5.2 Direct VariationDirect 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:

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

b) 3y + 4x = 8

Inverse of Multiplication___ __ 7 7

y = x Equation is in y=kx with k=

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 = kxUsing 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 = kxy = 7x

We now go further and find the value of y when x = 9 as follows:

y = 7xy = 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 itTime (x) Distance (y)

10s 2 mi15s 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)

Equation F(x)

Dependent Variable (y)

Ordered Pair (x, y)

-2 -7(-2) 14 (-2, 14)

-1 -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)

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) Equation

F(x)Dependent Variable (y)

Ordered Pair (x, y)

-2 2(-2) -4 (-2, -4)

-1 2(-1) -2 (-1, -2)0 2(0) 0 (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/algebra-functions/direct_inverse_variation/v/recognizing-direct-and-inverse-variation

https://www.khanacademy.org/math/algebra/algebra-functions/direct_inverse_variation/v/direct-and-inverse-variation

Class Work:

Pages: 302-303

Problems: As many as needed to master the concept