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5.2 Direct Variation Direct Variation: the relationship that can be represented by a function if the form: y = kx Constant of variation: the constant variable K is the coefficient of x on the y=kx equation. Inverse variation: the relationship that can be represented by the function: y= Joint Variation: the relationship that can be represented by the function: y = kxz Real World: 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 = 2x b) 3y + 4x = 8 Ex: (solution) If we can writer the equation in y = kx we have a direct variation. a)___ 7y = 2x Inverse of Multiplication __ 7 7 y= x Equation is in y=kx with k= b) 3y + 4x = 8 Isolate y, subtract 4x and divide by 3 y=− x+ Equation is not in y=kx 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? 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 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. 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. 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. Real World: Let’s solve it Real World: Let’s solve it Time (x) 10s 15s Distance (y) 2 mi 3 mi Using the direct variation equation and y = 2mi when x = 10s y = kx 2 = k(10) k= = Thus the direct variation eq: y = x 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. Ex: Graph f(x) = -7x Independent Variable(x) -2 Equation F(x) -7(-2) Dependent Variable (y) 14 Ordered Pair (x, y) (-2, 14) -1 0 -7(-1) -7(0) 7 0 (-1, 7) (0, 0) 1 -7(1) -7 (1, -7) 2 -7(2) -14 (2, -14) Now we must graph the ordered pairs (last column) Ordered Pair (x, y) (-2, 14) (-1, 7) (0, 0) (1, -7) (2, -14) Y = -7x YOU TRY IT: Graph y = 2X YOU TRY IT: (SOLUTION) Graph y = 2x Independent Variable(x) -2 Equation F(x) 2(-2) Dependent Variable (y) -4 Ordered Pair (x, y) (-2, -4) -1 0 2(-1) 2(0) -2 0 (-1, -2) (0, 0) 1 2(1) 2 (1, 2) 2 2(2) 4 (2, 4) Ordered Pair (x, y) (-2, -4) (-1, -2) (0, 0) (1, 2) (2, 4) Y = 2x VIDEOS: Graphs https://www.khanacademy.org/math/algebra/alge bra-functions/direct_inverse_variation/v/directand-inverse-variation https://www.khanacademy.org/math/algebra/algeb rafunctions/direct_inverse_variation/v/recognizingdirect-and-inverse-variation Class Work: Pages: 302-303 Problems: As many as needed to master the concept