Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (For help, go to Lessons 5-2 and 1-5.) Evaluate each function rule for x = –5. 1. y = x – 7 2. y = 7 – x 3. y = 2x + 5 4. y = – x + 3 Write in simplest form. 5. 6. 7. 8. 9. 10. 2 5 7 – 3 3 – 1 3 – 5 6 – 0 8 – (–4) 3 – 7 –1 – 2 0 – 5 –6 – (–4) –2 – 6 0 – 1 1 – 0 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Solutions 1. y = x – 7 for x = –5: y = –5 – 7 = –12 2. y = 7 – x for x = –5: y = 7 – (–5) = 12 3. y = 2x + 5 for x = –5: y = 2(–5) + 5 = –10 + 5 = –5 4. y = – x + 3 for x = –5: y = – (–5) + 3 = 2 + 3 = 5 5. = = 2 6. = – = – 7. = – = –3 8. = = 9. = = 10. = – = –1 2 5 2 5 7 – 3 3 – 1 4 2 3 – 5 6 – 0 2 6 1 3 8 – (–4) 3 – 7 12 4 –1 – 2 0 – 5 –3 –5 3 5 –6 – (–4) –2 – 6 –2 –8 1 4 0 – 1 1 – 0 1 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 For the data in the table, is the rate of change the same for each pair of consecutive mileage amounts? Fee for Miles Driven Miles Fee 100 $30 150 200 250 $42 $54 $66 Find the rate of change for each pair of consecutive mileage amounts. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) change in cost Cost depends on the change in number of miles number of miles. rate of change = 42 – 30 12 54 – 42 12 66 – 54 12 150–100 50 200–150 50 200–250 50 = The rate of change for each pair of consecutive mileage amounts is $12 per 50 miles. The rate of change is the same for all the data. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Below is a graph of the distance traveled by a motorcycle from its starting point. Find the rate of change. Explain what this rate of change means. Choose two points On the graph (0,0) and (400,20) 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) Using the points (0,0) and (400,20), find the rate of change. vertical change change in distance horizontal change change in time rate of change = 400 – 0 20 – 0 400 20 = Use two points. Divide the vertical change by the horizontal change. Simplify. The rate of change is 20 m/s. The motorcycle is traveling 20 meters each second. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Find the slope of each line. slope = rise run a. = – 3 2 = –2 4 – 1 0 – 2 The slope of the line is – . 3 2 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) slope = rise run b. Find the slope of the line. = 2 = –2 –1 –1 – 1 –2 – (–2) The slope of the line is 2. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Find the slope of each line through E(3, –2) and F(–2, –1). slope = y2 – y1 x2 – x1 = – 1 5 = –5 Substitute (–2, –1) for (x2, y2) and (3, –2) for (x1, y1). Simplify. –1 – (–2) –2 – 3 The slope of EF is – . 1 5 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Find the slope of each line. a. slope = y2 – y1 x2 – x1 2 – 2 1 – (–4) = 5 Substitute (1, 2) for (x2, y2) and (–4, 2) for (x1, y1). Simplify. = 0 The slope of the horizontal line is 0. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) slope = y2 – y1 x2 – x1 b. Find the slope of the line. = 5 Substitute (2, 1) for (x2, y2) and (2, –4) for (x1, y1). 1 – (–4) 2 – 2 Simplify. Division by zero is undefined. The slope of the vertical line in undefined. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 10. 2 11. 2 12. 13. – 14. –1 15. –1 16. 17. 18. – 19. – 20. 21. –5 22. 0 23. undefined 24. 0 25. undefined 26. 0 27. in./month 28. $5/person 29. 30 mi/hr 30. 1 2 pages 286–289  Exercises 1. 3; the temperature increases 3°F each hour. 2. 3.95; the cost per person is $3.95. 3. – gal/mi 4. ; there are 2 lb of carbon emissions for 3 h of television use. 5. –16 ; the skydiver descends 16 ft/s. 6. ; the cost of oregano is $1 for 4 ounces. 7. 8. –3 9. 3 4 1 15 3 2 2 3 2 3 2 3 1 4 5 9 9 10 1 2 5 3 6 5 2 3 1 2 2 5 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 31. 32. 33. –20 34. undefined 35. –3 36. 37. 1 6 38. 39. 40. a. C b. C greatest; A least; the slope 41. a. b. c. Answers may vary. Sample: = 42. 2 ; 3 43. No; for example the line passing through the points such as (1, 6) and (2, 5) has a slope of –1. 44. PQ: ; QR: 0; RS: 5; SP: –1 45. JK: – ; KL: 2; ML: – ;MJ: 2 2 3 1 4 2 3 y2 – y1 x2 – x1 y1 – y2 x2 – x1 2 5 3 5 1 2 1 2 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 46. a. b. c. d. equal 47. a. Answers may vary. Sample: (0, 0), (1, ) b. Answers may vary. Sample: (0, 0), (1, – ) 48. 0 49. –6 50. 6 51. 4 52. 12 53. 3 54–60. Counter examples may vary. 54. False; it can be neg. or undefined. 55. true 56. False; y = x + 2 57. true 58. False; y = x 59. False; y = 0x 60. true 61. a.111; $111/h b. 52 customers per h 3 4 2 1 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 62. Friend found instead of . 63. 0 64. – 65. 66. Yes; AB and BC have the same slope. 67. Yes; GH and HI have 68. No; DE and EF do not have the same slope. 69. No; PQ and QR do not have the same slope. 70. Yes; GH and HI have the same slope. 71. No; ST and TU do not 72. C 73. G 74. [2] = ; the slope is . = 12.5%; the grade is 12.5%. [1] finds grade only 75. A 76. c = 3.5n 77. p = 4.95q – 232 78. 0 79. 80. 81. 1 82. –3 83. –4 84. 85. 7 86. 5 87. –10 run rise rise run 1 2 n 2m 1 3 2d – b c – 2a 3 24 1 8 1 8 19 9 1 8 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 1. Find the rate of change for the data in the table. 5 $75 6 7 8 $90 $105 $120 Tickets Cost 2. Find the rate of change for the data in the graph. 3. Find the slope of the line. 4. Find the slope of the line through (3, –2) and (–2, 5). 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 5. State whether the slope is zero or undefined. a. b. 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 1. Find the rate of change for the data in the table. 2. Find the rate of change for the data in the graph. 3. Find the slope of the line. 5 $75 6 7 8 $90 $105 $120 Tickets Cost $15 per ticket 4 3 400 calories per hour 7 5 – 4. Find the slope of the line through (3, –2) and (–2, 5). 6-1

Rate of Change and Slope ALGEBRA 1 LESSON 6-1 5. State whether the slope is zero or undefined. a. b. undefined 6-1

Evaluate each expression. 1. 6a + 3 for a = 2 2. –2x – 5 for x = 3 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 (For help, go to Lessons 1-6 and 2-6.) Evaluate each expression. 1. 6a + 3 for a = 2 2. –2x – 5 for x = 3 3. x + 2 for x = 16 4. 0.2x + 2 for x = 15 Solve each equation for y. 5. y – 5 = 4x 6. y + 2x = 7 7. 2y + 6 = –8x 1 4 6-2

Slope-Intercept Form Solutions ALGEBRA 1 LESSON 6-2 Solutions 1. 6a + 3 for a = 2: 6(2) + 3 = 12 + 3 = 15 2. –2x – 5 for x = 3: –2(3) – 5 = –6 – 5 = –11 3. x + 2 for x = 16: (16) + 2 = 4 + 2 = 6 4. 0.2x + 2 for x = 15: 0.2(15) + 2 = 3 + 2 = 5 5. y – 5 = 4x 6. y + 2x = 7 y – 5 + 5 = 4x + 5 y + 2x – 2x = 7 – 2x y = 4x + 5 y = –2x + 7 7. 2y + 6 = –8x 2y + 6 – 6 = –8x – 6 2y = –8x – 6 = y = –4x – 3 1 4 1 4 2y y –8x – 6 2 6-2

What are the slope and y-intercept of y = 2x – 3? Slope-Intercept Form ALGEBRA 1 LESSON 6-2 What are the slope and y-intercept of y = 2x – 3? mx + b = 2x – 3 Use the Slope-Intercept form. The slope is 2; the y-intercept is –3. 6-2

Write an equation of the line with slope and y-intercept 4. Slope-Intercept Form ALGEBRA 1 LESSON 6-2 Write an equation of the line with slope and y-intercept 4. 2 5 2 5 Use the slope-intercept form. Substitute for m and 4 for b. y = mx + b y = x + 4 6-2

Write an equation for the line. Slope-Intercept Form ALGEBRA 1 LESSON 6-2 Write an equation for the line. Step 1 Find the slope. Two points on the line are (0, 1) and (3, –1). –1 – 1 3 – 0 slope = = – 2 3 2 3 y = mx + b y = – x + 1 Substitute – for m and 1 for b. Step 2 Write an equation in slope– intercept form. The y-intercept is 1. 6-2

The slope is . Use slope to plot a second point. Step 3 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 Graph y = x – 2. 1 3 Step 1 The y-intercept is –2. So plot a point at (0,–2). Step 2 The slope is . Use slope to plot a second point. 1 3 Step 3 Draw a line through the two points. 6-2

The base pay for a used car salesperson is $300 per week. Slope-Intercept Form ALGEBRA 1 LESSON 6-2 The base pay for a used car salesperson is $300 per week. The salesperson also earns 15% commission on sales made. The equation t = 300 + 0.15s relates total earnings t to sales s. Graph the equations. Step 1  Identify the slope and y-intercept. t = 300 + 0.15s t = 0.15s + 300  Rewrite the equation in slope intercept form. slope y-intercept 6-2

Step 3 Draw a line through the points. Slope-Intercept Form ALGEBRA 1 LESSON 6-2 (continued) Step 2  Plot two points. First plot a point at the y-intercept. Then use the slope to plot a second point. The slope is 0.15, which equals . Plot a second point 15 units above and 100 units to the right of the y-intercept. 15 100 Step 3  Draw a line through the points. 6-2

Slope-Intercept Form pages 294–296 Exercises 1. –2; 1 2. – ; 2 3. 1; – ALGEBRA 1 LESSON 6-2 pages 294–296  Exercises 1. –2; 1 2. – ; 2 3. 1; – 4. 5; 8 5. ; 1 6. –4; 0 7. –1; –7 8. –0.7; –9 9. – ; –5 10. y = x + 3 11. y = 3x + 12. y = x + 3 13. y = 1 14. y = –x – 6 15. y = – x + 5 16. y = 0.3x + 4 17. y = 0.4x + 0.6 18. y = –7x + 19. y = – x – 20. y = – x + 21. y = x + 1 2 5 4 3 9 22. y = – x + 1 23. y = x + 2 24. y = 2x – 2 25. y = x + 26. y = – x + 2.8 27. y = x – 28. y = x + 4 8 6-2

Slope-Intercept Form 29. y = x – 1 30. y = –5x + 2 31. y = 2x + 5 32. ALGEBRA 1 LESSON 6-2 29. y = x – 1 30. y = –5x + 2 31. y = 2x + 5 32. y = x + 4 33. y = –x + 2 34. y = 4x – 3 35. y = – x 36. y = x – 3 37. y = – x + 2 3 2 38. y = – x + 4 39. y = –0.5x + 2 40. a. t = 14c – 4 b. $80 4 5 6-2

Slope-Intercept Form 51. y = –2x 52. y = – x – 2 53. y = 5x – 6 ALGEBRA 1 LESSON 6-2 51. y = –2x 52. y = – x – 2 53. y = 5x – 6 41. –3; 2 42. – ; 0 43. 9; 44. 3; –9 45. ; 3 46. 9; –15 47. c; d 48. 2 – a; a 49. –3; –2n 50. y = 7 – 3x 54. y = – x – 55.   y = 6x – 8 56. The slope was used for the y-int., and the y-int. was used for the slope. 2 3 1 6-2

Slope-Intercept Form 68. A; slope in A = > = 2 = slope in B. ALGEBRA 1 LESSON 6-2 68. A; slope in A = > = 2 = slope in B. 69. y = 2x – 1 70. y = –4x + 7 71. y = – x + 8 72. y = x + 5 73. y = –x – 3 74. y = 3x – 6 75. a–b. c. Both; check students’ work. 57. a. b. h = – t + 12 c. 90 min 58. a. Slope represents the weight of a gallon of fuel. b. 2662 lb 59. no 60. yes 61. no 62. III 63. I 64. II 8 4 10 4.5 65. a. d = 7p b. 84 dog years 66. a. t = 5d + 15 b. t = 5d + 15 c. Answers may vary. Sample: no neg. charges and no neg. number of days 67. Answers may vary. Sample: Plot point (0, 5), then move up 3 and right 4. Plot (4, 8) and connect the two points. 1 2 15 6-2

Slope-Intercept Form 76. a. ; b. 2; –2 c. same slopes ALGEBRA 1 LESSON 6-2 1 4 1 4 76. a. ; b. 2; –2 c. same slopes 77. Check students’ work. 78. – 79. –5 80. 81. a. x = –2; x = 2 y = 3; y = –3 b. Rectangle; check students’ work. c. y = x OR y = – x; explanations may vary. 82. a. r = –15d + 265 b. r = –15d + 265 c. 18 days 83. D 84. G 85. A 86. [2] slope: = = 2 y-intercept: 3 equation: y = 2x + 3 [1] correct graph and minor error in finding function 87. – 88. – 89. 90. –1 91. 4.8 billion 3 2 3 2 5 – 3 1 – 0 1 2 3 4 9 7 5 6 9 2 6-2

Find the slope and y-intercept of each equation. Slope-Intercept Form ALGEBRA 1 LESSON 6-2 Find the slope and y-intercept of each equation. 1. y = x – 3 2. y = –2x Write an equation of a line with the given slope and y-intercept. 3. m = – , b = –4 4. m = 0.5, b = 1 3 4 5 5. Write an equation for the line. m = ; b = –3 3 4 y = x – 3 1 2 m = –2; b = 0 y = – x – 4 3 5 6. Graph y = –x + 3. y = 0.5x + 1 6-2

Solve each equation for y. 1. 3x + y = 5 2. y – 2x = 10 3. x – y = 6 Standard Form ALGEBRA 1 LESSON 6-3 (For help, go to Lessons 2-3 and 2-6.) Solve each equation for y. 1. 3x + y = 5 2. y – 2x = 10 3. x – y = 6 4. 20x + 4y = 8 5. 9y + 3x = 1 6. 5y – 2x = 4 Clear each equation of decimals. 7. 6.25x + 8.5 = 7.75 8. 0.4 = 0.2x – 5 9. 0.9 – 0.222x = 1 6-3

Standard Form Solutions 1. 3x + y = 5 2. y – 2x = 10 ALGEBRA 1 LESSON 6-3 Solutions 1. 3x + y = 5 2. y – 2x = 10 3x – 3x + y = 5 – 3x y – 2x + 2x = 10 + 2x y = –3x + 5 y = 2x + 10 3. x – y = 6 4. 20x + 4y = 8 x = 6 + y 4y = –20x + 8 x – 6 = y y = y = x – 6 y = –5x + 2 5. 9y + 3x = 1 6. 5y – 2x = 4 9y = –3x + 1 5y = 2x + 4 y = y = y = – x + y = x + 7. Multiply each term by 100: 100(6.25x)+100(8.5) = 100(7.75) Simplify: 625x + 850 = 775 8. Multiply each term by 10: 10(0.4) = 10(0.2x) – 10(5) 4 = 2x – 50 9. Multiply each term by 1000: 1000(0.9)–1000(0.222x)=1000(1) 900 – 222x = 1000 –20x + 8 4 –3x+ 1 9 2x+ 4 5 1 3 1 9 2 5 4 5 6-3

Find the x- and y-intercepts of 2x + 5y = 6. Standard Form ALGEBRA 1 LESSON 6-3 Find the x- and y-intercepts of 2x + 5y = 6. Step 1  To find the x-intercept, substitute 0 for y and solve for x. 2x + 5y = 6 2x + 5(0) = 6 2x = 6 x = 3 The x-intercept is 3. Step 2  To find the y-intercept, substitute 0 for x and solve for y. 2x + 5y = 6 2(0) + 5y = 6 5y = 6 y = The y-intercept is . 6 5 6-3

Graph 3x + 5y = 15 using intercepts. Standard Form ALGEBRA 1 LESSON 6-3 Graph 3x + 5y = 15 using intercepts. Step 1  Find the intercepts. 3x + 5y = 15 3x + 5(0) = 15 Substitute 0 for y. 3x = 15 Solve for x. x = 5 3(0) + 5y = 15 Substitute 0 for x. 5y = 15 Solve for y. y = 3 Step 2  Plot (5, 0) and (0, 3). Draw a line through the points. 6-3

0 • x + 1 • y = 4 Write in standard form. For all values of x, y = 4. ALGEBRA 1 LESSON 6-3 a. Graph y = 4 b. Graph x = –3. 0 • x + 1 • y = 4 Write in standard form. For all values of x, y = 4. 1 • x + 0 • y = –3 Write in standard form. For all values of y, x = –3. 6-3

Write y = x + 6 in standard form using integers. ALGEBRA 1 LESSON 6-3 Write y = x + 6 in standard form using integers. 2 3 y = x + 6 2 3 3y = 3( x + 6 ) Multiply each side by 3. 3y = 2x + 18 Use the Distributive Property. –2x + 3y = 18 Subtract 2x from each side. The equation in standard form is –2x + 3y = 18. 6-3

Define: Let = the hours mowing lawns. Standard Form ALGEBRA 1 LESSON 6-3 Write an equation in standard form to find the number of hours you would need to work at each job to make a total of $130. Define:  Let = the hours mowing lawns. Let = the hours delivering newspapers. x y Amount Paid per hour Job Mowing lawns $12 Delivering newspapers Relate: $12 per h   plus $5 per h equals   $130 mowing delivering $5 Write:  12 + 5 = 130 x y The equation standard form is 12x + 5y = 130. 6-3

Standard Form pages 301–303 Exercises 1. 18; 9 2. 3; –9 3. –6; 30 ALGEBRA 1 LESSON 6-3 pages 301–303  Exercises 1. 18; 9 2. 3; –9 3. –6; 30 4. ; –3 5. – ; – 6. –8; 12 7. 6; 4 8. ; –2 9. –5; 4 10. B 11. C 12. A 13. 14. 15. 16. 17. 18. 19. horizontal 20. vertical 21. horizontal 22. vertical 23. 24. 3 2 9 2 3 2 4 7 6-3

Standard Form 25. 26. 27. –3x + y = 1 28. 4x – y = 7 29. x – 2y = 6 ALGEBRA 1 LESSON 6-3 25. 26. 27. –3x + y = 1 28. 4x – y = 7 29. x – 2y = 6 30. –2x + 3y = 15 31. –3x – 4y = 16 32. –4x – 5y = 35 33. –14x + 4y = 1 34. 4x + 10y = 1 35. 3x + y = 0 36. a. Answers may vary. Sample: x = no. of cars; y = no. of vans or trucks b. 5x + 6.5y = 800 37. a. Answers may vary. Sample: x = time walking; y = time running b. 3x + 8y = 15 38. 39. 40. 41. 42. 6-3

Standard Form 47. a. 3x + 7y = 28 43. 51. y = – x – 3 b. 7 oz 44. ALGEBRA 1 LESSON 6-3 47. a. 3x + 7y = 28 b. 7 oz 48. 4.29x + 3.99y = 30 49. 50. 4 5 43. 44. 45. 51. y = – x – 3 52. y = x – 5 9 3 y = x + 10 4 5 46. y = x + 3 6 7 6-3

Sample: slope-intercept form when comparing the Standard Form ALGEBRA 1 LESSON 6-3 53. y = – x – 8 54. y = x – 16 11 1 9 2 3 55. Answers may vary. Sample: slope-intercept form when comparing the steepness of two lines; standard form when making quick graphs 56. Answers may vary. Sample: 0x + 0y = 0, no linear equation exists. 57. –3x instead of 3x 58. y = 2 59. y = –2 60. x = 1 61. x = –2 62. a. 200s + 150a = 1200 b. Answers may vary. Sample: s = $2.00, a = $5.33; s = $3.00, a = $4.00; s = $6.00, a = $0.00 s = $3.00 and a = $4.00 because they are whole dollar amounts, and adults pay more. 63. y = x + 4 5 6-3

three of the following: 1 Supreme, 35 Prestige; [3] correct equation Standard Form ALGEBRA 1 LESSON 6-3 64.   square 65. a. b. – ; – c. Answers may vary. Sample: The x- and y-intercepts of 2x + 3y = 18 are 3 times those of 2x + 3y = 6. 66. C 67. H 68. [2] y = x + 1; x – 4y = –4 [1] correct slope- intercept equation 69. [4] a. 48x + 56y = 2008 b. c. Answers will include three of the following: 1 Supreme, 35 Prestige; 8 Supreme, 29 Prestige; 15 Supreme, 23 Prestige; 22 Supreme, 17 Prestige; 29 Supreme, 11 Prestige; 36 Supreme, 5 Prestige [3] correct equation and graph [2] wrong equation, but a corresponding correct graph [1] Correct equation only 70. yes 71. no 72. no 73. 74. 75. 4 76. 2.25 77. –0.5 78. –21 1 4 1 36 1 12 2 3 2 3 6-3

Find each x- and y-intercepts of each equation. Standard Form ALGEBRA 1 LESSON 6-3 Find each x- and y-intercepts of each equation. 1. 3x + y = 12 2. –4x – 3y = 9 3. Graph 2x – y = 6 using x- and y-intercepts. For each equation, tell whether its graph is horizontal or vertical. 4. y = 3 5. x = –8 6. Write y = x – 3 in standard form using integers. x = – , y = –3 9 4 x = 4, y = 12 horizontal vertical 5 2 –5x + 2y = –6 or 5x – 2y = 6 6-3

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (For help, go to Lessons 6-1 and 1-7.) Find the rate of change of the data in each table. 1. 2. 3. Simplify each expression. 4. –3(x – 5) 5. 5(x + 2) 6. – (x – 6) 4 9 2 5 8 11 –2 –8 –14 x y –3 –5 –1 1 3 –4 10 7.5 2.5 –6 –11 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1. Use points (2, 4) and (5, –2). rate of change = = = –2 2. Use points (–3, –5) and (–1, –4). rate of change = = = = 3. Use points (10, 4) and (7.5, –1). rate of change = = = = 2 4. –3(x – 5) = –3x – (–3)(5) = –3x + 15 5. 5(x + 2) = 5x + 5(2) = 5x + 10 6. – (x – 6) = – x – (– )(6) = – x + 4 9 8 3 –5 + 4 –3 + 1 4 – (–2) 2 – 5 4 – (–1 ) 10 – 7.5 –5 – (–4) –3 – (–1) 4 + 1 2.5 5 –1 –2 6 –3 1 2 Solutions 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1 3 Graph the equation y – 2 = (x – 1). 1 3 The equation of a line that passes through (1, 2) with slope . Start at (1, 2). Using the slope, go up 1 unit and right 3 units to (4, 3). Draw a line through the two points. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Write the equation of the line with slope –2 that passes through the point (3, –3). y – y1 = m(x – x1) Substitute (3, –3) for (x1, y1) and –2 for m. y – (–3) = –2(x – 3) Simplify the grouping symbols. y + 3 = –2(x – 3) 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Write equations for the line in point-slope form and in slope-intercept form. The slope is – . 1 3 Step 1  Find the slope. = m y2 – y1 x2 – x1 4 – 3 –1 – 2 = – 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) Step 2  Use either point to write the the equation in point-slope form. Use (–1, 4). y – y1 = m(x – x1) y – 4 = – (x + 1) 1 3 Step 3  Rewrite the equation from Step 2 in slope– intercept form. y – 4 = – (x + 1) y – 4 = – x – y = – x + 3 1 3 2 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. –2 –1 –6 –3 = 2 3 6 2 –1 –3 4 –2 –6 x y –1( ) –2 –3( ) –6 –2( ) –4 The relationship is linear. The rate of change is 2. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) Step 2 Use the slope 2 and a point (2,4) to write an equation. y – y1 = m(x – x1) Use the point-slope form. y – 4 = 2(x – 2) Substitute (2, 4) for (x1, y1) and 2 for m. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. 1 2 = 1 – / –2 –1 1 2 x y 1 ( ) 1 2 ( ) 1 The relationship is not linear. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 pages 307–309  Exercises 1. 2. 3. 7. 8. 9. 4. 5. 6. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 10. y + 4 = 6(x – 3) 11. y – 2 = – (x – 4) 12. y – 2 = (x) 13. y + 7 = – (x + 2) 14. y = 1(x – 4) 15. y + 8 = –3(x – 5) 16. y – 2 = 0(x + 5) or y = 2 17. y + 8 = – (x – 1) 18. y – 1 = (x + 6) 19–30. Answers may vary for the point indicated by the equation. 19. y = 1(x + 1); y = x + 1 20. y – 5 = (x – 3); y = x 21. y + 2 = – (x – 4); y = – x + 22. y + 4 = –1(x – 6); y = –x + 2 23. y + 5 = (x + 1); y = x – 24. y + 4 = (x + 3); y = x – 3 25. y – 7 = 11(x – 2); y = 11x – 15 26. y – 6 = – (x + 2); y = – x + 4 27. y + 8 = – (x – 3); y = – x – 28. y – = (x – 1); 29. y – 2 = –1(x – ); y = –x + 30. y – 1.1 = (x – 0.2); y = x + 5 3 4 2 1 6 14 29 7 13 1.9 6.8 7.1 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 3 2 46. y – 2 = – (x – 2); 5x + 3y = 16 47. y – 1 = – (x + 7); x + 6y = –1 48. y – 4 = – (x + 8); 3x + 2y = –16 49. y – 4 = 2(x – 2); –2x + y = 0 50. y – 3 = –2(x – 5); 2x + y = 13 51. y – 1 = x; –x + 3y = 3 52. y – 4 = – (x + 2); 9x + 2y = –10 5 3 31. Yes; answers may vary. Sample: y – 9 = –2(x + 4) 32. Yes; answers may vary. Sample: y – 40 = 3(x – 5) 33. no 34. Yes; answers may vary. Sample: y – 75 = 10(x – 10) 35. no 36–53. Answers may vary for point indicated by the equation. 36. y – 2 = (x – 1) 37. y + 3 = (x – 1) 38. y = – (x – 5) 39. y – 4 = (x – 1); –3x + 2y = 5 40. y + 3 = 0(x – 6); y = –3 41. y + 2 = 2(x + 1); –2x + y = 0 42. y – 2 = 0(x – 0); y = 2 43. y – 6 = – (x + 6); x + 3y = 12 44. y – 3 = – (x – 2); 2x + 3y = 13 45. y + 3 = – (x – 5); 7x + 2y = 29 1 6 3 2 1 3 3 4 2 3 1 3 2 5 7 2 9 2 5 7 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 53. y – 2 = (x – 6); –3x + 5y = –8 54. a. y = – x + 1 b. about 4 atmospheres 55. y = –2.6x + 315.6 56. a. Answers may vary. Sample: y + 6 = 2(x + 4); chose slope and substituted into y – y1 = m(x – x1) b. infinite; infinite number of slopes 57. y-intercept changes 58. Yes; the point satisfies the equation. 59. Answers may vary. Sample: a. y = x + 1 b. –x + y = 1 c. y – 1 = 1(x – 0) 60. a. Answers may vary. Sample: y – 332 = (x – 0) b. 341 m/s c. 368 m/s 61. y = 7x + 16 62. y = 3 63. y = x – 64. a. 14.75 b. 57.5 c. –4 d. 100 3 5 1 33 2 14 65. 66. 9 67. 68. 4 69. 6 70. 71. 7 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 72. 73. 74. 75. 76. 5; 3; 8 77. ; ; 78. 0.07; 2.66; 2.73 79. –0.05; –3.35; –3.4 80. 17; 69; 86 81. 4; 5; 9 2 6 3 11 16 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1. Graph the equation y + 1 = –(x – 3). 2. Write an equation of the line with slope – that passes through the point (0, 4). 3. Write an equation for the line that passes through (3, –5) and (–2, 1) in Point-Slope form and Slope-Intercept form. 4. Is the relationship shown by the data linear? If so, model that data with an equation. 2 3 y – 4 = – (x – 0), or y = – x + 4 2 3 –10 –7 5 20 –3 –1 x y 6 5 7 y + 5 = – (x – 3); y = – x – 2 5 yes; y + 3 = (x – 0) 6-4

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (For help, go to Lessons 1-6 and 6-2.) What is the reciprocal of each fraction? 1. 2. 3. – 4. – What are the slope and y-intercept of each equation? 5. y = x + 4 6. y = x – 8 7. y = 6x 8. y = 6x – 2 1 2 4 3 2 5 7 5 5 3 5 3 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 1. The reciprocal of is or 2. 2. The reciprocal of is . 3. The reciprocal of – is – . 4. The reciprocal of – is – . 5. y = x + 4 6. y = x – 8 slope: slope: y-intercept: 4 y-intercept: –8 7. y = 6x 8. y = 6x – 2 slope: 6 slope: 6 y-intercept: 0 y-intercept: – 2 1 2 4 3 5 7 Solutions 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Are the graphs of y = –2x – 1 and 4x + 2y = 6 parallel? Write 4x + 2y = 6 in slope-intercept form. Then compare with y = –2x – 1. 2y = –4x + 6 Subtract 4x from each side. y = –2x + 3 Simplify. Divide each side by 2. 2y –4x + 6 2 2 = The lines are parallel. The equations have the same slope, –2, and different y-intercepts. 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Write an equation for the line that contains (–2, 3) and is parallel to y = x – 4. 5 2 Step 1 Identify the slope of the given line. y = x – 4 slope 5 2 Step 2  Write the equation of the line through (–2, 3) using slope-intercept form. y – y1 = m(x – x1) Use point-slope form. y – 3 = (x + 2) Substitute (–2, 3) for (x1, y1) and for m. 5 2 y – 3 = x + (2) Use the Distributive Property. 5 2 y – 3 = x + 5 Simplify. 5 2 y = x + 8 Add 3 to each side and simplify. 5 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Write an equation of the line that contains (6, 2) and is perpendicular to y = –2x + 7. Step 1 Identify the slope of the given line. y = – 2 x + 7 slope Step 2  Find the negative reciprocal of the slope. The negative reciprocal of –2 is . 1 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (continued) Step 3  Use the slope-intercept form to write an equation. y = mx + b 2 = (6) + b Substitute for m, 6 for x, and 2 for y. 1 2 1 2 2 = 3 + b Simplify. 2 – 3 = 3 + b – 3 Subtract 3 from each side. –1 = b Simplify. The equation is y = x – 1. 1 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 The line in the graph represents the street in front of a new house. The point is the front door. The sidewalk from the front door will be perpendicular to the street. Write an equation representing the sidewalk. Step 1  Find the slope m of the street. m = = = = – Points (0, 2) and (3, 0) are on the street. y2 – y1 x2 – x1 0 – 2 3 – 0 –2 3 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (continued) Step 2 Find the negative reciprocal of the slope. The negative reciprocal of – is . So the slope of the sidewalk is . The y-intercept is –3. 2 3 The equation for the sidewalk is y = x – 3. 3 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 pages 314–317  Exercises 1. 2. – 3. 1 4. 0 5. – 6. 7 7. no, different slopes 8. yes, same slopes and different y-intercepts 9. yes, same slopes and 10. no, different slopes 11. yes, same slopes and 12. yes, same slopes and 13. y = 6x 14. y = –3x + 9 15. y = –2x – 1 16. y = – x – 20 17. y = 0.5x – 9 18. y = – x + 19. – 20. 21. – 22. 5 1 2 3 4 23. 24. undefined 25. y = – x 26. y = –x + 10 27. y = 3x – 10 28. y = x + 29. y = – x + 24 30. y = – x + 2 31. y = x + 1 32. perpendicular 33. parallel 34. perpendicular 7 5 11 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 35. neither 36. parallel 37. perpendicular 38. parallel 39. neither 40. perpendicular 41. neither 42. Answers may vary. Sample: same x and y coefficients 43. y = – x – ; y = – x + 44. y = x + y = –3x + 7 45. y = – x y = 2x 46. y = x + ; y = x – 47. y = 4; y = 2 48. y = x; y = –x + 1 49. about 50. Answers may vary. Sample: same slope of – 51. Answers may vary. Sample: • – = –1 4 5 19 3 1 52. a. The screen is not square. b. The lines appear perpendicular. 53. Answers may vary. Sample: y = 4x + 1 54. No; the slopes are not equal. 55. No; the slopes are not neg. reciprocals. 56. yes; same slopes and different y-intercepts 57. False; the product of two positive numbers can’t be –1. 58. True; y = x + 2 and y = x + 3 are parallel. 2 21 / 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 59. False; all direct variations go through the point (0, 0). If they have the same slope, they are the same line, not parallel lines. 60. The slopes of AD and BC are both undefined, so they are parallel. The slopes of AB and CD are both so they are parallel. The quadrilateral is a parallelogram. 2 5 61. The slope of JK is . The slope of KL is –2. The slope of LM is . The slope of JM is –4. The quadrilateral is not a parallelogram. 62. The slopes of PQ and RS are both – .The slopes of QR and SP are both – .The quadrilateral is 1 6 3 63. The slopes of AB and CD are both .The slopes of BC and AD are both – . The product is –1, so the quadrilateral is a rectangle. 64. The slopes of KL and MN are both – .The slopes of LM and KN are both 5. The product is not –1, so the quadrilateral is not a rectangle. 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 The diagonal AC has a slope of . The diagonals are perpendicular. ABCD is a rhombus. 67. RP has a slope of . RQ has a slope of – . RP is the neg. reciprocal of RQ, so PQR is a right triangle. 68. parallel 69. perpendicular 65. The slopes of PQ and RS are both . The slopes of PS and QR are both –2. The product is –1, so the quadrilateral is a rectangle. 66. BC and AD both have a slope of zero. BC and AD are parallel. AB and CD both have a slope of . AB and CD are parallel. The diagonal BD has a slope of –2. 1 2 4 3 70. y = – x – ; y = x + 7 71. y = x – 3; y = –2x + 7 72. –1.5; 24 73. D 74. F 75. C 8 17 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 76. [2] Find slope of 2x + y = 3: y = –2x + 3, therefore slope is –2. Find x: = –2, = –2, –4 = –2 + 2x, x = –1 (OR equivalent explanation) [1] correct value of x but no work OR minor computation error in work 77. C 78. A 79. B 80. y = 3x + 4 81. y = –4x – 8 82. y + 3 = (x – 5) 83. y + 9 = – (x + 1) 84. y – 4 = – (x + 6) 85. y – 11 = (x – 7) 86. 7; 11; 15 87. –9; –17; –25 88. yes 2 – 6 1 – x – 4 89. yes 90. no 91. yes 3 4 2 5 1 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 3 2 1. Find the slope of a line parallel to 3x – 2y = 1. 2. Find the slope of a line perpendicular to 4x + 5y = 7. Tell whether the lines for each pair of equations are parallel, perpendicular, or neither. 3. y = 3x – 1, y = – x + 2 4. y = 2x + 5, 2x + y = –4 5. y = – x – 1, 2x + 3y = 6 6. Write an equation of the line that contains (2, 1) and is perpendicular to y = – x + 3. 5 4 1 3 perpendicular neither 2 3 parallel 1 2 y = 2x – 3 6-5

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 (For help, go to Lessons 1-9.) Use the data in each table to draw a scatter plot. 1 2 3 4 –3 8 9 5 –25 y x 1. 21 15 12 7 2. 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 Solutions 1. Graph the points:   (1, 2) (2, –3) (3, 8) (4, 9) (5, –25) 2. Graph the points:   (1, 21) (2, 15) (3, 12) (4, 9) (5, 7) 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 Make a scatter plot to represent the data. Draw a trend line and write an equation for the trend line. Use the equation to predict the time needed to travel 32 miles on a bicycle. Speed on a Bicycle Trip Step 1  Draw a scatter plot. Use a       straight edge to draw a trend line. Estimate two points on the line. Miles Time (min) 5 27 10 46 14 71 18 78 22 107 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 (continued) Step 2  Write an equation of the trend line. m = = = 5 y2 – y1 x2 – x1 120 – 27 25 – 5 93 20 y – y1 = m(x – x1)   Use point-slope for m. y – 27 = 5(x – 5) Substitute 5 for m and (5, 27) for (x1, y1). Step 3  Predict the time needed to travel 32 miles. y – 27 = 5(32 – 5) Substitute 32 for x. y – 27 = 5(27) Simplify within the parentheses. y – 27 =135 Multiply. y = 162 Add 27 to each side. The time needed to travel 32 miles is about 162 minutes. 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 1996 5086.6 1997 1998 1999 4930.0 4619.3 4266.8 Year No. of Crimes 1995 5275.9 [Source: Crime in the United States, 1999, FBI, Uniform Crime Reports] Use a graphing calculator to find the equation of the line of best fit for the data below. What is the correlation coefficient? U.S. Crime Rate (per 100,000 inhabitants) 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 (continued) Step 1  Use the EDIT feature of the screen on your graphing calculator. Let 95 correspond to 1995. Enter the data for years and then enter data for crimes. Step 2  Use the CALC feature in the screen. Find the equation for the line of best fit. The equation for the line of best fit is y = –248.55x + 28,945.07 for values a and b rounded to the nearest hundredth. The value of the correlation coefficient is –0.9854908654. 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 pages 320–324  Exercises 1–6. Trend lines may vary. Samples given. 1. y – 52.5 = 2(x – 91) 2. y – 100 =15.71(x – 5) 3. y – 16.4 = 0.64(x – 69.9) 4. y – 85 = – 15.25(x – 1) 5. y – 5= 0.66(x – 240) 6. y = 1.6x – 80; 40 in. 7. y = –1.06x + 92.31; –0.9701709306 8. y = 10.60x – 772.66; 0.990733298 9. y = –2.29x + 613.93; –0.8108238756 10. y = 2.64x + 70.51; 0.9900170523 11. y = 1.35x – 31.42; 0.999808967 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 12. a. b. Answers may vary. y = 3.25x – 1 c. Answers may vary. Sample: The slope is the approximate ratio of the circumference to the diameter. d. 14 cm 13. a. b. Answers may vary. Sample: y = 0.939x + 13,800 c. 143,800,000 d. Answers may vary. Sample: No, the year is too far in the future. 14. a. Check students’ work. b. 1 15. Answers may vary. Sample: pos. slope; as temp. increases, more students are absent. 16. a. y = 0.61x + 35.31 Sample: small set of data with weak correlation 17. y = 0.37x – 28.66; $12.04 billion 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 18. a. y = –16.70x + 297.57 b. –0.6681863355 c. No, the correlation coefficient is not close to 1 or –1, so the equation does not closely model the data. 19. a. (2, 3) and (6, 6); y = 0.75x + 1.5 b. y = 0.75x + 1.21 20. a. y = 4.82x – 29.65 b. 404 ft c. The speed is much faster than those speeds used to find the equation of a trend line. 21. B 22. H 23. [4] a. b–c. Answers may vary. Samples: b. Let 1960 = 60. Two points on line are (62, 18) and (90, 30). y = 0.429x – 8.6 c. y = 0.429(105) – 8.6; 36.4 million people 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 [3] appropriate methods but one computational error [2] incorrect points used correctly OR correct points used incorrectly; function written appropriately, given previous results. [1] correct function, without work shown 24. y + 3 = 5(x – 2) 25. y – 5 = –x 26. y – 4 = – (x + 1) 27. y + 4 = – (x – 3) 28. y + 1 = –2(x + 2) 29. y – 2 = (x + 1) 30. x > 1 31. x < 5 32. x > –1 33. x –5 34. x > 35. x < 3 2 3 1 4 < – 6-6

Scatter Plots and Equations of Lines ALGEBRA 1 LESSON 6-6 1. Graph the data in a scatter plot. Draw a trend line. 2. Write an equation for the trend line. 3. Predict the number of households in the U.S. in 2005. (Use 105 for x.) 4. Use a calculator to find the line of best fit for the data. 5. What is the correlation coefficient? Number of Households in the U.S. [Source: U.S. Census Bureau, Current Population Reports. From Statistical Abstract of the United States, 2000] 1980 80.8 1985 1990 1995 86.8 93.3 99.0 Year Households (millions) 1975 71.1 y – 86.8 = 1.3(x – 85) about 112.8 million households y = 1.366x – 29.91 0.9943767027 6-6

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 (For help, go to Lessons 1-5 and 5-3.) Simplify each expression. 1. |2 – 7| 2. |7 – 12| 3. |38 – 56| 4. |–24 + 12| Model each rule using a table of values. 5. y = 6 – x 6. y = |x| + 1 7. y = |x + 1| 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 Solutions 1. |2 – 7| = |–5| = 5 2. |7 – 12| = |–5| = 5 3. |38 – 56| = |–18| = 18 4. |–24 + 12| = |–12| = 12 5. 6. 7. 6 1 2 3 5 4 x y –1 –2 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 Describe how the graphs of y = |x| and y = |x| + 1 are the same and how they are different. The graphs are the same shape. The y-intercept of the first graph is 0. The y-intercept of the second graph is 1. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 Graph y = |x| – 3 by translating y = |x|. Start with a graph of y = |x|. Translate the graph down 3 units. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 Write an equation for each translation of y = |x|. a. 9 units down b. 13 units up The equation is y = |x| – 9. The equation is y = |x| + 13. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 Graph each equation by translating y = |x|. a. y = |x + 2.5| b. y = |x – 2.5| Start with a graph of y = |x|. Start with a graph of y = |x|. Translate the graph left 2.5 units. Translate the graph right 2.5 units. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 b. 7 units right Write an equation for each translation of y = |x|. a. 10 units left The equation is y = |x + 10|. The equation is y = |x – 7|. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 pages 327–329  Exercises 1. Answers may vary. Sample: same shape, shifted 3 units up 2. Answers may vary. shifted 3 units down 3. Answers may vary. shifted 7 units down 4. 5. 6. 7. 8. 9. 10. y = |x| + 9 11. y = |x| – 6 12. y = |x| + 0.25 13. y = |x| + 14. y = |x| + 5.90 15. y = |x| – 1 5 2 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 16. 17. 18. 19. 20. 21. 22. y = |x + 9| 23. y = |x – 9| 24. y = |x – | 25. y = |x + | 26. y = |x + 0.5| 27. y = |x – 8.2| 5 2 3 28. 29. 30. 31. 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 32. y = –|x| + 2 33. y = –|x + 2.25| 34. y = –|x| – 35. y = –|x – 4| 36. B 37. 38. 39. 40. 41. a. b. (2, 3) c. Answers may vary. Sample: The x-coordinate is the horizontal translation, and the y-coordinate is the vertical translation. d. Use (a, b) for the vertex. Graph y = x and y = –x above the vertex. 42. a. b. 3 2 x y –1 2 0 0 1 2 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 c. d. 43. a. y = |x|; y = 1 b. –1 and 1 c. y = – x or y = – x + 2 44. B 45. H 46. A 47. G 48. [4] a. and b. c. Part (a) graph is shifted 8 units up to get part (b) graph. [3] one incorrect graph but correct answer to part (c) based on graphs drawn OR incorrect answer based on correct graphs [2] both graphs incorrect but correct answer to part (c) based on graphs drawn [1] both graphs incorrect and no answer given for part (c) 49. y = 5000x – 413,000 50. y = 4000x – 313,000 51. 12 5 2 6 52. 4 1 1 4 53. 2.5 14 –2.0 10.7 1 2 1 2 6-7

Graphing Absolute Value Equations ALGEBRA 1 LESSON 6-7 1. Describe how the graphs of y = |x| and y = |x| – 9 are the same and how they are different. 2. Graph y = |x| – 2 by translating y = |x|. 3. Write an equation for the translation of y = |x|, 1.5 units down. 4. Graph y = |x – 5|. 5. Write an equation for the translation of y = |x|, 7 units left. The graphs are the same shape. The y-intercept of the first graph is 0. The y-intercept of the second graph is –9. y = |x| – 1.5 y = |x + 7| 6-7

Linear Equations and Their Graphs ALGEBRA 1 CHAPTER 6 1. False; a rate of change could also be negative or 0. 2. False; a vertical line has an undefined rate of change. 3. –5 4. – 5. 6. 7. 8. 9. y = – x + 10. y = x + 6 11. y = – x + 25 12. y = x – 13. –8; –6 14. ; –4 15. –12; 6 16. 1; 1 17. y + 7 = (x + 2) 18. y + 8 = 3(x – 4) 19. y – 3 = – x 20. y = –5(x – 9) 21–24. Samples given. 21. y – 9 = (x – 4) 22. y = (x + 1) 23. y + 8 = 0 1 4 8 7 3 5 9 2 13 6-A

Linear Equations and Their Graphs ALGEBRA 1 CHAPTER 6 24. y – 7 = –2x 25. D 26. y = 5x – 11 27. y = 6 28. y = x – 2 29. y = 2 30. Answers may vary. Sample: y = 0.5x + 2 31. a. y = 5x – 30 b. intercepts: 6; –30 32. y = |x| – 2 33. y = |x – | 34. 35. 36. a. y = 0.0436x + 15.34 (for 1967 = 67) b. For sample in (a): 20,100 municipalities 37. a. y = –0.197x + 31.95 (for 1967 = 67) b. For sample in (a): 10,300 school districts 3 4 1 2 6-A