# How to find the asymptotes of f(x) = (x+3) /( x^2 + 8x + 15) ?

Feb 28, 2016

Simplify $f \left(x\right)$ to see that there is a horizontal asymptote $y = 0$, vertical asymptote $x = - 5$ and hole (removable singularity) at $\left(- 3 , \frac{1}{2}\right)$

#### Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} + 8 x + 15}$

$= \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)}$

$= \frac{1}{x + 5}$

with exclusion $x \ne - 3$

This function has horizontal asymptote $y = 0$ since $\frac{1}{x + 5} \to 0$ as x->+-oo#

It has a vertical asymptote $x = - 5$, where the denominator is zero, but the numerator non-zero.

It has a hole (removable singularity) at $\left(- 3 , \frac{1}{2}\right)$ since the left and right limits exist at $\left(- 3 , \frac{1}{2}\right)$ and are equal.