# How do you find the asymptotes for f(x) = (2x^2 + 1) / (x^2 - 4)?

Oct 2, 2016

The vertical asymptotes are $x = 2$ and $x = - 2$. The horizontal asymptote is $y = 2$. There are no oblique asymptotes.

#### Explanation:

Vertical asymptotes occur where $x$ is undefined, i.e. the denominator equals zero.

${x}^{2} - 4 = 0$
$\left(x + 2\right) \left(x - 2\right) = 0 \textcolor{w h i t e}{a a a a a a a}$ Factor
$x + 2 = 0 \textcolor{w h i t e}{a a a} x - 2 = 0 \textcolor{w h i t e}{a a a}$ Set each factor equal to zero
$x = - 2 \textcolor{w h i t e}{a a a} x = 2 \textcolor{w h i t e}{a a a a a a a}$ Solve for x

The vertical asymptotes are $x = - 2$ and $x = 2$

Horizontal asymptotes are found by comparing the degree of the denominator to the degree of the numerator.

If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$.

If the degree of the denominator is equal to the degree of the numerator, the horizontal asymptote is the leading coefficient of the numerator divided by the leading coefficient of the denominator.

$f \left(x\right) = \frac{\textcolor{b l u e}{2} {x}^{\textcolor{red}{2}} + 1}{\textcolor{b l u e}{1} {x}^{\textcolor{red}{2}} - 4}$

In this example, the $\textcolor{red}{\mathrm{de} g r e e}$ is $\textcolor{red}{2}$ in both the denominator and numerator, so the horizontal asymptote is the leading coefficient of the numerator over the leading coefficient of the denominator
$y = \frac{\textcolor{b l u e}{2}}{\textcolor{b l u e}{1}} = 2$

If the degree of the denominator is less than the degree of the numerator, there is an oblique asymptote. This example does not have an oblique asymptote.