1. Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Introduction to Graphing 3.1Reading Graphs, Plotting Points, and Scaling Graphs 3.2Graphing Linear Equations 3.3Graphing and Intercepts 3.4Rates 3.5Slope 3.6Slope--Intercept Form 3.7Point--Slope Form 3

3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Reading Graphs, Plotting Points, and Scaling Graphs Problem Solving with Bar, Circle, and Line Graphs Points and Ordered Pairs Numbering the Axes Appropriately 3.1

4. Slide 3- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A bar graph is a convenient way of showing comparisons. In every bar graph certain categories, such as body weight in the example, are paired with certain numbers.

5. Slide 3- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A blood-alcohol level of 0.08% or higher makes driving illegal in the United States. The bar graph shows how many drinks a person of a certain weight would need to consume in 1 hr to achieve a blood- alcohol level of 0.08% a) Approximately how many drinks would a 180- pound person have consumed if he or she had a blood-alcohol level of 0.08%? b) What can be concluded about the weight of someone who can consume 2 drinks in an hour without reaching a blood-alcohol level of 0.08%? Example

6. Slide 3- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) Approximately how many drinks would a 180- pound person have consumed if he or she had a blood-alcohol level of 0.08%? Go to the top of the bar that is above the body weight 180 lb. Then we move horizontally from the top of the bar to the vertical scale listing numbers of drinks. It appears that approximately 4 drinks will give a 180-lb person a blood-alcohol level of 0.08%.

7. Slide 3- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution b) What can be concluded about the weight of someone who can consume 2 drinks in an hour without reaching a blood-alcohol level of 0.08%? By moving up the vertical scale to the number 2, and then moving horizontally, we see that the first bar to reach a height of 2 corresponds to a weight of 100 lb. Thus an individual should weigh over 100 pounds if he or she wishes to consume 2 drinks in an hour without exceeding a blood-alcohol level of 0.08%.

8. Slide 3- 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The circle graph shows the favorite colors of Americans and the percentage that prefers each color. There are approximately 292 million Americans, and three quarters of them live in cities. Assuming that color preference is unaffected by geographical location, how many urban Americans choose blue as their favorite color? Example

9. Slide 3- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1. Familarize. The problem involves percents. We are told that three quarters of all Americans live in cities. Since there are about 292 million Americans, this amounts to The chart indicates that 44% of the U.S. population prefers blue. We let b = the number of urban Americans who select blue as their favorite color.

10. Slide 3- 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Rewording: What is 44% of 219 million? Translating: b = 0.44 219,000,000 3. Carry out. b = 0.44 219,000,000 b = 96,360,000 4. Check. The check is left to the student. 5. State. About 96,360,000 urban Americans choose blue as their favorite color.

11. Slide 3- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Points and Ordered Pairs To graph, or plot, points we use two perpendicular number lines called axes. The point at which the axes cross is called the origin. Arrows on the axes indicate the positive directions. Consider the pair (2, 3). The numbers in such a pair are called the coordinates. The first coordinate in this case is 2 and the second coordinate is 3.

12. Slide 3- 12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Points and Ordered Pairs continued To plot the point (2, 3) we start at the origin, move horizontally to the 2, move up vertically 3 units, and then make a “dot”. (2, 3)

13. Slide 3- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Plot the point (  4, 3). Solution Starting at the origin, we move 4 units in the negative horizontal direction. The second number, 3, is positive, so we move 3 units in the positive vertical direction (up). 4 units left 3 units up Example

14. Slide 3- 14 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the coordinates of points A, B, C, D, E, F, and G. Solution Point A is 5 units to the right of the origin and 3 units above the origin. Its coordinates are (5, 3). The other coordinates are as follows: B: (  2, 4) C: (  3,  4) D: (3,  2) E: (2, 3) F: (  3, 0) G: (0, 2) A B C D E F G Example

15. Slide 3- 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Numbering the Axes Appropriately Often it is necessary to graph a range of x-values and /or y-values that is too large to be displayed if each square of the grid is one unit wide and one unit high. Use a grid 10 square wide and 10 squares high to plot (  24, 460) and (38, 85) and (10,  170). Solution The x-values range from a low of  24 to a high of 38. The span is 38  (  24) = 62 units Example

16. Slide 3- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Because 62 is not a multiple of 10, we round up to the next multiple of 10, which is 70. Dividing 70 by 10 we would have a scale of 7 which is not convenient so we round up to 10. The y-values span from  170 to 460, the vertical squares must span from 460  (  170) = 630. For convenience, we round 630 up to 700, and then divide 700 by 10 = 70. Using 70 as the scale. Solution (continued)

17. Slide 3- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Combine the work from the x and y-values to create the grid and plot the points.

18. Slide 3- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The horizontal and vertical axes divide the plane into four regions, or quadrants.

19. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Linear Equations Solutions of Equations Graphing Linear Equations Applications 3.2

20. Slide 3- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether each of the following pairs is a solution of 4y + 3x = 18: a) (2, 3); b)(1, 5). Solution a) We substitute 2 for x and 3 for y. 4y + 3x = 18 43 + 32 | 18 12 + 6 | 18 18 = 18 True b) We substitute 1 for x and 5 for y. 4y + 3x = 18 45 + 31 | 18 20 + 3 | 18 23 = 18 False Since 18 = 18 is true, the pair (2, 3) is a solution. Since 23 = 18 is false, the pair (1, 5) is not a solution. Example

21. Slide 3- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Graph a Linear Equation 1.Select a value for one coordinate and calculate the corresponding value of the other coordinate. Form an ordered pair. This pair is one solution of the equation. 2.Repeat step (1) to find a second ordered pair. A third ordered pair can be used as a check. 3.Plot the ordered pairs and draw a straight line passing through the points. The line represents all solutions of the equation.

22. Slide 3- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph y =  4x + 1 Solution We select convenient values for x and compute y, and form an ordered pair. If x = 2, then y =  4(2) + 1 =  7 and (2,  7) is a solution. If x = 0, then y =  4(0) + 1 = 1 and (0, 1) is a solution. If x =  2, then y =  4(  2) + 1 = 9 and (  2, 9) is a solution. Example

23. Slide 3- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Results are often listed in a table. (1) Choose x. (2) Compute y. (3) Form the pair (x, y). (4) Plot the points. xy(x, y) 2 77(2,  7) 01(0, 1) 22 9 (  2, 9)

24. Slide 3- 24 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that all three points line up. If they didn’t we would know that we had made a mistake. Finally, use a ruler or other straightedge to draw a line. Every point on the line represents a solution of y =  4x + 1 Solution (continued)

25. Slide 3- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Select some convenient x-values and compute y-values. If x = 6, then 6 + 2y = 6  y = 0 If x = 0, then 0 + 2y = 6  y = 3 If x = 2, then 2 + 2y = 6  y = 2 Graph x + 2y = 6 xy(x, y) 60(6, 0) 03(0, 3) 22(2, 2) Example

26. Slide 3- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 4y = 3x Solution Begin by solving for y. To graph we can select values of x that are multiples of 4. This will allow us to avoid fractions when corresponding y-values are computed. Example

27. Slide 3- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Select some convenient x-values and compute y-values. If x = 0, then ¾ (0) = 0 If x = 4, then ¾ (4) = 3 If x =  4, then ¾ (  4) =  3 xy(x, y) 00(0, 0) 43(4, 3) 44 33(  4,  3) Solution (continued)

28. Slide 3- 28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications The cost c, in dollars, of shipping a FedEx Priority Overnight package weighing 1 lb or more a distance of 1001 to 1400 mi is given by c = 2.8w + 21.05 where w is the package’s weight in pounds. Graph the equation and then use the graph to estimate the cost of shipping a 10 ½-pound package. Example

29. Slide 3- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Select values for w and then calculate c. c = 2.8w + 21.05 If w = 2, then c = 2.8(2) + 21.05 = 26.65 If w = 4, then c = 2.8(4) + 21.05 = 32.25 If w = 8, then c = 2.8(8) + 21.05 = 43.45 wc 226.65 432.25 843.45

30. Slide 3- 30 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Weight (in pounds) Mail cost (in dollars) Solution (continued) Plot the points. To estimate an 10 ½ pound package, we locate the point on the line that is above 10 ½ and then find the value on the c-axis that corresponds to that point. The cost of shipping an 10 ½ pound package is about $51.00. 10 ½ pounds

31. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing and Intercepts Intercepts Using Intercepts to Graph Graphing Horizontal or Vertical Lines 3.3

32. Slide 3- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For the graph shown below, a) give the coordinates of any x-intercepts and b) give the coordinates of any y-intercepts. Solution a) The x-intercepts are (  2, 0) and (2, 0). b) The y-intercept is (  4, 0). Example

33. Slide 3- 33 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Find Intercepts To find the y-intercept(s) of an equation’s graph, replace x with 0 and solve for y. To find the x-intercept(s) of an equation’s graph, replace y with 0 and solve for x.

34. Slide 3- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the y-intercept and the x-intercept of the graph of 5x + 2y = 10. Solution To find the y-intercept, we let x = 0 and solve for y: 5 0 + 2y = 10 2y = 10 y = 5 The y-intercept is (0, 5). To find the x-intercept, we let y = 0 and solve for x. 5x + 2 0 = 10 5x = 10 x = 2 The x-intercept is (2, 0). Replacing x with 0 Replacing y with 0 Example

35. Slide 3- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 5x + 2y = 10 using intercepts. Solution We found the intercepts in the previous example. Before drawing the line, we plot a third point as a check. If we let x = 4, then 5 4 + 2y = 10 20 + 2y = 10 2y =  10 y =  5 We plot (4,  5), (0, 5) and (2, 0). 5x + 2y = 10 x-intercept (2, 0) y-intercept (0, 5) Example

36. Slide 3- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 3x  4y = 8 using intercepts. Solution To find the y-intercept, we let x = 0. This amounts to ignoring the x-term and then solving.  4y = 8 y =  2 The y-intercept is (0,  2). To find the x-intercept, we let y = 0. This amounts to ignoring the y-term and then solving. 3x = 8 x = 8/3 The x-intercept is (8/3, 0). Example

37. Slide 3- 37 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Graph 3x  4y = 8 using intercepts. Solution Find a third point. If we let x = 4, then 3 4  4y = 8 12  4y = 8  4y =  4 y = 1 We plot (0,  2); (8/3, 0); and (4, 1). x-intercept y-intercept 3x  4y = 8

38. Slide 3- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph y = 2 Solution We regard the equation y = 2 as 0 x + y = 2. No matter what number we choose for x, we find that y must equal 2. y = 2 Choose any number for x. y must be 2. xy(x, y) 02(0, 2) 42(4, 2) 44 2 (  4, 2) Example

39. Slide 3- 39 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued y = 2 Solution When we plot the ordered pairs (0, 2), (4, 2) and (  4, 2) and connect the points, we obtain a horizontal line. Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y-intercept (0, 2). y = 2 (  4, 2) (0, 2) (4, 2)

40. Slide 3- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley xy(x, y) 22 4 (  2, 4) 22 1 (  2, 1) 22 44(  2,  4) x must be  2. Graph x =  2 Solution We regard the equation x =  2 as x + 0 y =  2. We make up a table with all  2 in the x-column. x =  2 Any number can be used for y. Example

41. Slide 3- 41 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued x =  2 Solution When we plot the ordered pairs (  2, 4), (  2, 1), and (  2,  4) and connect them, we obtain a vertical line. Any ordered pair of the form (  2, y) is a solution. The line is parallel to the y-axis with x-intercept (  2, 0). x =  2 (  2,  4) (  2, 4) (  2, 1)

42. Slide 3- 42 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

43. Slide 3- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write an equation for the graph. Solution Note that every point on the horizontal line passing through (0,  3) has  3 as the y-coordinate. The equation of the line is y =  3. Example

44. Slide 3- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write an equation for the graph. Solution Note that every point on the vertical line passing through (4, 0) has 4 as the x-coordinate. The equation of the line is x = 4. Example

45. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rates Rates of Change Visualizing Rates 3.4

46. Slide 3- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rate A rate is a ratio that indicates how two quantities change with respect to each other.

47. Slide 3- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley On March 4, Nichole rented a mini-van with a full tank of gas and 10,324 mi on the odometer. On March 9, she returned the mini-van with 10,609 mi on the odometer. If the rental agency charged Nichole $126 for the rental and needed 15 gal of gas to fill up the gas tank, find the following rates: a) The car’s average rate of gas consumption, in miles per gallon. b) The average cost of the rental, in dollars per day. c) The car’s rate of travel, in miles per day. Example

48. Slide 3- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley On March 4, Nichole rented a mini-van with a full tank of gas and 10,324 mi on the odometer. On March 9, she returned the mini-van with 10,609 mi on the odometer. If the rental agency charged Nichole $126 for the rental and needed 15 gal of gas to fill up the gas tank, find the following rates: Solution a) The car’s average rate of gas consumption, in miles per gallon. Rate, in miles per gallon Example

49. Slide 3- 49 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley On March 4, Nichole rented a mini-van with a full tank of gas and 10,324 mi on the odometer. On March 9, she returned the mini-van with 10,609 mi on the odometer. If the rental agency charged Nichole $126 for the rental and needed 15 gal of gas to fill up the gas tank, find the following rates: Solution (continued) b) The average cost of the rental, in dollars per day. Rate, in dollars per day Example

50. Slide 3- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) c) The car’s rate of travel, in miles per day. Rate, in miles per day On March 4, Nichole rented a mini-van with a full tank of gas and 10,324 mi on the odometer. On March 9, she returned the mini-van with 10,609 mi on the odometer. If the rental agency charged Nichole $126 for the rental and needed 15 gal of gas to fill up the gas tank, find the following rates: Example

51. Slide 3- 51 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Wanda’s Hair Salon has a graph displaying data from a recent day of work. a) What rate can be determined from the graph? b) What is that rate? Solution a) We can find the rate Number of haircuts per hour. b) 12345 Example

52. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope Rate and Slope Horizontal and Vertical Lines Applications 3.5

53. Slide 3- 53 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope The slope of the line containing points (x 1, y 1 ) and (x 2, y 2 ) is given by Change in y Change in x

54. Slide 3- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph the line containing the points (  4, 5) and (4,  1) and find the slope. Solution Change in y =  6 Change in x = 8 Example

55. Slide 3- 55 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the slope of the line y = 3 Solution Consider the points (  3, 3) and (2, 3), which are on the line. A horizontal line has slope 0. (  3, 3) (2, 3) Example

56. Slide 3- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the slope of the line x = 2 Solution Consider the points (2, 4) and (2,  2), which are on the line. The slope of a vertical line is undefined. (2, 4) (2,  2) Example

57. Slide 3- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications of Slope Some applications use slope to measure the steepness. For examples, numbers like 2%, 3%, and 6% are often used to represent the grade of a road, a measure of a road’s steepness. That is, a 3% grade means that for every horizontal distance of 100 ft, the road rises or drops 3 ft.

58. Slide 3- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the slope (or grade) of the treadmill. 0.42 ft 5.5 ft Solution The grade of the treadmill is 7.6%. ** Reminder: Grade is slope expressed as a percent. Example

59. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope-Intercept Form Using the y-intercept and the Slope to Graph a Line Equations in Slope-Intercept Form Graphing and Slope-Intercept Form Parallel and Perpendicular Lines 3.6

60. Slide 3- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Draw a line that has slope 2/3 and y-intercept (0,  1). Solution We plot (0,  1) and from there move up 2 units and to the right three units. This locates the point (3, 1). We plot the point and draw a line passing through the two points. y-intercept up 2 right 3 (0,  1) (3, 1) Example

61. Slide 3- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Slope-Intercept Equation The equation y = mx + b is called the slope-intercept equation. The equation represents a line of slope m with y-intercept (0, b).

62. Slide 3- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the slope and the y-intercept of each line whose equation is given. a)b) 3x + y = 7c) 4x  5y = 10 Solution a) The slope is. The y-intercept is (0,  2). b) We first solve for y to find an equivalent form of y = mx + b. 3x + y = 7 y =  3x + 7 The slope is  3. The y-intercept is (0, 7). Example

63. Slide 3- 63 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley c) We rewrite the equation in the form y = mx + b. 4x  5y = 10 The slope is 4/5. The y-intercept is (0,  2).

64. Slide 3- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A line has slope  3/7 and y- intercept (0, 8). Find an equation of the line. Solution We use the slope-intercept equation, substituting  3/7 for m and 8 for b: The desired equation is Example

65. Slide 3- 65 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph: Solution The slope is 4/3 and the y-intercept is (0,  2). We plot (0,  2), then move up 4 units and to the right 3 units. We could also move down 4 units and to the left 3 units. Then draw the line. up 4 units right 3 down 4 left 3 (  3,  6) (3, 2) (0,  2) Example

66. Slide 3- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph: 3x + 4y = 12 Solution Rewrite the equation in slope-intercept form. Example

67. Slide 3- 67 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution The slope is  3/4 and the y-intercept is (0, 3). We plot (0, 3), then move down 3 units and to the right 4 units. An alternate approach would be to move up 3 units and to the left 4 units. up 3 down 3 left 4 right 4 (0, 3) (4, 0) (  4, 6)

68. Slide 3- 68 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parallel and Perpendicular Lines Two lines are parallel if they lie in the same plane and do not intersect no matter how far they are extended.

69. Slide 3- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope and Parallel Lines Two lines are parallel if they have the same slope or if both lines are vertical.

70. Slide 3- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether the graphs of and 3x  2y =  5 are parallel. Solution Remember that parallel lines extend indefinitely without intersecting. Thus, two lines with the same slope but different y-intercepts are parallel. Example

71. Slide 3- 71 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) The line has slope 3/2 and y-intercept 3. We need to rewrite 3x  2y =  5 in slope-intercept form: 3x  2y =  5  2y =  3x  5 The slope is 3/2 and the y-intercept is 5/2. Both lines have slope 3/2 and different y-intercepts, the graphs are parallel.

72. Slide 2- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope and Perpendicular Lines Two lines are perpendicular if the product of their slopes is –1 or if one line is vertical and the other is horizontal. Thus, if one line has slope m the slope of a line perpendicular to it is –1/m.

73. Slide 2- 73 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Consider the line given by the equation 5y = 3x + 10. a) Find an equation for a parallel line passing through (0, 6). b) Find an equation for a perpendicular line passing through (0, 6). Example

74. Slide 2- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Both parts (a) and (b) require find the slope of the line given by 5y = 3x + 10. Solve for y to find the slope-intercept form: The slope is 3/5.

75. Slide 2- 75 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley a) The slope of any parallel line will be 3/5. The slope-intercept form yields: b) The slope of any perpendicular line will be –5/3 (the opposite of the reciprocal of 3/5). The slope-intercept form yields:

76. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Point-Slope Form Writing Equations in Point-Slope Form Graphing and Point-Slope Form Estimations and Predictions Using Two Points 3.7

77. Slide 3- 77 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Point-Slope Equation The equation is called the point-slope equation for the line with slope m that contains the point (x 1, y 1 ).

78. Slide 3- 78 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write a point-slope equation for the line with slope 2/3 that contains the point (4, 9). Solution We substitute 2/3 for m, and 4 for x 1, and 9 for x 2 : Example

79. Slide 3- 79 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write the slope-intercept equation for the line with slope 3 and point (4, 3). Solution There are two parts to this solution. First, we write an equation in point-slope form: Example

80. Slide 3- 80 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Next, we find an equivalent equation of the form y = mx + b: Using the distributive law Adding 3 to both sides to get the slope-intercept form

81. Slide 3- 81 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing and Point-Slope Form When we know a line’s slope and a point that is on the line, we can draw the graph, much like we did in Section 3.6.

82. Slide 3- 82 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Since y  3 = 2(x  1) is in point- slope form, we know that the line has slope 2 and passes through the point (1, 3). We plot (1, 3) and then find a second point by moving up 2 units and to the right 1 unit. Then the line can be drawn. Graph y  3 = 2(x  1) (3, 1) up 2 right 1 (1, 3) Example

83. Slide 3- 83 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find an equivalent equation: The line passes through (  2,  3) and has slope  4/3. Graph down 4 right 3 (  2,  3) Example

84. Slide 3- 84 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Estimations and Predictions Using Two Points It is possible to use line graphs to estimate real-life quantities that are not already known. To do so, we calculate the coordinates of an unknown point by using two points with known coordinates. When the unknown point is located between the two points, this process is called interpolation. Sometimes a graph passing through the known points is extended to predict future values. Making predictions in this manner is called extrapolation.

85. Slide 3- 85 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Aerobic exercise. A person’s target heart rate is the number of beats per minute that bring the most aerobic benefit to his or her heart. The target heart rate for a 20-year-old is 150 beats per minute and for a 60-year- old, 120 beats per minute. a) Graph the given data and calculate the target heart rate for a 46-year-old. b) Calculate the target heart rate for a 70-year-old. Example

86. Slide 3- 86 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) We draw the axes and label, using a scale that will permit us to view both the given and the desired data. The given information allows us to then plot (20, 150) and (60, 120).

87. Slide 3- 87 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued We determine the slope of the line. Use one point and write the equation of the line.

88. Slide 3- 88 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued a) To calculate the target heart rate for a 46-year-old, we substitute 46 for x in the slope-intercept equation: The graph confirms the target heart rate.

89. Slide 3- 89 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) To calculate the target heart rate for a 70-year-old, we substitute 70 for x in the slope-intercept equation: The graph confirms the target heart rate.