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

Feb 28, 2016

Vertical asymptote at $x = 6$.
Horizontal asymptote at $y = 0$.

#### Explanation:

By factorizing the denominator as a difference of 2 squares, we get that $h \left(x\right) = \frac{x + 6}{\left(x + 6\right) \left(x - 6\right)} = \frac{1}{x - 6}$.

Hence a vertical asymptote occurs at points leading to division by zero in the denominator, ie when $x = 6$.

Note that $x = - 6$ is a removable singularity and not an asymptote.

Horizontal asymptotes occur at $y = {\lim}_{x \to \pm \infty} h \left(x\right) = 0$.

The plotted graph verifies the existence of these asymptotes.

graph{(x+6)/(x^2-36) [-3.71, 16.29, -5.6, 4.4]}