1. Copyright © Cengage Learning. All rights reserved. Graphs; Equations of Lines; Functions; Variation 3

2. Copyright © Cengage Learning. All rights reserved. Section 3.5 Slope-Intercept Form

3. 3 Objectives  Find the slope and y-intercept given a linear equation. 2.Write an equation of the line that has a given slope and passes through a specified point. Use the slope-intercept form to write an equation of the line that passes through two given points. Graph a linear equation using the slope and y-intercept. 1 1 2 2 3 3 4 4

4. 4 Objectives 5.Determine whether two linear equations define lines that are parallel, perpendicular, or neither parallel nor perpendicular. 6.Write an equation of the line passing through a specified point and parallel or perpendicular to a given line.  Write an equation of a line representing real- world data. 5 5 6 6 7 7

5. 5 Find the slope and y-intercept given a linear equation 1.

6. 6 Equation of a line—Slope-intercept Form Since the y-intercept of the line shown in Figure 3-37 is the point (0, b), we can write its equation by substituting 0 for x 1 and b for y 1 in the point-slope equation and simplifying. y – y 1 = m(x – x 1 ) y – b = m(x – 0) y – b = mx (1) y = mx + b Figure 3-37

7. 7 Equation of a Line—Point-slope Form Because Equation 1 displays the slope m and the y-coordinate b of the y-intercept, it is called the slope-intercept form of the equation of a line. Slope-Intercept Form of the Equation of a Line The equation of the line with slope m and y-intercept (0, b) is y = mx + b

8. 8 Example Find the slope and y-intercept of the line with the given equation. a. b. 3x – 4y = 9 Solution: a. Since the equation is solved for y, we can identify the slope and the y-intercept from the equation. The slope is, and the y-intercept is (0, 5).

9. 9 Example 1 – Solution b. To write the equation 3x – 4y = 9 in slope-intercept form, we solve it for y. Subtract 3x from both sides. Divide both sides by –4. cont’d

10. 10 Example 1 – Solution We can read the slope and the y-intercept from the equation because it is now in y = mx + b form The slope is. The y-intercept is cont’d

11. 11 Write an equation of the line that has a given slope and passes through a specified point 2.

12. 12 Equation of a Line—Point-slope Form If we know the slope of a line and the coordinates of one point on the line, we can write the equation of the line.

13. 13 Example 2 Write the slope-intercept equation of the line that has a slope of 5 and passes through the point (–2, 9). Solution: We are given that m = 5 and that the pair (–2, 9) satisfies the equation. To find b, we can substitute –2 for x, 9 for y, and 5 for m in the equation y = mx + b. y = mx + b 9 = 5(–2) + b Substitute 9 for y, 5 for m, and –2 for x.

14. 14 Example – Solution 9 = –10 + b 19 = b Since m = 5 and b = 19, the equation in slope-intercept form is y = 5x + 19. Multiply. Add 10 to both sides. cont’d

15. 15 Use the slope-intercept form to write an equation of the line that passes through two given points 3.

16. 16 Example Use the slope-intercept form to write an equation of the line that passes through the points (–3, 4) and (5, –8). Solution: We first find the slope of the line. This is the formula for slope. Substitute –8 for y 2, 4 for y 1, 5 for x 2, and – 3 for x 1. –8 – 4 = –12, 5 – (–3) = 8 Simplify.

17. 17 Example – Solution Then we write an equation of the line that has a slope of and passes through either of the given points, say (–3, 4). To do so, we substitute for m, –3 for x, 4 for y, and solve the equation y = mx + b for b. Substitute. cont’d

18. 18 Example – Solution Since and an equation is or more simply, Subtract from both sides and simplify. cont’d

19. 19 Graph a linear equation using the slope and y-intercept 4.

20. 20 Graphing a Line Using the Slope-intercept Form To graph y = (4/3)x – 2, we note that b = –2 and m = 4/3

21. 21 Graphing a Line Using the Slope-intercept Form Since the slope of the line is, we can locate another point on the line by starting at the point (0, –2) and counting 3 units to the right and 4 units up. The change in x is 3, and the corresponding change in y is 4. The line joining the two points is the graph of the equation.

22. 22 Example Find the slope and the y-intercept of the line with the equation 2(x – 3) = –3(y + 5). Then graph the line. Solution: We write the equation in the form y = mx + b to find the slope m and the y-intercept (0, b). 2(x – 3) = –3(y + 5) 2x – 6 = –3y – 15 2x + 3y – 6 = –15 Use the distributive property to remove parentheses. Add 3y to both sides.

23. 23 Example – Solution 3y – 6 = –2x – 15 3y = –2x – 9 The slope is and the y-intercept is (0, –3). Subtract 2x from both sides. Add 6 to both sides. Divide both sides by 3. cont’d

24. 24 Example – Solution To draw the graph, we plot the y-intercept (0, –3) and then locate a second point on the line by moving 3 units to the right and 2 units down. We draw a line through the two points to obtain the graph shown in Figure 3-39. Figure 3-39 cont’d

25. 25 Determine whether two linear equations define lines that are parallel, perpendicular, or neither parallel nor perpendicular 5.

26. 26 Parallel Lines We know that different lines having the same slope are parallel. Also recall that if the slopes of two lines are negative reciprocals, the lines are perpendicular. We will use these facts in the next example.

27. 27 Example Show that the lines represented by 4x + 8y = 16 and x = 6 – 2y are parallel. Solution: We solve each equation for y to see whether their slopes are equal and then determine if the lines are different. 4x + 8y = 16 x = 6 – 2y 8y = –4x + 16 2y = –x + 6 y = x + 2 y = x + 3

28. 28 Example – Solution Since the slope of each line is the lines are possibly parallel, but we need to know whether the lines are distinct. Since the values of b in these equations are different, the lines are distinct and, therefore, parallel. cont’d

29. 29 Write an equation of the line passing through a specified point and parallel or perpendicular to a given line 6.

30. 30 Parallel Lines We will now use the slope properties of parallel and perpendicular lines to write more equations of lines.

31. 31 Example Write an equation of the line passing through (–3, 2) and parallel to the line y = 8x – 5. Solution: Since y = 8x – 5 is written in slope-intercept form, the slope of its graph is the coefficient of x, which is 8. The desired equation is to have a graph that is parallel to the line, thus its slope must be 8 as well.

32. 32 Example 7 – Solution To find the equation of the line, we substitute –3 for x, 2 for y, and 8 for m in the slope-intercept equation and solve for b. y = mx + b 2 = 8(–3) + b 2 = –24 + b 26 = b Since the slope of the desired line is 8 and the y-intercept is (0, 26), the equation is y = 8x + 26. Substitute. Multiply. Add 24 to both sides. cont’d

33. 33 Write an equation of a line representing real-world data 7.

34. 34 Write an equation of a line representing real- world data As machinery wears out, it becomes worth less. Accountants often estimate the decreasing value of aging equipment with linear depreciation, a method based on linear equations.

35. 35 Example – Linear Depreciation A company buys a $12,500 computer with an estimated life of 6 years. The computer can then be sold as scrap for an estimated salvage value of $500. If y represents the value of the computer after x years of use and y and x are related by the equation of a line, a. Find an equation of the line and graph it. b. Find the value of the computer after 2 years. c. Find the economic meaning of the y-intercept of the line. d. Find the economic meaning of the slope of the line.

36. 36 Example – Solution To find an equation of the line, we calculate its slope and then use the slope-intercept form to find its equation. When the computer is new, its age x is 0 and its value y is the purchase price of $12,500. We can represent this information as the point with coordinates (0, 12,500). When it is six years old, x = 6 and y = 500, its salvage value. We can represent this information as the point with coordinates (6, 500).

37. 37 Example – Solution Since the line passes through these two points as shown in Figure 3-40, the slope of the line is The slope is –2,000. Figure 3-40 cont’d

38. 38 Example – Solution To find an equation of the line, we substitute –2,000 for m, 0 for x, and 12,500 for y in the slope-intercept form and solve for b. y = mx + b 12,500 = –2,000(0) + b 12,500 = b The current value of the computer is related to its age by the equation y = –2,000x + 12,500. cont’d

39. 39 Example – Solution To find the value of the computer after 2 years, we substitute 2 for x in the equation y = –2,000x + 12,500. y = –2,000x + 12,500 y = –2,000(2) + 12,500 = –4,000 + 12,500 = 8,500 In 2 years, the computer will be worth $8,500. cont’d

40. 40 Example – Solution Since the y-intercept of the graph is (0, 12,500), the y-coordinate of the y-intercept is the computer’s original purchase price. cont’d

41. 41 Example – Solution Each year, the value decreases by $2,000, shown by the slope of the line, –2,000. The slope of the depreciation line is called the annual depreciation rate. cont’d