1. 2 Graphs and Functions © 2008 Pearson Addison-Wesley. All rights reserved Sections 2.4–2.5

2. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-2 Linear Functions 2.4 Graphing Linear Functions ▪ Standard From Ax + By = C ▪ Slope ▪ Average Rate of Change ▪ Linear Models

3. Linear Functions Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-3 A linear function has the form: f(x) = mx + b where m and b are real numbers and m is the slope which comes from the equation for the given points (x 1,y 1 ) and (x 2,y 2 ).

4. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-4 2.4 Finding Slopes with the Slope Formula Find the slope of the line through the given points. (a) (–2, 4), (2, –6) (b) (–3, 8), (5, 8) (c) (–4, –10), (–4, 10) the slope is undefined

5. Forms for the Equations of a Line 1. uses slope, m and given point (x 1, y 1 ) 2.Slope-intercept form:y = mx + b uses slope, m and y-intercept = b 3.Standard form:Ax +By = C Slope is m = -A/B and (0, C/B) and (C/A,0) 4.Horizontal form:y = b Slope is zero, y-interecept (0,b) 5.Vertical form:x = a slope is undefined, not a function. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-5 Point-slope form: y – y 1 = m(x – x 1 )

6. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-6 2.4 Example 6 Finding the Slope from an Equation (page 222) Find the slope of the line y = 5x – 3. This equation is in slope-intercept form. You can identify the slope, m from its form. m = 5

7. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-7 2.4 Graphing a Line Using a Point and the Slope Graph the line passing through (–2, –3) and having slope. Plot the point (–2, –3). The slope is, so a change of 3 units horizontally produces a change of 4 units vertically. This gives the point (1, 1). Join the points with a straight line to complete the graph.

8. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-8 2.4 Graphing a Linear Function Using Intercepts Graph. Give the domain and range. Since this equation is in slope-intercept form, pull the slope from the equation, m = 3/2 and the y-intercept, (0,6). Use the slope to move from the point (0,6) up 3 units and to the right 2 units. (0,6) (2,9) Domain: Range:

9. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-9 2.4 Graphing a Linear Function Using Intercepts Corresponding calculator graph:

10. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-10 2.4 Example 2 Graphing a Horizontal Line (page 219) Graph f(x) = 2. Give the domain and range. Since f(x) always equals 2, the value of y can never be 0. So, there is no x-intercept. The line is parallel to the x-axis. The y-intercept is 2. Domain: Range: {2}

11. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-11 2.4 Example 3 Graphing a Vertical Line (page 219) Graph x = 5. Give the domain and range. Since x always equals 5, the value of x can never be 0. So, there is no y-intercept. The line is parallel to the y-axis. The relation is not a function. The x-intercept is 5. Domain: {5} Range:

12. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-12 2.4 Example 4 Graphing Ax + By = C with C = 0 (page 220) Graph 3x + 4y = 0. Give the domain and range. Find the intercepts. x-intercept ( x,0) 3x + 4(0) = 0 x = 0 y-intercept (0, y ) 3(0) + 4y = 0 y = 0 There is only one intercept, the origin, (0, 0). Choose x = 4 to find a second point. 3(4) + 4y = 0 12 + 4y = 0 y = –3

13. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-13 2.4 Example 4 Graphing Ax + By = C with C = 0 (cont.) Graph the points (0, 0) and (4, –3) and join with a straight line. Domain: Range:

14. Average Rate of Change Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-14 We know that the slope of a line is the ratio of vertical change in y to the horizontal change in x. Thus, the slope gives the average rate of change in y per unit of change in x, where y = f(x). If f is a linear function we get the slope formula. Average rate of change of f on a given interval, [a,b] is

15. Types of Slopes The slope indicates what is happening to a line on the graph: a)If m > 0, the line will rise from left to right, i.e. it is increasing, b)If m < 0, the line will fall from left to right, i.e. it is decreasing, c)If m = 0, then there is no change in the line and it is horizontal, i.e. it is constant d)If m is undefined, the line is a vertical line and not a function. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-15

16. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-16 2.4 Example 8 Interpreting Slope as Average Rate of Change (page 223) In 1997, sales of VCRs numbered 16.7 million. In 2002, estimated sales of VCRs were 13.3 million. Find the average rate of change in VCR sales, in millions, per year. Graph as a line segment, and interpret the result. The average rate of change per year is

17. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-17 2.4 Example 8 Interpreting Slope as Average Rate of Change (cont.) In 1997, sales of VCRs numbered 16.7 million. In 2002, estimated sales of VCRs were 13.3 million. Find the average rate of change in VCR sales, in millions, per year. Graph as a line segment, and interpret the result. The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope. Sales of VCRs decreased by an average of 0.68 million each year from 1997 to 2002.

18. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-18 2.4 Example 9 Writing Linear Cost, Revenue, and Profit Functions (page 224) Assume that the cost to produce an item is a linear function and all items produced are sold. The fixed cost is $2400, the variable cost per item is $120, and the item sells for $150. Write linear functions to model (a) cost, (b) revenue, and (c) profit. (a) Since the cost function is linear, it will have the form C(x) = mx + b with m = 120 and b = 2400. C(x) = 120x + 2400 (b) The revenue function is R(x) = px with p = 150. R(x) = 150x

19. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-19 2.4 Example 9 Writing Linear Cost, Revenue, and Profit Functions (cont.) (c) The profit is the difference between the revenue and the cost. = 150x – (120x + 2400) P(x) = R(x) – C(x) = 30x – 2400 (d) How many items must be sold for the company to make a profit? To make a profit, P(x) must be positive. 30x – 2400 > 030x > 2400x > 80 The company must sell at least 81 items to make a profit.

20. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-20 2.4 Example 9 Writing Linear Cost, Revenue, and Profit Functions (cont.) Graphing calculator solution Define Y 1 as 30x – 2400 and graph the line. Then find the x-intercept. The graph shows that y-values for x less than 80 are negative, y-values for x greater than 80 are positive. So, at least 81 items must be sold for the company to make a profit.

21. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. R-21 Equations of Lines; Curve Fitting 2.5 Point-Slope Form ▪ Slope-Intercept Form ▪ Vertical and Horizontal Lines ▪ Parallel and Perpendicular Lines ▪ Modeling Data ▪ Modeling Data ▪ Solving Linear Equations in One Variable by Graphing

22. Forms for the Equations of a Line 1. uses slope, m and given point (x 1, y 1 ) 2.Slope-intercept form:y = mx + b uses slope, m and y-intercept = b 3.Standard form:Ax +By = C Slope is m = -A/B and (0, C/B) and (C/A,0) 4.Horizontal form:y = b Slope is zero, y-interecept (0,b) 5.Vertical form:x = a slope is undefined, not a function. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-22 Point-slope form: y – y 1 = m(x – x 1 )

23. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-23 Find an equation of the line through (3, –5) having slope –2. 2.5 Example 1 Using the Point-Slope Form (Given a Point and the Slope) (page 232) Point-slope form: y – y 1 = m(x – x 1 ) x 1 = 3, y 1 = –5, m = –2

24. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-24 Find an equation of the line through (–4, 3) and (5, –1). 2.5 Example 2 Using the Point-Slope Form (Given Two Points) (page 233) First, find the slope: Use either point for (x 1, y 1 ) Point-slope form

25. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-25 Find an equation of the line through (–4, 3) and (5, –1). 2.5 Example 2 Using the Point-Slope Form (Given Two Points) (page 233) Verify using (5, –1) for (x 1, y 1 ) : Point-slope form

26. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-26 Find the slope and y-intercept of the line with equation 3x – 4y = 12. 2.5 Example 3 Find the Slope and y -intercept From an Equation of a Line (page 234) Write the equation in slope-intercept form: The slope is and the y-intercept is –3.

27. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-27 Find an equation of the line through (–2, 4) and (2, 2). Then graph the line using the slope- intercept form. 2.5 Example 4 Using the Slope-Intercept Form (Given Two Points (page 234) First, find the slope: The equation is Substitute for m and the coordinates of one of the points (say, (2, 2)) for x and y into the slope-intercept form y = mx + b, then solve for b:

28. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-28 2.5 Example 4 Using the Slope-Intercept Form (Given Two Points (cont.)

29. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-29 Use the graph to (a) find the slope, y-intercept, and x-intercept, and (b) write the equation of the function. 2.5 Example 5 Finding an Equation From a Graph (page 235) The line rises 5 units each time the x-value increases by 2 units. The slope is.

30. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-30 2.5 Example 5(a) Finding an Equation From a Graph (cont.) The graph intersects the y-axis at (0, 5) and the x-axis at (–2, 0). The y-intercept is 5. The x-intercept is –2.

31. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-31 2.5 Example 5(b) Finding an Equation From a Graph (cont.) Slope, y-intercept 5

32. Parallel and Perpendicular Lines Parallel Lines: Two distinct non-vertical lines are parallel if and only if they have the same slopes. In other words, if line l 1 has slope of m 1 and l 2 has slope of m 2 and l 1 is parallel to l 2, then Perpendicular Lines: The product of the slopes of perpendicular lines is -1. In other words, if line l 1 has slope of m 1 and l 2 has slope of m 2 and l 1 is perpendicular to l 2, then Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-32

33. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-33 Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is parallel to the line 3x – 2y = 5. 2.5 Example 6(a) Finding Equations of Parallel and Perpendicular Lines (page 236) Write the equation in slope-intercept form to find the slope: The slope is. Parallel lines have the same slope, so the slope of the line whose equation is to be found is.

34. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-34 Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is parallel to the line 3x – 2y = 5. 2.5 Example 6(a) Finding Equations of Parallel and Perpendicular Lines (page 236) Point-slope form

35. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-35 Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is perpendicular to the line 3x – 2y = 5. 2.5 Example 6(b) Finding Equations of Parallel and Perpendicular Lines (page 236) Write the equation in slope-intercept form to find the slope: The slope is. The slopes of perpendicular lines are negative reciprocals, so the slope of the line whose equation is to be found is.

36. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-36 Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is perpendicular to the line 3x – 2y = 5. 2.5 Example 6(b) Finding Equations of Parallel and Perpendicular Lines (page 236) Point-slope form

37. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-37 Average annual tuition and fees for in-state students at public 4-year colleges are shown in the table for selected years and in the graph below, with x = 0 representing 1996, x = 4 representing 2000, etc. 2.5 Example 7 Finding an Equation of a Line That Models Data (page 238)

38. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-38 Find an equation that models the data. Use the data for 1998 and 2004. 2.5 Example 7(a) Finding an Equation of a Line That Models Data (page 238) 1998 is represented by x = 2 and 2004 is represented by x = 8. Find the slope:

39. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-39 2.5 Example 7(a) Finding an Equation of a Line That Models Data (cont.) Use either point, (2, 3486) or (8, 5148) for (x 1, y 1 )

40. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-40 2.5 Example 7(b) Finding an Equation of a Line That Models Data (page 238) Use the equation from part (a) to predict the cost of tuition and fees in 2008. For 2008, x = 12. According to the model, average tuition and fees will be $6256 in 2008.

41. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-41 The table and graph illustrate how the percent of women in the civilian labor force has changed from 1960 to 2005. 2.5 Example 8 Finding an Equation of a Line That Models Data (page 239)

42. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-42 Use the points (1965, 39.3) and (1995, 58.9) to find a linear equation that models the data. 2.5 Example 8(a) Finding an Equation of a Line That Models Data (page 239) Find the slope: Use either point for (x 1, y 1 ).

43. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-43 Use the equation to estimate the percent for 2005. How does the result compare to the actual figure of 59.3%? 2.5 Example 8(b) Finding an Equation of a Line That Models Data (page 239) Let x = 2005. Solve for y: The model estimates about 65.4% in 2005. This is 6.1% more than the actual figure of 59.3%.