1. Unit 1

2. Warm-Up – X.X

3. Vocabulary – X.X Holder Holder 2 Holder 3 Holder 4

4. Notes – X.X – LESSON TITLE. Holder

5. Examples X.X

6. Unit 1 Section 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.6-4.7 Holder

7. Warm-Up – 4.1

8. ANSWER y = – x + 1 2 5 2 9. 6x + 4y = 16 Prerequisite Skills SKILLS CHECK Write the equation so that y is a function of x. y = – x + 4 3 2 ANSWER 10. x + 2y = 5 ANSWER y = 2x – 2 11. –12x + 6y = –12

9. 3. y = x + 6; when x = 0, 2, 4, 6, and 8 Prerequisite Skills SKILLS CHECK Graph the function on a coordinate plane and give the Input/ Output table. ANSWER InputOutput 06 28 410 612 814

10. Vocabulary – 4.1 Domain Set of INPUTS to a function Sometimes these are considered the X variables AKA the “independent” variable Range Set of OUTPUTS ofa function Sometimes these are considered the Yvariables AKA the “dependent” variable Quadrants 4 regions of a coordinate plane Function A numerical relationship where ONE input has EXACTLY ONE output

11. Notes – 4.1 – Plot Pts. and Graphs To plot points, move along the X axis first, and then the Y axis You have to run before you jump (or drop!). Domain = Inputs Range = Outputs Usually in Table format Quadrants labeled With Roman Numerals

12. Examples 4.1

13. GUIDED PRACTICE for Example 1 1. Use the coordinate plane in Example 1 to give the coordinates of points C, D, and E. C = (0,2) D = (3,1) E = (-2,-3) C.C. SOLUTION

14. GUIDED PRACTICE for Example 1 y -coordinate of any point on the x -axis is 0 2. What is the y -coordinate of any point on the x -axis ? SOLUTION

15. Make a table by substituting the domain values into the function. STEP 1 SOLUTION Graph the function y = 2x – 1 with domain – 2, – 1, 0, 1, and 2. Then identify the range of the function. EXAMPLE 3 Graph a function

16. STEP 2 List the ordered pairs: (– 2, – 5),(– 1, – 3), (0, – 1), (1, 1), (2, 3). Then graph the function. EXAMPLE 3 Graph a function Identify the range. The range consists of the y -values from the table: – 5, – 3, – 1, 1, and 3.

17. GUIDED PRACTICE for Examples 2 and 3 7. Graph the function y = – x + 2 with domain – 6, –3, 0, 3, and 6. Then identify the range of the function. 1 3 STEP 1 Make a table by substituting the domain values into the function.

18. GUIDED PRACTICE for Examples 2 and 3 y = – (6) + 2 = 06 y = – (3) + 2 = 13 y = – (0) + 2 = 20 y = – (– 3) + 2 = 3– 3 y = – (– 6) + 2 = 4– 6 y = – x + 2x 1 3 1 3 1 3 1 3 1 3 1 3

19. GUIDED PRACTICE for Examples 2 and 3 STEP 2 List the ordered pairs: (– 6, 4),(– 3, 3), (0, 2), (3, 1), (6,0). Then graph the function. STEP 3 Identify the range. The range consists of the y -values from the table: 0, 1, 2, 3 and 4.

20. Graph a function represented by a table EXAMPLE 4 VOTING In 1920 the ratification of the 19 th amendment to the United States Constitution gave women the right to vote. The table shows the number (to the nearest million) of votes cast in presidential elections both before and since women were able to vote.

21. Graph a function represented by a table EXAMPLE 4 Years before or since 1920 – 12– 8– 404812 Votes (millions) 15 1927293740 Explain how you know that the table represents a function. a.a. Graph the function represented by the table. b.b. Describe any trend in the number of votes cast. c.c.

22. Graph a function represented by a table EXAMPLE 4 SOLUTION The table represents a function because each input has exactly one output. a.a. To graph the function, let x be the number of years before or since 1920. Let y be the number of votes cast (in millions). b.b. The graph of the function is shown.

23. Graph a function represented by a table EXAMPLE 4 SOLUTION In the three election years before 1920, the number of votes cast was less than 20 million. In 1920, the number of votes cast was greater than 20 million. The number of votes cast continued to increase in the three election years since 1920. c.c.

24. Warm-Up – 4.2 You may use a calculator on every assignment from this point forward unless otherwise told not to!

25. Lesson 4.2, For use with pages 215-222 1. Graph y = –x – 2 with domain –2, –1, 0, 1, and 2. ANSWER

26. Lesson 4.2, For use with pages 215-222 2. 3x + 4y = 16 Rewrite the equation so y is a function of x. ANSWER 3 4 y = – x + 4 3. –6x – 2y = –12 ANSWER y = –3x + 6

27. Vocabulary – 4.2 Linear Equation –The graph of the solutions to the function form a STRAIGHT LINE!

28. Notes – 4.2 – Graph Linear Equations Standard Form of a Linear Equation looks like this: Ax + By = C, where A, B, and C are real numbers and A and B are not both = 0 There are several ways to sketch graphs of linear equations, but the most common is THIS 1.GET Y BY ITSELF!!! 2.Build a table with at least 3 values 3.Sketch the graph The graphs of y = constant and x = constant are special cases of linear equations. We’ll check those out in a minute!

29. Examples 4.2

30. Substitute 3 for x and 4 for y. Simplify. Write original equation. Check whether each ordered pair is a solution of the equation. SOLUTION Which ordered pair is a solution of 3x – y = 7? EXAMPLE 1 Standardized Test Practice (3, 4) A (1, –4) B (5, –3) C (–1, –2) D Test (3, 4) : 3(3) – 4 = ? 7 3x – y =3x – y =7 5 =7

31. Simplify. Write original equation. Standardized Test Practice EXAMPLE 1 Test (1, – 4): 3x – y =3x – y =7 3(1) – (– 4) = ? 7 Substitute 1 for x and – 4 for y. So, (3, 4) is not a solution, but (1, – 4) is a solution of 3x – y = 7. ANSWER The correct answer is B. AB DC 7 =7

32. Solve the equation for y. SOLUTION EXAMPLE 2 Graph an equation Graph the equation – 2x + y = – 3. – 2x + y = – 3 y = 2x – 3 STEP 1 Make a table by choosing a few values for x and finding the values of y. x– 2– 1012 y– 7– 5– 3– 11 STEP 2

33. Graph an equation EXAMPLE 2 Plot the points. Notice that the points appear to lie on a line. STEP 3

34. Graph (a) y = 2 and (b) x = – 1. Graph y = b and x = a EXAMPLE 3 y = 2 x = -1

35. GUIDED PRACTICE for Examples 2 and 3 Graph the equation 2. y + 3x = – 2 Solve the equation for y. SOLUTION y + 3x = – 2 y = – 3x – 2 STEP 1

36. GUIDED PRACTICE for Examples 2 and 3 Make a table by choosing a few values for x and finding the values of y. x– 2– 1012 y41– 2– 5– 8 Plot the points. Notice that the points appear to lie on a line. Connect the points by drawing a line through them. Use arrows to indicate that the graph goes on without end. STEP 3 STEP 4 STEP 2

37. GUIDED PRACTICE for Examples 2 and 3 3. y = 2.5 SOLUTION For every value of x, the value of y is 2.5. The graph of the equation y = 2.5 is a horizontal line 2.5 units above the x -axis. 4. x = – 4 SOLUTION For every value of y, the value of x is – 4. The graph of the equation x = – 4 is a vertical line 4 units to the left of the y -axis.

38. SOLUTION STEP 1 Make a table. x 02468 y43210 EXAMPLE 4 Graph a linear function 1 2 Graph the function y = –x + 4 with domain x > 0. Then identify the range of the function. –

39. STEP 2 STEP 3 Connect the points with a ray because the domain is restricted. STEP 4 Identify the range. From the graph, you can see that all points have a y- coordinate of 4 or less, so the range of the function is y ≤ 4. EXAMPLE 4 Graph a linear function Plot the points.

40. GUIDED PRACTICE for Example 4 5. Graph the function y = – 3x + 1 with domain x < 0. Then identify the range of the function. – SOLUTION STEP 1 Make a table. x 0– 1– 2– 3– 4 y1471013

41. GUIDED PRACTICE for Example 4 STEP 2 STEP 3 Connect the points with a ray because the domain is restricted. STEP 4 Identify the range. From the graph, you can see that all points have a y- coordinate of 1 or more, so the range of the function is y  1. – Plot the points.

42. Warm-Up – 4.3

43. Daily Homework Quiz For use after Lesson 4.2 1. Graph y + 2x = 4 ANSWER

44. Daily Homework Quiz For use after Lesson 4.2 2. The distance in miles an elephant walks in t hours is given by d = 5t. The elephant walks for 2.5 hours. Graph the function and identify its domain and range. domain : 0 < t < 2.5 range : 0 < d < 12.5 ANSWER – – – –

45. Vocabulary – 4.3 X-intercept Where a graph crosses the X axis Y-intercept Where a graph crosses the Y axis

46. Notes – 4.3 – Graph using Intercepts. The primary reason to use the Standard Form of a linear equation is b/c it does make finding the x and y intercepts VERY easy! To find the X-intercept of a function Set Y=0 and solve for X To find the Y-intercept of a function Set X=0 and solve for Y Since you only need two points to make a line Graph the X and Y intercepts and connect them!

47. Examples 4.3

48. Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept of the graph of 2x + 7y = 28. Find the intercepts of the graph of an equation EXAMPLE 1 Solve for x. 2x + 7(0) = 28 x == 14 28 2 2x + 7y = 28

49. 2(0) + 7y = 28 Find the intercepts of the graph of an equation EXAMPLE 1 To find the y- intercept, substitute 0 for x and solve for y. Write original equation. Substitute 0 for x. Solve for y. ANSWER The x- intercept is ( 14,0). The y- intercept is (0,4). 2x +7y = 28 y = 28 7 = 4

50. Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept of the graph of the equation. Solve for x. 3x + 2(0) = 6 x = 2 3x + 2y = 6 GUIDED PRACTICE for Example 1 1. 3x + 2y = 6

51. 3(0) + 2y = 6 Find the intercepts of the graph of an equation EXAMPLE 1 To find the y- intercept, substitute 0 for x and solve for y. Write original equation. Substitute 0 for x. Solve for y. ANSWER The x- intercept is ( 2,0). The y- intercept is (0, 3). 3x +2y = 6 y =3 GUIDED PRACTICE for Example 1

52. SOLUTION STEP 1 Use intercepts to graph an equation EXAMPLE 2 Graph the equation x + 2y = 4. x + 2y = 4 x =  x- intercept 4 Find the intercepts. x + 2(0) =4 0 + 2y = 4 y =  y- intercept 2 x + 2y = 4 X intercept = (4,0) Y intercept = (0,2)

53. Use intercepts to graph an equation EXAMPLE 2 STEP 2 Plot points. The x- intercept is 4, so plot the point (4, 0). The y- intercept is 2, so plot the point (0, 2). Draw a line through the points.

54. EVENT PLANNING Solve a multi-step problem EXAMPLE 4 You are helping to plan an awards banquet for your school, and you need to rent tables to seat 180 people. Tables come in two sizes. Small tables seat 4 people, and large tables seat 6 people. This situation can be modeled by the equation. 4x + 6y = 180 where x is the number of small tables and y is the number of large tables. Find the intercepts of the graph of the equation.

55. SOLUTION STEP 1 Solve a multi-step problem EXAMPLE 4 Give four possibilities for the number of each size table you could rent. Graph the equation. Find the intercepts. 4x + 6(0) = 180 x =  x -intercept 45 4x + 6y =180 4(0) + 6y = 180 y =  y -intercept 30 4x + 6y = 180

56. Solve a multi-step problem EXAMPLE 4 Since x and y both represent numbers of tables, neither x nor y can be negative. So, instead of drawing a line, draw the part of the line that is in Quadrant I. STEP 2 Graph the equation. The x- intercept is 45, so plot the point (45, 0). The y- intercept is 30, so plot the point (0, 30).

57. Solve a multi-step problem EXAMPLE 4 STEP 3 Find the number of tables. For this problem, only whole - number values of x and y make sense. You can see that the line passes through the points (0, 30),(15,20),(30, 10), and (45, 0).

58. Solve a multi-step problem EXAMPLE 4 So, four possible combinations of tables that will seat 180 people are : 0 small and 30 large, 15 small and 20 large, 30 small and 10 large,and 45 small and 0 large.

59. Warm-Up – 4.4 1) Do pages 10-13 from the “Classified” ads packet as a group. 2) You have ~15 minutes to work on this.

60. Vocabulary – 4.4 Rate of Change Ratio of How much something changed over how long did it take to change. Slope –The STEEPNESS of a line –Same thing as UNIT RATE!!!! –Same thing as Rate of Change!!!! –HOW FAST SOMETHING IS CHANGING!!! Rise –Vertical or UP/DOWN change –Change in Y’s Run –Horizontal or LEFT/RIGHT change –Change in X’s

61. Notes – 4.4–Slope and Rate of Change NOTES  Slope = RISE RUN  4 Kinds of Slope 1.Positive = Slants UP 2.Negative = Slants DOWN 3.Zero = Horizontal line 4.No Slope = Vertical line  Finding slope with a Graph 1.Draw a right triangle connecting the points 2.Calculate RISE and RUN 3.Use Slope = RISE/RUN

62. Notes – Continued NOTES - CONTINUED  Finding slope with a Table  Slope = Change in Y = How much change = 2 ND – 1 ST Change in X = How long did it take = 2 ND – 1 ST  SAME as UNIT RATE and RATE OF CHANGE!!!  Pay attention to positives/negatives!!  Finding slope Using Coordinates  Variable for slope is usually m. 1.Slope = m = Y 2 – Y 1 = RISE X 2 – X 1 = RUN 2.Plug in what you know and solve for what you don’t!

63. Examples 4.4

64. EXAMPLE 2 Find a negative slope Find the slope of the line shown. m = y 2 – y 1 x 2 – x 1 Let (x 1, y 1 ) = (3, 5) and (x 2, y 2 ) = (6, –1). –1 – 5 6 – 3 = – 6 3 = = –2–2 Write formula for slope. Substitute. Simplify.

65. EXAMPLE 3 Find the slope of a horizontal line Find the slope of the line shown. Let (x 1, y 1 ) = (– 2, 4) and (x 2, y 2 ) = (4, 4). m = y 2 – y 1 x 2 – x 1 4 – 4 4 – (– 2) = 0 6 = = 0 Write formula for slope. Substitute. Simplify.

66. EXAMPLE 4 Find the slope of a vertical line Find the slope of the line shown. Let (x 1, y 1 ) = (3, 5) and (x 2, y 2 ) = (3, 1). m = y 2 – y 1 x 2 – x 1 Write formula for slope. 1 – 5 3 – 3 = Substitute. Division by zero is undefined. ANSWER Because division by zero is undefined, the slope of a vertical line is undefined. – 4 0 =

67. EXAMPLE 2 Write an equation from a graph GUIDED PRACTICE for Examples 2, 3 and 4 Find the slope of the line that passes through the points. 4. (5, 2) and (5, – 2) m = y 2 – y 1 x 2 – x 1 Let (x 1, y 1 ) = (5, 2) and (x 2, y 2 ) = (5, –2). –2 – 2 5 – 5 = – 5 0 = Write formula for slope. Substitute. Division by zero is undefined. ANSWER The slope is undefined. SOLUTION

68. EXAMPLE 5 Find a rate of change INTERNET CAFE The table shows the cost of using a computer at an Internet cafe for a given amount of time. Find the rate of change in cost with respect to time. Time ( hours ) 246 Cost ( dollars ) 71421

69. EXAMPLE 5 Find a rate of change Rate of change = change in cost change in time 14 –7 4 – 2 = 7 2 = 3.5 = ANSWER The rate of change in cost is $3.50 per hour. SOLUTION

70. Time ( minute ) 306090 Distance ( miles ) 1.534.5 GUIDED PRACTICE for Example 5 SOLUTION The table shows the distance a person walks for exercise. Find the rate of change in distance with respect to time. 7. EXERCISE

71. EXAMPLE 5 Find a rate of change Rate of change = change in distance change in time 3 – 1.5 60 – 30 = = 0.05 ANSWER The rate of change in cost is $0.05 mile / minute. GUIDED PRACTICE for Example 5

72. EXAMPLE 6 Use a graph to find and compare rates of change COMMUNITY THEATER A community theater performed a play each Saturday evening for 10 consecutive weeks. The graph shows the attendance for the performances in weeks 1, 4, 6, and 10. Describe the rates of change in attendance with respect to time.

73. SOLUTION EXAMPLE 6 Use a graph to find and compare rates of change Find the rates of change using the slope formula. Weeks 1–4: 232 – 124 4 – 1 = 108 3 = 36 people per week Weeks 4–6: 204 – 232 6 – 4 = – 28 2 = – 14 people per week Weeks 6–10: 72 – 204 10 – 6 = – 132 4 = – 33 people per week ANSWER Attendance increased during the early weeks of performing the play. Then attendance decreased, slowly at first, then more rapidly.

74. EXAMPLE 7 Interpret a graph COMMUTING TO SCHOOL A student commutes from home to school by walking and by riding a bus. Describe the student’s commute in words.

75. EXAMPLE 7 Interpret a graph The first segment of the graph is not very steep, so the student is not traveling very far with respect to time. The student must be walking. The second segment has a zero slope, so the student must not be moving. He or she is waiting for the bus. The last segment is steep, so the student is traveling far with respect to time. The student must be riding the bus. SOLUTION

76. Warm-Up – 4.5

77. Lesson 4.5, For use with pages 243-250 1. Rewrite 5x + y = 8 so y is a function of x. 2. Find the slope of the line that passes through (–5, 6) and (0, 8). ANSWER y = –5x + 8 5 2

78. 3. Find the intercepts of the graph of the function a = 20t – 600. ANSWER Lesson 4.5, For use with pages 243-250 a- intercept: – 600, t- intercept: 30 4. Find the slope of -2x + y = 1. BUILD A TABLE! 4. What is slope?? ANSWER Slope = 2

79. Vocabulary – 4.5 Parallel Lines –Lines that never intersect –Lines that have the SAME SLOPE! slope-intercept form –Linear equation where y = mx + b –m = slope of the line –b = y-intercept

80. Notes – 4.5 – Slope-Intercept Form Slope Intercept Form of an Equation –  y = mx + b  To use the slope intercept form 1.Solve the equation so that Y is by itself 2.The coefficient of X is the slope. 3.The constant number is the Y intercept.  To graph a function using the slope intercept form 1.Graph the Y intercept 2.Use the slope = rise/run to find the next point 3.Graph the second point and connect the two points BrainPop: Slope and Intercept

81. Examples 4.5

82. SOLUTION EXAMPLE 1 Identify slope and y-intercept Identify the slope and y- intercept of the line with the given equation. y = 3x + 41. 3x + y = 22. The equation is in the form y = m x + b. So, the slope of the line is 3, and the y- intercept is 4. a. b.Rewrite the equation in slope-intercept form by solving for y.

83. EXAMPLE 1 Identify slope and y-intercept 3x + y = 2= 2 Write original equation. y = – 3x + 2 Subtract 3x from each side. ANSWER The line has a slope of – 3 and a y- intercept of 2.

84. SOLUTION EXAMPLE 1 Identify slope and y-intercept Identify the slope and y- intercept of the line with the given equation. The equation is in the form y = mx + b. So, the slope of the line is 5, and the y- intercept is –3. y = 5x – 31. GUIDED PRACTICE for Example 1

85. EXAMPLE 1 Identify slope and y-intercept Identify the slope and y- intercept of the line with the given equation. 3x – 3y = 122. GUIDED PRACTICE for Example 1

86. EXAMPLE 1 Identify slope and y-intercept ANSWER The line has a slope of –1 and a y- intercept of –4. GUIDED PRACTICE for Example 1 SOLUTION Rewrite the equation in slope-intercept form by solving for y. 3x – 3y = 12 Write original equation. Divide 3 by equation. 3x – 12 = 3y Rewrite original equation. y 3x + 12 = 3 y x – 4 = Simplify.

87. EXAMPLE 1 Identify slope and y-intercept Identify the slope and y- intercept of the line with the given equation. x + 4y = 63. GUIDED PRACTICE for Example 1

88. EXAMPLE 1 Identify slope and y-intercept GUIDED PRACTICE for Example 1 ANSWER The line has a slope of and a y- intercept of 1. 4 1 – 2 1 SOLUTION Rewrite the equation in slope-intercept form by solving for y. x + 4y = 6 Write original equation. Divide 3 by equation. 4y4y = – x + 6 Rewrite original equation. Simplify. – x 4 + 6 4 = – x + 6 = 4 y

89. SOLUTION EXAMPLE 2 Graph an equation using slope-intercept form Graph the equation 2x + y = 3. STEP 1 Rewrite the equation in slope-intercept form. y – 2x + 3 =

90. EXAMPLE 2 Graph an equation using slope-intercept form Identify the slope and the y- intercept. STEP 2 STEP 3 Plot the point that corresponds to the y- intercept, (0, 3). STEP 4 Use the slope to locate a second point on the line. Draw a line through the two points. = – 2 m and = 3 b

91. ESCALATORS EXAMPLE 3 Change slopes of lines To get from one floor to another at a library, you can take either the stairs or the escalator. You can climb stairs at a rate of 1.75 feet per second, and the escalator rises at a rate of 2 feet per second. You have to travel a vertical distance of 28 feet. The equations model the vertical distance d (in feet) you have left to travel after t seconds. Stairs: d = – 1.75t + 28 Escalator: d = – 2t + 28

92. SOLUTION EXAMPLE 3 Change slopes of lines a. Graph the equations in the same coordinate plane. b. How much time do you save by taking the escalator ? a. Draw the graph of d = – 1.75t + 28 using the fact that the d- intercept is 28 and the slope is – 1.75. Similarly, draw the graph of d = – 2t + 28. The graphs make sense only in the first quadrant.

93. EXAMPLE 3 Change slopes of lines The equation d = – 1.75t + 28 has a t- intercept of 16. The equation d = – 2t + 28 has a t- intercept of 14. So, you save 16 – 14 = 2 seconds by taking the escalator. b.

94. SOLUTION EXAMPLE 2 Graph an equation using slope-intercept form 4. Graph the equation y = – 2x + 5. STEP 1 Identify the slope and the y- intercept. GUIDED PRACTICE for Examples 2 and 3 STEP 2 Plot the point that corresponds to the y- intercept, (0, 5). = – 2 m and = 5 b STEP 3 Use the slope to locate a second point on the line. Draw a line through the two points.

95. EXAMPLE 5 Identify parallel lines Determine which of the lines are parallel. Find the slope of each line. Line a: m = – 1 – 0 – 1 – 2 – 3 – (–1 ) 0 – 5 = – 1 – 3 1 3 = Line b: m = – 2 – 5 = 2 5 = Line c: m = – 5 – (–3) – 2 – 4 – 2 – 6 = 1 3 = ANSWER Line a and line c have the same slope, so they are parallel.

96. EXAMPLE 5 Identify parallel lines GUIDED PRACTICE for Examples 4 and 5 Determine which lines are parllel: line a through (-1, 2) and (3, 4); line b through (3, 4) and ( 5, 8); line c through (-9, -2) and (-1, 2). 7. SOLUTION Line a: m = 4 – 2 3 –(1) 8 – 4 5 – 3 = 2 3+1 1 2 = Line b: m = – 4 – 2 = Line c: m = 2 – (–2) – 1 – (– 9) 2+2 –1+ 9 = 4 8 = Find the slope of each line. = 2 = 1 2

97. Warm-Up – 4.6 and 4.7

98. Lesson 4.6, For use with pages 253-259 1.4x – y = –8 2.-9x - 3y = 21 ANSWER y=4x+8 Write the equation in slope intercept form and sketch the graph y = -3x - 7

99. Lesson 4.6, For use with pages 253-259 1.Slope = 1 and y-int = 2 ANSWER Y = x + 2 Write the equation and sketch the graph

100. 3. You are traveling by bus. After 4.5 hours, the bus has traveled 234 miles. Use the formula d = rt where d is distance, r is a rate, and t is time to find the average rate of speed of the bus. ANSWER Lesson 4.6, For use with pages 253-259 52 mi/h Find the unit rate.

101. Vocabulary – 4.6-4.7 Direct Variation –Linear equation where y = kx Constant of Variation –In a Direct Variation, the letter k is the constant of variation –It’s the unit rate, the rate of change and … –SAME AS SLOPE!! Function Notation –Different way of writing functions –F(x) means the “the function of x” Family of Functions –A group of functions with similar characteristics (e.g. their graphs are all linear) Parent Linear Function –Simplest form of a family of functions –F(x) = x is the parent linear function

102. Notes – 4.6-4.7 – Direct Variations and Graphing Linear Functions A Direct variation has the form  y = kx  k = the constant of variation and is AKA  THE SLOPE!  A direct variation graph ALWAYS goes through the origin.  A direct variation is ALWAYS PROPORTIONAL!  There are two ways to find the constant, k 1.Find the UNIT RATE 2.If they give you a coordinate (x,y), 1.Plug in the numbers to y = kx 2.Solve for k.

103. Notes – 4.6-4.7 – Direct Variations and Graphing Linear Functions – cont.  A “function” is usually written as  F(x) and we read it as “F of x”  OR y = F(x)  To evaluate functions  Plug in what you know and …..???

104. Examples 4.6-4.7

105. EXAMPLE 1 Identify direct variation equations Tell whether the equation represents direct variation. If so, identify the constant of variation. 2x – 3y = 0 a.a. – x + y = 4 b.b.

106. EXAMPLE 1 Identify direct variation equations ANSWER Because the equation 2x – 3y = 0 can be rewritten in the form y = ax, it represents direct variation. The constant of variation is. 2 3 SOLUTION To tell whether an equation represents direct variation, try to rewrite the equation in the form y = ax. Write original equation. Subtract 2x from each side. – 3y = – 2x y = 2 3 x Simplify. 2x – 3y = 0

107. EXAMPLE 1 Identify direct variation equations – x + y = 4 Write original equation. Add x to each side. y = x + 4 ANSWER Because the equation – x + y = 4 cannot be rewritten in the form y = ax, it does not represent direct variation. b.b.

108. GUIDED PRACTICE for Example 1 Tell whether the equation represents direct variation. If so, identify the constant of variation. 1. – x + y = 1

109. GUIDED PRACTICE for Example 1 ANSWER Because the equation – x + y = 1 cannot be rewritten in the form y = ax, it does not represent direct variation. SOLUTION To tell whether an equation represents direct variation, try to rewrite the equation in the form y = ax. Write original equation. – x + y = 1 y = x + 1 Subtract Add x each side.

110. GUIDED PRACTICE for Example 1 2. 2x + y = 0 2x + y = 0 Write original equation. y = – 2x Subtract 2x from each side. SOLUTION ANSWER Because the equation 2x + y = 0 can be rewritten in the form y = ax, it represents direct variation. The constant of variation is. – 2

111. GUIDED PRACTICE for Example 1 3. 4x – 5y = 0 To tell whether an equation represents direct variation, try to rewrite the equation in the form y = ax. 4x – 5y = 0 Write original equation. 4x = 5y Subtract Add 5y each side. y = 4 5 x Simplify. ANSWER Because the equation 4x – 5y = 0 can be rewritten in the form y = ax, it represents direct variation. The constant of variation is. 4 5 SOLUTION

112. EXAMPLE 2 Graph direct variation equations Graph the direct variation equation. SOLUTION a.a. y = x 2 3 y = – 3x b.b. a.a. Plot a point at the origin. The slope is equal to the constant of variation, or Find and plot a second point, then draw a line through the points. 2 3

113. EXAMPLE 2 Graph direct variation equations Plot a point at the origin. The slope is equal to the constant of variation, or – 3. Find and plot a second point, then draw a line through the points. b.b.

114. EXAMPLE 3 Write and use a direct variation equation The graph of a direct variation equation is shown. SOLUTION y = ax y = ax Write direct variation equation. Substitute. 2 = a (– 1) Write the direct variation equation. a.a. Find the value of y when x = 30. b.b. Because y varies directly with x, the equation has the form y = ax. Use the fact that y = 2 when x = – 1 to find a. a.a. Solve for a. – 2 = a

115. EXAMPLE 3 Graph direct variation equations ANSWER A direct variation equation that relates x and y is y = – 2x. b. When x = 30, y = – 2(30) = – 60.

116. SOLUTION GUIDED PRACTICE for Examples 2 and 3 4. Graph the direct variation equation. y = 2x Plot a point at the origin. The slope is equal to the constant of variation, or Find and plot a second point, then draw a line through the points.

117. SOLUTION GUIDED PRACTICE for Examples 2 and 3 Because y varies directly with x, the equation has the form y = ax. Use the fact that y = 6 when x = 4 to find a. y = axy = ax 6 = a (4) 6 4 a = 3 2 = Solve for a. Write direct variation equation. Substitute. 5.The graph of a direct variation on equation passes through the point ( 4,6 ). Write the direct variation equation and find the value of y when x =24.

118. GUIDED PRACTICE for Examples 2 and 3 ANSWER A direct variation equation that relates x and y is y = x when x = 24. y = (24) = 36 3 2 3 2

119. SOLUTION Standardized Test Practice EXAMPLE 1 Substitute -3 for x. Write original function. f (x) 3x – 15 = = (– 3) 3(– 3) – 15 f ANSWER The correct answer is A. A BC D Simplify. = 24

120. GUIDED PRACTICE for Example 1 1. Evaluate the function h(x) = – 7x when x = 7. Substitute 7 for x. Write original function. h(x) = – 7x (7) = – 7(7) h = – 49 Simplify. SOLUTION

121. Find an x-value EXAMPLE 2 Write original function. Substitute 6 for f(x). 8 x = Solve for x. ANSWER When x = 8, f(x) = 6. 6 2x – 10= = f(x) 6. For the function f(x) 2x – 10, find the value of x so that = = 2x – 10 f(x)f(x)

122. Graph a function The gray wolf population in central Idaho was monitored over several years for a project aimed at boosting the number of wolves. The number of wolves can be modeled by the function f(x) = 37x + 7 where x is the number of years since 1995. Graph the function and identify its domain and range. GRAY WOLF EXAMPLE 3

123. SOLUTION To graph the function, make a table. xf(x) 037(0) + 7 = 7 137(1) + 7 = 44 237(2) + 7 = 81 The domain of the function is x 0. From the graph or table, you can see that the range of the function is f(x) 7. > = > = Graph a function EXAMPLE 3