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Rate of Change and Slope

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 = 2 x + 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.

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Rate of Change and Slope

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  1. 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

  2. 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 1 6-1

  3. 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 $42 200 $54 250 $66 Rate of Change and Slope ALGEBRA 1 LESSON 6-1 Find the rate of change for each pair of consecutive mileage amounts. 6-1

  4. change in costCost depends on the change in number of milesnumber of miles. rate of change = 42 – 3012 54 – 4212 66 – 5412 150–100 50 200–150 50 200–250 50 = = = Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) 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

  5. 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) Rate of Change and Slope ALGEBRA 1 LESSON 6-1 6-1

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

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

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

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

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

  11. y2 – y1 x2 – x1 slope = Substitute (2, 1) for (x2, y2) and (2, –4) for (x1, y1). 1 – (–4) 2 – 2 = 5 0 = Simplify. Rate of Change and Slope ALGEBRA 1 LESSON 6-1 (continued) Find the slope of the line. b. Division by zero is undefined. The slope of the vertical line in undefined. 6-1

  12. 9 10 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 23 5 9 1 4 1 2 5 3 6 5 2 3 1 2 2 5 6-1

  13. y1 – y2 x2 – x1 y2 – y1 x2 – x1 1 2 1 2 1 2 Rate of Change and Slope ALGEBRA 1 LESSON 6-1 31. 32. 33. –20 34. undefined 35. –3 36. 37. 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 1 6 2 3 2 3 1 4 2 5 3 5 6-1

  14. 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 2 3 2 3 3 4 1 2 Rate of Change and Slope ALGEBRA 1 LESSON 6-1 6-1

  15. rise run 1 2 1 3 1 8 3 24 1 8 19 9 1 8 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 the same slope. 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 have the same slope. 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 n 2m 2d – b c – 2a 6-1

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

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

  18. Tickets Cost 5 $75 6 $90 7 $105 8 $120 7 5 – 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. $15 per ticket 4 3 400 calories per hour 4. Find the slope of the line through (3, –2) and (–2, 5). 6-1

  19. 5. State whether the slope is zero or undefined. a. b. Rate of Change and Slope ALGEBRA 1 LESSON 6-1 undefined 0 6-1

  20. 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 = 4x6.y + 2x = 7 7. 2y + 6 = –8x 1 4 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 (For help, go to Lessons 1-6 and 2-6.) 6-2

  21. 1 4 1 4 2y y Slope-Intercept Form 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 = 4x6.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 –8x – 6 2 6-2

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

  23. 2 5 Use the slope-intercept form. Substitute for m and4 for b. y = mx + b 2 5 2 5 y = x + 4 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 Write an equation of the line with slope and y-intercept 4. 6-2

  24. 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 = – y = mx + b Step 2 Write an equation in slope– intercept form. The y-intercept is 1. 2 3 y = – x + 1 Substitute – for mand 1 for b. 2 3 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

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

  26. 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. slopey-intercept Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

  27. 15 100 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 theright of the y-intercept. 6-2

  28. 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 + 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 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 + 9 2 2 3 3 4 1 2 5 4 1 2 1 2 2 3 2 3 2 5 5 4 1 2 1 3 1 5 2 5 3 4 1 4 5 4 2 9 8 3 2 3 1 2 2 9 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

  29. 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 38. y = – x + 4 39. y = –0.5x + 2 40. a. t = 14c – 4 b. $80 2 3 4 5 3 2 2 5 2 3 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

  30. 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 51. y = –2x 52. y = – x – 2 53. y = 5x – 6 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. 1 2 1 2 2 3 1 3 3 2 2 3 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

  31. 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 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. 8 4 10 4.5 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. 1 2 2 15 1 4 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 6-2

  32. 5 – 3 1 – 0 9 7 5 6 9 2 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 1 4 1 4 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 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 3 2 3 2 1 2 3 4 6-2

  33. 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 1 2 3 4 y = x – 3 3 4 m = ; b = –3 3 5 3 5 y = – x – 4 Slope-Intercept Form ALGEBRA 1 LESSON 6-2 5. Write an equation for the line. m = –2; b = 0 6. Graph y = –x + 3. y = 0.5x + 1 6-2

  34. 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

  35. 2x+ 4 5 2 5 4 5 1 3 Standard Form 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 = yy = 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) Simplify: 4 = 2x – 50 9. Multiply each term by 1000: 1000(0.9)–1000(0.222x)=1000(1) Simplify: 900 – 222x = 1000 –20x + 8 4 –3x+ 1 9 1 9 6-3

  36. Find the x- and y-intercepts of 2x + 5y = 6. 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 5 Standard Form ALGEBRA 1 LESSON 6-3 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. 6-3

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

  38. Standard Form 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

  39. 2 3 3y = 3( x + 6 ) Multiply each side by 3. 2 3 y = x + 6 Standard Form ALGEBRA 1 LESSON 6-3 2 3 Write y = x + 6 in standard form using integers. 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

  40. 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. x Define:  Let = the hours mowing lawns. Let = the hours delivering newspapers. y x y Write:  12 + 5 = 130 Standard Form ALGEBRA 1 LESSON 6-3 Amount Paid per hour Job Mowing lawns $12 Delivering newspapers Relate: $12 per h   plus $5 per h equals   $130 mowing delivering $5 The equation standard form is 12x + 5y = 130. 6-3

  41. Standard Form 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

  42. Standard Form 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

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

  44. 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 16 11 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 53.y = – x – 8 54.y = x – 1 9 2 3 3 5 Standard Form ALGEBRA 1 LESSON 6-3 6-3

  45. 1 36 1 12 Standard Form ALGEBRA 1 LESSON 6-3 [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 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 1 4 2 3 2 3 6-3

  46. 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. 9 4 x = – , y = –3 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. Standard Form ALGEBRA 1 LESSON 6-3 x = 4, y = 12 horizontal vertical 5 2 –5x + 2y = –6 or 5x – 2y = 6 6-3

  47. 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) x y x y x y 2 4 –3 –5 10 4 5 –2 –1 –4 7.5 –1 8 –8 1 –3 5 –6 11 –14 3 –2 2.5 –11 4 9 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (For help, go to Lessons 6-1 and 1-7.) 6-4

  48. Solutions 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 – (–2) 2 – 5 6 –3 –5 – (–4) –3 – (–1) 1 2 –5 + 4 –3 + 1 –1 –2 4 – (–1 ) 10 – 7.5 4 + 1 2.5 5 2.5 4 9 4 9 4 9 4 9 8 3 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 6-4

  49. 1 3 The equation of a line that passes through (1, 2) with slope . 1 3 Start at (1, 2). Using the slope, go up1 unitand right 3 units to (4, 3). Draw a line through the two points. Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Graph the equation y – 2 = (x – 1). 6-4

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

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