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=
𝒌
𝒙
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
𝟐
𝟕
y= x
Equation is in y=kx with k=
b) 3y + 4x = 8
Isolate y, subtract 4x and
divide by 3
y=−
𝟒
𝟖
x+
𝟑
𝟑
Equation is not in y=kx
𝟐
𝟕
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=
𝟐
𝟏𝟎
=
𝟏
𝟓
𝟏
𝟓
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