I removed the other version because of some updates :o)

1. 11-1 Graphing Linear Equations Course 3

2. Warm Up Solve each equation for y. 1. 6 y – 12 x = 24 2. – 2 y – 4 x = 20 3. 2 y – 5 x = 16 4. 3 y + 6 x = 18 y = 2 x + 4 y = – 2 x – 10 y = – 2 x + 6 Course 3 11-1 Graphing Linear Equations y = x + 8 5 2

3. Learn to identify and graph linear equations. Course 3 11-1 Graphing Linear Equations

4. WHAT IS A LINEAR EQUATION? A linear equation is an equation whose solutions fall on a line on the coordinate plane. All solutions of a particular linear equation fall on the line, and all the points on the line are solutions of the equation. Course 3 11-1 Graphing Linear Equations

5. HOW DO I KNOW IF AN EQUATION IS LINEAR? If an equation is linear, a constant change in the x -value corresponds to a constant change in the y-value. 3 3 3 2 2 2 The graph shows an example where each time the x -value increases by 3, the y -value increases by 2. Course 3 11-1 Graphing Linear Equations

6. Make a table, graph the equation, and tell whether it is linear. A. y = 3 x – 1 Graphing Equations – 7 3 (–2) – 1 3 (–1) – 1 3 (0) – 1 3 (1) – 1 3 (2) – 1 – 4 – 1 2 5 (–2, –7) (–1, –4) (0, –1) (1, 2) (2, 5) 2 1 0 – 1 – 2 ( x , y ) y 3 x – 1 x Course 3 11-1 Graphing Linear Equations

7. Each time x increases by 1 unit, y increases by 3 units. y = 3 x – 1 Course 3 11-1 Graphing Linear Equations (2, 5) (1, 2) (0, -1) (-1, -4) (-2, -7) ( x , y )

8. Make a table, graph the equation, and tell whether it is linear. B. y = x 3 Graphing Equations – 8 (–2) 3 (–1) 3 (0) 3 (1) 3 (2) 3 – 1 0 1 8 (–2, –8) (–1, –1) (0, 0) (1, 1) (2, 8) 2 1 0 – 1 – 2 ( x , y ) y x 3 x Course 3 11-1 Graphing Linear Equations

9. The equation y = x 3 is not a linear equation because its graph is not a straight line. Also notice that as x increases by a constant of 1 unit, the change in y is not constant. +7 +1 +1 +7 8 1 0 – 1 – 8 y 2 1 0 – 1 – 2 x Course 3 11-1 Graphing Linear Equations

10. Fill in the table below. Then, graph the equation and tell whether it is linear. A. y = 2 x + 1 Try This – 3 2 (–2) + 1 2 (–1) + 1 2 (0) + 1 2 (1) + 1 2 (2) + 1 – 1 1 3 5 (–3, –3) (–2, –1) (–1, 1) (0, 3) (2, 5) 2 1 0 – 1 – 2 ( x , y ) y 2 x + 1 x Course 3 11-1 Graphing Linear Equations

11. The equation y = 2 x + 1 is a linear equation. Each time x increase by 1 unit, y increases by 2 units. Course 3 11-1 Graphing Linear Equations

12. Complete the first two problems on page 85

14. More Graphing Equations Graph the equation and tell whether it is linear. C. y = – 3 x 4 Course 3 11-1 Graphing Linear Equations

15. Additional Example 1 Continued The equation y = – is a linear equation. 3 x 4 Course 3 11-1 Graphing Linear Equations

16. Graph the equation and tell whether it is linear. D. y = 2 More Graphing Equations For any value of x, y = 2. 2 2 2 2 2 2 2 2 2 2 (–2, 2) (–1, 2) (0, 2) (1, 2) (2, 2) 2 1 0 – 1 – 2 ( x , y ) y 2 x Course 3 11-1 Graphing Linear Equations

17. Additional Example 1D Continued The equation y = 2 is a linear equation because the points form a straight line. As the value of x increases, the value of y has a constant change of 0. Course 3 11-1 Graphing Linear Equations

18. Try This Graph the equation and tell whether it is linear. C. y = x – 8 – 6 0 4 8 (–8, –8) (–6, –6) (0, 0) (4, 4) (8, 8) 8 4 0 – 6 – 8 ( x , y ) y x Course 3 11-1 Graphing Linear Equations

19. Try This : Example 1C Continued The equation y = x is a linear equation because the points form a straight line. Each time the value of x increases by 1, the value of y increases by 1. Course 3 11-1 Graphing Linear Equations

20. Application In an amusement park ride, a car travels according to the equation D = 1250 t where t is time in minutes and D is the distance in feet the car travels. Graph the relationship between time and distance. How far has each person traveled? 3 min Colette 2 min Greg 1 min Ryan Time Rider Course 3 11-1 Graphing Linear Equations

21. Continued The distances are: Ryan, 1250 ft; Greg, 2500 ft; and Collette, 3750 ft. (3, 3750) 3750 1250( 3 ) 3 (2, 2500) 2500 1250( 2 ) 2 (1, 1250) 1250 1250( 1 ) 1 ( t , D) D D =1250 t t Course 3 11-1 Graphing Linear Equations

22. Continued x y This is a linear equation because when t increases by 1 unit, D increases by 1250 units. 1250 2500 1 2 3750 5000 3 4 Time (min) Distance (ft) Course 3 11-1 Graphing Linear Equations

23. Sports Application A lift on a ski slope rises according to the equation a = 130 t + 6250, where a is the altitude in feet and t is the number of minutes that a skier has been on the lift. Five friends are on the lift. What is the altitude of each person if they have been on the ski lift for the times listed in the table? Draw a graph that represents the relationship between the time on the lift and the altitude. Course 3 11-1 Graphing Linear Equations

24. Additional Example 2 Continued Course 3 11-1 Graphing Linear Equations

25. Additional Example 2 Continued Course 3 11-1 Graphing Linear Equations

26. The altitudes are: Anna, 6770 feet; Tracy, 6640 feet; Kwani, 6510 feet; Tony, 6445 feet; George, 6380 feet. This is a linear equation because when t increases by 1 unit, a increases by 130 units. Note that a skier with 0 time on the lift implies that the bottom of the lift is at an altitude of 6250 feet. Additional Example 2 Continued Course 3 11-1 Graphing Linear Equations

28. 11-2 Slope of a Line Course 3

29. Warm Up Evaluate each equation for x = –1, 0, and 1. 1. y = 3 x 2. y = x – 7 3. y = 2 x + 5 4. y = 6 x – 2 – 3, 0, 3 – 8, –7, –6 3, 5, 7 – 8, –2, 4 Course 3 11-2 Slope of a Line

30. Learn to find the slope of a line and use slope to understand and draw graphs. Course 3 11-2 Slope of a Line

31. What is slope? vertical change horizontal change change in y change in x = This ratio is often referred to as , or “rise over run,” where rise indicates the number of units moved up or down and run indicates the number of units moved to the left or right. Slope can be positive, negative, zero, or undefined. rise run Course 3 11-2 Slope of a Line

32. Course 3 11-2 Slope of a Line

33. Course 3 11-2 Slope of a Line

34. GET ON YOUR FEET! Course 3 11-2 Slope of a Line

35. Finding Slope from a Graph Choose two points on the line: (0, 1) and (3, –4). Guess by looking at the graph: – 5 3 rise run = – 5 3 = – 5 3 Course 3 11-2 Slope of a Line

36. How to find the slope of a line (without a visual)… If you have two points ( x 1 , y 1 ) & ( x 2 , y 2 ) Use the following formula: y 2 – y 1 x 2 – x 1 Course 3 11-2 Slope of a Line

37. Find the slope of the line that passes through (–2, –3) and (4, 6). Finding Slope, Given Two Points Let ( x 1 , y 1 ) be (–2, –3) and ( x 2 , y 2 ) be (4, 6). Substitute 6 for y 2 , –3 for y 1 , 4 for x 2 , and –2 for x 1 . 6 – (–3) 4 – (–2) 9 6 = The slope of the line that passes through (–2, –3) and (4, 6) is . 3 2 = y 2 – y 1 x 2 – x 1 3 2 = Course 3 11-2 Slope of a Line

38. Find the slope of the line that passes through (–4, –6) and (2, 3). Try This : Example 1 Let ( x 1 , y 1 ) be (–4, –6) and ( x 2 , y 2 ) be (2, 3). Substitute 3 for y 2 , –6 for y 1 , 2 for x 2 , and –4 for x 1 . 3 – (–6) 2 – (–4) 9 6 = The slope of the line that passes through (–4, –6) and (2, 3) is . 3 2 = y 2 – y 1 x 2 – x 1 3 2 = Course 3 11-2 Slope of a Line

39. Use the graph of the line to determine its slope. Try This : Example 2 Course 3 11-2 Slope of a Line

40. Try This : Example 2 Continued Choose two points on the line: (1, 1) and (0, –1). Guess by looking at the graph: Use the slope formula. Let (1, 1) be (x 1 , y 1 ) and (0, –1) be (x 2 , y 2 ). = 2 1 2 rise run = 2 1 = 2 = y 2 – y 1 x 2 – x 1 – 2 – 1 = – 1 – 1 0 – 1 Course 3 11-2 Slope of a Line

42. Class Assignment/Homework Workbook 11-2 Complete #1-6 Course 3 11-2 Slope of a Line

44. Something you need to remember: Parallel lines have the same slope . The slopes of two perpendicular lines are negative reciprocals of each other. Course 3 11-2 Slope of a Line

46. Parallel and Perpendicular Lines by Slope line 1: (–6, 4) and (2, –5) line 2: (–1, –4) and (8, 4) slope of line 1: slope of line 2: Line 1 has a slope equal to – and line 2 has a slope equal to , – and are negative reciprocals of each other, so the lines are perpendicular. 9 8 8 9 8 9 9 8 = y 2 – y 1 x 2 – x 1 – 9 8 = – 5 – 4 2 – (–6) 4 – (–4) 8 – (–1) = y 2 – y 1 x 2 – x 1 8 9 = 9 8 = – Course 3 11-2 Slope of a Line

47. Parallel and Perpendicular Lines by Slope line 1: (0, 5) and (6, –2) line 2: (–1, 3) and (5, –4) slope of line 1: slope of line 2: Both lines have a slope equal to – , so the lines are parallel. 7 6 = y 2 – y 1 x 2 – x 1 – 7 6 = – 2 – 5 6 – 0 = y 2 – y 1 x 2 – x 1 7 6 = – – 7 6 = 7 6 = – – 4 – 3 5 – (–1) Course 3 11-2 Slope of a Line

48. Try This line 1: (1, 1) and (2, 2) line 2: (1, –2) and (2, -1) Line 1 has a slope equal to 1 and line 2 has a slope equal to –1. 1 and –1 are negative reciprocals of each other, so the lines are perpendicular. slope of line 1: slope of line 2: = 1 = –1 = y 2 – y 1 x 2 – x 1 1 1 = 2 – 1 2 – 1 = y 2 – y 1 x 2 – x 1 – 1 1 = – 1 – (–2) 2 – (1) Course 3 11-2 Slope of a Line

49. Try This line 1: (–8, 2) and (0, –7) line 2: (–3, –6) and (6, 2) slope of line 1: slope of line 2: Line 1 has a slope equal to – and line 2 has a slope equal to , – and are negative reciprocals of each other, so the lines are perpendicular. 9 8 8 9 8 9 9 8 = y 2 – y 1 x 2 – x 1 – 9 8 = – 7 – 2 0 – (–8) 2 – (–6) 6 – (–3) = y 2 – y 1 x 2 – x 1 8 9 = 9 8 = – Course 3 11-2 Slope of a Line

50. Graphing a Line Using a Point and the Slope 2. Use the slope to count your units. Place another point once you get to the end. 3. Continue until you have enough point and then draw a line. 1. Plot the point given to you. Course 3 11-2 Slope of a Line

51. Graphing a Line Using a Point and the Slope Graph the line passing through (3, 1) with slope 2. Plot the point (3, 1). Then move 2 units up and right 1 unit and plot the point (4, 3). Use a straightedge to connect the two points. The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit. 2 1 Course 3 11-2 Slope of a Line

52. Additional Example 4 Continued 1 2 (3, 1) Course 3 11-2 Slope of a Line

53. Try This : Example 4 Graph the line passing through (1, 1) with slope 2. Plot the point (1, 1). Then move 2 units up and right 1 unit and plot the point (2, 3). Use a straightedge to connect the two points. The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit. 2 1 Course 3 11-2 Slope of a Line

54. Try This : Example 4 Continued 1 2 (1, 1) Course 3 11-2 Slope of a Line

56. Warm Up Find the slope of the line that passes through each pair of points. 1. (3, 6) and (-1, 4) 2. (1, 2) and (6, 1) 3. (4, 6) and (2, -1) 4. (-3, 0) and (-1, 1) Course 3 11-3 Using Slopes and Intercepts 1 2 - 1 5 7 2 1 2

57. 11-3 Using Slopes and Intercepts Course 3

58. Learn to use slopes and intercepts to graph linear equations. Course 3 11-3 Using Slopes and Intercepts

59. As you watch the video, take notes on your handout. Insert Lesson Title Here Course 3 11-3

60. One way to graph a linear equation easily is by finding the x -intercept and the y -intercept . The x -intercept is the value of x where the line crosses the x -axis ( y = 0). The y -intercept is the value of y where the line crosses the y -axis ( x = 0). Course 3 11-3 Using Intercepts

61. Find the x -intercept and y -intercept of the line 4 x – 3 y = 12. Use the intercepts to graph the equation. Example 1 Find the x -intercept ( y = 0). 4 x – 3 y = 12 4 x – 3 (0) = 12 4 x = 12 x = 3 The x -intercept is 3. 4 x 4 12 4 = Course 3 11-3 Using Intercepts

62. Example 1 Continued Find the y -intercept ( x = 0). 4 x – 3 y = 12 4 (0) – 3 y = 12 – 3 y = 12 y = –4 The y -intercept is –4. -3 y -3 12 -3 = Course 3 11-3 Using Intercepts

63. 4 x – 3 y = 12 Crosses the x -axis at the point (3, 0) Crosses the y -axis at the point (0, –4) Course 3 11-3 Using Intercepts

64. Find the x -intercept and y -intercept of the line 8 x – 6 y = 48. Use the intercepts to graph the equation. Try This Find the x -intercept ( y = 0). 8 x – 6 y = 48 8 x – 6 (0) = 48 8 x = 48 x = 6 The x -intercept is 6 so the point is (6, 0). 8 x 8 48 8 = Course 3 11-3 Using Intercepts

65. Try This Find the y -intercept ( x = 0). 8 x – 6 y = 48 8(0) – 6 y = 48 – 6 y = 48 y = –8 The y -intercept is –6 so the point is (0, -8). -6 y -6 48 -6 = Course 3 11-3 Using Intercepts

66. Try This : Example 1 Continued The graph of 8 x – 6 y = 48 is the line that crosses the x -axis at the point (6, 0) and the y -axis at the point (0, –8). Course 3 11-3 Using Slopes and Intercepts

67. In an equation written in slope-intercept form , y = mx + b , m is the slope and b is the y -intercept. y = m x + b Slope y -intercept Course 3 11-3 Using Slopes and Intercepts

68. Using the Slope-Intercept Form 1. Isolate the y so that you equation is in y = m x +b form. 2. Slope will always be “m” (the number in front of x ). 3. The y-intercept will always be “b” (the number by itself). Write this point as (0,b) For an equation such as y = x – 6, write it as y = x + (–6) to read the y -intercept, –6. The point would be (0,-6) Helpful Hint Course 3 11-3 Using Slopes and Intercepts

69. Example 2 Write each equation in slope-intercept form, and then find the slope and y -intercept. A. 2 x + y = 3 2 x + y = 3 – 2 x –2 x Subtract 2x from both sides. y = 3 – 2 x y = –2 x + 3 The equation is in slope-intercept form. m = –2 b = 3 The slope of the line is –2, and the y -intercept is 3. Course 3 11-3 Using Slopes and Intercepts

70. More Examples B. 5 y = 3 x 5 y = 3 x Divide both sides by 5 to solve for y. The equation is in slope-intercept form. b = 0 The slope of the line is , and the y -intercept is 0. = x 3 5 5 y 5 y = x + 0 3 5 m = 3 5 3 5 Course 3 11-3 Using Slopes and Intercepts

71. More Examples C. 4 x + 3 y = 9 4 x + 3 y = 9 Subtract 4x from both sides. b = 3 – 4 x –4 x 3 y = –4 x + 9 Divide both sides by 3. The equation is in slope-intercept form. y =- x + 3 4 3 m =- 4 3 The slope of the line 4 x+ 3 y = 9 is – , and the y -intercept is 3. 4 3 = + – 4 x 3 3 y 3 9 3 Course 3 11-3 Using Slopes and Intercepts

72. Try This Write each equation in slope-intercept form, and then find the slope and y-intercept. A. 4 x + y = 4 – 4 x –4 x Subtract 4x from both sides. y = 4 – 4 x Rewrite to match slope-intercept form. y = –4 x + 4 The equation is in slope-intercept form. m = –4 b = 4 The slope of the line 4 x + y = 4 is –4, and the y -intercept is 4. Course 3 11-3 Using Slopes and Intercepts

73. Try This B. 7y = 2x 7 y = 2 x Divide both sides by 7 to solve for y. The equation is in slope-intercept form. b = 0 = x 2 7 7 y 7 y = x + 0 2 7 m = 2 7 The slope of the line 7 y = 2 x is , and the y -intercept is 0. 2 7 Course 3 11-3 Using Slopes and Intercepts

74. Try This C. 5 x + 4 y = 8 5 x + 4 y = 8 Subtract 5x from both sides. Rewrite to match slope-intercept form. b = 2 – 5 x –5 x 4 y = 8 – 5 x 5 x + 4 y = 8 Divide both sides by 4. The equation is in slope-intercept form. y =- x + 2 5 4 The slope of the line 5 x + 4 y = 8 is – , and the y -intercept is 2. 5 4 = + – 5 x 4 4 y 4 8 4 m =- 5 4 Course 3 11-3 Using Slopes and Intercepts

75. Additional Example 3: Entertainment Application A video club charges $8 to join, and $1.25 for each DVD that is rented. The linear equation y = 1.25 x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y -intercept. y = 1.25 x + 8 The equation is in slope-intercept form. b = 8 m =1.25 Course 3 11-3 Using Slopes and Intercepts

76. Additional Example 3 Continued The slope of the line is 1.25, and the y -intercept is 8. The line crosses the y -axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right. Course 3 11-3 Using Slopes and Intercepts

77. Try This : Example 3 A salesperson receives a weekly salary of $500 plus a commission of 5% for each sale. Total weekly pay is given by the equation S = 0.05 c + 500. Graph the equation using the slope and y -intercept. y = 0.05 x + 500 The equation is in slope-intercept form. b = 500 m =0.05 Course 3 11-3 Using Slopes and Intercepts

78. Try This : Example 3 Continued The slope of the line is 0.05, and the y -intercept is 500. The line crosses the y -axis at the point (0, 500) and moves up 0.05 units for every 1 unit it moves to the right. x y 500 1000 1500 2000 10,000 5000 15,000 Course 3 11-3 Using Slopes and Intercepts

79. Additional Example 4: Writing Slope-Intercept Form Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form. Find the slope. The slope is –2. Choose either point and substitute it along with the slope into the slope-intercept form. y = mx + b 4 = –2 (–1) + b 4 = 2 + b Substitute –1 for x, 4 for y, and –2 for m. Simplify. = –2 4 – (–4) – 1 – 3 = y 2 – y 1 x 2 – x 1 8 – 4 = Course 3 11-3 Using Slopes and Intercepts

80. Additional Example 4 Continued Solve for b . Subtract 2 from both sides. Write the equation of the line, using –2 for m and 2 for b . 4 = 2 + b – 2 –2 2 = b y = –2 x + 2 Course 3 11-3 Using Slopes and Intercepts

81. Try This : Example 4 Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form. Find the slope. The slope is 4. Choose either point and substitute it along with the slope into the slope-intercept form. y = mx + b 2 = 4 (1) + b 2 = 4 + b Substitute 1 for x, 2 for y, and 4 for m. Simplify. = 4 6 – 2 2 – 1 = y 2 – y 1 x 2 – x 1 4 1 = Course 3 11-3 Using Slopes and Intercepts

82. Try This : Example 4 Continued Solve for b . Subtract 4 from both sides. Write the equation of the line, using 4 for m and –2 for b . 2 = 4 + b – 4 –4 – 2 = b y = 4 x – 2 Course 3 11-3 Using Slopes and Intercepts

83. Clear your desk except for a pencil and your calculator.

84. Warm Up Problem of the Day Lesson Presentation 11-4 Point-Slope Form Course 3

85. Warm Up Write the equation of the line that passes through each pair of points in slope-intercept form. 1. (0, –3) and (2, –3) 2. (5, –3) and (5, 1) 3. (–6, 0) and (0, –2) 4. (4, 6) and (–2, 0) y = –3 x = 5 y = x + 2 Course 3 11-4 Point-Slope Form y = – x – 2 1 3

86. Problem of the Day Without using equations for horizontal or vertical lines, write the equations of four lines that form a square. Possible answer: y = x + 2, y = x – 2, y = – x + 2, y = – x – 2 Course 3 11-4 Point-Slope Form

87. Learn to find the equation of a line given one point and the slope. Course 3 11-4 Point-Slope Form

88. Vocabulary point-slope form Insert Lesson Title Here Course 3 11-4 Point-Slope Form

89. Point on the line ( x 1 , y 1 ) Point-slope form y – y 1 = m ( x – x 1 ) slope The point-slope of an equation of a line with slope m passing through ( x 1 , y 1 ) is y – y 1 = m ( x – x 1 ). Course 3 11-4 Point-Slope Form

90. Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 7 = 3( x – 4) Additional Example 1: Using Point-Slope Form to Identify Information About a Line y – y 1 = m ( x – x 1 ) y – 7 = 3 ( x – 4 ) m = 3 ( x 1 , y 1 ) = (4, 7) The line defined by y – 7 = 3( x – 4) has slope 3, and passes through the point (4, 7). The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation. Course 3 11-4 Point-Slope Form

91. B. y – 1 = ( x + 6) Additional Example 1B: Using Point-Slope Form to Identify Information About a Line y – y 1 = m ( x – x 1 ) ( x 1 , y 1 ) = (–6, 1) Rewrite using subtraction instead of addition. 1 3 1 3 y – 1 = ( x + 6) y – 1 = [ x – ( –6 )] 1 3 m = 1 3 The line defined by y – 1 = ( x + 6) has slope , and passes through the point (–6, 1). 1 3 1 3 Course 3 11-4 Point-Slope Form

92. Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 5 = 2 ( x – 2) Try This : Example 1 y – y 1 = m ( x – x 1 ) y – 5 = 2 ( x – 2 ) m = 2 ( x 1 , y 1 ) = (2, 5) The line defined by y – 5 = 2( x – 2) has slope 2, and passes through the point (2, 5). The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation. Course 3 11-4 Point-Slope Form

93. B. y – 2 = (x + 3) Try This : Example 1B ( x 1 , y 1 ) = (–3, 2) Rewrite using subtraction instead of addition. y – y 1 = m ( x – x 1 ) 2 3 2 3 y – 2 = ( x + 3 ) y – 2 = [ x – (–3)] 2 3 m = 2 3 The line defined by y – 2 = ( x + 3) has slope , and passes through the point (–3, 2). 2 3 2 3 Course 3 11-4 Point-Slope Form

94. Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 4 passing through (5, -2) Additional Example 2: Writing the Point-Slope Form of an Equation y – y 1 = m (x – x 1 ) The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4( x – 5). Substitute 5 for x 1 , –2 for y 1 , and 4 for m. [ y – ( –2 )] = 4 ( x – 5 ) y + 2 = 4( x – 5) Course 3 11-4 Point-Slope Form

95. B. the line with slope –5 passing through (–3, 7) Additional Example 2: Writing the Point-Slope Form of an Equation y – y 1 = m ( x – x 1 ) The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5( x + 3). Substitute –3 for x 1 , 7 for y 1 , and –5 for m. y – 7 = -5[ x – ( –3 )] y – 7 = –5( x + 3) Course 3 11-4 Point-Slope Form

96. Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 2 passing through (2, –2) Try This : Example 2A y – y 1 = m (x – x 1 ) The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2( x – 2). Substitute 2 for x 1 , –2 for y 1 , and 2 for m. [ y – ( –2 )] = 2 ( x – 2 ) y + 2 = 2( x – 2) Course 3 11-4 Point-Slope Form

97. B. the line with slope -4 passing through (-2, 5) Try This : Example 2B y – y 1 = m ( x – x 1 ) The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4( x + 2). Substitute –2 for x 1 , 5 for y 1 , and –4 for m. y – 5 = –4[ x – ( –2 )] y – 5 = –4( x + 2) Course 3 11-4 Point-Slope Form

98. A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet. Additional Example 3: Entertainment Application As x increases by 30, y increases by 20, so the slope of the line is or . The line passes through the point (0, 18). 20 30 2 3 Course 3 11-4 Point-Slope Form

99. Additional Example 3 Continued y – y 1 = m ( x – x 1 ) y – 18 = 100 y = 118 The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward. Substitute 0 for x 1 , 18 for y 1 , and for m. 2 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x . Substitute 150 for x to find the value of y . 2 3 y – 18 = ( 150 ) 2 3 y – 18 = ( x – 0 ) 2 3 Course 3 11-4 Point-Slope Form

100. Try This: Example 3 A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet. As x increases by 45, y increases by 15, so the slope of the line is or . The line passes through the point (0, 15). 15 45 1 3 Course 3 11-4 Point-Slope Form

101. Try This : Example 3 Continued y – y 1 = m ( x – x 1 ) y – 15 = 100 y = 115 The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward. Substitute 0 for x 1 , 15 for y 1 , and for m. 1 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x . Substitute 300 for x to find the value of y . 1 3 y – 15 = ( 300 ) 1 3 y – 15 = ( x – 0 ) 1 3 Course 3 11-4 Point-Slope Form

102. Warm Up Problem of the Day Lesson Presentation 11-5 Direct Variation Course 3

103. Warm Up Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. 1. y – 3 = – ( x – 9) 2. y + 2 = ( x – 5) 3. y – 9 = –2( x + 4) 4. y – 5 = – ( x + 7) (–4, 9), –2 Course 3 11-5 Direct Variation 1 7 2 3 1 4 (9, 3), – 1 7 (5, –2), 2 3 (–7, 5), – 1 4

104. Problem of the Day Where do the lines defined by the equations y = –5 x + 20 and y = 5 x – 20 intersect? (4, 0) Course 3 11-5 Direct Variation

105. Learn to recognize direct variation by graphing tables of data and checking for constant ratios. Course 3 11-5 Direct Variation

106. Vocabulary direct variation constant of proportionality Insert Lesson Title Here Course 3 11-5 Direct Variation

107. Course 3 11-5 Direct Variation

108. Course 3 11-5 Direct Variation The graph of a direct-variation equation is always linear and always contains the point (0, 0). The variables x and y either increase together or decrease together. Helpful Hint

109. Determine whether the data set shows direct variation. A. Additional Example 1A: Determining Whether a Data Set Varies Directly Course 3 11-5 Direct Variation

110. Make a graph that shows the relationship between Adam’s age and his length. Additional Example 1A Continued Course 3 11-5 Direct Variation

111. You can also compare ratios to see if a direct variation occurs. 81 264 81 ≠ 264 The ratios are not proportional. The relationship of the data is not a direct variation. Additional Example 1A Continued 22 3 27 12 = ? Course 3 11-5 Direct Variation

112. Determine whether the data set shows direct variation. B. Additional Example 1B: Determining Whether a Data Set Varies Directly Course 3 11-5 Direct Variation

113. Make a graph that shows the relationship between the number of minutes and the distance the train travels. Additional Example 1B Continued Plot the points. The points lie in a straight line. (0, 0) is included. Course 3 11-5 Direct Variation

114. You can also compare ratios to see if a direct variation occurs. The ratios are proportional. The relationship is a direct variation. = = = Compare ratios. Additional Example 1B Continued 25 10 50 20 75 30 100 40 Course 3 11-5 Direct Variation

115. Determine whether the data set shows direct variation. A. Try This : Example 1A 0 3 5 Number of Baskets 40 30 20 Distance (ft) Kyle's Basketball Shots Course 3 11-5 Direct Variation

116. Make a graph that shows the relationship between number of baskets and distance. Try This : Example 1A Continued Number of Baskets Distance (ft) 2 3 4 20 30 40 1 5 Course 3 11-5 Direct Variation

117. You can also compare ratios to see if a direct variation occurs. Try This : Example 1A 60 150 150  60. The ratios are not proportional. The relationship of the data is not a direct variation. 5 20 3 30 = ? Course 3 11-5 Direct Variation

118. Determine whether the data set shows direct variation. B. Try This : Example 1B Cup (c) Ounces (oz) 4 3 2 1 32 24 16 8 Ounces in a Cup Course 3 11-5 Direct Variation

119. Make a graph that shows the relationship between ounces and cups. Try This : Example 1B Continued Plot the points. The points lie in a straight line. (0, 0) is included. Number of Cups Number of Ounces 2 3 4 8 16 24 1 32 Course 3 11-5 Direct Variation

120. You can also compare ratios to see if a direct variation occurs. Try This : Example 1B Continued The ratios are proportional. The relationship is a direct variation. Compare ratios. Course 3 11-5 Direct Variation = 1 8 = = 2 16 3 24 4 32

121. Find each equation of direct variation, given that y varies directly with x. A. y is 54 when x is 6 Additional Example 2A: Finding Equations of Direct Variation y = kx 54 = k  6 9 = k y = 9 x y varies directly with x. Substitute for x and y. Solve for k. Substitute 9 for k in the original equation. Course 3 11-5 Direct Variation

122. B. x is 12 when y is 15 Additional Example 2B: Finding Equations of Direct Variation y = kx 15 = k  12 y varies directly with x. Substitute for x and y. Solve for k. = k 5 4 Substitute for k in the original equation. 5 4 y = k 5 4 Course 3 11-5 Direct Variation

123. C. y is 8 when x is 5 Additional Example 2C: Finding Equations of Direct Variation y = kx 8 = k  5 y varies directly with x. Substitute for x and y. Solve for k. = k 8 5 Substitute for k in the original equation. 8 5 y = k 8 5 Course 3 11-5 Direct Variation

124. Find each equation of direct variation, given that y varies directly with x. A. y is 24 when x is 4 Try This : Example 2A y = kx 24 = k  4 6 = k y = 6 x y varies directly with x. Substitute for x and y. Solve for k. Substitute 6 for k in the original equation. Course 3 11-5 Direct Variation

125. B. x is 28 when y is 14 Try This : Example 2B y = kx 14 = k  28 y varies directly with x. Substitute for x and y. Solve for k. = k 1 2 Substitute for k in the original equation. 1 2 y = k 1 2 Course 3 11-5 Direct Variation

126. C. y is 7 when x is 3 Try This : Example 2C y = kx 7 = k  3 y varies directly with x. Substitute for x and y. Solve for k. = k 7 3 Substitute for k in the original equation. 7 3 y = k 7 3 Course 3 11-5 Direct Variation

127. Mrs. Perez has $4000 in a CD and $4000 in a money market account. The amount of interest she has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation. Additional Example 3: Money Application Course 3 11-5 Direct Variation

128. Additional Example 3 Continued A. interest from CD and time The second and third pairs of data result in a common ratio. In fact, all of the nonzero interest from CD to time ratios are equivalent to 17. The variables are related by a constant ratio of 17 to 1, and (0, 0) is included. The equation of direct variation is y = 17 x , where x is the time, y is the interest from the CD, and 17 is the constant of proportionality. interest from CD time = 17 1 interest from CD time = = 17 34 2 = = = 17 interest from CD time = = 17 1 34 2 51 3 68 4 Course 3 11-5 Direct Variation

129. Additional Example 3 Continued B. interest from money market and time 19 ≠ 18.5 If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. interest from money market time = = 19 19 1 interest from money market time = =18.5 37 2 Course 3 11-5 Direct Variation

130. Mr. Ortega has $2000 in a CD and $2000 in a money market account. The amount of interest he has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation. Try This : Example 3 Course 3 11-5 Direct Variation

131. Try This : Example 3 Continued 50 50 4 45 40 3 40 30 2 15 12 1 0 0 0 Money Market ($) from CD ($) Time (mo) Interest from Interest Course 3 11-5 Direct Variation

132. Try This : Example 3 Continued The second and third pairs of data do not result in a common ratio. If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. A. interest from CD and time interest from CD time = 12 1 interest from CD time = = 15 30 2 Course 3 11-5 Direct Variation

133. Try This: Example 3 Continued B. interest from money market and time 15 ≠ 20 If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. interest from money market time = = 15 15 1 interest from money market time = =20 40 2 Course 3 11-5 Direct Variation

134. Warm Up Problem of the Day Lesson Presentation 11-6 Graphing Inequalities in Two Variables Course 3

135. Warm Up Find each equation of direct variation, given that y varies directly with x . 1. y is 18 when x is 3. 2. x is 60 when y is 12. 3. y is 126 when x is 18. 4. x is 4 when y is 20. y = 6 x y = 7 x y = 5 x Course 3 11-6 Graphing Inequalities in Two Variables y = x 1 5

136. Problem of the Day The circumference of a pizza varies directly with its diameter. If you graph that direct variation, what will the slope be?  Course 3 11-6 Graphing Inequalities in Two Variables

137. Learn to graph inequalities on the coordinate plane. Course 3 11-6 Graphing Inequalities in Two Variables

138. Vocabulary boundary line linear inequality Insert Lesson Title Here Course 3 11-6 Graphing Inequalities in Two Variables

139. A graph of a linear equation separates the coordinate plane into three parts: the points on one side of the line, the points on the boundary line , and the points on the other side of the line. Course 3 11-6 Graphing Inequalities in Two Variables

140. Course 3 11-6 Graphing Inequalities in Two Variables

141. When the equality symbol is replaced in a linear equation by an inequality symbol, the statement is a linear inequality . Any ordered pair that makes the linear inequality true is a solution. Course 3 11-6 Graphing Inequalities in Two Variables

142. Graph each inequality. A. y < x – 1 Additional Example 1A: Graphing Inequalities First graph the boundary line y = x – 1. Since no points that are on the line are solutions of y < x – 1, make the line dashed . Then determine on which side of the line the solutions lie. (0, 0) y < x – 1 Test a point not on the line. Substitute 0 for x and 0 for y. 0 < 0 – 1 ? 0 < –1 ? Course 3 11-6 Graphing Inequalities in Two Variables

143. Additional Example 1A Continued Since 0 < –1 is not true, (0, 0) is not a solution of y < x – 1. Shade the side of the line that does not include (0, 0). Course 3 11-6 Graphing Inequalities in Two Variables

144. B. y  2 x + 1 Additional Example 1B: Graphing Inequalities First graph the boundary line y = 2 x + 1. Since points that are on the line are solutions of y  2 x + 1, make the line solid . Then shade the part of the coordinate plane in which the rest of the solutions of y  2 x + 1 lie. (0, 4) Choose any point not on the line. Substitute 0 for x and 4 for y. y ≥ 2 x + 1 4 ≥ 0 + 1 ? Course 3 11-6 Graphing Inequalities in Two Variables

145. Additional Example 1B Continued Since 4  1 is true, (0, 4) is a solution of y  2 x + 1. Shade the side of the line that includes (0, 4). Course 3 11-6 Graphing Inequalities in Two Variables

146. C. 2 y + 5 x < 6 Additional Example 1C: Graphing Inequalities First write the equation in slope-intercept form. 2 y < –5 x + 6 2 y + 5 x < 6 Subtract 5x from both sides. Divide both sides by 2. y < – x + 3 5 2 Then graph the line y = – x + 3. Since points that are on the line are not solutions of y < – x + 3, make the line dashed. Then determine on which side of the line the solutions lie. 5 2 5 2 Course 3 11-6 Graphing Inequalities in Two Variables

147. Additional Example 1C Continued (0, 0) Choose any point not on the line. Since 0 < 3 is true, (0, 0) is a solution of y < – x + 3. Shade the side of the line that includes (0, 0). 5 2 y < – x + 3 5 2 0 < 0 + 3 ? 0 < 3 ? Course 3 11-6 Graphing Inequalities in Two Variables

148. Graph each inequality. A. y < x – 4 Try This : Example 1A First graph the boundary line y = x – 4. Since no points that are on the line are solutions of y < x – 4, make the line dashed . Then determine on which side of the line the solutions lie. (0, 0) y < x – 4 Test a point not on the line. Substitute 0 for x and 0 for y. 0 < 0 – 4 ? 0 < –4 ? Course 3 11-6 Graphing Inequalities in Two Variables

149. Try This : Example 1A Continued Since 0 < –4 is not true, (0, 0) is not a solution of y < x – 4. Shade the side of the line that does not include (0, 0). Course 3 11-6 Graphing Inequalities in Two Variables

150. B. y > 4 x + 4 Try This : Example 1B First graph the boundary line y = 4 x + 4. Since points that are on the line are solutions of y  4 x + 4, make the line solid . Then shade the part of the coordinate plane in which the rest of the solutions of y  4 x + 4 lie. (2, 3) Choose any point not on the line. Substitute 2 for x and 3 for y. y ≥ 4 x + 4 3 ≥ 8 + 4 ? Course 3 11-6 Graphing Inequalities in Two Variables

151. Try This : Example 1B Continued Since 3  12 is not true, (2, 3) is not a solution of y  4 x + 4. Shade the side of the line that does not include (2, 3). Course 3 11-6 Graphing Inequalities in Two Variables

152. C. 3 y + 4 x  9 Try This : Example 1C First write the equation in slope-intercept form. 3 y  –4 x + 9 3 y + 4 x  9 Subtract 4x from both sides. Divide both sides by 3. y  – x + 3 4 3 4 3 Then graph the line y = – x + 3. Since points that are on the line are solutions of y  – x + 3, make the line solid. Then determine on which side of the line the solutions lie. 4 3 Course 3 11-6 Graphing Inequalities in Two Variables

153. Try This : Example 1C Continued (0, 0) Choose any point not on the line. Since 0  3 is not true, (0, 0) is not a solution of y  – x + 3. Shade the side of the line that does not include (0, 0). 4 3 y  – x + 3 4 3 0  0 + 3 ? 0  3 ? Course 3 11-6 Graphing Inequalities in Two Variables

154. A successful screenwriter can write no more than seven and a half pages of dialogue each day. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write a 200-page screenplay in 30 days? Additional Example 2: Career Application First find the equation of the line that corresponds to the inequality. In 0 days the writer writes 0 pages. point (0, 0) point (1, 7.5) In 1 day the writer writes no more than 7 pages. 1 2 Course 3 11-6 Graphing Inequalities in Two Variables

155. Additional Example 2 Continued With two known points, find the slope. y  7.5 x + 0 The y-intercept is 0. No more than means  . Graph the boundary line y = 7.5 x . Since points on the line are solutions of y  7.5 x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y  7.5 x lie. Course 3 11-6 Graphing Inequalities in Two Variables m = 7.5 – 0 1 – 0 7.5 1 = = 7.5

156. (2, 2) Choose any point not on the line. y  7.5 x Substitute 2 for x and 2 for y. Since 2  15 is true, (2, 2) is a solution of y  7.5 x . Shade the side of the line that includes point (2, 2). Additional Example 2 Continued 2  7.5  2 ? 2  15 ? Course 3 11-6 Graphing Inequalities in Two Variables

157. The point (30, 200) is included in the shaded area, so the writer should be able to complete the 200 page screenplay in 30 days. Additional Example 2 Continued Course 3 11-6 Graphing Inequalities in Two Variables

158. A certain author can write no more than 20 pages every 5 days. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write 140 pages in 20 days? Try This : Example 2 First find the equation of the line that corresponds to the inequality. In 0 days the writer writes 0 pages. point (0, 0) point (5, 20) In 5 days the writer writes no more than 20 pages. Course 3 11-6 Graphing Inequalities in Two Variables

159. Try This : Example 2 Continued With two known points, find the slope. y  4 x + 0 The y-intercept is 0. No more than means  . Graph the boundary line y = 4 x . Since points on the line are solutions of y  4 x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y  4 x lie. 20 - 0 5 - 0 m = = 20 5 = 4 Course 3 11-6 Graphing Inequalities in Two Variables

160. (5, 60) Choose any point not on the line . y  4 x Substitute 5 for x and 60 for y. Since 60  20 is not true, (5, 60) is not a solution of y  4 x . Shade the side of the line that does not include (5, 60). Try This : Example 2 Continued 60  4  5 ? 60  20 ? Course 3 11-6 Graphing Inequalities in Two Variables

161. The point (20, 140) is not included in the shaded area, so the writer will not be able to write 140 pages in 20 days. Try This : Example 2 Continued x y 200 180 160 140 120 100 80 60 40 20 Pages 5 10 15 20 25 30 35 40 45 50 Days Course 3 11-6 Graphing Inequalities in Two Variables

162. Warm Up Problem of the Day Lesson Presentation 11-7 Lines of Best Fit Course 3

163. Warm Up Answer the questions about the inequality 5 x + 10 y > 30. 1. Would you use a solid or dashed boundary line? 2. Would you shade above or below the boundary line? 3. What are the intercepts of the graph? dashed above (0, 3) and (6, 0) Course 3 11-7 Lines of Best Fit

164. Problem of the Day Write an inequality whose positive solutions form a triangular region with an area of 8 square units. ( Hint : Sketch such a region on a coordinate plane.) Possible answer: y < – x + 4 Course 3 11-7 Lines of Best Fit

165. Learn to recognize relationships in data and find the equation of a line of best fit. Course 3 11-7 Lines of Best Fit

167. Plot the data and find a line of best fit. Additional Example 1: Finding a Line of Best Fit Plot the data points and find the mean of the x - and y -coordinates. Course 3 11-7 Lines of Best Fit x m = = 6 4 + 7 + 3 + 8 + 8 + 6 6 y m = = 4 4 + 5 + 2 + 6 + 7 + 4 6 2 3 4 7 6 2 5 4 y 6 8 8 3 7 4 x 2 3 ( x m , y m )= 6 , 4

168. Course 3 11-7 Lines of Best Fit The line of best fit is the line that comes closest to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line. Remember!

169. Additional Example 1 Continued Course 3 11-7 Lines of Best Fit Draw a line through 6 , 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line. 2 3

170. Find the slope. y – y 1 = m ( x – x 1 ) Use point-slope form. Substitute. Additional Example 1 Continued Course 3 11-7 Lines of Best Fit y – 4 = ( x – 6 ) 2 3 2 3 y – 4 = x – 4 2 3 2 3 2 3 y = x + 2 3 The equation of a line of best fit is . 2 3 y = x + 2 3 2 3 1 3 m = = = 6 – 4 8 – 6 1 2 2 3

171. Plot the data and find a line of best fit. Try This : Example 1 Plot the data points and find the mean of the x - and y -coordinates. ( x m , y m ) = (2 , 1 ) Course 3 11-7 Lines of Best Fit x m = = 2 – 1 + 0 + 2 + 6 + –3 + 8 6 y m = = 1 – 1 + 0 + 3 + 7 + –7 + 4 6 4 – 7 7 3 0 – 1 y 8 – 3 6 2 0 – 1 x

172. Try This : Example 1 Continued Draw a line through ( 2 , 1 ) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line. Course 3 11-7 Lines of Best Fit

173. Find the slope. y – y 1 = m ( x – x 1 ) Use point-slope form. Substitute. Try This : Example 1 Continued Course 3 11-7 Lines of Best Fit y – 1 = ( x – 2 ) 9 8 y – 1 = x – 9 8 9 4 The equation of a line of best fit is . y = x – 9 8 5 4 m = = 10 – 1 10 – 2 9 8 y = x – 9 8 5 4

174. Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in 2006. Additional Example 2: Sports Application Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107). ( x m , y m ) = ( 5 , 103 ) Course 3 11-7 Lines of Best Fit x m = = 5 0 + 2 + 4 + 7 + 12 5 107 106 103 101 98 Distance (ft) 2002 1997 1994 1992 1990 Year y m = = 103 98 + 101 + 103 + 106 + 107 5

175. Additional Example 2 Continued Draw a line through ( 5 , 103 ) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line. Course 3 11-7 Lines of Best Fit

176. Find the slope. y – y 1 = m ( x – x 1 ) Use point-slope form. y – 103 = 0.8 ( x – 5 ) Substitute. y – 103 = 0.8 x – 4 y = 0.8 x + 99 The equation of a line of best fit is y = 0.8 x + 99. Since 1990 represents year 0, 2006 represents year 16. Additional Example 2 Continued Course 3 11-7 Lines of Best Fit m = = 0.8 107 - 103 10 - 5

177. Substitute. y = 12.8 + 99 y = 0.8 (16) + 99 The equation predicts a winning distance of about 112 feet for the year 2006. y = 111.8 Additional Example 2 Continued Course 3 11-7 Lines of Best Fit

178. Predict the winning weight lift in 2010. Try This : Example 2 Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170). ( x m , y m ) = ( 6 , 132 ) Course 3 11-7 Lines of Best Fit x m = = 6 0 + 5 + 7 + 8 + 10 5 y m = = 132 100 + 120 + 130 + 140 + 170 5 170 140 130 120 100 Lift (lb) 2000 1998 1997 1995 1990 Year

179. Try This : Example 2 Continued Draw a line through ( 5 , 132 ) the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line. Course 3 11-7 Lines of Best Fit Years since 1990 weight (lb) 0 100 120 140 160 180 2 4 6 8 10 200

180. Find the slope. y – y 1 = m ( x – x 1 ) Use point-slope form. y – 132 = 4 ( x – 5 ) Substitute. y – 132 = 4 x – 20 y = 4 x + 112 The equation of a line of best fit is y = 4 x + 112. Since 1990 represents year 0, 2010 represents year 20. Try This : Example 2 Continued Course 3 11-7 Lines of Best Fit m = = 4 140 – 132 7 – 5

181. Substitute. y = 192 y = 4 (20) + 112 The equation predicts a winning lift of about 192 lb for the year 2010. Try This : Example 2 Continued Course 3 11-7 Lines of Best Fit