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Relations and Functions

Relations and Functions. ALGEBRA 2 LESSON 2-1. (For help, go to Skills Handbook page 848 and Lesson 1-2.). Graph each ordered pair on the coordinate plane. 1. (–4, –8) 2.  (3, 6) 3.  (0, 0) 4.  (–1, 3) 5.  (–6, 5). Evaluate each expression for x = –1, 0, 2, and 5.

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Relations and Functions

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  1. Relations and Functions ALGEBRA 2 LESSON 2-1 (For help, go to Skills Handbook page 848 and Lesson 1-2.) Graph each ordered pair on the coordinate plane. 1. (–4, –8) 2. (3, 6) 3. (0, 0) 4. (–1, 3) 5. (–6, 5) Evaluate each expression for x = –1, 0, 2, and 5. 6. x + 2 7. –2x + 3 8. 2x2 + 1 9. |x – 3| 2-1

  2. Relations and Functions ALGEBRA 2 LESSON 2-1 Solutions 1. 2. 3. 4. 5. 2-1

  3. Relations and Functions ALGEBRA 2 LESSON 2-1 Solutions (continued) 6.x + 2 for x = –1, 0, 2, and 5: –1 + 2 = 1; 0 + 2 = 2; 2 + 2 = 4; 5 + 2 = 7 7. –2x + 3 for x = –1, 0, 2, and 5: –2(–1) + 3 = 2 + 3 = 5; –2(0) + 3 = 0 + 3 = 3; –2(2) + 3 = –4 + 3 = –1; –2(5) + 3 = –10 + 3 = –7 8. 2x2 + 1 for x = –1, 0, 2, and 5: 2 • (–1)2 + 1 = 2 • 1 + 1 = 2 + 1 = 3; 2 • 02 + 1 = 2 • 0 + 1 = 0 + 1 = 1; 2 • 22 + 1 = 2 • 4 + 1 = 8 + 1 = 9; 2 • 52 + 1 = 2 • 25 + 1 = 50 + 1 = 51 9. |x – 3| for x = –1, 0, 2, and 5: |–1 – 3| = |–4| = 4; |0 – 3| = |–3| = 3; |2 – 3| = |–1| = 1; |5 – 3| = |2| = 2 2-1

  4. Graph and label each ordered pair. Relations and Functions ALGEBRA 2 LESSON 2-1 Graph the relation {(–3, 3), (2, 2), (–2, –2), (0, 4), (1, –2)}. 2-1

  5. Relations and Functions ALGEBRA 2 LESSON 2-1 Write the ordered pairs for the relation. Find the domain and range. {(–4, 4), (–3, –2), (–2, 4), (2, –4), (3, 2)} The domain is {–4, –3, –2, 2, 3}. The range is {–4, –2, 2, 4}. 2-1

  6. Pair the domain elements with the range elements. Relations and Functions ALGEBRA 2 LESSON 2-1 Make a mapping diagram for the relation {(–1, 7), (1, 3), (1, 7), (–1, 3)}. 2-1

  7. Relations and Functions ALGEBRA 2 LESSON 2-1 Determine whether the relation is a function. The element –3 of the domain is paired with both 4 and 5 of the range. The relation is not a function. 2-1

  8. Relations and Functions ALGEBRA 2 LESSON 2-1 Use the vertical-line test to determine whether the graph represents a function. If you move an edge of a ruler from left to right across the graph, keeping the edge vertical as you do so, you see that the edge of the ruler never intersects the graph in more than one point in any position. Therefore, the graph does represent a function. 2-1

  9. 9 1 – 2 9 1 – x c.ƒ(x) = 9 –1 ƒ(2) = = = –9 Relations and Functions ALGEBRA 2 LESSON 2-1 Find ƒ(2) for each function. a.ƒ(x) = –x2 + 1 ƒ(2) = –22 + 1 = –4 + 1 = –3 b.ƒ(x) = |3x| ƒ(2) = |3 • 2| = |6| = 6 2-1

  10. Relations and Functions ALGEBRA 2 LESSON 2-1 Pages 59–61 Exercises 1. 2. 3. 4. 5. (–2, –2), (–1, 1), (1, 1), (1, 0), (3, 3), (3, –2); domain {–2, –1, 1, 3}, range {–2, 0, 1, 3} 6. (–2, 3), (0, 1), (2, –1), (3, –2); domain {–2, 0, 2, 3}, range {–2, –1, 1, 3} 7. (–2, 0), (–1, 2), (0, 3), (1, 2), (2, 0); domain {–2, –1, 0, 1, 2}, range {0, 2, 3} 8. 2-1

  11. Relations and Functions ALGEBRA 2 LESSON 2-1 9. 10. 11. 12. not a function 13. function 14. function 15. function 16. function 17. not a function 18. function 19. function 20. not a function 21. function 22. –7, –3, 4, 11 23. 13, 7, –3.5, –14 24. 4.5, 6.5, 10, 13.5 25. –2, –4, –7.5, –11 26. 21, 13, –1, –15 27. –13, –9, –2, 5 28. 29 , 17 , –3 , –24 29. – , – , , 1 3 1 3 2 3 2 3 23 6 13 6 3 4 11 3 2-1

  12. x 39.37 Relations and Functions ALGEBRA 2 LESSON 2-1 9 2 7 2 7 4 30. – , – , – , 0 31.y = , where y is the number of meters and x is the number of inches; 1.50 m 32. domain {2, 3, 4, 5}, range {4, 5, 6, 7} 33. domain {–4, –3, –2, –1}, range {1, 2, 3, 4} 34. domain { – , 0, 2}, range { –2, – , , 2} 35. domain { – , , , } range { – , } 36. domain {2, 4, 8}, range {4, 8, 16}; function 37. domain {–2, –1, 0, 9}, range {2, 5, 7}; not a function 3 2 1 2 3 2 5 2 1 2 1 2 1 2 1 2 1 2 2-1

  13. < < < < – – – – > – Relations and Functions ALGEBRA 2 LESSON 2-1 38. domain {all real numbers}, range {y 0}; function 39. domain {–3.2 x 3.2}, range {–1 y 1}; not a function 40. A 41. B 42. C 43. yes 44. no 45. no 46.v(s) = s3; 2460.375 cm3 47.v(r) = r 3; about 4849 cm3 48. No; since 2 in the domain maps to 1 and 4, it is not a function. 49. Check students’ work. 50. –3 51. – 52. –4x – 14 53. 7 54. 55. Yes; each x is paired with a unique y. 56. No; each positive x is paired with two y values. 2 3 3 4 2-1

  14. Relations and Functions ALGEBRA 2 LESSON 2-1 57. Yes; each x is paired with a unique y. 58. Function; if the domain and range are interchanged, it is not a function because 6 would be paired with both 2.5 and 3.0. 59. Domain {all integers}, range {all even integers}; function, each integer pairs to a unique even integer. 60. Domain {all integers}, range {all integers}; function, each integer pairs with its opposite. 61. Domain {all integers}, range {all even integers}; not a function, each nonzero integer pairs with a pos. and neg. even integer. 62. C 63. G 64.[2]ƒ(5); ƒ(5) = –2(5) + 3 = –7 and g(–2) = 4(–2) – 3 = –11, and –7 > –11. [1] only includes answer ƒ(5) or g(–2) 65.[4] Yes; S(c) = 6c2; S(2.5) = 6(2.5)2 = 37.5; the surface area is 37.5 cm2. [3] minor computational error [2] includes only function [1] no work shown 2-1

  15. < < – – = / 6 11 1 11 2 11 4 11 7 11 3 11 Relations and Functions ALGEBRA 2 LESSON 2-1 66. 67. 68. 69. 0 70. 71. 72. 73. –6, 14 74. –2 b 2 75.x 6 76. 11.1% increase 77. 50% increase 78. 25% decrease 2-1

  16. Relations and Functions ALGEBRA 2 LESSON 2-1 1. Write the ordered pairs for the relation. Find the domain and range. 2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function. 3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function. 2-1

  17. Relations and Functions ALGEBRA 2 LESSON 2-1 4. Determine whether the graph represents a function. 5. Find ƒ(–5) for each function. a. ƒ(x) = 5x + 35 b. ƒ(x) = x2 – x 2-1

  18. Relations and Functions ALGEBRA 2 LESSON 2-1 1. Write the ordered pairs for the relation. Find the domain and range. 2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function. 3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function. {(–3, 2), (–2, 0), (–1, –1), (2, 1), (2, 3), (4, 1)}; D = {–3, –2, –1, 2, 4], R = {–1, 0, 1, 2, 3} function (–4, 2) or (–4, 6) 2-1

  19. Relations and Functions ALGEBRA 2 LESSON 2-1 4. Determine whether the graph represents a function. 5. Find ƒ(–5) for each function. a. ƒ(x) = 5x + 35 b. ƒ(x) = x2 – x not a function 10 30 2-1

  20. Linear Equations ALGEBRA 2 LESSON 2-2 (For help, go to Lesson 1-2.) Evaluate each expression for x = –2, 0, 1, and 4. 1.x + 7 2.x – 2 3. 3x + 1 4.x – 8 2 3 3 5 1 2 2-2

  21. Linear Equations ALGEBRA 2 LESSON 2-2 Solutions 1.x + 7 for x = –2, 0, 1, 4: (–2) + 7 = – + = = or 5 ; (0) + 7 = 0 + 7 = 7; (1) + 7 = + 7 = 7 or ; (4) + 7 = + = = or 9 2 3 2 3 4 3 21 3 –4 + 21 3 17 3 2 3 2 3 2 3 2 3 23 3 2 3 2 3 8 3 21 3 8 + 21 3 29 3 2 3 2-2

  22. –6 + (–10) 5 3 + (–10) 5 12 + (–10) 5 Linear Equations ALGEBRA 2 LESSON 2-2 Solutions (continued) 2.x – 2 for x = –2, 0, 1, 4: (–2) – 2 = – + – = = – or –3 ; (0) – 2 = 0 – 2 = –2; (1) – 2 = + – = = – = – or –1 ; (4) – 2 = + – = = 3 5 3 5 6 5 10 5 16 5 1 5 3 5 3 5 3 5 10 5 7 5 7 5 2 5 3 5 12 5 10 5 2 5 3. 3x + 1 for x = –2, 0, 1, 4: 3(–2) + 1 = –6 + 1 = –5; 3(0) + 1 = 0 + 1 = 1; 3(1) + 1 = 3 + 1 = 4; 3(4) + 1 = 12 + 1 = 13 2-2

  23. Linear Equations ALGEBRA 2 LESSON 2-2 Solutions (continued) 4.x – 8 for x = –2, 0, 1, 4: (–2) – 8 = –1 – 8 = –9; (0) – 8 = 0 – 8 = –8; (1) – 8 = – 8 = –7 or – ; (4) – 8 = 2 – 8 = –6 1 2 1 2 1 2 1 2 15 2 1 2 1 2 1 2 2-2

  24. Plot the points (0, 2) and (3, –2) and then draw the line through these two points. Linear Equations ALGEBRA 2 LESSON 2-2 4 3 Graph the equation y = – x + 2. If x = 0, then y = 2. If x = 3, then y = –2. 2-2

  25. Set x or y equal to zero to find each intercept. 10x + 5y = 40 10x + 5y = 40 10x + 5(0) = 40 10(0) + 5y = 40 10x = 40 5y = 40 x = 4 y = 8 Linear Equations ALGEBRA 2 LESSON 2-2 The equation 10x + 5y = 40 models how you can give $.40 change if you have only dimes and nickels. The variable x is the number of dimes, and y is the number of nickels. Graph the equation. Describe the domain and the range. Explain what the x- and y-intercepts represent. 2-2

  26. The x-intercept is (4, 0), which means that the change can be given using 4 dimes and 0 nickels. The y-intercept is (0, 8), which means that the change can be given using 0 dimes and 8 nickels. The possible solutions for this situation are limited to those points on the line segment connecting (0, 8) and (4, 0) whose x- and y-coordinates are whole numbers. Linear Equations ALGEBRA 2 LESSON 2-2 (continued) Use the intercepts to graph the equation. The number of dimes and the number of nickels must each be a whole number. Therefore, the domain is {0, 1, 2, 3, 4} and the range is {0, 2, 4, 6, 8}. 2-2

  27. y2 –y1 x2 – x1 Slope = Use the slope formula. =    Substitute (–2, 7) for (x1, y1) and (8, –6) for (x2, y2). 13 10 = – Simplify. –6 – 7 8 – (–2) 13 10 The slope of the line is – . Linear Equations ALGEBRA 2 LESSON 2-2 Find the slope of the line through the points (–2, 7) and (8, –6). 2-2

  28. Linear Equations ALGEBRA 2 LESSON 2-2 Write in standard form an equation of the line with slope 3 through the point (–1, 5). y – y1 = m(x – x1) Use the point-slope equation. y – 5 = 3[x – (–1)] Substitute 3 for m, 5 for y1, and –1 for x1. y – 5 = 3[(x + 1)] Simplify. y – 5 = 3x + 3 Distributive Property 3x – y = –8 Write in standard form. 2-2

  29. y2 – y1 x2 – x1 y – y1 = (x – x1) Substitute the slope formula for m. y – (–3) = (x – 4) Substitute: x1 = 4, y1 = –3, x2 = 5, y2 = –1. –1 – (–3) 5 – 4 Linear Equations ALGEBRA 2 LESSON 2-2 Write in point-slope form an equation of the line through (4, –3) and (5, –1). y – y1 = m(x – x1) Write the point-slope equation. y + 3 = 2(x – 4) Simplify. You can also use (5, –1) for (x1, y1) and (4, –3) for (x2, y2). This gives the equation y + 1 = 2(x – 5). Both equations define the same line. 2-2

  30. 7 2 y = x + 4 Write in the slope-intercept form. 7 2 The slope of the line is . Linear Equations ALGEBRA 2 LESSON 2-2 Find the slope of –7x + 2y = 8. –7x + 2y = 8 Add 7x to both sides. 2-2

  31. 1 4 m = – Find the negative reciprocal of 4. y = mx + bUse slope-intercept form. 1 4 1 4 y = – x + bSlope is – . 1 4 –3 = – (5) + bSubstitute (5, –3) for (x, y). 5 4 –3 = – + bSimplify. 7 4 – = bSolve for b. 1 4 7 4 y = – x – Write the equation. Linear Equations ALGEBRA 2 LESSON 2-2 Write an equation of the line through (5, –3) and perpendicular to y = 4x + 1. Graph both lines. 2-2

  32. Linear Equations ALGEBRA 2 LESSON 2-2 Pages 67–71 Exercises 1. 2. 3. 4. 5. 6. 2-2

  33. > > – – Linear Equations ALGEBRA 2 LESSON 2-2 7. 8. 9. a. y = 0.23x domain {x | x 0} range {y | y 0} 10. a. b. The x-intercept is the point that represents selling 1000 caps and 0 sweatshirts in order to raise the $4500. The y-intercept represents selling 0 caps and 360 sweatshirts to raise the $4500. b.x-intercept (0, 0), y-intercept (0, 0); when no miles have been driven, there is no cost. c. 0.23 represents a cost of $.23 per mile driven. 2-2

  34. 4 11 Linear Equations ALGEBRA 2 LESSON 2-2 4 3 11. –1 12. –2 13. 3 14. 15. – 16. 1 17. undefined 18. 0 19. 20. 3x – y = –2 21.x – y = 22. x + y = – 23.y = –2 24.x + y = 2 25. 5x – y = –2 26.y – 3 = –1(x + 10) 27.y – 0 = (x – 1) 28. y – 10 = – (x + 4) 29. y + 1 = – (x – 0) 30.y – 11 = 1(x – 7) 31.y – 9 = – (x – 1) 32. –5 33. 34. – 35. – 36. 37. 0 5 6 19 3 7 5 3 5 12 5 3 2 1 5 1 2 A B 5 4 A B 5 2 17 5 2-2

  35. 38.y = –3x – 5 39.y = x + 5 2 13 2 Linear Equations ALGEBRA 2 LESSON 2-2 40.y = 10 41.x = 1 42. 43. 44. 2-2

  36. Linear Equations ALGEBRA 2 LESSON 2-2 1 4 45. 46. 47. 48. 49. 50. 51. 52. 53. – 54. , (0, 4), (–6, 0) 55. –1, (0, 1000), (1000, 0) 56. , 0, – , , 0 57. 5, (0, –1), , 0 58. undefined slope, no y-intercept, (–3, 0) 59. 0, (0, 0), all pts. on x-axis 2 3 R S T S T R 1 5 1 3 2 3 2-2

  37. 7 10 5 13 Linear Equations ALGEBRA 2 LESSON 2-2 1 2 5 2 3 2 1 2 60. – , 0, – , (–5, 0) 61. –0.8, (0, 0.4), (0.5, 0) 62. – , 0, , , 0 63. – 64. 65. – 66.y = x + 3 67. y = 3x + 2 68. y = – x – 69.a. I, II; III; graphs I and II show constant rate of change. b. I and III c. II 70. Vertical lines cannot be graphed by this method. 71.y = –1  72.y = 2x + 1  73.y = x +   C B C A A B 5 6 10 3 7 5 3 4 2-2

  38. Linear Equations ALGEBRA 2 LESSON 2-2 3 2 74.y = – x – 1 75. 3x – 2y = 2 76. 9x + 3y = 2 77. 3x + 12y = –4 78.a–e. Check students’ work. The polygon is a rectangle. 79. yes 80. no 81. a. b.y = 3x + 6 c.y = – x + d. They are perpendicular. 82. The equation of the line connecting (1, 3) to (–2, 6) is y = –x + 4. The equation of the line connecting (1, 3) to (3, 5) is y = x + 2. The slopes are negative reciprocals so the lines are perpendicular. Therefore by def. of a right triangle it is a right triangle. 1 3 8 3 2-2

  39. Linear Equations ALGEBRA 2 LESSON 2-2 83. The slope of the line connecting (2, 5) to (4, 8) is , (2, 5) to (5, 3) is – , (4, 8) to (7, 6) is – , and (5, 3) to (7, 6) is . Since the adjacent sides’ slopes are negative reciprocals they are perpendicular. By the def. of a rectangle, it is a rectangle. 84.p: y = 4x + 16 q: y = – x + r : y = 4 85. A 86. G 87. C 88. B 89. C 90. B 91. domain {–2, 1, 2, 3, 4}, range {–2, –1, 2, 3}; not a function 92. domain {all reals}, range {all reals}; function 93. domain {–3, 0, 1, 7}, range {–10, –5, –1, 3}; not a function 1 4 13 4 3 2 2 3 2 3 3 2 2-2

  40. Linear Equations ALGEBRA 2 LESSON 2-2 94. Commutative Prop. of Add. 95. multiplicative inverses 96. Distributive Prop. 97. additive inverses, additive identity 98. studio: $6.26; designer: $9.39 2-2

  41. 4 7 1 5 y – 8 = – (x + 3) 1 4 1 2 y = x – 3 2 2 3 – ; Linear Equations ALGEBRA 2 LESSON 2-2 1. Find the slope of the line through the points (–5, –1) and (2, 3). 2. Write an equation in standard form for the line with slope 3 through (9, –4). 3. Write in point-slope form an equation of the line through the points (–3, 8) and (7, 6). Use (–3, 8) as the point for the equation. 4. Write the equation 3x – 12y = 6 in slope-intercept form. 5. What is the slope of a line perpendicular to y = x – 7? What is the slope of a line parallel to y = x – 7? 3x – y = 31 2 3 2 3 2-2

  42. 6 15 9 24 6.7.8.9. 1 4 2 8 2 5 12 36 20 24 30 36 Direct Variation ALGEBRA 2 LESSON 2-3 (For help, go to Lesson 1-3 and Skills Handbook page 844.) Solve each equation for y. 1. 12y = 3x2. 12y = 5x3.y = 15 4. 0.9y = 27x5. 5y = 35 3 4 Tell whether each equation is true. 2-3

  43. 1. 12y = 3x y = = = 2. 12y = 5x y = = x 3x 12 3 • x 3 • 4 x 4 5x 12 3.y = 15 y = 15 = = 20 4. 0.9y = 27x y = = 30x 3 4 60 3 27x 0.9 5 12 4 3 Direct Variation ALGEBRA 2 LESSON 2-3 Solutions 5. 5y = 35 y = 7 1 4 2 8 6. 1(8) 4(2) 8 = 8 true 2-3

  44. Solutions (continued) = / 12 36 8. 9(36) 24(12) 324 288 false 7. 2(15) 5(6) 30 = 30 true 2 5 20 24 30 36 6 15 9 24 9. 20(36) 24(30) 720 = 720 true Direct Variation ALGEBRA 2 LESSON 2-3 2-3

  45. a. = and are both equal to –3, but = 3. 15 5 3 –1 –6 2 y x b. x –1 2 5 y 3 –6 15 x 7 9 –4 y 14 18 –8 14 7 18 9 –8 –4 = = = = 2, so y does vary directly with x. y x Direct Variation ALGEBRA 2 LESSON 2-3 For each function, determine whether y varies directly with x. If so, find the constant of variation and write the equation. Since the three ratios are not all equal, y does not vary directly with x. The constant of variation is 2. The equation is y = 2x. 2-3

  46. 5 2 5x = –2y is equivalent to y = – x, so y varies directly with x. 5 2 The constant of variation is – . Direct Variation ALGEBRA 2 LESSON 2-3 For each function, tell whether y varies directly with x. If so, find the constant of variation. a. 3y = 7x + 7 Since you cannot write the equation in the form y = kx, y does not vary directly with x. b. 5x = –2y 2-3

  47. Direct Variation ALGEBRA 2 LESSON 2-3 The perimeter of a square varies directly as the length of a side of the square. The formula P = 4s relates the perimeter to the length of a side. a. Find the constant of variation. The equation P = 4s has the form of a direct variation equation with k = 4. b. Find how long a side of the square must be for the perimeter to be 64 cm. P = 4sUse the direct variation. 64 = 4sSubstitute 64 for P. 16 = sSolve for s. The sides of the square must have length 16 cm. 2-3

  48. y1 x1 y2 x2 = Write a proportion. 15 27 y 18 Substitute. = 15 • 18 27 y = Solve for y. Direct Variation ALGEBRA 2 LESSON 2-3 Suppose y varies directly with x, and y = 15 when x = 27. Find y when x = 18. Let (x1, y1) = (27, 15) and let (x2, y2) = (18, y). 15(18) = 27(y) Write the cross products. y = 10 Simplify. 2-3

  49. Direct Variation ALGEBRA 2 LESSON 2-3 Pages 74–77 Exercises 2 7 10 7 1. yes; k = 2, y = 2x 2. yes; k = –3, y = –3x 3. no 4. yes; k = , y = x 5. yes; k = 7, y = 7x 6. no 7. yes; k = –2, y = –2x 8. no 9. yes; k = 12 10. yes; k = 6 11. yes; k = –2 12. no 13. no 14. yes; k = –5 15. yes; k = 6 16. no 17. k = ; – 18.k = – ; 19.k = –1; 5 20.k = 2; –10 21.k = – ; 21 22. k = – ; 1 23.a.k = b. s = h c. 23 ft 1 in. 5 3 25 3 1 3 1 3 17 4 1 4 1 4 1 4 13 36 13 36 2-3

  50. Direct Variation ALGEBRA 2 LESSON 2-3 24. –3 25. 4 26. 10.5 27. 28. 681.8 mi/h 29. yes; k = , y = x 30. no 31. no 32. yes; k = 1.3, y = 1.3x 33.y = 2x 34.y = x 35.y = – x 36.y = –500x 37. y = x 38. y = – x 39. y = x 40.y = – x 41. 9 42. 6 43. 90 44. –140 45. 1.2 46. No; y = 1.7x does not contain the point (9, –9). 47. Yes; y = – x contains the point (15, –12 ). 48. Yes; y = x contains the point (6 , 22 ). 7 3 9 2 5 3 3 5 1 9 2 3 2 3 5 6 2 7 1 2 14 3 7 2 1 2 3 4 2-3

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