1. 12-5 Direct Variation Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

2. Warm Up Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. 1. y – 3 = – (x – 9) 2. y + 2 = (x – 5) 3. y – 9 = –2(x + 4) 4. y – 5 = – (x + 7) (–4, 9), –2 Course 3 12-5 Direct Variation 1 7 2 3 1 4 (9, 3), – 1 7 (5, –2), 2 3 (–7, 5), – 1 4

3. Problem of the Day Where do the lines defined by the equations y = –5x + 20 and y = 5x – 20 intersect? (4, 0) Course 3 12-5 Direct Variation

4. Learn to recognize direct variation by graphing tables of data and checking for constant ratios. Course 3 12-5 Direct Variation

5. Vocabulary direct variation constant of proportionality Insert Lesson Title Here Course 3 12-5 Direct Variation

6. Course 3 12-5 Direct Variation

7. Course 3 12-5 Direct Variation The graph of a direct-variation equation is always linear and always contains the point (0, 0). The variables x and y either increase together or decrease together. Helpful Hint

8. Determine whether the data set shows direct variation. Additional Example 1A: Determining Whether a Data Set Varies Directly Course 3 12-5 Direct Variation

9. Make a graph that shows the relationship between Adam’s age and his length. The graph is not linear. Additional Example 1A Continued Course 3 12-5 Direct Variation

10. You can also compare ratios to see if a direct variation occurs. 22 3 27 12 = ? 81 264 81 ≠ 264 The ratios are not proportional. The relationship of the data is not a direct variation. Additional Example 1A Continued Course 3 12-5 Direct Variation

11. Determine whether the data set shows direct variation. Additional Example 1B: Determining Whether a Data Set Varies Directly Course 3 12-5 Direct Variation

12. Make a graph that shows the relationship between the number of minutes and the distance the train travels. Additional Example 1B Continued Plot the points. The points lie in a straight line. Course 3 12-5 Direct Variation (0, 0) is included.

13. You can also compare ratios to see if a direct variation occurs. The ratios are proportional. The relationship is a direct variation. 25 10 50 20 75 30 100 40 === Compare ratios. Additional Example 1B Continued Course 3 12-5 Direct Variation

14. Determine whether the data set shows direct variation. Check It Out: Example 1A Kyle's Basketball Shots Distance (ft)203040 Number of Baskets530 Course 3 12-5 Direct Variation

15. Make a graph that shows the relationship between number of baskets and distance. The graph is not linear. Check It Out: Example 1A Continued Number of Baskets Distance (ft) 2 3 4 203040 1 5 Course 3 12-5 Direct Variation

16. You can also compare ratios to see if a direct variation occurs. Check It Out: Example 1A Continued 5 20 3 30 = ? 60 150 150  60. The ratios are not proportional. The relationship of the data is not a direct variation. Course 3 12-5 Direct Variation

17. Determine whether the data set shows direct variation. Check It Out: Example 1B Ounces in a Cup Ounces (oz)8162432 Cup (c)1234 Course 3 12-5 Direct Variation

18. Make a graph that shows the relationship between ounces and cups. Check It Out: Example 1B Continued Number of Cups Number of Ounces 2 3 4 81624 1 32 Course 3 12-5 Direct Variation Plot the points. The points lie in a straight line. (0, 0) is included.

19. You can also compare ratios to see if a direct variation occurs. Check It Out: Example 1B Continued Course 3 12-5 Direct Variation The ratios are proportional. The relationship is a direct variation. Compare ratios. = 1 8 == 2 16 3 24 4 32

20. Find each equation of direct variation, given that y varies directly with x. y is 54 when x is 6 Additional Example 2A: Finding Equations of Direct Variation y = kx 54 = k  6 9 = k y = 9x y varies directly with x. Substitute for x and y. Solve for k. Substitute 9 for k in the original equation. Course 3 12-5 Direct Variation

21. x is 12 when y is 15 Additional Example 2B: Finding Equations of Direct Variation y = kx 15 = k  12 y varies directly with x. Substitute for x and y. Solve for k. = k 5 4 Substitute for k in the original equation. 5 4 y = x 5 4 Course 3 12-5 Direct Variation

22. Find each equation of direct variation, given that y varies directly with x. y is 24 when x is 4 Check It Out: Example 2A y = kx 24 = k  4 6 = k y = 6x y varies directly with x. Substitute for x and y. Solve for k. Substitute 6 for k in the original equation. Course 3 12-5 Direct Variation

23. x is 28 when y is 14 Check It Out: Example 2B y = kx 14 = k  28 y varies directly with x. Substitute for x and y. Solve for k. = k 1 2 Substitute for k in the original equation. 1 2 y = x 1 2 Course 3 12-5 Direct Variation

24. Mrs. Perez has $4000 in a CD and $4000 in a money market account. The amount of interest she has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation. Additional Example 3: Money Application Course 3 12-5 Direct Variation

25. Additional Example 3 Continued interest from CD and time interest from CD time = 17 1 = = 17 interest from CD time 34 2 The second and third pairs of data result in a common ratio. In fact, all of the nonzero interest from CD to time ratios are equivalent to 17. The variables are related by a constant ratio of 17 to 1, and (0, 0) is included. The equation of direct variation is y = 17x, where x is the time, y is the interest from the CD, and 17 is the constant of proportionality. = = = 17 interest from CD time = 17 1 34 2 51 3 68 4 Course 3 12-5 Direct Variation

26. Additional Example 3 Continued interest from money market and time interest from money market time = = 19 19 1 interest from money market time = =18.5 37 2 19 ≠ 18.5 If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. Course 3 12-5 Direct Variation

27. Mr. Ortega has $2000 in a CD and $2000 in a money market account. The amount of interest he has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation. Check It Out: Example 3 Course 3 12-5 Direct Variation InterestInterest from Time (mo)from CD ($)Money Market ($) 000 11215 23040 3 45 450

28. Check It Out: Example 3 Continued interest from CD time = 12 1 interest from CD time = = 15 30 2 The second and third pairs of data do not result in a common ratio. If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. A. interest from CD and time Course 3 12-5 Direct Variation

29. Check It Out: Example 3 Continued B. interest from money market and time interest from money market time = = 15 15 1 interest from money market time = =20 40 2 15 ≠ 20 If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included. Course 3 12-5 Direct Variation

30. Lesson Quiz: Part I Find each equation of direct variation, given that y varies directly with x. 1. y is 78 when x is 3. 2. x is 45 when y is 5. 3. y is 6 when x is 5. y = 26x Insert Lesson Title Here y = x 1 9 6 5 Course 3 12-5 Direct Variation

31. Lesson Quiz: Part II 4. The table shows the amount of money Bob makes for different amounts of time he works. Determine whether there is a direct variation between the two sets of data. If so, find the equation of direct variation. Insert Lesson Title Here direct variation; y = 12x Course 3 12-5 Direct Variation