# How do you find the asymptotes for f(x) = (x+3 )/(x^2 + 8x + 15)?

Nov 13, 2015

Factor the denominator and simplify, finding that $f \left(x\right)$ has a horizontal asymptote $y = 0$ and vertical asymptote $x = - 5$

#### Explanation:

$f \left(x\right) = \frac{x + 3}{x + 8 x + 15} = \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)}$

$= \frac{1}{x + 5}$ with exclusion $x \ne 3$

As $x \to \pm \infty$, $\frac{1}{x + 5} \to 0$, so $f \left(x\right)$ has a horizontal asymptote $y = 0$.

When $x = - 5$, the denominator is zero and the numerator is non-zero, so $f \left(x\right)$ has a vertical asymptote $x = - 5$