# Direct Variation

### Learning Objectives

- Solve a direct variation problem
- Use a constant of variation to describe the relationship between two variables

*e*= 0.16

*s*tells us her earnings,

*e*, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.

s, sales prices |
e = 0.16s |
Interpretation |
---|---|---|

$4,600 | e = 0.16(4,600) = 736 |
A sale of a $4,600 vehicle results in $736 earnings. |

$9,200 | e = 0.16(9,200) = 1,472 |
A sale of a $9,200 vehicle results in $1472 earnings. |

$18,400 | e = 0.16(18,400) = 2,944 |
A sale of a $18,400 vehicle results in $2944 earnings. |

**direct variation**. Each variable in this type of relationship

**varies directly**with the other. The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[/latex] is used for direct variation. The value

*k*is a nonzero constant greater than zero and is called the

**constant of variation**. In this case,

*k*= 0.16 and

*n*= 1.

### A General Note: Direct Variation

If*x*and

*y*are related by an equation of the form

[latex]y=k{x}^{n}[/latex]

then we say that the relationship is**direct variation**and

*y*

**varies directly**with the

*n*th power of

*x*. In direct variation relationships, there is a nonzero constant ratio [latex]k=\frac{y}{{x}^{n}}[/latex], where

*k*is called the

**constant of variation**, which help defines the relationship between the variables.

### How To: Given a description of a direct variation problem, solve for an unknown.**
**

- Identify the input,
*x*, and the output,*y*. - Determine the constant of variation. You may need to divide
*y*by the specified power of*x*to determine the constant of variation. - Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.

### Example: Solving a Direct Variation Problem

The quantity*y*varies directly with the cube of

*x*. If

*y*= 25 when

*x*= 2, find

*y*when

*x*is 6.

Answer:
The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing *y* by the cube of *x*.

[latex]\begin{array}{l} k=\frac{y}{{x}^{3}} \\ =\frac{25}{{2}^{3}}\\ =\frac{25}{8}\end{array}[/latex]

Now use the constant to write an equation that represents this relationship.[latex]y=\frac{25}{8}{x}^{3}[/latex]

Substitute*x*= 6 and solve for

*y*.

[latex]\begin{array}{l}y=\frac{25}{8}{\left(6\right)}^{3}\hfill \\ \text{ }=675\hfill \end{array}[/latex]

#### Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.### Q & A

**Do the graphs of all direct variation equations look like Example 1?**

*No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).*

### Try It

The quantity*y*varies directly with the square of

*x*. If

*y*= 24 when

*x*= 3, find

*y*when

*x*is 4.

Answer: [latex]\frac{128}{3}[/latex]

## Licenses & Attributions

### CC licensed content, Original

- Revision and Adaptation.
**Provided by:**Lumen Learning**License:**CC BY: Attribution.

### CC licensed content, Shared previously

- Question ID 91391.
**Authored by:**Jenck, Michael.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC-BY + GPL. - College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected]. - Direct Variation Applications .
**Authored by:**James Sousa (Mathispower4u.com) .**License:**CC BY: Attribution.