1. 5.2 Direct Variation Direct Variation: the relationship that can be represented by a function if the form: Constant of variation: the constant variable K is the coefficient of x on the y=kx equation. y = kx Inverse variation: the relationship that can be represented by the function: Joint Variation: the relationship that can be represented by the function: y = kxz

2. Real World:

3. Identifying a Direct Variation: If the equation can be written in y = kx we have a direct variation. Ex: Does the equation represent a direct variation? a) 7y = 2xb) 3y + 4x = 8

4. Ex: (solution) If we can writer the equation in y = kx we have a direct variation. a) 7y = 2x b) 3y + 4x = 8 Inverse of Multiplication ___ __ 7 7 Isolate y, subtract 4x and divide by 3 Equation is not in y=kx

5. WRITING DIRECT VARIATION EQUATIONS: To write a direct variation equation we must first find the constant of variation k using ordered pairs given. Ex: Suppose y varies directly with x, and y = 35 when x = 5. What direct variation equation relates x and y? What is the value of y when x = 9?

6. AGAIN: To write a direct variation equation we must first find the constant of variation k using ordered pairs given. From the problem, we are given the following: y = 35 when x = 5. That is: (5, 35) Since we have “varies directly” we must have an equation on the form: y = kx Using the equation and info given, we have: 35 = k(5) k = 35/5 = 7

7. Once we know the constant of variation (K = 7) we can now write the direct variation equation as follows: y = kx y = 7x We now go further and find the value of y when x = 9 as follows: y = 7x y = 7(9) Thus: y = 63 when x = 9.

8. YOU TRY IT: Suppose y varies directly with x, and y = 10 when x = -2. Write a direct variation equation and find the value of y when x = - 15.

9. YOU TRY IT (SOLUTION): Given: y = 10, x = - 2 Varies Directly equation: y = kx To find the constant of variation (k): y = kx 10 = k(-2) K = - 5 Therefore our equation is: y = -5x Using this equation to find y when x = -15 y = -5x y = -5(-15)  y = 75.

10. Real World: Let’s solve it

11. Time (x)Distance (y) 10s2 mi 15s3 mi Using the direct variation equation and y = 2mi when x = 10s y = kx 2 = k(10)

12. GRAPHING DIRECT VARIATIONS: To graph a direct variation equation we must go back to tables: Independent Variable(x) Equation F(x)Dependent Variable (y) Ordered Pair (x, y) Remember: the Independent variable(x) is chosen by you if you are not given any x values.

13. Ex: Graph f(x) = -7x Independent Variable(x) Equation F(x) Dependent Variable (y) Ordered Pair (x, y) -2-7(-2)14(-2, 14) -7(-1)7(-1, 7) 0-7(0)0(0, 0) 1-7(1)-7(1, -7) 2-7(2)-14(2, -14) Now we must graph the ordered pairs (last column)

14. Ordered Pair (x, y) (-2, 14) (-1, 7) (0, 0) (1, -7) (2, -14) Y = -7x

15. YOU TRY IT: Graph y = 2X

16. YOU TRY IT: (SOLUTION) Graph y = 2x Independent Variable(x) Equation F(x) Dependent Variable (y) Ordered Pair (x, y) -22(-2)-4(-2, -4) 2(-1)-2(-1, -2) 02(0)0(0, 0) 12(1)2(1, 2) 22(2)4(2, 4)

17. Ordered Pair (x, y) (-2, -4) (-1, -2) (0, 0) (1, 2) (2, 4) Y = 2x

18. VIDEOS: Graphs https://www.khanacademy.org/math/algebra/algeb ra- functions/direct_inverse_variation/v/recognizing- direct-and-inverse-variation https://www.khanacademy.org/math/algebra/alge bra-functions/direct_inverse_variation/v/direct- and-inverse-variation

19. Class Work: Pages: 302-303 Problems: As many as needed to master the concept