Math Core

A rational function is a fraction whose numerator and denominator are both polynomials, like $f(x) = \dfrac{2x - 4}{x + 1}$ . Its graph can break apart, shoot off toward infinity, and level off near a line it never quite reaches. Once you know where to look for these features, you can sketch any rational function quickly.

Definition

Rational function

A rational function has 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 every real number except the zeros of $q(x)$ .

Transformations of $y = \dfrac{1}{x}$

The simplest rational function is the parent function $y = \dfrac{1}{x}$ , the inverse variation graph from the last lesson. It has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 0$ .

The transformations you know from other parent functions work here too. In

y = \frac{a}{x - h} + k,

the graph of $y = \dfrac{1}{x}$ is stretched by $a$ (and reflected if $a < 0$ ), shifted right $h$ units and up $k$ units. The asymptotes move with it: the vertical asymptote becomes $x = h$ and the horizontal asymptote becomes $y = k$ .

Worked example: A shifted hyperbola

Graph $y = \dfrac{2}{x - 3} + 1$ .

Here $a = 2$ , $h = 3$ and $k = 1$ . Draw the asymptotes first: $x = 3$ and $y = 1$ . Then plot a couple of points on each side of $x = 3$ :

$x$	1	2	4	5
$y$	0	$-1$	3	2

Since $a > 0$ , the branches sit in the upper right and lower left of the new "center" $(3, 1)$ .

y = 2/(x − 3) + 1 with asymptotes x = 3 and y = 1 (dashed).Open in grapher →

Features of a general rational function

Most rational functions aren't written in shifted form. To graph $f(x) = \dfrac{p(x)}{q(x)}$ , factor both polynomials and then find these features.

Holes. If a factor $(x - c)$ appears in both the numerator and the denominator, it cancels. The graph looks just like the simplified function but with a single missing point (a hole) at $x = c$ . To find the $y$ -coordinate of the hole, plug $c$ into the simplified function.

Vertical asymptotes. After canceling, any remaining zero of the denominator gives a vertical asymptote. Near it, the function values grow without bound in the positive or negative direction.

$x$ -intercepts. After canceling, the zeros of the numerator are the $x$ -intercepts.

$y$ -intercept. Evaluate $f(0)$ , if $0$ is in the domain.

Horizontal asymptote. This describes what happens as $x$ gets very large or very negative. It depends only on the leading terms, so compare the degree of the numerator ( $n$ ) with the degree of the denominator ( $d$ ).

Horizontal asymptote rules

For $f(x) = \dfrac{a x^n + \cdots}{b x^d + \cdots}$ :

degrees	horizontal asymptote
$n < d$	$y = 0$
$n = d$	$y = \dfrac{a}{b}$ , the ratio of leading coefficients
$n > d$	none

Why? For huge $x$ , the leading terms swamp everything else. If $n = d$ , the fraction behaves like $\dfrac{a x^n}{b x^n} = \dfrac{a}{b}$ . If $n < d$ , the denominator grows faster and the fraction shrinks toward $0$ . If $n > d$ , the numerator wins and the function keeps growing. (When $n = d + 1$ , polynomial division shows the graph approaches a slanted line, called a slant asymptote.)

Worked example: Finding every feature

Graph $f(x) = \dfrac{2x - 4}{x + 1}$ .

Nothing cancels, since $2x - 4 = 2(x - 2)$ shares no factor with $x + 1$ .
Vertical asymptote: $x + 1 = 0$ , so $x = -1$ .
$x$ -intercept: $2x - 4 = 0$ , so $x = 2$ .
$y$ -intercept: $f(0) = \dfrac{-4}{1} = -4$ .
Horizontal asymptote: both degrees are $1$ , so $y = \dfrac{2}{1} = 2$ .

f(x) = (2x − 4)/(x + 1), with asymptotes x = −1 and y = 2 and intercepts (2, 0) and (0, −4).Open in grapher →

Worked example: A hole and an asymptote

Find the holes, asymptotes and intercepts of $g(x) = \dfrac{x^2 - 1}{x^2 - x - 2}$ .

Factor:

g(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 1)}.

The factor $(x + 1)$ cancels, so there is a hole at $x = -1$ . The simplified function is $\dfrac{x - 1}{x - 2}$ , and $\dfrac{-1 - 1}{-1 - 2} = \dfrac{2}{3}$ , so the hole is at $\left(-1, \dfrac{2}{3}\right)$ .

Vertical asymptote: $x = 2$ (the remaining zero of the denominator).
$x$ -intercept: $x = 1$ .
$y$ -intercept: $g(0) = \dfrac{-1}{-2} = \dfrac{1}{2}$ .
Horizontal asymptote: degrees are equal (both $2$ ) with leading coefficients $1$ and $1$ , so $y = 1$ .

g(x) = (x² − 1)/(x² − x − 2). The curve has a hole at (−1, 2/3), where x = −1 is not in the domain.Open in grapher →

Common mistake

Don't find vertical asymptotes from the unfactored denominator. In $g(x)$ above, $x = -1$ makes the denominator zero, but it is a hole, not an asymptote, because the factor cancels. Always factor and cancel first, then read off the asymptotes.

Worked example: Comparing degrees

Find the horizontal asymptote, if any.

$h(x) = \dfrac{3x}{x^2 + 1}$ : degree $1$ over degree $2$ , so $y = 0$ .
$k(x) = \dfrac{6x^2 - 5}{2x^2 + x}$ : equal degrees, so $y = \dfrac{6}{2} = 3$ .
$m(x) = \dfrac{x^2 + 1}{x - 1}$ : degree $2$ over degree $1$ , so there is no horizontal asymptote. (Dividing gives $x + 1 + \dfrac{2}{x - 1}$ , so the graph follows the slant line $y = x + 1$ .)

Tip

A graph can cross its horizontal asymptote in the middle; the asymptote only describes the far left and far right. For instance, $h(x) = \dfrac{3x}{x^2 + 1}$ passes through $(0, 0)$ , right on its asymptote $y = 0$ . A graph can never cross a vertical asymptote, because the function is undefined there.

Practice

Practice 1

For $f(x) = \dfrac{3}{x - 2} + 1$ , the vertical asymptote is $x = c$ . What is $c$ ?

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

Practice 2

For $f(x) = \dfrac{3}{x - 2} + 1$ , the horizontal asymptote is $y = c$ . What is $c$ ?

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

Practice 3

The graph of $f(x) = \dfrac{2x + 1}{x - 3}$ has horizontal asymptote $y = c$ . What is $c$ ?

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

Practice 4

What is the $x$ -intercept of $f(x) = \dfrac{x - 4}{x + 2}$ ? Give the $x$ -value.

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

Practice 5

Find all vertical asymptotes of $f(x) = \dfrac{x + 1}{x^2 - x - 6}$ . Enter the $x$ -values.

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

Practice 6

What is the horizontal asymptote of $f(x) = \dfrac{4x^2 - 1}{2x^2 + x - 3}$ ?

Practice 7

The graph of $f(x) = \dfrac{x^2 - 9}{x - 3}$ is a line with one hole. Give the coordinates of the hole as an ordered pair.

Enter a point like (2, -3)

Practice 8

Which statement about $f(x) = \dfrac{x - 1}{x^2 - 1}$ is true?

Transformations of y=1xy = \dfrac{1}{x}y=x1​

Features of a general rational function

Practice

Transformations of $y = \dfrac{1}{x}$