1. 1 Copyright © 2011 Pearson Education, Inc.

2. Equations and Inequalities in Two Variables; Functions CHAPTER 3.1Graphing Linear Equations 3.2The Slope of a Line 3.3The Equation of a Line 3.4Graphing Linear Inequalities 3.5Introduction to Functions and Function Notation 3

3. Copyright © 2011 Pearson Education, Inc. The Slope of a Line 3.2 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.

4. Slide 3- 4 Copyright © 2011 Pearson Education, Inc. Example 1a 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

5. Slide 3- 5 Copyright © 2011 Pearson Education, Inc. Example 1b 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

6. Slide 3- 6 Copyright © 2011 Pearson Education, Inc. 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 line gets.

7. Slide 3- 7 Copyright © 2011 Pearson Education, Inc. Slope: The ratio of the vertical change (change in y) to the horizontal change (change in x) between any two points on a line between those points.

8. Slide 3- 8 Copyright © 2011 Pearson Education, Inc. Example 2 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)

9. Slide 3- 9 Copyright © 2011 Pearson Education, Inc. Graphs of Equations in Slope-Intercept Form The graph of an equation in the form y = mx + b, (slope-intercept form) is a line with slope m and y- intercept (0, b). The following rules indicate how m affects the graph. If m > 0, the line slants upward from left to right. If m < 0, the line slants downward from left to right. The greater the absolute value of m, the steeper the line. m > 0 m < 0

10. Slide 3- 10 Copyright © 2011 Pearson Education, Inc. Example 3 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).

11. Slide 3- 11 Copyright © 2011 Pearson Education, Inc. continued We begin at (0, –4) and then rise 2 and run 5. m = y-intercept: (0, –4) rise 2 run 5

12. Slide 3- 12 Copyright © 2011 Pearson Education, Inc. 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 (x 1, y 1 ) (x2, y 2 )

13. Slide 3- 13 Copyright © 2011 Pearson Education, Inc. Find the slope of the line connecting (4, 6) and (−2, 8). Solution Example 4 Using, replace the variables with their corresponding values and then simplify.

14. Slide 3- 14 Copyright © 2011 Pearson Education, Inc. Graph the line connecting the given points and find its slope. (3, 8) and (−2, 8) Solution Example 5a Because the y-coordinates are the same, the graphs is a horizontal line.

15. Slide 3- 15 Copyright © 2011 Pearson Education, Inc. Graph the line connecting the given points and find its slope. (6, 1) and (6, −4) Solution Example 5b Because the x-coordinates are the same, the graphs is a vertical line. Graph the line connecting the given points and find its slope. (6, 1) and (6, −4)

16. Slide 3- 16 Copyright © 2011 Pearson Education, Inc. Slopes of Horizontal and Vertical Lines Two points with different x-coordinates and the same y- coordinates, (x 1, c) and (x 2, c), will form a horizontal line with slope 0 and equation y = c. Two points with the same x-coordinates and different y- coordinates, (c, y 1 ) and (c, y 2 ), will form a vertical line with undefined slope and equation x = c.

17. Slide 2 - 17 Copyright © 2011 Pearson Education, Inc. Example 6 The following graph shows the hourly wage earned by Henry each of the five years after his hire date. An analysis determines that the red line reasonably describes the trend shown in the data. Find the slope of that line. Understand We are to find the slope on a line that passes through or near a set of data points.

18. Slide 2 - 18 Copyright © 2011 Pearson Education, Inc. continued Execute Plan To find the slope of the line, we use the slope formula with two data points that are on the line. We will use (3, 15) and (5, 18). Answer The slope of the line is, which means that Henry’s wage increased about $1.50 per hour each year. Check We could use a different pair of data points on the line and see if we get the same slope. We will leave this to the viewer.

19. Slide 3- 19 Copyright © 2011 Pearson Education, Inc. What is the slope of the line 3x + y = 6? a) m = 3 b) m =  3 c) d) m = 2 3.2

20. Slide 3- 20 Copyright © 2011 Pearson Education, Inc. What is the slope of the line 3x + y = 6? a) m = 3 b) m =  3 c) d) m = 2 3.2

21. Slide 3- 21 Copyright © 2011 Pearson Education, Inc. 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 3.2

22. Slide 3- 22 Copyright © 2011 Pearson Education, Inc. 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 3.2

23. Slide 3- 23 Copyright © 2011 Pearson Education, Inc. What is the slope of the line through the given points (5, 6) and (6,  3)? a) m =  9 b) m = 9 c) d) 3.2

24. Slide 3- 24 Copyright © 2011 Pearson Education, Inc. What is the slope of the line through the given points (5, 6) and (6,  3)? a) m =  9 b) m = 9 c) d) 3.2

25. Copyright © 2011 Pearson Education, Inc. The Equation of a Line 3.3 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.

26. Slide 3- 26 Copyright © 2011 Pearson Education, Inc. 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.

27. Slide 3- 27 Copyright © 2011 Pearson Education, Inc. Example 1 A line with a slope of 4 crosses the y-axis at the point (0, 5). Write the equation of the line in slope-intercept form. Solution Use y = mx + b m = 4 b = 5 from the point given (0, 5) which is the y-intercept. y = 4x + 5

28. Slide 3- 28 Copyright © 2011 Pearson Education, Inc. 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, calculate the slope using then use y = mx + b.

29. Slide 3- 29 Copyright © 2011 Pearson Education, Inc. Example 2 Write the equation of the line passing through (0,  6) and (3, 6) in slope-intercept form. Solution The y-intercept is (0,  6). Find the slope: y = mx + b y = 4x  6

30. Slide 3- 30 Copyright © 2011 Pearson Education, Inc. 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 ).

31. Slide 3- 31 Copyright © 2011 Pearson Education, Inc. Example 3a Write the equation of a line with a slope of 5 that passes 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.

32. Slide 3- 32 Copyright © 2011 Pearson Education, Inc. Example 3b 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.

33. Slide 3- 33 Copyright © 2011 Pearson Education, Inc. continued Points (4,  2) and (  4, 4), y – y 1 = m(x – x 1 ) Replace m with -3/4, x 1 with 4 and y 1 with -2. Simplify. Subtract 2 from both sides to isolate y.

34. Slide 3- 34 Copyright © 2011 Pearson Education, Inc. Equations can also be written in standard form, which is Ax + By = C, where A, B, and C are real numbers. It should also be written so that the x-term is first and with a positive coefficient and, if possible, A, B, and C are all integers.

35. Slide 3- 35 Copyright © 2011 Pearson Education, Inc. Example 4 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 ( ).

36. Slide 3- 36 Copyright © 2011 Pearson Education, Inc. 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.

37. Slide 3- 37 Copyright © 2011 Pearson Education, Inc. Parallel Lines Nonvertical parallel lines have equal slopes and different y-intercepts. Vertical lines are parallel. y = 2x + 1 y = 2x – 3

38. Slide 3- 38 Copyright © 2011 Pearson Education, Inc. Example 5 Write the equation of the line in slope-intercept form that passes through (1, –5) and parallel to the graph of y = –3x + 4. 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.

39. Slide 3- 39 Copyright © 2011 Pearson Education, Inc. Perpendicular Lines The slope of a line perpendicular to a line with a slope of will be Horizontal and vertical lines are perpendicular.

40. Slide 3- 40 Copyright © 2011 Pearson Education, Inc. Example 6 Write the equation of a line in standard form that passes through (7, 1) and is perpendicular to the graph of 7x – 2y = –2. Solution Determine the slope of the line 7x – 2y = –2. Slope of perpendicular line:

41. Slide 3- 41 Copyright © 2011 Pearson Education, Inc. continued slope = ; point (7, 1) y – y 1 = m(x – x 1 ) Multiply both sides by 7, simplify and distribute. Rearrange the x and y terms, and add 7 to both sides to put the equation in Standard Form.

42. Slide 3- 42 Copyright © 2011 Pearson Education, Inc. 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 3.3

43. Slide 3- 43 Copyright © 2011 Pearson Education, Inc. 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 3.3

44. Slide 3- 44 Copyright © 2011 Pearson Education, Inc. 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) 3.3

45. Slide 3- 45 Copyright © 2011 Pearson Education, Inc. 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) 3.3

46. Slide 3- 46 Copyright © 2011 Pearson Education, Inc. What is the relationship between the two lines? 5x – 3y = 11 3x + 5y = 8 a) parallel b) perpendicular c) neither 3.3

47. Slide 3- 47 Copyright © 2011 Pearson Education, Inc. What is the relationship between the two lines? 5x – 3y = 11 3x + 5y = 8 a) parallel b) perpendicular c) neither 3.3

48. Copyright © 2011 Pearson Education, Inc. Introduction to Functions and Function Notation 3.5 1.Identify the domain and range of a relation and determine whether a relation is a function. 2.Find the value of a function. 3.Graph functions.

49. Slide 3- 49 Copyright © 2011 Pearson Education, Inc. Relation: A set of ordered pairs. Domain: The set containing initial values of a relation; its input values; the first coordinates in ordered pairs. Range: The set containing all values that are paired to domain values in a relation; its output values; the second coordinates in ordered pairs.

50. Slide 3- 50 Copyright © 2011 Pearson Education, Inc. Function: A relation in which each value in the domain is assigned to 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.

51. Slide 3- 51 Copyright © 2011 Pearson Education, Inc. 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

52. Slide 3- 52 Copyright © 2011 Pearson Education, Inc. Example 1 Identify the domain and range of the relation, then determine if it is a function. BirthdateFamily member 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. Domain: {March 1, April 17, Sept 3, Oct 9} Range: {Donna, Dennis, Catherine, Denise, Nancy}

53. Slide 3- 53 Copyright © 2011 Pearson Education, Inc. Determining the Domain and Range of a Graph The domain is a set containing the first coordinate (x-coordinate) of every point on the graph. The range is a set containing the second coordinate (y-coordinate) of every point on the graph.

54. Slide 3- 54 Copyright © 2011 Pearson Education, Inc. Vertical Line Test To determine whether a graphical relation is a function, draw or imagine vertical lines through each value in the domain. If each vertical line intersects the graph at only one point, the relation is a function. If any vertical line intersects the graph more than once, the relation is not a function.

55. Slide 3- 55 Copyright © 2011 Pearson Education, Inc. Example 2 For each graph, identify 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

56. Slide 3- 56 Copyright © 2011 Pearson Education, Inc. 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”

57. Slide 3- 57 Copyright © 2011 Pearson Education, Inc. 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 then evaluate or simplify.

58. Slide 3- 58 Copyright © 2011 Pearson Education, Inc. Example 3 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

59. Slide 3- 59 Copyright © 2011 Pearson Education, Inc. Example 4 Use the graph to find the indicated value of the function. a. f(0)b. f(2)c. f(8) Solution a. When x = 0, y = 0, so f(0) = 0. b. When x = 2, y = 2, so f(2) = 2. c. When x = 8, y = 4, so f(8) = 4.

60. Slide 3- 60 Copyright © 2011 Pearson Education, Inc. 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

61. Slide 3- 61 Copyright © 2011 Pearson Education, Inc. Example 5 Graph: f(x) = 2x + 1. Give the domain and range. Solution We could make a table of values or use the fact that the slope is 2 and the y-intercept is 1. Domain: Range: xf(x)f(x) 01 13 25 f(x) = 2x + 1

62. Slide 3- 62 Copyright © 2011 Pearson Education, Inc. Example 7 Graph. f(x) = 2x 2 Solution We create a table of ordered pairs, plot the points, and connect with a smooth curve. xf(x)f(x) 22 8 11 2 00 12 28

63. Slide 3- 63 Copyright © 2011 Pearson Education, Inc. Is the relation a function? a) yes b) no 3.5

64. Slide 3- 64 Copyright © 2011 Pearson Education, Inc. Is the relation a function? a) yes b) no 3.5

65. Slide 3- 65 Copyright © 2011 Pearson Education, Inc. For the function f(x) = 2x 2 + x – 4 find f(1). a) 3 b) –4 c) 2 d)  1 3.5

66. Slide 3- 66 Copyright © 2011 Pearson Education, Inc. For the function f(x) = 2x 2 + x – 4 find f(1). a) 3 b) –4 c) 2 d)  1 3.5