# How do you find the vertical, horizontal and slant asymptotes of: f(x) = (4x)/(x^2-25)?

Sep 18, 2016

The VA's are $x = - 5$ and $x = 5$. The HA is $y = 0$.

#### Explanation:

$f \left(x\right) = \frac{4 x}{{x}^{2} - 25}$

Factor the denominator.

$f \left(x\right) = \frac{4 x}{\left(x + 5\right) \left(x - 5\right)}$

The vertical asymptotes can be found by setting the denominator equal to zero and solving for x. f(x) is undefined when the denominator = 0.

$\left(x + 5\right) \left(x - 5\right) = 0$
$x = - 5$ and $x = 5$ These are the VA's.

The horizontal asymptote is found by comparing the degree of the numerator to the degree of the denominator.
If the degree of the numerator is less than the degree of the numerator, the HA is y=0.
If the degrees are equal, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
*If the degree of the numerator is greater, there is a slant asymptote.

In $f \left(x\right) = \frac{4 {x}^{\textcolor{red}{1}}}{{x}^{\textcolor{b l u e}{2}} - 25}$, the degree of the numerator is $\textcolor{red}{1}$ and the degree of the denominator is $\textcolor{b l u e}{2}$. Thus, the HA is $y = 0$.

There are no slant asymptotes because the degree of the numerator is not greater than the degree of the denominator.