inc. Chapter 2 Test 4. The midpoint has coordinates  2  3 1  4   1 5  ,     , . 2 2  2 2 (b) This is the graph of a function because no vertical line intersects the graph in more than one point. The domain of the function is (– , – 1)  (–1, ). The range is (– , 0)  (0, ). As x is getting 5. Use the point-slope form with ( x1 , y1 )  (2,1) and m  53 . larger on the intervals  , 1 and y  y1  m( x  x1 ) y  1  53 [ x  (2)] 1,   , the value of y is decreasing, so y  1  53 ( x  2)  5  y  1  3( x  2)  5 y  5  3 x  6  5 y  3 x  11  3x  5 y  11  3 x  5 y  11 6. Solve 3x – 5y = –11 for y. 3x  5 y  11 5 y  3 x  11 y  53 x  11 5 the function is decreasing on these intervals. (The function is never increasing or constant.) 10. Point A has coordinates (5, –3). (a) The equation of a vertical line through A is x = 5. (b) The equation of a horizontal line through A is y = –3. Therefore, the linear function is f ( x)  53 x  11 . 5 7. (a) The center is at (0, 0) and the radius is 2, so the equation of the circle is x2  y 2  4 . (b) The center is at (1, 4) and the radius is 1, so the equation of the circle is ( x  1) 2  ( y  4) 2  1 2 2 8. x  y  4 x  10 y  13  0 Complete the square on x and y to write the equation in standard form: x 2  y 2  4 x  10 y  13  0 11. The slope of the graph of y  3 x  2 is –3. (a) A line parallel to the graph of y  3 x  2 has a slope of –3. Use the point-slope form with ( x1 , y1 )  (2, 3) and m  3. y  y1  m( x  x1 ) y  3  3( x  2) y  3  3 x  6  y  3x  9 (b) A line perpendicular to the graph of y  3 x  2 has a slope of 13 because  3 13  1. y  3  13 ( x  2) 3  y  3  x  2  3 y  9  x  2  3 y  x  7  y  13 x  73  x  4 x     y  10 y    13  x  4 x  4   y  10 y  25  13  4  25 2 2 261 2 2  x  22   y  52  16 The circle has center (−2, 5) and radius 4. 12. (a) 2,   (b) 0, 2 (c) , 0 (d) ,   (e) ,   (f)  1,   13. To graph f  x   x  2  1 , we translate the graph of y  x , 2 units to the right and 1 unit down. 9. (a) This is not the graph of a function because some vertical lines intersect it in more than one point. The domain of the relation is [0, 4]. The range is [– 4, 4]. Copyright © 2017 Pearson Education, Inc. 262 14. Chapter 2 Graphs and Functions f ( x)   x  1 To get y = 0, we need 0  x  1  1  1  x  0. To get y = 1, we need 1  x  1  2  0  x  1. Follow this pattern to graph the step function. (c) Reflect f(x), across the x-axis. (d) Reflect f( x), across the y-axis. 15. if x  2 3 f ( x)   1 2  2 x if x  2 For values of x with x < –2, we graph the horizontal line y = 3. For values of x with x  2, we graph the line with a slope of  12 and a y-intercept of (0, 2). Two points on this line are (–2, 3) and (0, 2). 16. (a) Shift f(x), 2 units vertically upward. (e) Stretch f(x), vertically by a factor of 2. 17. Starting with y  x , we shift it to the left 2 units and stretch it vertically by a factor of 2. The graph is then reflected over the x-axis and then shifted down 3 units. 18. 3x 2  2 y 2  3 (a) Replace y with –y to obtain 3x 2  2( y ) 2  3  3x 2  2 y 2  3. The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. (b) Shift f(x), 2 units horizontally to the left. (b) Replace x with –x to obtain 3( x) 2  2 y 2  3  3 x 2  2 y 2  3. The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. (c) The graph is symmetric with respect to the x-axis and with respect to the y-axis, so it must also be symmetric with respect to the origin. Copyright © 2017 Pearson Education, Inc. Chapter 2 Test 19. (g) g  x   2 x  1  g 0  2 0  1  0  1  1. Therefore,  f  g 0  f  g 0 f ( x)  2 x 2  3x  2, g ( x)  2 x  1 (a) ( f  g )( x)  f ( x)  g ( x)    2 x 2  3x  2   2 x  1 2  f 1  2  12  3  1  2  2 1  3 1  2  23 2 1  2 x  3x  2  2 x  1  2×2  x  1 f  f ( x) 2 x 2  3x  2  (b)   ( x)  g ( x) 2 x  1 g For exercises 20 and 21, f  x   x  1 and (c) We must determine which values solve the equation 2 x  1  0. 2 x  1  0  2 x  1  x  12 g  x   2 x  7. 20. and the domain is (d)   The domain and range of g are (, ) , while the domain of f is [0, ) . We need to find the values of x which fit the domain of f: 2 x  6  0  x  3 . So, the domain of f  g is [3, ) .  1 , . 2 f ( x)  2 x 2  3x  2 f  x  h  2  x  h  3  x  h  2 2    2 x 2  2 xh  h 2  3 x  3h  2 2 2  2 x  4 xh  2h  3 x  3h  2 f ( x  h)  f ( x )  (2 x 2  4 xh  2h 2  3 x  3h  2)  (2 x 2  3 x  2)  2 x 2  4 xh  2h 2  3x 3h  2  2 x 2  3x  2 2  4 xh  2h  3h f ( x  h)  f ( x) 4 xh  2h 2  3h  h h h(4 x  2h  3)  h  4 x  2h  3 (e) ( f  g )(1)  f (1)  g (1)  (2  12  3  1  2)  ( 2  1  1)  (2  1  3  1  2)  (2  1  1)  (2  3  2)  (2  1)  1  (1)  0 (f) ( fg )(2)  f (2)  g (2)  (2  22  3  2  2)  ( 2  2  1)  (2  4  3  2  2)  ( 2  2  1)  (8  6  2)  (4  1)  4(3)  12  f  g   f  g  x   f 2 x  7  (2 x  7)  1  2 x  6 Thus, 12 is excluded from the domain,  , 12 263 21.  g  f   g  f  x   g  x  1   2 x 1  7 The domain and range of g are (, ) , while the domain of f is [0, ) . We need to find the values of x which fit the domain of f: x  1  0  x  1 . So, the domain of g  f is [1, ) . 22. (a) C(x) = 3300 + 4.50x (b) R(x) = 10.50x (c) P ( x)  R( x)  C ( x)  10.50 x  (3300  4.50 x)  6.00 x  3300 (d) P( x)  0 6.00 x  3300  0 6.00 x  3300 x  550 She must produce and sell 551 items before she earns a profit. Copyright © 2017 Pearson Education, Inc.