# How do you find the horizontal asymptote for (x-3) / (x+5)?

Mar 22, 2018

$y = 1$

#### Explanation:

There are two ways of solving this.
1. Limits:

$y = {\lim}_{x \to \pm \infty} \frac{a x + b}{c x + d} = \frac{a}{c}$, therefore horizontal asymptote occurs when $y = \frac{1}{1} = 1$

2. Inverse:
Let's take the inverse of $f \left(x\right)$, this is because the $x$ and $y$ asymptotes of $f \left(x\right)$ will be the $y$ and $x$ asymptotes for ${f}^{-} 1 \left(x\right)$

$x = \frac{y - 3}{y + 5}$
$x y + 5 x = y - 3$
$x y - y = - 5 x - 3$
$y \left(x - 1\right) = - 5 x - 3$
$y = {f}^{-} 1 \left(x\right) = - \frac{5 x + 3}{x - 1}$

The vertical asymptote is the same as the horizontal asymptote of $f \left(x\right)$

The vertical asymptote of ${f}^{-} 1 \left(x\right)$ is $x = 1$, therefore the horizontal asymptote of $f \left(x\right)$ is $y = 1$