Math Core

Drive a fixed distance faster and the trip takes less time. Share a pizza among more friends and each slice gets smaller. When multiplying one quantity by some factor divides the other by that same factor, the two quantities vary inversely, and the equation connecting them is your first rational function.

Direct versus inverse variation

You already know direct variation, $y = kx$ . There the ratio $\dfrac{y}{x}$ stays constant: double $x$ and $y$ doubles too.

Inverse variation flips that relationship. Here the product $xy$ stays constant. If a 120-mile trip is driven at speed $r$ for time $t$ , then $rt = 120$ no matter how fast you go:

speed $r$ (mph)	20	30	40	60
time $t$ (hours)	6	4	3	2

Every column multiplies to $120$ . Doubling the speed from $20$ to $40$ cuts the time in half, from $6$ to $3$ . Solving $rt = 120$ for $t$ gives $t = \dfrac{120}{r}$ , where the variable sits in the denominator.

Definition

Inverse variation

Two variables $x$ and $y$ vary inversely if there is a nonzero constant $k$ with

y = \frac{k}{x} \qquad\text{or equivalently}\qquad xy = k.

The number $k$ is the constant of variation. You also say " $y$ is inversely proportional to $x$ ."

The two forms say the same thing. Use $xy = k$ to find $k$ from a data point, and $y = \dfrac{k}{x}$ to predict new values.

Finding the constant and making predictions

Every inverse variation problem follows the same two-step plan:

Use one known pair $(x, y)$ to find $k = xy$ .
Write $y = \dfrac{k}{x}$ and substitute the new value.

Worked example: Finding k and predicting

$y$ varies inversely with $x$ , and $y = 12$ when $x = 5$ . Write the equation and find $y$ when $x = 15$ .

First find the constant: $k = xy = 5 \cdot 12 = 60$ . So the equation is

y = \frac{60}{x}.

When $x = 15$ , $y = \dfrac{60}{15} = 4$ .

Check the pattern: $x$ tripled (from $5$ to $15$ ), so $y$ should be divided by $3$ (from $12$ to $4$ ). It was.

Because $x_1 y_1 = k = x_2 y_2$ , you can skip writing the equation and go straight to $x_1 y_1 = x_2 y_2$ . In the example, $5 \cdot 12 = 15 \cdot y_2$ gives $y_2 = 4$ in one line.

Worked example: Negative values

$y$ is inversely proportional to $x$ , and $y = -4$ when $x = 6$ . Find $x$ when $y = 8$ .

The constant is $k = 6 \cdot (-4) = -24$ . Using $xy = -24$ with $y = 8$ :

8x = -24 \quad\Longrightarrow\quad x = -3.

The constant can be negative. Then $x$ and $y$ always have opposite signs.

Is it inverse variation?

To test a table, multiply each pair. If every product is the same nonzero number, the data vary inversely. To test an equation, try to rewrite it as $xy = k$ .

$xy = 18$ : inverse variation with $k = 18$ .
$y = \dfrac{-7}{x}$ : inverse variation with $k = -7$ .
$\dfrac{y}{x} = 3$ : this is $y = 3x$ , which is direct variation.
$y = \dfrac{5}{x} + 1$ : not inverse variation. The $+1$ ruins the constant product.

Common mistake

" $y$ goes down as $x$ goes up" is not enough to call a relationship inverse variation. The line $y = 10 - x$ decreases, but $x = 1, y = 9$ gives a product of $9$ while $x = 5, y = 5$ gives $25$ . Only a constant product counts.

The graph of $y = \dfrac{k}{x}$

The graph of an inverse variation is a hyperbola with two separate pieces called branches. Below are $y = \dfrac{6}{x}$ and the two axes, which the curve approaches but never touches.

y = 6/x. The branches get close to the x-axis and the y-axis but never reach them.Open in grapher →

Three features to notice:

No point at $x = 0$ . You cannot divide by zero, so $x = 0$ is not in the domain. As $x$ gets close to $0$ from the right, $\dfrac{6}{x}$ grows without bound. The line $x = 0$ is a vertical asymptote.
No point with $y = 0$ . A fraction $\dfrac{6}{x}$ is never zero. As $x$ grows large, $\dfrac{6}{x}$ shrinks toward $0$ . The line $y = 0$ is a horizontal asymptote.
Where the branches sit. When $k > 0$ , the branches lie in Quadrants I and III (where $x$ and $y$ have the same sign). When $k < 0$ , they lie in Quadrants II and IV.

The next lesson shifts and stretches this basic shape to graph many more rational functions.

Other kinds of inverse variation

Sometimes a quantity varies inversely with a power of another. The brightness of a light, for example, is inversely proportional to the square of your distance from it:

B = \frac{k}{d^2}.

The method is identical: find $k$ from one data point, then substitute.

Worked example: Inverse square

The intensity of a light is $36$ units at a distance of $2$ meters and varies inversely with the square of the distance. What is the intensity at $6$ meters?

Find $k$ : $36 = \dfrac{k}{2^2}$ , so $k = 36 \cdot 4 = 144$ . At $d = 6$ :

B = \frac{144}{6^2} = \frac{144}{36} = 4 \text{ units}.

The distance tripled, so the intensity was divided by $3^2 = 9$ .

Tip

Before computing, predict how the answer should change. If $y = \dfrac{k}{x}$ and $x$ is multiplied by $c$ , then $y$ is divided by $c$ . If $y = \dfrac{k}{x^2}$ , then $y$ is divided by $c^2$ . A quick prediction catches most arithmetic slips.