1. Chapter 5 - Linear Functions Algebra I

2. Table of Contents 5.6 - Direct Variation 5.6 5.7- Slope – Intercept Form 5.7 5.8 – Point – Slope Form 5.8 5.9 – Slopes of Parallel and Perpendicular Lines 5.9 5.10 – Transforming Linear Functions (Skip 15-16) 5.10

3. 5.6 - Direct Variation Algebra I

4. A recipe for paella calls for 1 cup of rice to make 5 servings. In other words, a chef needs 1 cup of rice for every 5 servings. The equation y = 5x describes this relationship. In this relationship, the number of servings varies directly with the number of cups of rice. 5.6 A direct variation is a special type of linear relationship that can be written in the form y = kx, where k is a nonzero constant called the constant of variation. Just Read Top until Definitions (then copy) Algebra 1 (bell work)

5. Tell whether the equation represents a direct variation. If so, identify the constant of variation. y = 3x This equation represents a direct variation because it is in the form of y = kx. The constant of variation is 3. 5.6Example 1Identifying Direct Variations from Equations 3x + y = 8 –3x y = –3x + 8 This equation is not a direct variation because it cannot be written in the form y = kx. –4x + 3y = 0 +4x +4x 3y = 4x This equation represents a direct variation because it is in the form of y = kx. The constant of variation is

6. What happens if you solve y = kx for k? y = kx So, in a direct variation, the ratio is equal to the constant of variation. Another way to identify a direct variation is to check whether is the same for each ordered pair (except where x = 0). 5.6

7. Tell whether the relationship is a direct variation. Explain. Method 1 Write an equation. y = 3x This is direct variation because it can be written as y = kx, where k = 3. 5.6Example 2Identifying Direct Variations from Ordered Pairs Method 2 Find for each ordered pair. This is a direct variation because is the same for each ordered pair.

8. Method 1 Write an equation. y = x – 3 This is not a direct variation because it cannot be written as y = kx. Tell whether the relationship is a direct variation. Explain. 5.6

9. Math Joke Parent: How is your math homework going? Child: Well, m is negative, so at least I’m on the down slope 5.6

10. The value of y varies directly with x, and y = 3, when x = 9. Find y when x = 21. Method 1 Find the value of k and then write the equation. y = kx 3 = k(9) The equation is y = x. When x = 21, y = (21) = 7. 5.6Writing and Solving Direct Variation EquationsExample 3 Method 2 Use a proportion. 9y = 63 y = 7

11. The value of y varies directly with x, and y = 4.5 when x = 0.5. Find y when x = 10. Method 1 Find the value of k and then write the equation. y = kx 4.5 = k(0.5) The equation is y = 9x. When x = 10, y = 9(10) = 90. 9 = k 5.6 Method 2 Use a proportion. 0.5y = 45 y = 90

12. HW pg. 339 5.6- – 2-9 (9 No Graph), 36, 39, 47-55 (Odd) – Ch: 17-20, 37, 38, 40

13. 5.7 - Slope – Intercept Form Algebra I

14. Graph the line given the slope and y-intercept. y intercept = 4 5.7Example 1Graphing by Using Slope and y-intercept

15. Graph the line given the slope and y-intercept. slope = 4; y-intercept = 5.7

16. Graph the line given the slope and y-intercept. slope = 2, y-intercept = –3 5.7

17. slope = y-intercept = 1 Graph the line given the slope and y-intercept. 5.7

18. Write the equation that describes the line in slope-intercept form. slope = ; y-intercept = 4 y = mx + b y = x + 4 slope = ; y-intercept = –6 y = mx + b 5.7 Example 2Writing Linear Equations in Slope-Intercept Form

19. slope = –9; y-intercept = y = mx + b Write the equation that describes the line in slope-intercept form. y = –9x + slope = 3; y-intercept = y = mx + b 5.7

20. slope = 2; (3, 4) is on the line y = mx + b 4 = 2(3) + b Step 1 Find the y-intercept.–2 = b 4 = 6 + b –6 Write the equation that describes the line in slope-intercept form. Step 2 Write the equation. y = mx + b y = 2x + (–2) y = 2x – 2 5.7 A line has a slope of 8 and (3, –1) is on the line Step 1 Find the y-intercept. y = mx + b –1 = 8(3) + b –25 = b –1 = 24 + b –24 Step 2 Write the equation. y = mx + b y = 8x + (–25) y = 8x – 25

21. Math Joke Q: Why was the scientist smiling as he worked? A: His graph had a great plot! 5.7

22. Write the equation in slope -intercept form. Then graph the line described by the equation. 2y + 3x = 6 Step 1 : Solve for Y= 2y + 3x = 6 –3x –3x 2y = –3x + 6 5.7Example 3Using Slope-Intercept From to GraphDay 2

23. Write the equation in slope-intercept form. Then graph the line described by the equation. 5.7

24. 6x + 2y = 10 Step 1 – Solve for Y = 6x + 2y = 10 –6x –6x 2y = –6x + 10 Write the equation in slope-intercept form. Then graph the line described by the equation. 5.7

25. A closet organizer charges a $100 initial consultation fee plus $30 per hour. The cost as a function of the number of hours worked is graphed below. a. Write an equation that represents the cost as a function of the number of hours. Costis $30 for each hour plus $100 y = 30 x + 100 An equation is y = 30x + 100. 5.7Example 4Application

26. HW pg.348 5.7- (Due Friday b/c of quiz Thursday) – Day 1: 5-9, 14-22, 45-50 – Day 2: 10-12, 23-31, 35 – Ch: 36

27. 5.8 – Point-Slope Form Algebra I

28. 5.8Algebra 1 (bell work)

29. Write an equation in point-slope form for the line with the given slope that contains the given point. A.B. C. 5.8Example 1Writing Linear Equations in Point-Slope Form

30. Write an equation in point-slope form for the line with the given slope that contains the given point. a. b. slope = 0; (3, –4) y – (–4) = 0(x – 3) y + 4 = 0(x – 3) 5.8Optional

31. 5.8 A. B. Example 2Using Point-Slope Form to Graph

32. Write an equation in slope- intercept form for the line with slope 3 that contains (–1, 4). Step 1 Write the equation in point-slope form: y – 4 = 3[x – (–1)] Step 2 Write the equation in slope- intercept form by solving for y. y – 4 = 3(x + 1) y – 4 = 3x + 3 + 4 + 4 y = 3x + 7 y – y 1 = m(x – x 1 ) 5.8 Example 3 Writing Linear Equations in Slope-Intercept Form Write an equation in slope- intercept form for the line with slope that contains (–3, 1). Step 1 Write the equation in point-slope form: y – y 1 = m(x – x 1 ) Step 2 Write the equation in slope- intercept form by solving for y. +1

33. Math Joke Q: What did the horizontal line say to the vertical line? A: It was nice running into you 5.9

34. Write an equation in slope-intercept form for the line through the two points. (2, –3) and (4, 1) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the point-slope form. y – y 1 = m(x – x 1 ) y – (–3) = 2(x – 2) Step 3 Write the equation in slope-intercept form. y = 2x – 7 –3 y + 3 = 2(x – 2) y + 3 = 2x – 4 5.8Example 3Writing Linear Equations in Slope-Intercept Form

35. Write an equation in slope-intercept form for the line through the two points. (0, 1) and (–2, 9) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the point-slope form. y – y 1 = m(x – x 1 ) y – 1 = –4(x – 0) Step 3 Write the equation in slope-intercept form. y = –4x + 1 + 1 +1 y – 1 = –4(x – 0) y – 1 = –4x 5.8Optional

36. The cost to stain a deck is a linear function of the deck’s area. The cost to stain 100, 250, 400 square feet are shown in the table. Write an equation in slope-intercept form that represents the function. Then find the cost to stain a deck whose area is 75 square feet. Step 1 Choose any two ordered pairs from the table to find the slope. 5.8Skip 15_16Example 5Application

37. Step 2 Substitute the slope and any ordered pair from the table into the point- slope form. y – 150 = 1.25(x – 100) y – y 1 = m(x – x 1 ) Step 3 Write the equation in slope-intercept form by solving for y. y – 150 = 1.25(x – 100) y – 150 = 1.25x – 125 y = 1.25x + 25 5.8

38. Step 4 Find the cost to stain an area of 75 sq. ft. y = 1.25x + 25 y = 1.25(75) + 25 = 118.75 The cost of staining 75 sq. ft. is $118.75. 5.8

39. What if…? At a newspaper the costs to place an ad for one week are shown. Write an equation in slope-intercept form that represents this linear function. Then find the cost of an ad that is 21 lines long. Step 1 Choose any two ordered pairs from the table to find the slope. Step 2 Substitute the slope and any ordered pair from the table into the point- slope form. y – y 1 = m(x – x 1 ) y – 17.25 = 2.25(x – 5) 5.8

40. Step 3 Write the equation in slope-intercept form by solving for y. y – 17.25 = 2.25(x – 5) y – 17.25 = 2.25x – 11.25 y = 2.25x + 6 Step 4 Find the cost for an ad that is 21 lines long. y = 2.25x + 6 y = 2.25(21) + 6 = 53.25 The cost of the ad 21 lines long is $53.25. 5.8

41. HW pg. 355 5.8- – 1-15, 50, 59, 60 – Ch: 34, 35, 36, 40-43, 51, 53

42. 5.9 – Slopes of Parallel and Perpendicular Lines Algebra I

43. To sell at a particular farmers’ market for a year, there is a $100 membership fee. Then you pay $3 for each hour that you sell at the market. However, if you were a member the previous year, the membership fee is reduced to $50. The red line shows the total cost if you are a new member. The blue line shows the total cost if you are a returning member. 5.9Algebra 1 (bell work) Just Read These two lines are parallel. Parallel lines are lines in the same plane that have no points in common. In other words, they do not intersect.

44. 5.9

45. The lines described by and both have slope These lines are parallel. The lines described by y = x and y = x + 1 both have slope 1. These lines are parallel. 5.9Example 1Identifying Parallel Lines

46. Write all equations in slope-intercept form to determine the slope. y = 2x – 3 slope-intercept form slope-intercept form Identify which lines are parallel. 2x + 3y = 8 –2x – 2x 3y = –2x + 8 y + 1 = 3(x – 3) y + 1 = 3x – 9 – 1 y = 3x – 10 5.9

47. y = 2x – 3 y + 1 = 3(x – 3) 5.9Identify which lines are parallel.

48. Write all equations in slope-intercept form to determine the slope. Slope-intercept form y = 3x Slope-intercept form –3x + 4y = 32 +3x 4y = 3x + 32 y – 1 = 3(x + 2) y – 1 = 3x + 6 + 1 y = 3x + 7 5.9Optional

49. The lines described by –3x + 4y = 32 and y = + 8 have the same slope, but they are not parallel lines. They are the same line. –3x + 4y = 32 y – 1 = 3(x + 2) y = 3x The lines described by y = 3x and y – 1 = 3(x + 2) represent parallel lines. They each have slope 3. 5.9

50. Math Joke Q: What did the lines do at the gym to avoid each other? A: They used the parallel bars. 5.9

51. Show that JKLM is a parallelogram. Use the ordered pairs and the slope formula to find the slopes of MJ and KL. MJ is parallel to KL because they have the same slope. JK is parallel to ML because they are both horizontal. Since opposite sides are parallel, JKLM is a parallelogram. 5.9Skip (15_16)Example 2Geometry Application

52. Perpendicular lines are lines that intersect to form right angles (90 °). The slopes of perpendicular lines are opposite or negative reciprocals 5.9

53. Helpful Hint If you know the slope of a line, the slope of a perpendicular line will be the "opposite reciprocal.” 5.9

54. Identify which lines are perpendicular: y = 3; x = –2; y = 3x;. The graph described by y = 3 is a horizontal line, and the graph described by x = –2 is a vertical line. These lines are perpendicular. y = 3 x = –2 y =3x The slope of the line described by y = 3x is 3. The slope of the line described by is These lines are perpendicular because the product of their slopes is –1. 5.9Example 3Identifying Perpendicular Lines

55. Identify which lines are perpendicular: y = –4; y – 6 = 5(x + 4); x = 3; y = The graph described by x = 3 is a vertical line, and the graph described by y = –4 is a horizontal line. These lines are perpendicular. The slope of the line described by y – 6 = 5(x + 4) is 5. The slope of the line described by y = is y = –4 x = 3 y – 6 = 5(x + 4) These lines are perpendicular because the product of their slopes is –1. 5.9

56. Show that ABC is a right triangle. slope of Therefore, ABC is a right triangle because it contains a right angle. If ABC is a right triangle, AB will be perpendicular to AC. AB is perpendicular to AC because 5.9Skip (15_16)Example 4Geometry Application

57. Write an equation in slope-intercept form for the line that passes through (4, 10) and is parallel to the line described by y = 3x + 8. Step 1 Find the slope of the line. y = 3x + 8 The parallel line also has a slope of 3. Step 2 Write the equation in point-slope form. y – y 1 = m(x – x 1 ) y – 10 = 3(x – 4) Step 3 Write the equation in slope-intercept form. y – 10 = 3(x – 4) y – 10 = 3x – 12) y = 3x – 2 5.9Example 4 Writing Equations of Parallel and Perpendicular Lines Day 2

58. Write an equation in slope-intercept form for the line that passes through (2, –1) and is perpendicular to the line described by y = 2x – 5. Step 1 Find the slope of the line. y = 2x – 5 The perpendicular line has a slope of because Step 2 Write the equation in point-slope form. y – y 1 = m(x – x 1 ) Step 3 Write the equation in slope-intercept form. 5.9

59. Write an equation in slope-intercept form for the line that passes through (5, 7) and is parallel to the line described by y = x – 6. Step 1 Find the slope of the line. Step 2 Write the equation in point-slope form. y = x –6 The parallel line also has a slope of. y – y 1 = m(x – x 1 ) Step 3 Write the equation in slope-intercept form. 5.9

60. Write an equation in slope-intercept form for the line that passes through (–5, 3) and is perpendicular to the line described by y = 5x. Step 1 Find the slope of the line. y = 5x Step 2 Write the equation in point-slope form. The perpendicular line has a slope of because y – y 1 = m(x – x 1 ) Step 3 Write in slope-intercept form. 5.9

61. HW pg. 365 5.9- – Day 1: 1-3, 5, 6, 18-21, 48, 61, 63 – Day 2: 8, 17, 25, 27, 35, 37, 46, 47 – Ch: 50, 51

62. 5.10 - Transforming Linear Functions Algebra I

63. A family of functions is a set of functions whose graphs have basic characteristics in common. For example, all linear functions form a family because all of their graphs are the same basic shape. A parent function is the most basic function in a family. For linear functions, the parent function is f(x) = x. The graphs of all other linear functions are transformations of the graph of the parent function, f(x) = x. A transformation is a change in position or size of a figure. 5.10

64. The graphs of g(x) = x + 3, h(x) = x – 2, and k(x) = x – 4, are vertical translations of the graph of the parent function, f(x) = x. A translation is a type of transformation that moves every point the same distance in the same direction. You can think of a translation as a “slide.” 5.10

65. Graph f(x) = 2x and g(x) = 2x – 6. Then describe the transformation from the graph of f(x) to the graph of g(x). The graph of g(x) = 2x – 6 is the result of translating the graph of f(x) = 2x 6 units down. f(x) = 2x g(x) = 2x −6 f(x) = 2x g(x) = 2x – 6 5.10

66. Graph f(x) = x + 4 and g(x) = x – 2. Then describe the transformation from the graph of f(x) to the graph of g(x). The graph of g(x) = x – 2 is the result of translating the graph of f(x) = x + 4 6 units down. f(x) = x + 4 g(x) = x −2 f(x) = x + 4 g(x) = x – 2 5.10

67. Math Joke Q: Why did the student need directions during math class? A: He was lost in translation 5.10

68. The graphs of g(x) = 3x, h(x) = 5x, and k(x) = are rotations of the graph f(x) = x. A rotation is a transformation about a point. You can think of a rotation as a “turn.” The y-intercepts are the same, but the slopes are different. 5.10

69. Graph f(x) = x and g(x) = 5x. Then describe the transformation from the graph of f(x) to the graph of g(x). The graph of g(x) = 5x is the result of rotating the graph of f(x) = x about (0, 0). The graph of g(x) is steeper than the graph of f(x). g(x) = 5x f(x) = x g(x) = 5x f(x) = x 5.10

70. Graph f(x) = 3x – 1 and g(x) = x – 1. Then describe the transformation from the graph of f(x) to the graph of g(x). The graph of g(x) is the result of rotating the graph of f(x) about (0, –1). The graph of g(x) is less steep than the graph of f(x). f(x) = 3x – 1 g(x) = x – 1 f(x) = 3x – 1 g(x) = x – 1 5.10

71. The diagram shows the reflection of the graph of f(x) = 2x across the y-axis, producing the graph of g(x) = –2x. A reflection is a transformation across a line that produces a mirror image. You can think of a reflection as a “flip” over a line. 5.10

72. Graph f(x) = 2x + 2. Then reflect the graph of f(x) across the y-axis. Write a function g(x) to describe the new graph. f(x) = 2x + 2 To find g(x), multiply the value of m by –1. f(x) = 2x + 2, m = 2. 2(–1) = –2 g(x) = –2x + 2 f(x)f(x) g(x)g(x) g(x)g(x)f(x)f(x) 5.10

73. Graph. Then reflect the graph of f(x) across the y-axis. Write a function g(x) to describe the new graph. To find g(x), multiply the value of m by –1. f(x) = x + 2, m =. (–1) = – g(x) = – x + 2 g(x)g(x) f(x)f(x) f(x)f(x) g(x)g(x) 5.10

74. HW pg. 373 5.10- – 3-19 (Odd), 28, 35 - 37, 45 – B: 38, 48,