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2. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Linear Equations and Inequalities CHAPTER 4.1The Rectangular Coordinate System 4.2Graphing Linear Equations 4.3Graphing Using Intercepts 4.4Slope-Intercept Form 4.5Point-Slope Form 4.6Graphing Linear Inequalities 4.7Introduction to Functions and Function Notation 4

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Rectangular Coordinate System 4.1 1.Determine the coordinates of a given point. 2.Plot points in the coordinate plane. 3.Determine the quadrant for a given coordinate. 4.Determine whether the graph of a set of data points is linear.

4. Slide 4- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Determine the coordinates of a given point.

5. Slide 4- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Axis: A number line used to locate a point in a plane. the origin Positive numbers are to the right and up from the origin. Negative numbers are to the left and down from the origin. The notation for writing ordered pairs is: (horizontal coordinate, vertical coordinate).

6. Slide 4- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identifying the Coordinates of a Point To determine the coordinates of a given point in the rectangular system: 1. Follow a vertical line from the point to the x-axis (horizontal axis). The number at this position on the x-axis is the first coordinate. 2. Follow a horizontal line from the point to the y-axis (vertical axis). The number at this position on the y- axis is the second coordinate.

7. Slide 4- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the coordinates for each point shown. A B D F E C We can get to point A by moving to the right 1 and then up 4. A: (1, 4) B: (  3, 2) C: (  2,  2) D: (4, 0) E: (0,  2) F: (4,  3) Left 3, up 2 Left 2, down 2 Right 4 Down 2 Right 4, down 3

8. Slide 4- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Plot points in the coordinate plane.

9. Slide 4- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Plotting a Point To graph or plot a point given its coordinates: 1. Beginning at the origin, (0, 0) move to the right or left along the x-axis the amount indicated by the first coordinate. 2. From that position on the x-axis, move up or down the amount indicated by the second coordinate. 3.Draw a dot to represent the point described by the coordinates.

10. Slide 4- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Plot the point described by the coordinates. a. (4,  3) b. (  2, 4) c. (0, 3) (4, –3 ): Begin at the origin and move to the right 4, then down 3. (4,  3) (–2, 4): Begin at the origin and move to the left 2, then up 4. (  2, 4) (0, 3): Begin at the origin and move up 3. (0, 3)

11. Slide 4- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Determine the quadrant for a given coordinate.

12. Slide 4- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The quadrants are numbered using Roman numerals. Note that the signs of the coordinates determine the quadrant in which a point lies. Quadrant I (+, +) Quadrant II ( , +) Quadrant III ( ,  ) Quadrant IV (+,  )

13. Slide 4- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identifying Quadrants To determine the quadrant for a given ordered pair, consider the signs of the coordinates. (+, +) means the point is in quadrant I (–, +) means the point is in quadrant II (–, –) means the point is in quadrant III (+, –) means the point is in quadrant IV

14. Slide 4- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example State the quadrant in which each point is located. a. (–14, 45) Answer Quadrant II (upper left), because the first coordinate is negative and the second coordinate is positive. b. (0, –15 ) Answer Since the x-coordinate is 0, this point is on the y-axis and is not in a quadrant.

15. Slide 4- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 4 Determine whether the graph of a set of data points is linear.

16. Slide 4- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In many problems data are listed as ordered pairs. If we plot each pair of data as an ordered pair we can see whether the points form a straight line. If they form a straight line they are linear. Points that do not form a straight line are nonlinear.

17. Slide 4- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The following points show the path of an object over time. Plot the points with the time along the horizontal axis and the distance along the vertical axis. Then state whether the path is linear or nonlinear. Time (hours) Distance (miles) 00 162 2124 3186 4248 The data points form a straight line; the path of the object is linear.

18. Slide 4- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identify the coordinates of the point shown below. a) (–4, –3 ) b) (–3, –4) c) (4, –3) d) (–3, 4)

19. Slide 4- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identify the coordinates of the point shown below. a) (–4, –3 ) b) (–3, –4) c) (4, –3) d) (–3, 4)

20. Slide 4- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In which quadrant is the point (5, –2) located? a) I b) II c) III d) IV

21. Slide 4- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In which quadrant is the point (5, –2) located? a) I b) II c) III d) IV

22. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Linear Equations 4.2 1.Determine whether a given pair of coordinates is a solution to a given equation with two unknowns. 2.Find solutions for an equation with two unknowns. 3.Graph linear equations.

23. Slide 4- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Determine whether a given pair of coordinates is a solution to a given equation with two unknowns.

24. Slide 4- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Checking a Potential Solution for an Equation with Two Variables To determine whether a given ordered pair is a solution for an equation with two variables: 1. Replace the variables in the equation with the corresponding coordinates. 2. Verify that the equation is true.

25. Slide 4- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether the ordered pair is a solution for the equation. (  3, 5); y = 4x – 3 Solution y = 4x – 3 5 = 4(–3) – 3 Replace x with –3 and y with 5. 5 ? –12 – 3 5  –15 Because the equation is not true, (–3, 5) is not a solution for y = 4x – 3.

26. Slide 4- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Find solutions for an equation with two unknowns.

27. Slide 4- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding Solutions to Equations with Two Variables To find a solution to an equation in two variables: 1. Choose a value for one of the variables (any value). 2. Replace the corresponding variable with your chosen value. 3. Solve the equation for the value of the other variable.

28. Slide 4- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find three solutions for the equation. 4x + y = 5 Solution To find a solution we replace one of the variables with a chosen value and then solve for the other variable. Choose x = 0 4x + y = 5 4(0) + y = 5 y = 5 Solution (0, 5) Choose x = 1 4x + y = 5 4(1) + y = 5 4 + y = 5 y = 1 Solution (1, 1) Choose x = 2 4x + y = 5 4(2) + y = 5 8 + y = 5 y = –3 Solution (2,  3)

29. Slide 4- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued The solutions can be summarized in a table: Keep in mind that there are an infinite number of correct solutions for a given equation in two variables. xyOrdered Pair 05(0, 5) 11(1, 1) 2 33(2,  3)

30. Slide 4- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Graph linear equations.

31. Slide 4- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations in two variables have an infinite number of solutions. Because of this fact, there is no way that all solutions to an equation can be found. However, all the solutions can be represented using a graph. The graph of the solutions of every linear equation will be a straight line.

32. Slide 4- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Linear Equations To graph a linear equation: 1. Find at least two solutions to the equation. 2. Plot the solutions as points in the rectangular coordinate system. 3. Connect the points to form a straight line.

33. Slide 4- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph 4x + y = 5. Solution We found three solutions to this equation in the previous example. xyOrdered Pair 05(0, 5) 11(1, 1) 2 33(2,  3) Plot each point and then connect the points to form a straight line. 4x + y = 5

34. Slide 4- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph y = 2. Solution y is equal to a constant To establish ordered pairs, we can rewrite the equation as 0x + y = 2 y is always 2 no matter what we choose for x. xyOrdered Pair 02(0, 2) 12(1, 2) 22(3, 2) y = 2

35. Slide 4- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal Lines The graph of y = c where, c is a real-number constant, is a horizontal line parallel to the x-axis that passes through the y-axis at a point with coordinates (0, c).

36. Slide 4- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph x = 2. Solution x is equal to a constant. To establish ordered pairs, we can rewrite the equation as x + 0y = 2. x is always 2 no matter what we choose for y. xyOrdered Pair 20(2, 0) 21(2, 1) 22(2, 2) x = 2

37. Slide 4- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Vertical Lines The graph of x = c where, c is a real-number constant, is a vertical line parallel to the y-axis that passes through the x-axis at a point with coordinates (c, 0).

38. Slide 4- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which ordered pair is a solution of the equation 3x – 2y = –6? a) (  4, 3) b) (1, 5) c) (2, 0) d) (0, 3)

39. Slide 4- 39 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which ordered pair is a solution of the equation 3x – 2y = –6? a) (  4, 3) b) (1, 5) c) (2, 0) d) (0, 3)

40. Slide 4- 40 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 4x – y = –4 a)b) c)d)

41. Slide 4- 41 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 4x – y = –4 a)b) c)d)

42. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Using Intercepts 4.3 1.Given an equation, find the coordinates of the x - and y - intercepts. 2.Graph linear equations using intercepts.

43. Slide 4- 43 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Given an equation, find the coordinates of the x - and y - intercepts.

44. Slide 4- 44 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Notice that the line intersects the x-axis as (4, 0) and the y-axis at (0, –2). These points are called intercepts. The point where a graph intersects the x-axis is called the x-intercept. The point where a graph intersects the y-axis is called the y-intercept. x-intercept (4, 0) y-intercept (0,  2)

45. Slide 4- 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the x- and y-intercepts To find an x-intercept:To find a y-intercept: 1. Replace y with 0 in the given equation. 1. Replace x with 0 in the given equation. 2. Solve for x.2. Solve for y.

46. Slide 4- 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the equation 7x – 2y = 14, find the x- and y-intercepts. Solution For the x-intercept, replace y with 0 and solve for x. 7x – 2y = 14 7x – 2(0) = 14 7x = 14 x = 2 x-intercept: (2, 0) For the y-intercept, replace x with 0 and solve for y. 7x – 2y = 14 7(0) – 2y = 14 –2y = 14 y = –7 y-intercept: (0, –7)

47. Slide 4- 47 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the equation, find the x- and y- intercepts. Solution For the y-intercept, replace x with 0 and solve for y. y-intercept: (0, 0) For the x-intercept, replace y with 0 and solve for x. x-intercept: (0, 0)

48. Slide 4- 48 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Intercepts for y = mx If an equation can be written in the form y = mx, where m is a real number other than 0, then the x- and y-intercepts are at the origin, (0, 0).

49. Slide 4- 49 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the equation y = 3x + 4, find the coordinates of the x- and y-intercepts. Solution x-intercept: y = 3x + 4 0 = 3x + 4 –4 = 3x x-intercept: y-intercept: y = 3x + 4 y = 3(0) + 4 y = 4 y-intercept: (0, 4)

50. Slide 4- 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The y-intercept for y = mx + b If an equation is in the form y = mx + b, where m and b are real numbers, then the y-intercept will be (0, b).

51. Slide 4- 51 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Intercepts for y = c The graph of an equation in the form y = c, where c is a real-number constant, has no x-intercept and the y-intercept is (0, c). Intercepts for x = c The graph of an equation in the form x = c, where c is a real-number constant, has no y-intercept and the x-intercept is (c, 0).

52. Slide 4- 52 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Graph linear equations using intercepts.

53. Slide 4- 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph using the x- and y-intercepts: 7x – 2y = 14 Solution The intercepts were found in a previous example to be (2, 0) and (0, –7). We plot the points and connect them with a line. It is helpful to find a third solution and verify that all three points can be connected to form a straight line. 7x – 2y = 14

54. Slide 4- 54 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Choose x = 3 as a check. 7x – 2y = 14 7(3) – 2y = 14 –2y = –7 y = 3.5 Table of Solutions xy x-intercept:20 y-intercept:0 22 check33.5 7x – 2y = 14

55. Slide 4- 55 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph using x- and y-intercepts. (0, 0) is the x- and y-intercept. We need to find one more point and a third point as a check. Choose x = 4.Choose x =  4.

56. Slide 4- 56 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Plot the solutions and connect the points to form the line. Table of Solutions xy x- and y- intercept: 00 solution43 check 44 33

57. Slide 4- 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph using x- and y-intercepts. (0, 2) is the y-intercept. We need to find one more point and the x-intercept. Choose x = 4.Let y = 0.

58. Slide 4- 58 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Plot the solutions and connect the points to form the line. Table of Solutions xy x-intercept  2.6 0 y-intercept02 check45

59. Slide 4- 59 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the x-intercept for 3x – y = 6? a) (2, 0) b) (0, –6) c) (–6, 0) d) (0, 2)

60. Slide 4- 60 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the x-intercept for 3x – y = 6? a) (2, 0) b) (0, –6) c) (–6, 0) d) (0, 2)

61. Slide 4- 61 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the y-intercept for y = 2x – 6? a) (3, 0) b) (0, –6) c) (–6, 0) d) (0, 3)

62. Slide 4- 62 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the y-intercept for y = 2x – 6? a) (3, 0) b) (0, –6) c) (–6, 0) d) (0, 3)

63. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope-Intercept Form 4.4 1.Compare lines with different slopes. 2.Graph equations in slope-intercept form. 3.Find the slope of a line given two points on the line.

64. Slide 4- 64 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Compare lines with different slopes.

65. Slide 4- 65 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph each of the following on the same grid. y = xy = 3xy = 4x Solution Complete a table of values. If x isy = xy = 3xy = 4x 0000 1134 2268 y = x y = 3x y = x y = 3x y = x y = 4x

66. Slide 4- 66 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph each of the following on the same grid. Solution Complete a table of values. If x is y =  x 0000 11 1 22 2

67. Slide 4- 67 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If the coefficient of m increases, the graphs get steeper. Because the coefficient m affects how steep a line is, m is called the slope of the line. If a slope of the line is a fraction between 0 and 1, then the smaller the fraction is, the less inclined or flatter the lines get.

68. Slide 4- 68 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of y = mx Given an equation of the form y = mx, the graph of the equation is a line passing through the origin and having the following characteristics: If m > 0, then the graph is a line that slants uphill from left to right. If m < 0, then the graph is a line that slants downhill from left to right. The greater the absolute value of m, the steeper the line. m > 0 m < 0

69. Slide 4- 69 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Graph equations in slope-intercept form.

70. Slide 4- 70 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope: The ratio of the vertical change between any two points on a line to the horizontal change between these points.

71. Slide 4- 71 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Equations in Slope-Intercept Form To graph an equation in slope-intercept form, y = mx + b, 1. Plot the y-intercept, (0, b). 2. Plot a second point by rising the number of units indicated by the numerator of the slope, m, then running the number of units indicated by the denominator of the slope, m. 3. Draw a straight line through the two points. Note: You can check by locating additional points using the slope. Every point you locate using the slope should be on the line.

72. Slide 4- 72 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the equation determine the slope and the y-intercept. Then graph the equation. Solution m = y-intercept: (0, 3) Plot the y-intercept and then use the slope to find other points. rise  2 run 3 (3, 1)

73. Slide 4- 73 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the equation  2x + 5y =  20, determine the slope and the y-intercept. Then graph the equation. Solution Write the equation in slope-intercept form by isolating y.  2x + 5y =  20 5y = 2x  20 The slope is and the y-intercept is (0, –4).

74. Slide 4- 74 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued We begin at (0, –4) and then rise 2 and run 5. m = y-intercept: (0, –4) rise 2 run 5

75. Slide 4- 75 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Find the slope of a line given two points on the line.

76. Slide 4- 76 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Slope Formula Given two points (x 1, y 1 ) and (x 2, y 2 ), where x 2  x 1, the slope of the line connecting the two points is given by the formula Rise: y 2 – y 1 Run: x 2 – x 1

77. Slide 4- 77 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the slope of the line connecting the given points. a. (4, 6) and (–2, 8) Solution b. (3, 8) and (–2, 8) Solution

78. Slide 4- 78 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Zero Slope Two points with different x-coordinates and the same y- coordinates, (x 1, c) and (x 2, c), will form the line with a slope 0 (a horizontal line) and equation y = c. Undefined Slope Two points with the same x-coordinates and different y- coordinates, (c, y 1 ) and (c, y 2 ), will form a line with a slope that is undefined (a vertical line) and equation x = c.

79. Slide 4- 79 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the slope of the line 3x + y = 6? a) m = 3 b) m =  3 c) d) m = 2

80. Slide 4- 80 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the slope of the line 3x + y = 6? a) m = 3 b) m =  3 c) d) m = 2

81. Slide 4- 81 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley How would you graph the line y = 3x – 2? a) Plot (0, –2), down 2, right 3 b) Plot (0, –2), down 3, right 1 c) Plot (0, 2), up 3, left 2 d) Plot (0, –2), up 3, right 1

82. Slide 4- 82 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley How would you graph the line y = 3x – 2? a) Plot (0, –2), down 2, right 3 b) Plot (0, –2), down 3, right 1 c) Plot (0, 2), up 3, left 2 d) Plot (0, –2), up 3, right 1

83. Slide 4- 83 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the slope of the line through the given points (5, 6) and (6,  3)? a) m =  9 b) m = 9 c) d)

84. Slide 4- 84 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the slope of the line through the given points (5, 6) and (6,  3)? a) m =  9 b) m = 9 c) d)

85. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Point-Slope Form 4.5 1.Use slope-intercept form to write the equation of a line. 2.Use point-slope form to write the equation of a line. 3.Write the equation of a line parallel to a given line. 4.Write the equation of a line perpendicular to a given line.

86. Slide 4- 86 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Use slope-intercept form to write the equation of a line.

87. Slide 4- 87 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If we are given the y-intercept (0, b) of a line, to write the equation of the line, we will need either the slope or another point so that we can calculate the slope.

88. Slide 4- 88 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A line with a slope of 4 crosses the y-axis at the point (0, 5). Write the equation of the line. Solution Use y = mx + b m = 4 b = 5 from the point (0, 5) y = 4x + 5

89. Slide 4- 89 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the equation of the line connecting the given points in slope-intercept form. (0,  6), (3, 6) Solution The y-intercept is (0,  6). Find the slope: y = mx + b y = 4x  6

90. Slide 4- 90 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equation of a Line Given Its y-Intercept To write the equation of a line given its y-intercept, (0, b), and its slope, m, use the slope-intercept form of the equation, y = mx + b. If given a second point and not the slope, we must first calculate the slope using then use y = mx + b.

91. Slide 4- 91 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Use point-slope form to write the equation of a line.

92. Slide 4- 92 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley You can use the point-slope form to write the equation of a line given any two points on the line. Using the Point-Slope Form of the Equation of a Line To write the equation of a line given its slope and any point, (x 1, y 1 ), on the line, use the point-slope form of the equation of a line, y – y 1 = m(x – x 1 ). If given a second point (x 2, y 2 ), and not the slope, we first calculate the slope using then use y – y 1 = m(x – x 1 ).

93. Slide 4- 93 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the equation of a line with a slope of 5 and passing through the point (3, 12). Write the equation in slope-intercept form. Solution Begin with the point-slope formula. Replace m = 5, x 1 = 3, y 1 = 12 y – y 1 = m(x – x 1 ) y – 12 = 5(x – 3) y – 12 = 5x – 15 Simplify. y = 5x – 3 Add 15 to both sides to isolate y.

94. Slide 4- 94 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the equation of a line passing through the points (4,  2) and (  4, 4). Write the equation in slope- intercept form. Solution Calculate the slope. Use the point-slope form, then isolate y to write the slope-intercept form.

95. Slide 4- 95 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Points (4,  2) and (  4, 4), We can use either point for (x 1, y 1 ). y – y 1 = m(x – x 1 ) Substitute. Simplify. Subtract 2 from both sides to isolate y.

96. Slide 4- 96 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations can also be written in standard form, Ax + By = C, where A, B, and C are real numbers.

97. Slide 4- 97 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A line connects the points (2, 6) and (–4, 3). Write the equation of the line in the form Ax + By = C, where A, B, and C are integers and A > 0. Solution Find the slope: Use point-slope form: y – y 1 = m(x – x 1 ) Distribute to clear ( ).

98. Slide 4- 98 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Multiply both sides by the LCD, 2. Subtract x from both sides to get x and y together. Add 12 to both sides to get the constant terms together. Multiply by –1 so that the coefficient of x is positive.

99. Slide 4- 99 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Write the equation of a line parallel to a given line.

100. Slide 4- 100 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parallel Lines The slopes of parallel lines are equal. y = 2x + 1 y = 2x – 3

101. Slide 4- 101 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the equation of a line that passes through (1, –5) and parallel to y = –3x + 4. Write the equation in slope-intercept form. Solution In y = –3x + 4, the slope is –3, so the slope of the line parallel will also be –3. Use point-slope form. y – y 1 = m(x – x 1 ) y – (  5) = –3(x – 1) y + 5 = –3x + 3 y = –3x – 2 y 1 =  5, x 1 = 1 and m = –3 Simplify. Subtract 5 from both sides to isolate y.

102. Slide 4- 102 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 4 Write the equation of a line perpendicular to a given line.

103. Slide 4- 103 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Perpendicular Lines The slope of a line perpendicular to a line with a slope of will be Horizontal and vertical lines are perpendicular.

104. Slide 4- 104 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the equation of a line that passes through (7, 1) and is perpendicular to 7x – 2y = –2. Write the equation in slope-intercept form. Solution Determine the slope of the line 7x – 2y = –2. Slope of perpendicular line:

105. Slide 4- 105 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued slope = ; point (7, 1) y – y 1 = m(x – x 1 ) Simplify. Add 1 to both sides to isolate y.

106. Slide 4- 106 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write the equation of the line in slope- intercept form given m =  2 and the point (4,  5). a) y =  2x – 3 b) y = 2x + 3 c) y =  2x + 3 d) y =  2x – 5

107. Slide 4- 107 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write the equation of the line in slope- intercept form given m =  2 and the point (4,  5). a) y =  2x – 3 b) y = 2x + 3 c) y =  2x + 3 d) y =  2x – 5

108. Slide 4- 108 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the equation of the line connecting the points (4,  3) and (  1, 7)? a) y =  2x + 5 b) y = 2x + 5 c) y =  2x – 5 d)

109. Slide 4- 109 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the equation of the line connecting the points (4,  3) and (  1, 7)? a) y =  2x + 5 b) y = 2x + 5 c) y =  2x – 5 d)

110. Slide 4- 110 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the relationship between the two lines? 5x – 3y = 11 3x + 5y = 8 a) parallel b) perpendicular c) neither

111. Slide 4- 111 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the relationship between the two lines? 5x – 3y = 11 3x + 5y = 8 a) parallel b) perpendicular c) neither

112. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Linear Inequalities 4.6 1.Determine whether an ordered pair is a solution for a linear inequality with two variables. 2.Graph linear inequalities.

113. Slide 4- 113 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Determine whether an ordered pair is a solution for a linear inequality with two variables.

114. Slide 4- 114 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Checking an Ordered Pair To determine whether an ordered pair is a solution for an inequality, replace the variables with the corresponding coordinates and see if the resulting inequality is true. If so, the ordered pair is a solution. Linear inequalities have the same form as linear equations, except that they contain an inequality symbol instead of an equal sign. Remember that a solution for a linear equation in two variables is an ordered pair that makes the equation true.

115. Slide 4- 115 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether (3, –5) is a solution for y  3x + 4. Solution y  3x + 4 – 5  3(3) + 4 –5  9 + 4 –5  13 This statement is true, so (3, –5) is a solution. Replace x with 3 and y with  5.

116. Slide 4- 116 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Graph linear inequalities.

117. Slide 4- 117 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing linear inequalities is very much like graphing linear equations. Graphing Linear Inequalities To graph a linear inequality: 1. Graph the related equation. The related equation has an equal sign in place of the inequality symbol. If the inequality symbol is  or , then draw a solid line. If the inequality symbol is, then draw a dashed line. 2. Choose an ordered pair on one side of the boundary line and test this ordered pair in the inequality. If the ordered pair satisfies the inequality, then shade the region that contains it. If the ordered pair does not satisfy the inequality, then shade the region on the other side of the boundary line.

118. Slide 4- 118 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph: y > 4x + 1 Solution Graph the related equation y = 4x + 1. (0, 1) and (  1,  3) Select a test point: (0, 0) y > 4x + 1 0 > 4(0) + 1 0 > 1 False Shade other side of the boundary line.

119. Slide 4- 119 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph: x + 4y   8 Solution Graph the related equation. (0,  2) and (8, 0) Select a test point: (0, 0) x – 4y  8 0 – 4(0)  8 0  8 True Shade the region that contains the point.

120. Slide 4- 120 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which point is a solution to the inequality 3x + 2y  12? a) (0,  3) b) (0, 0) c) (2, 5) d) (  4, 5)

121. Slide 4- 121 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which point is a solution to the inequality 3x + 2y  12? a) (0,  3) b) (0, 0) c) (2, 5) d) (  4, 5)

122. Slide 4- 122 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 2x + 3y  12. a)b) c)d)

123. Slide 4- 123 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 2x + 3y  12. a)b) c)d)

124. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Functions and Function Notation 4.7 1.Identify the domain and range of a relation. 2.Identify functions and their domains and ranges. 3.Find the value of a function. 4.Graph linear functions.

125. Slide 4- 125 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Identify the domain and range of a relation.

126. Slide 4- 126 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relation: A set of ordered pairs. Domain: The set of all input values for a relation. Range: The set of all output values for a relation.

127. Slide 4- 127 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine the domain and range of the relation {(3,  2), (4, 0), (5, 6), (6, 9)}. Solution Domain—set of all x-values {3, 4, 5, 6} Range—set of all y-values {  2, 0, 6, 9}

128. Slide 4- 128 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determining Domain and Range To determine the domain of a relation given its graph, answer the question: What are all the x-values that have a corresponding y-value? To determine the range, answer the question: What are all the y-values that have a corresponding x-value?

129. Slide 4- 129 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine the domain and range of the relation. Solution The graph begins at (  2, 0). Values along the x-axis begin at  2 and continue infinitely, so the domain is {x|x   2}. The y-values begin at 0 and continue infinitely, so the range is {y|y  0}.

130. Slide 4- 130 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Identify functions and their domains and ranges.

131. Slide 4- 131 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function: A relation in which every value in the domain is paired with exactly one value in the range. Domain Range 02 14 26 38 410 Each element in the domain has a single arrow pointing to an element in the range.

132. Slide 4- 132 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Every function is a relation, but not every relation is a function. If any value in the domain is assigned to more than one value in the range, then the relation is not a function. Domain Range 02 14 26 10 12 not a function

133. Slide 4- 133 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether the relation is a function. DomainRange March 1Donna April 17Dennis Sept. 3Catherine October 9Denise Nancy The relation is not a function because an element in the domain, Sept. 3, is assigned to two names in the range.

134. Slide 4- 134 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Vertical Line Test To determine whether a relation is a function from its graph, perform a vertical line test: 1. Draw or imagine vertical lines through each point in the domain. 2. If each vertical line intersects the graph at only one point, then the graph is the graph of a function. 3. If any vertical line intersects the graph at two different points, then the graph is not the graph of a function.

135. Slide 4- 135 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For each graph, determine the domain and range. Then state whether each relation is a function. a.b. Domain: {x|x  1} Range: all real numbers Not a function Domain: all real numbers Range: {y   1} Function

136. Slide 4- 136 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Find the value of a function.

137. Slide 4- 137 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When written as an equation, the notation for a function is a modification of an equation in two variables. y = 3x + 4 could be written as f(x) = 3x + 4 f(x) is read as “a function in terms of x” or “f of x”

138. Slide 4- 138 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the Value of a Function Given a function f(x), to find f(a), where a is a real number in the domain of f, replace x in the function with a and calculate the value.

139. Slide 4- 139 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the function f(x) = 3x – 5, find the following. a. f(2)b. f(  4)c. f(a) Solution a. f(2) = 3x – 5 = 3(2) – 5 = 6 – 5 = 1 b. f(  4) = 3x – 5 = 3(  4) – 5 =  12 – 5 =  17 c. f(a) = 3x – 5 = 3(a) – 5 = 3a – 5

140. Slide 4- 140 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 4 Graph linear functions.

141. Slide 4- 141 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We create the graph of a function the same way that we create the graph of an equation in two variables. Slope-intercept form: y = mx + b Linear function: f(x) = mx + b

142. Slide 4- 142 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph: f(x) = 2x + 1 Solution We could make a table of values or use the fact that the slope is 2 and the y-intercept is 1. xf(x)f(x) 01 13 25 f(x) = 2x + 1

143. Slide 4- 143 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Is the relation a function? a) yes b) no

144. Slide 4- 144 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Is the relation a function? a) yes b) no

145. Slide 4- 145 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For the function f(x) = 2x 2 + x – 4 find f(1). a) 3 b) –4 c) 2 d)  1

146. Slide 4- 146 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For the function f(x) = 2x 2 + x – 4 find f(1). a) 3 b) –4 c) 2 d)  1