# How do you find the asymptotes for h(x) = (x+6)/( x^2 - 36)?

Dec 14, 2015

Find the x-value where the equation becomes undefined.

#### Explanation:

When an equation is undefined, that means that the denominator equals zero.

Set the denominator equal to zero and solve:

${x}^{2} - 36 = 0$
${x}^{2} = 36$
$\left(x = - 6\right)$ $\left(x = 6\right)$

The equation is undefined when $x = 6$ and $x = - 6$, so there are vertical asymptotes at $x = 6$ and $x = - 6$.

Also, since the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at $y = 0$.