Rational Functions

A rational function is a quotient of polynomial functions, $f(x)=\frac{p(x)}{q(x)}$ , with $q(x)\ne 0$ . The denominator creates domain restrictions that can produce holes or vertical asymptotes.

Factoring before simplifying is essential because cancellation can change the visible formula without restoring excluded input values.

On this page (8 sections)

Domain restrictions and vertical asymptotes

f(x)=\frac{p(x)}{q(x)},\qquad q(x)\ne0

The numerator and denominator are polynomials, and the denominator is not the zero polynomial. Exclude every zero of the original denominator before simplifying.

\frac{x+1}{x-2}=1+\frac3{x-2},\qquad x\ne2

This form makes the vertical asymptote $x=2$ and horizontal asymptote $y=1$ visible. Just left of 2, the fraction becomes large and negative; just right of 2, it becomes large and positive.

Two branches separated by an excluded input

$y=(x+1)/(x-2)=1+3/(x-2)$

Grey dashed lines mark the stated asymptotes.

The asymptotes are

x=2

and

y=1

Diagram explained

The left branch falls without bound as inputs approach 2 from below; the right branch rises without bound as inputs approach 2 from above. The curve is never joined across the asymptote. Reference points give its exact values.

Exact reference values for this graph
Input	Output
$-1$	$0$
$0$	$-1/2$
$1$	$-2$
$3$	$4$
$5$	$2$
$8$	$3/2$

The horizontal intercept is $(-1,0)$ , because the numerator vanishes at $x=-1$ . The vertical intercept is $(0,-1/2)$ . An intercept must belong to the original domain.

Cancelled factors and removable holes

For $(x^2-1)/(x-1)$ , factor before cancelling:

\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1,\qquad x\ne1

The graph is the line $y=x+1$ with a hole at $(1,2)$ . The simplified expression predicts the approached height but does not restore the excluded input.

Cancellation does not always leave a hole

\frac{x-1}{(x-1)^2}=\frac1{x-1},\qquad x\ne1

Here a denominator factor remains after cancellation, so $x=1$ is still a vertical asymptote. A removable hole occurs when the fully reduced expression has a finite value at the excluded input.

Horizontal asymptotes

In $1+3/(x-2)$ , the fraction approaches zero as $x$ grows in either direction, so the horizontal asymptote is $y=1$ .

Find the range, not just the asymptote

The function never equals $1$ , since $3/(x-2)$ cannot be zero. Every other real output can occur: solving $y=(x+1)/(x-2)$ for $x$ gives:

x=\frac{2y+1}{y-1},\qquad y\ne1

This value never equals 2, so it is a permitted input. The range is $(-\infty,1)\cup(1,\infty)$ .

For a general rational function, compare numerator and denominator degrees to study end behaviour. A detailed treatment of asymptotes belongs to the later advanced-functions work; the rewritten example here establishes the idea directly.

Solving a rational equation

To solve $(x+1)/(x-2)=2$ , retain the restriction $x\ne2$ .

Clear the denominator on its permitted domain

$x+1=2(x-2)$
Multiply by x − 2, which is non-zero on the stated domain.
$x+1=2x-4\quad\Longrightarrow\quad x=5$
Expand and collect terms.
$\frac{5+1}{5-2}=\frac63=2$
Check the candidate in the original equation.

Excluded inputs are never solutions, even if a later polynomial equation happens to produce them.

Reading asymptotes correctly

A vertical asymptote of a rational function occurs at an excluded input where values are unbounded. A horizontal asymptote instead concerns end behaviour and may be crossed at a finite input.

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

This function has horizontal asymptote $y=0$ and crosses it at $(0,0)$ . The earlier example $(x+1)/(x-2)$ does not cross its horizontal asymptote, but that is a feature of that function, not a universal rule.

Practice questions

Find the domain restriction for $\frac{x+1}{x-2}$ .
What happens at $x=1$ for $\frac{x^2-1}{x-1}$ ?
Solve $\frac{x+1}{x-2}=2$ .

Check your answers

$x\ne2$ , because the original denominator is zero at 2.
The reduced rule is $x+1$ , but $x=1$ remains excluded. The graph has a removable hole at $(1,2)$ .
$x+1=2x-4$ gives $x=5$ . It is permitted and makes the original fraction equal 2.

Sources and further reading

OpenStax: Rational functions and asymptotes. Definitions, conventions and related examples.
OpenStax Precalculus 2e. Functions, transformations, inverse functions and function families.
Wolfram Language: FunctionDomain. Domain conventions and symbolic domain checks.
Wolfram Language: FunctionRange. Range conventions and symbolic range checks.