Finding the $x$-intercepts of Rational Functions

Recall that a rational function is defined as the ratio of two real polynomials with the condition that the polynomial in the denominator is not a zero polynomial.

$f(x) = \dfrac{P(x)}{Q(x)}$, where $Q(x) \neq 0$

An example of a rational function is:

$f(x) = \dfrac{x + 1}{2x^2 - x - 1}$

Rational functions can be graphed on the coordinate plane. We can use algebraic methods to calculate their $x$-intercepts (also known as zeros or roots), which are points where the graph intersects the $x$-axis. Rational functions can have zero, one, or multiple $x$-intercepts. 

For any function, the $x$-intercepts are $x$-values for which the function has a value of zero: $f(x) = 0$.

In the case of rational functions, the $x$-intercepts exist when the numerator is equal to $0$. For $f(x) = \frac{P(x)}{Q(x)}$, if $P(x) = 0$, then $f(x) = 0$.

In order to solve rational functions for their $x$-intercepts, set the polynomial in the numerator equal to zero, and solve for $x$ by factoring where applicable.

Example 1

Find the $x$-intercepts of this function:

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

Set the numerator of this rational function equal to zero and solve for $x$: 

$\begin {aligned} 0 &=x^2 - 3x + 2 \\&= (x - 1)(x - 2) \end {aligned}$

Solutions for this polynomial are $x = 1$ or $x= 2$. This means that this function has $x$-intercepts at $1$ and $2$.

Example 2

Find the $x$-intercepts of the function:

$f(x) = \dfrac {1}{x}$ 

Here, the numerator is a constant, and therefore, cannot be set equal to $0$. Thus, this function does not have any $x$-intercepts.

Example 3

Find the roots of:

 $g(x) = \dfrac{x^3 - 2x}{2x^2 - 10}$

Factoring the numerator, we have:

 $\begin {aligned} 0&=x^3 - 2x \\&= x(x^2 - 2) \end {aligned}$

Given the factor $x$, the polynomial equals $0$ when $x=0$. 

Let the second factor equal zero, and solve for $x$:

$x^2 - 2 = 0 \\ x^2 = 2 \\ x = \pm \sqrt{2}$

Thus there are three roots, or $x$-intercepts: $0$, $-\sqrt{2}$ and $\sqrt{2}$. These can be observed in the graph of the function below.

Graph of $g(x) = \frac{x^3 - 2x}{2x^2 - 10}$

$x$-intercepts exist at $x = -\sqrt{2}, 0, \sqrt{2}$.