1. Chapter 3 Graphs and Functions

2. § 3.1 Graphing Equations

3. Martin-Gay, Intermediate Algebra, 5ed 33 Vocabulary Ordered pair – a sequence of 2 numbers where the order of the numbers is important Axis – horizontal or vertical number line Origin – point of intersection of two axes Quadrants – regions created by intersection of 2 axes Location of a point residing in the rectangular coordinate system created by a horizontal (x-) axis and vertical (y-) axis can be described by an ordered pair. Each number in the ordered pair is referred to as a coordinate

4. Martin-Gay, Intermediate Algebra, 5ed 44 To graph the point corresponding to a particular ordered pair (a,b), you must start at the origin and move a units to the left of right (right if a is positive, left if a is negative), then move b units up or down (up if b is positive, down if b is negative). Graphing an Ordered Pair

5. Martin-Gay, Intermediate Algebra, 5ed 55 Note that the order of the coordinates is very important, since (-4, 2) and (2, -4) are located in different positions. x-axis y-axis (5, 3) 5 units right 3 units up (0, 5) (-6, 0) (2, -4) (-4, 2) (0, 0) Quadrant I Quadrant IV Quadrant III Quadrant II origin Graphing an Ordered Pair

6. Martin-Gay, Intermediate Algebra, 5ed 66 Paired data are data that can be represented as an ordered pair A scatter diagram is the graph of paired data as points in the rectangular coordinate system An order pair is a solution of an equation in two variables if replacing the variables by the appropriate values of the ordered pair results in a true statement. Vocabulary

7. Martin-Gay, Intermediate Algebra, 5ed 77 Determine whether (3, – 2) is a solution of 2x + 5y = – 4. Let x = 3 and y = – 2 in the equation. 2x + 5y = – 4 2(3) + 5( – 2) = – 4 Replace x with 3 and y with –2. 6 + ( – 10) = – 4 Compute the products. – 4 = – 4 True So (3, -2) is a solution of 2x + 5y = – 4 Solutions of an Equation Example:

8. Martin-Gay, Intermediate Algebra, 5ed 88 Determine whether ( – 1, 6) is a solution of 3x – y = 5. Let x = – 1 and y = 6 in the equation. 3x – y = 5 3( – 1) – 6 = 5 Replace x with – 1 and y with 6. – 3 – 6 = 5 Compute the product. – 9 = 5 False So ( – 1, 6) is not a solution of 3x – y = 5 Solutions of an Equation Example:

9. Martin-Gay, Intermediate Algebra, 5ed 99 Linear Equation in Two Variables A linear equation in two variables is an equation that can be written in the form Ax + By = C where A and B are not both 0. This is called standard form. Linear Equations

10. Martin-Gay, Intermediate Algebra, 5ed 10 Graph the linear equation 2x – y = -4. We find two ordered pair solutions (and a third solution as a check on our computations) by choosing a value for one of the variables, x or y, then solving for the other variable. We plot the solution points, then draw the line containing the 3 points. Example: Graphing Linear Equations Continued.

11. Martin-Gay, Intermediate Algebra, 5ed 11 Graph the linear equation 2x – y = – 4. Let x = 1. Then 2x – y = – 4 becomes 2(1) – y = – 4 Replace x with 1. 2 – y = – 4 Simplify the left side. – y = – 4 – 2 = – 6 Subtract 2 from both sides. y = 6 Multiply both sides by – 1. So one solution is (1, 6) Example continued: Graphing Linear Equations Continued.

12. Martin-Gay, Intermediate Algebra, 5ed 12 Graph the linear equation 2x – y = – 4. For the second solution, let y = 4. Then 2x – y = – 4 becomes 2x – 4 = – 4 Replace y with 4. 2x = – 4 + 4 Add 4 to both sides. 2x = 0 Simplify the right side. x = 0 Divide both sides by 2. So the second solution is (0, 4) Graphing Linear Equations Example continued: Continued.

13. Martin-Gay, Intermediate Algebra, 5ed 13 Graph the linear equation 2x – y = – 4. For the third solution, let x = – 3. Then 2x – y = – 4 becomes 2(– 3) – y = – 4 Replace x with – 3. – 6 – y = – 4 Simplify the left side. – y = – 4 + 6 = 2 Add 6 to both sides. y = – 2 Multiply both sides by – 1. So the third solution is (– 3, – 2) Graphing Linear Equations Example continued: Continued.

14. Martin-Gay, Intermediate Algebra, 5ed 14 Now we plot all three of the solutions (1, 6), (0, 4) and (– 3, – 2). x y (1, 6) (0, 4) (– 3, – 2) And then we draw the line that contains the three points. Graphing Linear Equations Example continued:

15. Martin-Gay, Intermediate Algebra, 5ed 15 Since the equation is solved for y, we should choose values for x. To avoid fractions, we should select values of x that are multiples of 4 (the denominator of the fraction). Graph the linear equation y = x + 3. Graphing Linear Equations Example: Continued.

16. Martin-Gay, Intermediate Algebra, 5ed 16 Let x = 4. y = 3 + 3 = 6 Simplify the right side. So one solution is (4, 6) Graph the linear equation y = x + 3. Then y = x + 3 becomes y = (4) + 3 Replace x with 4. Graphing Linear Equations Example continued: Continued.

17. Martin-Gay, Intermediate Algebra, 5ed 17 For the second solution, let x = 0. y = 0 + 3 = 3 Simplify the right side. So a second solution is (0, 3) Graph the linear equation y = x + 3. Then y = x + 3 becomes y = (0) + 3 Replace x with 0. Graphing Linear Equations Example continued: Continued.

18. Martin-Gay, Intermediate Algebra, 5ed 18 For the third solution, let x = – 4. y = – 3 + 3 = 0 Simplify the right side. So the third solution is ( – 4, 0) Graph the linear equation y = x + 3. Then y = x + 3 becomes y = ( – 4) + 3 Replace x with – 4. Example continued: Graphing Linear Equations Continued.

19. Martin-Gay, Intermediate Algebra, 5ed 19 Now we plot all three of the solutions (4, 6), (0, 3) and (– 4, 0). x y (4, 6) (0, 3) (– 4, 0) And then we draw the line that contains the three points. Graphing Linear Equations Example continued:

20. Martin-Gay, Intermediate Algebra, 5ed 20 Intercepts Intercepts of axes (where graph crosses the axes) Since all points on the x-axis have a y-coordinate of 0, to find x-intercept, let y = 0 and solve for x Since all points on the y-axis have an x-coordinate of 0, to find y-intercept, let x = 0 and solve for y

21. Martin-Gay, Intermediate Algebra, 5ed 21 Find the y-intercept of 4 = x – 3y Let x = 0. Then 4 = x – 3y becomes 4 = 0 – 3y Replace x with 0. 4 = – 3y Simplify the right side. = y Divide both sides by – 3. So the y-intercept is (0, ) Intercepts Example:

22. Martin-Gay, Intermediate Algebra, 5ed 22 Find the x-intercept of 4 = x – 3y Let y = 0. Then 4 = x – 3y becomes 4 = x – 3(0) Replace y with 0. 4 = x S implify the right side. So the x-intercept is (4,0) Intercepts Example:

23. Martin-Gay, Intermediate Algebra, 5ed 23 Graph the linear equation 4 = x – 3y by plotting intercepts. We previously found that the y-intercept is (0, ) and the x-intercept is (4, 0). Plot both of these points and then draw the line through the 2 points. Note: You should still find a 3 rd solution to check your computations. Example: Graph by Plotting Intercepts Continued.

24. Martin-Gay, Intermediate Algebra, 5ed 24 Graph the linear equation 4 = x – 3y. Along with the intercepts, for the third solution, let y = 1. Then 4 = x – 3y becomes 4 = x – 3(1) Replace y with 1. 4 = x – 3 Simplify the right side. 4 + 3 = x Add 3 to both sides. 7 = x Simplify the left side. So the third solution is (7, 1) Example continued: Graph by Plotting Intercepts Continued.

25. Martin-Gay, Intermediate Algebra, 5ed 25 x y And then we draw the line that contains the three points. Now we plot the two intercepts (0, ) and (4, 0) along with the third solution (7, 1). (4, 0) (7, 1) (0, ) Graph by Plotting Intercepts Example continued:

26. Martin-Gay, Intermediate Algebra, 5ed 26 Graphing Nonlinear Equations x y = x 22 4 11 1 00 11 24 x y 5 5 55 55 Example:

27. Martin-Gay, Intermediate Algebra, 5ed 27 Graphing Nonlinear Equations xy = x 22 2 11 1 00 11 22 x y 5 5 55 55 Example:

28. Martin-Gay, Intermediate Algebra, 5ed 28 Graphing Nonlinear Equations x y = x 00 11 42 93 x y 5 5 55 55 Example:

29. Martin-Gay, Intermediate Algebra, 5ed 29 § 3.2 Introduction to Functions

30. Martin-Gay, Intermediate Algebra, 5ed 30 Relations Equations in two variables define relations between the two variables. There are other ways to describe relations between variables. Set to set Ordered pairs A set of ordered pairs is also called a relation.

31. Martin-Gay, Intermediate Algebra, 5ed 31 Some relations are also functions. A function is a set of order pairs that assigns to each x-value exactly one y-value. Functions

32. Martin-Gay, Intermediate Algebra, 5ed 32 Given the relation {(4,9), (–4,9), (2,3), (10, –5)}, is it a function? Since each element of the domain is paired with only one element of the range, it is a function. Note: It’s okay for a y-value to be assigned to more than one x-value, but an x-value cannot be assigned to more than one y-value (has to be assigned to ONLY one y-value). Functions Example:

33. Martin-Gay, Intermediate Algebra, 5ed 33 Is the relation y = x 2 – 2x a function? Since each element of the domain (the x- values) would produce only one element of the range (the y-values), it is a function. Functions Example:

34. Martin-Gay, Intermediate Algebra, 5ed 34 Is the relation x 2 – y 2 = 9 a function? Since each element of the domain (the x- values) would correspond with 2 different values of the range (both a positive and negative y-value), the relation is NOT a function. Functions Example:

35. Martin-Gay, Intermediate Algebra, 5ed 35 Relations and functions can also be described by graphing their ordered pairs. Graphs can be used to determine if a relation is a function. If an x-coordinate is paired with more than one y- coordinate, a vertical line can be drawn that will intersect the graph at more than one point. If no vertical line can be drawn so that it intersects a graph more than once, the graph is the graph of a function. This is the vertical line test. Vertical Line Test

36. Martin-Gay, Intermediate Algebra, 5ed 36 Use the vertical line test to determine whether the graph to the right is the graph of a function. x y Since no vertical line will intersect this graph more than once, it is the graph of a function. Vertical Line Test Example:

37. Martin-Gay, Intermediate Algebra, 5ed 37 Use the vertical line test to determine whether the graph to the right is the graph of a function. x y Since no vertical line will intersect this graph more than once, it is the graph of a function. Vertical Line Test Example:

38. Martin-Gay, Intermediate Algebra, 5ed 38 Use the vertical line test to determine whether the graph to the right is the graph of a function. Since vertical lines can be drawn that intersect the graph in two points, it is NOT the graph of a function. x y Vertical Line Test Example:

39. Martin-Gay, Intermediate Algebra, 5ed 39 Since the graph of a linear equation is a line, all linear equations are functions, except those whose graph is a vertical line Note: An equation of the form y = c is a horizontal line and IS a function. An equation of the form x = c is a vertical line and IS NOT a function. Vertical Line Test

40. Martin-Gay, Intermediate Algebra, 5ed 40 Recall that a set of ordered pairs is also called a relation. The domain is the set of x-coordinates of the ordered pairs. The range is the set of y-coordinates of the ordered pairs. Domain and Range

41. Martin-Gay, Intermediate Algebra, 5ed 41 Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)} Domain is the set of all x-values, {4, –4, 2, 10} Range is the set of all y-values, {9, 3, –5} Domain and Range Example:

42. Martin-Gay, Intermediate Algebra, 5ed 42 Find the domain and range of the function graphed to the right. Use interval notation. x y Domain is [ – 3, 4] Domain Range is [ – 4, 2] Range Domain and Range Example:

43. Martin-Gay, Intermediate Algebra, 5ed 43 Find the domain and range of the function graphed to the right. Use interval notation. x y Domain is (– ,  ) Domain Range is [– 2,  ) Range Domain and Range Example:

44. Martin-Gay, Intermediate Algebra, 5ed 44 Input (Animal) Polar Bear Cow Chimpanzee Giraffe Gorilla Kangaroo Red Fox Output (Life Span) 20 15 10 7 Find the domain and range of the following relation. Domain and Range Example:

45. Martin-Gay, Intermediate Algebra, 5ed 45 Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox} Range is {20, 15, 10, 7} Domain and Range Example continued:

46. Martin-Gay, Intermediate Algebra, 5ed 46 Specialized notation is often used when we know a relation is a function and it has been solved for y. For example, the graph of the linear equation y = –3x + 2 passes the vertical line test, so it represents a function. We often use letters such as f, g, and h to name functions. We can use the function notation f(x) (read “f of x”) and write the equation as f(x) = –3x + 2. Note: The symbol f(x) is a specialized notation that does NOT mean f x (f times x). Function Notation

47. Martin-Gay, Intermediate Algebra, 5ed 47 When we want to evaluate a function at a particular value of x, we substitute the x-value into the notation. For example, f(2) means to evaluate the function f when x = 2. So we replace x with 2 in the equation. For our previous example when f(x) = – 3x + 2, f(2) = – 3(2) + 2 = – 6 + 2 = – 4. When x = 2, then f(x) = – 4, giving us the order pair (2, – 4). Function Notation

48. Martin-Gay, Intermediate Algebra, 5ed 48 Given that g(x) = x 2 – 2x, find g(– 3). Then write down the corresponding ordered pair. g(– 3) = (– 3) 2 – 2(– 3) = 9 – (– 6) = 15. The ordered pair is (– 3, 15). Function Notation Example:

49. Martin-Gay, Intermediate Algebra, 5ed 49 Given the graph of the following function, find each function value by inspecting the graph. f(5) = 8 x y f(x) f(4) = 3 f(  5) =  1 1 f(  6) = – 7– 7 Function Notation Example:

50. Martin-Gay, Intermediate Algebra, 5ed 50 § 3.3 Graphing Linear Functions

51. Martin-Gay, Intermediate Algebra, 5ed 51 Graph the linear function f(x) = x + 3. Graphing Linear Functions Example: Let x = 4. = 3 + 3 = 6 Simplify the right side. So one solution is (4, 6) Then f(4) = (4) + 3 Replace x with 4. Continued.

52. Martin-Gay, Intermediate Algebra, 5ed 52 For the second solution, let x = 0. = 0 + 3 = 3 Simplify the right side. So a second solution is (0, 3) Graphing Linear Functions Example continued: Continued. Graph the linear function f(x) = x + 3. Then f(0) = (0) + 3 Replace x with 0.

53. Martin-Gay, Intermediate Algebra, 5ed 53 For the third solution, let x = – 4. = – 3 + 3 = 0 Simplify the right side. So a third solution is (– 4, 0) Graphing Linear Functions Example continued: Continued. Graph the linear function f(x) = x + 3. Then f(– 4) = ( – 4 ) + 3 Replace x with – 4.

54. Martin-Gay, Intermediate Algebra, 5ed 54 Now we plot all three of the solutions (4, 6), (0, 3) and (– 4, 0). x y (4, 6) (0, 3) (– 4, 0) And then we draw the line that contains the three points. Graphing Linear Functions Example continued:

55. Martin-Gay, Intermediate Algebra, 5ed 55 Graphing Linear Functions x f (x) = x 22 22 11 1 00 11 22 x y 5 5 55 55 Example:

56. Martin-Gay, Intermediate Algebra, 5ed 56 Intercepts Intercepts of axes (where graph crosses the axes) Since all points on the x-axis have a y-coordinate of 0, to find x-intercept, let y = 0 and solve for x Since all points on the y-axis have an x-coordinate of 0, to find y-intercept, let x = 0 and solve for y

57. Martin-Gay, Intermediate Algebra, 5ed 57 Find the y-intercept of 4 = x – 3y Let x = 0. Then 4 = x – 3y becomes 4 = 0 – 3y Replace x with 0. 4 = – 3y Simplify the right side. = y Divide both sides by – 3. So the y-intercept is (0, ) Intercepts Example:

58. Martin-Gay, Intermediate Algebra, 5ed 58 Find the x-intercept of 4 = x – 3y Let y = 0. Then 4 = x – 3y becomes 4 = x – 3(0) Replace y with 0. 4 = x S implify the right side. So the x-intercept is (4,0) Intercepts Example:

59. Martin-Gay, Intermediate Algebra, 5ed 59 Graph the linear equation 4 = x – 3y by plotting intercepts. We previously found that the y-intercept is (0, ) and the x-intercept is (4, 0). Plot both of these points and then draw the line through the 2 points. Note: You should still find a 3 rd solution to check your computations. Example: Graph by Plotting Intercepts Continued.

60. Martin-Gay, Intermediate Algebra, 5ed 60 Graph the linear equation 4 = x – 3y. Along with the intercepts, for the third solution, let y = 1. Then 4 = x – 3y becomes 4 = x – 3(1) Replace y with 1. 4 = x – 3 Simplify the right side. 4 + 3 = x Add 3 to both sides. 7 = x Simplify the left side. So the third solution is (7, 1) Example continued: Graph by Plotting Intercepts Continued.

61. Martin-Gay, Intermediate Algebra, 5ed 61 x y And then we draw the line that contains the three points. Now we plot the two intercepts (0, ) and (4, 0) along with the third solution (7, 1). (4, 0) (7, 1) (0, ) Graph by Plotting Intercepts Example continued:

62. Martin-Gay, Intermediate Algebra, 5ed 62 Graph y = 3 Note that this line can be written as y = 0·x + 3 The y-intercept is (0, 3), but there is no x-intercept! (Since an x-intercept would be found by letting y = 0, and 0  0·x + 3, there is no x-intercept) Every value we substitute for x gives a y-coordinate of 3. The graph will be a horizontal line through the point (0,3) on the y-axis Graphing Horizontal Lines Continued. Example:

63. Martin-Gay, Intermediate Algebra, 5ed 63 x y (0, 3) Graphing Horizontal Lines Example continued:

64. Martin-Gay, Intermediate Algebra, 5ed 64 Graph x = – 3 This equation can be written x = 0·y – 3 When y = 0, x = – 3, so the x-intercept is ( – 3,0), but there is no y-intercept Any value we substitute for y gives an x-coordinate of –3. So the graph will be a vertical line through the point ( – 3,0) on the x-axis Graphing Vertical Lines Continued. Example:

65. Martin-Gay, Intermediate Algebra, 5ed 65 x y (-3, 0) Graphing Vertical Lines Example continued:

66. Martin-Gay, Intermediate Algebra, 5ed 66 Vertical lines Appear in the form of x = c, where c is a real number x-intercept is at (c, 0), no y-intercept unless c = 0 (y-axis) Horizontal lines Appear in the form of y = c, where c is a real number y-intercept is at (0, c), no x-intercept unless c = 0 (x-axis) Vertical and Horizontal Lines

67. Martin-Gay, Intermediate Algebra, 5ed 67 § 3.4 The Slope of a Line

68. Martin-Gay, Intermediate Algebra, 5ed 68 Slope Slope of a Line

69. Martin-Gay, Intermediate Algebra, 5ed 69 Find the slope of the line through (4, -3) and (2, 2) If we let (x 1, y 1 ) be (4, -3) and (x 2, y 2 ) be (2, 2), then Note: If we let (x 1, y 1 ) be (2, 2) and (x 2, y 2 ) be (4, -3), then we get the same result. Slope Example:

70. Martin-Gay, Intermediate Algebra, 5ed 70 Slope-Intercept Form of a line y = mx + b has a slope of m and has a y-intercept of (0, b). This form is useful for graphing, since you have a point and the slope readily visible. Slope-Intercept Form

71. Martin-Gay, Intermediate Algebra, 5ed 71 Find the slope and y-intercept of the line –3x + y = -5. First, we need to solve the linear equation for y. By adding 3x to both sides, y = 3x – 5. Once we have the equation in the form of y = mx + b, we can read the slope and y-intercept. slope is 3 y-intercept is (0, – 5) Slope-Intercept Form Example:

72. Martin-Gay, Intermediate Algebra, 5ed 72 Find the slope and y-intercept of the line 2x – 6y = 12. First, we need to solve the linear equation for y. – 6y = – 2x + 12 Subtract 2x from both sides. y = x – 2 Divide both sides by – 6. Since the equation is now in the form of y = mx + b, slope is y-intercept is (0, – 2) Slope-Intercept Form Example:

73. Martin-Gay, Intermediate Algebra, 5ed 73 For any 2 points, the y values will be equal to the same real number. The numerator in the slope formula = 0 (the difference of the y-coordinates), but the denominator  0 (two different points would have two different x- coordinates). So the slope = 0. Slope of a Horizontal Line

74. Martin-Gay, Intermediate Algebra, 5ed 74 For any 2 points, the x values will be equal to the same real number. The denominator in the slope formula = 0 (the difference of the x-coordinates), but the numerator  0 (two different points would have two different y- coordinates), So the slope is undefined (since you can’t divide by 0). Slope of a Vertical Line

75. Martin-Gay, Intermediate Algebra, 5ed 75 If a line moves up as it moves from left to right, the slope is positive. If a line moves down as it moves from left to right, the slope is negative. Horizontal lines have a slope of 0. Vertical lines have undefined slope (or no slope). Summary of Slope of Lines

76. Martin-Gay, Intermediate Algebra, 5ed 76 Two lines that never intersect are called parallel lines. Parallel lines have the same slope unless they are vertical lines, which have no slope. Vertical lines are also parallel. Parallel Lines

77. Martin-Gay, Intermediate Algebra, 5ed 77 Find the slope of a line parallel to the line passing through (0,3) and (6,0) So the slope of any parallel line is also –½ Parallel Lines Example:

78. Martin-Gay, Intermediate Algebra, 5ed 78 Two lines that intersect at right angles are called perpendicular lines Two nonvertical perpendicular lines have slopes that are negative reciprocals of each other The product of their slopes will be –1 Horizontal and vertical lines are perpendicular to each other Perpendicular Lines

79. Martin-Gay, Intermediate Algebra, 5ed 79 Find the slope of a line perpendicular to the line passing through (-1,3) and (2,-8) So the slope of any perpendicular line is Perpendicular Lines Example:

80. Martin-Gay, Intermediate Algebra, 5ed 80 Determine whether the following lines are parallel, perpendicular, or neither. –5x + y = –6 and x + 5y = 5 First, we need to solve both equations for y. In the first equation, y = 5x – 6 Add 5x to both sides. In the second equation, 5y = –x + 5 Subtract x from both sides. y = x + 1 Divide both sides by 5. The first equation has a slope of 5 and the second equation has a slope of, so the lines are perpendicular. Parallel and Perpendicular Lines Example:

81. Martin-Gay, Intermediate Algebra, 5ed 81 § 3.5 Equations of Lines

82. Martin-Gay, Intermediate Algebra, 5ed 82 Slope-Intercept Form of a line y = mx + b has a slope of m and has a y-intercept of (0, b). This form is useful for graphing, since you have a point and the slope readily visible. Slope-Intercept Form

83. Martin-Gay, Intermediate Algebra, 5ed 83 Find the slope and y-intercept of the line – 3x + y = – 5. First, we need to solve the linear equation for y. By adding 3x to both sides, y = 3x – 5. Once we have the equation in the form of y = mx + b, we can read the slope and y-intercept. slope is 3 y-intercept is (0, – 5) Slope-Intercept Form Example:

84. Martin-Gay, Intermediate Algebra, 5ed 84 Slope-Intercept Form Example: Find the equation of the line with slope and y-intercept (0,  5). slope: y-intercept:

85. Martin-Gay, Intermediate Algebra, 5ed 85 The slope-intercept form uses, specifically, the y- intercept in the equation. The point-slope form allows you to use ANY point, together with the slope, to form the equation of the line. m is the slope (x 1, y 1 ) is a point on the line Point-Slope Form

86. Martin-Gay, Intermediate Algebra, 5ed 86 Find an equation of a line with slope – 2, through the point (– 11, – 12). Write the equation in standard form. First we substitute the slope and point into the point- slope form of an equation. y – (– 12) = – 2(x – (– 11)) y + 12 = – 2x – 22 Use distributive property. 2x + y + 12 = – 22 Add 2x to both sides. 2x + y = – 34 Subtract 12 from both sides. Point-Slope Form Example:

87. Martin-Gay, Intermediate Algebra, 5ed 87 Find the equation of the line through (– 4, 0) and (6, – 1). Write the equation in standard form. First find the slope. Point-Slope Form Example: Continued.

88. Martin-Gay, Intermediate Algebra, 5ed 88 Now substitute the slope and one of the points into the point-slope form of an equation. Clear fractions by multiplying both sides by 10. Use distributive property. Add x to both sides. Point-Slope Form Example continued:

89. Martin-Gay, Intermediate Algebra, 5ed 89 Find the equation of the line passing through points (2, 5) and ( – 4, 3). Write the equation in slope- intercept form. First find the slope. Point-Slope Form Example: Continued.

90. Martin-Gay, Intermediate Algebra, 5ed 90 Point-Slope Form Example continued:

91. Martin-Gay, Intermediate Algebra, 5ed 91 Remember that: nonvertical parallel lines have equal slopes. nonvertical perpendicular lines have slopes that are negative reciprocals of each other. if you rewrite linear equations into slope- intercept form, you can easily determine slope to compare lines. Horizontal and Vertical Lines

92. Martin-Gay, Intermediate Algebra, 5ed 92 Find an equation of a line that contains the point (– 2, 4) and is parallel to the line x + 3y = 6. Write the equation in standard form. First, we need to find the slope of the given line. 3y =  x + 6 Subtract x from both sides. y = x + 2 D ivide both sides by 3. Since parallel lines have the same slope, we use the slope of for our new equation, together with the given point. Horizontal and Vertical Lines Example: Continued.

93. Martin-Gay, Intermediate Algebra, 5ed 93 Multiply by 3 to clear fractions. Use distributive property. Add x to both sides. Add 12 to both sides. Horizontal and Vertical Lines Example continued:

94. Martin-Gay, Intermediate Algebra, 5ed 94 Find an equation of a line that contains the point (3,  5) and is perpendicular to the line 3x + 2y = 7. Write the equation in standard form. First, we need to find the slope of the given line. 2y =  3x + 7 Subtract 3x from both sides. y = x + Divide both sides by 2. Since perpendicular lines have slopes that are negative reciprocals of each other, we use the slope of for our new equation, together with the given point. Horizontal and Vertical Lines Example: Continued.

95. Martin-Gay, Intermediate Algebra, 5ed 95 Multiply by 3 to clear fractions. Use distributive property. Subtract 2x from both sides. Subtract 15 from both sides. Horizontal and Vertical Lines Example continued:

96. Martin-Gay, Intermediate Algebra, 5ed 96 § 3.6 Graphing Piecewise- Defined Functions and Shifting and Reflecting Graphs of Functions

97. Martin-Gay, Intermediate Algebra, 5ed 97 Graph each “piece” separately. Graph Graphing Piecewise-Defined Functions Continued. x f (x) = 3x – 1 0– 1 (closed circle) –1– 4 –2– 7 x f (x) = x + 3 1 4 2 5 3 6 Values  0. Values > 0. Example:

98. Martin-Gay, Intermediate Algebra, 5ed 98 Graphing Piecewise-Defined Functions x y x f (x) = x + 3 1 4 2 5 3 6 x f (x) = 3x – 1 0– 1 (closed circle) –1– 4 –2– 7 (0, –1) (–1, 4) (–2, 7) Open circle (0, 3) (3, 6) Example continued:

99. Martin-Gay, Intermediate Algebra, 5ed 99 Vertical and Horizontal Shifting Vertical Shifts (Upward or Downward) Let k be a Positive Number Graph ofSame As Moved g(x) = f(x) + kf(x)f(x)k units upward g(x) = f(x)  k f(x)f(x)k units downward Horizontal Shifts (To the Left or Right) Let h be a Positive Number Graph ofSame As Moved g(x) = f(x  h) f(x)f(x)h units to the right g(x) = f(x + h)f(x)f(x)h units to the left

100. Martin-Gay, Intermediate Algebra, 5ed 100 x y 5 5 55 55 Begin with the graph of f(x) = x 2. Shift the original graph downward 3 units. (0, –3) Vertical and Horizontal Shifting Example:

101. Martin-Gay, Intermediate Algebra, 5ed 101 Begin with the graph of f(x) = |x|. Shift the original graph to the left 2 units. (0, –2) x y 5 5 55 55 Vertical and Horizontal Shifting Example:

102. Martin-Gay, Intermediate Algebra, 5ed 102 Reflections of Graphs Reflection about the x-axis The graph of g(x) = – f(x) is the graph of f(x) reflected about the x-axis.

103. Martin-Gay, Intermediate Algebra, 5ed 103 Reflect the original graph about the x-axis. Reflections of Graphs x y 5 5 55 55 (4, –2) (4, 2) Example:

104. Martin-Gay, Intermediate Algebra, 5ed 104 § 3.7 Graphing Linear Inequalities

105. Martin-Gay, Intermediate Algebra, 5ed 105 Linear Inequalities in Two Variables Linear Inequality in Two Variables Written in the form Ax + By < C A, B, C are real #s, A & B are not both 0 Could use (>, ,  ) in place of < An ordered pair is a solution of the linear inequality if it makes the inequality a true statement.

106. Martin-Gay, Intermediate Algebra, 5ed 106 Graphing a Linear Inequality in Two Variables 1)Graph the boundary line found by replacing the inequality sign with an equal sign. If the sign is, graph a dashed line. If the sign is  or , graph a solid line. 2)Choose a test point not on the boundary line and substitute it into original inequality. 3)If a true statement is obtained in Step 2, shade the half-plane containing the point. If a false statement is obtained, shade the half-plane that does NOT contain the point. Linear Inequalities in Two Variables

107. Martin-Gay, Intermediate Algebra, 5ed 107 Graph 7x + y > –14 x y Pick a point not on the graph: (0,0) Graph 7x + y = –14 as a dashed line. Test it in the original inequality. 7(0) + 0 > – 14, 0 > – 14 True, so shade the side containing (0,0). (0, 0) Linear Inequalities in Two Variables Example:

108. Martin-Gay, Intermediate Algebra, 5ed 108 Graph 3x + 5y  –2 x y Pick a point not on the graph: (0,0), but just barely Graph 3x + 5y = –2 as a solid line. Test it in the original inequality. 3(0) + 5(0) > – 2, 0 > – 2 False, so shade the side that does not contain (0,0). (0, 0) Linear Inequalities in Two Variables Example:

109. Martin-Gay, Intermediate Algebra, 5ed 109 Graph 3x < 15 x y Pick a point not on the graph: (0,0) Graph 3x = 15 as a dashed line. Test it in the original inequality. 3(0) < 15, 0 < 15 True, so shade the side containing (0,0). (0, 0) Linear Inequalities in Two Variables Example:

110. Martin-Gay, Intermediate Algebra, 5ed 110 Note that although all of our examples allowed us to select (0, 0) as our test point, that will not always be true. If the boundary line contains (0,0), you must select another point that is not contained on the line as your test point. Warning! Linear Inequalities in Two Variables