Math Core

Divide one polynomial by another and you get a rational function. Unlike polynomials, rational functions can break apart, blow up near certain inputs, and settle down toward lines as $x$ grows. Those lines, the asymptotes, are the skeleton of the graph: find them first and the rest of the sketch nearly draws itself.

Definition

Rational function

A rational function is a function of the form $f(x) = \dfrac{p(x)}{q(x)}$ , where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. Its domain is all real numbers except the zeros of $q$ .

Vertical asymptotes and holes

Every zero of the denominator is excluded from the domain, but excluded inputs don't all look the same on the graph. What happens depends on whether the numerator is also zero there, so the first step is always to factor both polynomials and cancel common factors.

Holes and vertical asymptotes

Write $f$ in lowest terms by canceling common factors.

A factor that cancels completely leaves a hole: a single missing point. Its $y$ -coordinate is the value of the simplified function there.
A zero of the denominator that remains after canceling gives a vertical asymptote. Near it, $|f(x)|$ grows without bound.

The reason is size. Near a vertical asymptote $x = c$ , the denominator gets close to $0$ while the numerator stays near some nonzero number, and dividing by something tiny produces something huge. At a hole, the canceled factor was making both parts tiny at the same rate, so the quotient stays finite.

Behavior near a vertical asymptote

To see whether the graph shoots up or down on each side of $x = c$ , check the sign of $f(x)$ for $x$ just to the left and just to the right of $c$ . We write $x \to c^-$ for " $x$ approaches $c$ from the left" and $x \to c^+$ for "from the right."

Multiplicity matters here too. If the remaining factor is $(x - c)^k$ with $k$ odd, it changes sign at $c$ , so the two sides of the asymptote go in opposite directions. If $k$ is even, both sides go the same way. Compare $y = \dfrac{1}{x - 1}$ , which goes down on the left and up on the right, with $y = \dfrac{1}{(x - 1)^2}$ , which goes up on both sides.

y = 1/(x − 1) (odd power: opposite directions) and y = 1/(x − 1)² (even power: both up), with asymptote x = 1.Open in grapher →

Worked example: A hole and an asymptote

Find the domain, holes, vertical asymptotes and intercepts of $f(x) = \dfrac{x^2 - 4}{x^2 + x - 6}$ , and describe the behavior near each vertical asymptote.

Factor:

f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x + 3)} = \frac{x + 2}{x + 3}, \quad x \ne 2.

Domain: all real numbers except $2$ and $-3$ .
Hole: $x - 2$ cancels, so there is a hole at $x = 2$ . The simplified function gives $\dfrac{2 + 2}{2 + 3} = \dfrac{4}{5}$ , so the hole is at $\left(2, \dfrac{4}{5}\right)$ .
Vertical asymptote: $x = -3$ .
Intercepts: $x$ -intercept at $x = -2$ ; $y$ -intercept $f(0) = \dfrac{2}{3}$ .

Near $x = -3$ the numerator $x + 2$ is close to $-1$ . As $x \to -3^+$ , the denominator is a small positive number, so $f(x) \to -\infty$ . As $x \to -3^-$ , the denominator is a small negative number, so $f(x) \to \infty$ .

f(x) = (x² − 4)/(x² + x − 6) with asymptotes x = −3 and y = 1. There is a hole at (2, 4/5).Open in grapher →

Horizontal asymptotes

A horizontal asymptote $y = L$ describes the far ends of the graph: $f(x) \to L$ as $x \to \infty$ or $x \to -\infty$ . Just like end behavior of polynomials, it depends only on the leading terms.

Horizontal asymptote rules

Let $n$ be the degree of the numerator and $d$ the degree of the denominator, with leading coefficients $a$ and $b$ .

degrees	horizontal asymptote
$n < d$	$y = 0$
$n = d$	$y = \dfrac{a}{b}$
$n > d$	none

To see why, divide the numerator and denominator by the highest power of $x$ in the denominator. For example,

\frac{6x^2 - 5}{2x^2 + x} = \frac{6 - \dfrac{5}{x^2}}{2 + \dfrac{1}{x}}.

As $x \to \pm\infty$ , the terms $\dfrac{5}{x^2}$ and $\dfrac{1}{x}$ shrink to $0$ , leaving $\dfrac{6}{2} = 3$ .

Crossing a horizontal asymptote

A graph can never cross a vertical asymptote, because $f$ isn't defined there. But a horizontal asymptote only describes the ends, so the graph can cross it in the middle. To find where, solve $f(x) = L$ .

Worked example: Where the graph crosses its asymptote

Analyze $f(x) = \dfrac{2x^2 - 3x - 2}{x^2 + 1}$ and find where it crosses its horizontal asymptote.

The denominator $x^2 + 1$ is never $0$ , so there are no vertical asymptotes and the domain is all real numbers.
Equal degrees, so the horizontal asymptote is $y = \dfrac{2}{1} = 2$ .
$x$ -intercepts: $2x^2 - 3x - 2 = (2x + 1)(x - 2) = 0$ , so $x = -\dfrac{1}{2}$ and $x = 2$ .
$y$ -intercept: $f(0) = -2$ .

Now solve $f(x) = 2$ :

2x^2 - 3x - 2 = 2(x^2 + 1) \quad\Longrightarrow\quad -3x - 2 = 2 \quad\Longrightarrow\quad x = -\frac{4}{3}.

The graph crosses $y = 2$ at $\left(-\dfrac{4}{3}, 2\right)$ , rises a little above it, and then approaches it from above as $x \to -\infty$ .

f(x) = (2x² − 3x − 2)/(x² + 1) crosses its horizontal asymptote y = 2 at x = −4/3.Open in grapher →

Slant asymptotes

When the numerator's degree is exactly one more than the denominator's, there is no horizontal asymptote, but the graph still settles down, this time toward a slanted line.

Slant (oblique) asymptote

If $n = d + 1$ , divide: $f(x) = mx + b + \dfrac{r(x)}{q(x)}$ , where the remainder $r(x)$ has smaller degree than $q(x)$ . The line $y = mx + b$ is a slant asymptote, because the leftover fraction $\to 0$ as $x \to \pm\infty$ .

Worked example: Finding a slant asymptote

Find all asymptotes and intercepts of $f(x) = \dfrac{x^2 - x - 2}{x - 3}$ .

Divide by $x - 3$ with synthetic division using $3$ :

	$1$	$-1$	$-2$
$3$		$3$	$6$
	$1$	$2$	$4$

So $f(x) = x + 2 + \dfrac{4}{x - 3}$ .

Slant asymptote: $y = x + 2$ .
Vertical asymptote: $x = 3$ (the numerator is $4 \ne 0$ there).
$x$ -intercepts: $x^2 - x - 2 = (x - 2)(x + 1)$ , so $x = 2$ and $x = -1$ .
$y$ -intercept: $f(0) = \dfrac{-2}{-3} = \dfrac{2}{3}$ .

For large positive $x$ the fraction $\dfrac{4}{x - 3}$ is positive, so the graph sits just above the line; for large negative $x$ it sits just below.

f(x) = (x² − x − 2)/(x − 3) with vertical asymptote x = 3 and slant asymptote y = x + 2.Open in grapher →

Common mistake

Always cancel common factors before you name vertical asymptotes. In the first example, $x = 2$ makes the denominator $0$ , but it's a hole, not an asymptote. On the other hand, canceling never changes the horizontal or slant asymptote, since those depend only on the far ends of the graph.

Putting it all together

Worked example: A complete sketch

Sketch $f(x) = \dfrac{3x}{x^2 - x - 2}$ .

Factor: $f(x) = \dfrac{3x}{(x - 2)(x + 1)}$ . Nothing cancels.

Vertical asymptotes $x = 2$ and $x = -1$ ; horizontal asymptote $y = 0$ (degree $1$ over degree $2$ ).
The only intercept is $(0, 0)$ , which lies on the horizontal asymptote, so the graph crosses it there.

The numbers $-1$ , $0$ and $2$ split the line into four intervals. One test value in each gives the sign:

interval	$x$ below $-1$	$-1$ to $0$	$0$ to $2$	$x$ above $2$
test $x$	$-2$	$-0.5$	$1$	$3$
sign of $f$	$-$	$+$	$-$	$+$

So the graph is below the axis on the far left and comes up toward $0$ , drops to $-\infty$ as $x \to -1^-$ , comes down from $+\infty$ just right of $-1$ , passes through the origin, falls to $-\infty$ as $x \to 2^-$ , and comes down from $+\infty$ just right of $2$ toward $y = 0$ .

f(x) = 3x/((x − 2)(x + 1)) with vertical asymptotes x = −1 and x = 2 and horizontal asymptote y = 0.Open in grapher →

Tip

A sign chart is the fastest way to decide which way each branch goes near a vertical asymptote. You'll use exactly the same chart in the next lesson to solve rational inequalities.

Practice

Practice 1

Find all vertical asymptotes of $f(x) = \dfrac{x + 2}{x^2 - 2x - 15}$ . Enter the $x$ -values.

Separate answers with commas, e.g. 2, -5

Practice 2

The graph of $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$ has one hole. Give its coordinates.

Enter a point like (2, -3)

Practice 3

The graph of $f(x) = \dfrac{6x^3 - x}{3 - 2x^3}$ has horizontal asymptote $y = c$ . Find $c$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

For $f(x) = \dfrac{x - 1}{x + 2}$ , what happens to $f(x)$ as $x \to -2^+$ ?

Practice 5

Find the slant asymptote of $f(x) = \dfrac{2x^2 + 3x - 1}{x + 2}$ . Enter it as $y = \ldots$

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

At what $x$ -value does the graph of $f(x) = \dfrac{x^2 + 4x}{x^2 - 1}$ cross its horizontal asymptote?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Which function has a vertical asymptote $x = 1$ , a horizontal asymptote $y = 2$ , and an $x$ -intercept at $x = 3$ ?

Practice 8

The graph of $f(x) = \dfrac{ax^2 + 5}{2x^2 - x}$ has horizontal asymptote $y = 4$ . Find $a$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.