# How do you find the asymptotes for f(x)=x^2/(x+5)?

Aug 2, 2017

The vertical asymptote is $x = - 5$
The slant asymptote is $y = x - 5$
There is no horizontal asymptote

#### Explanation:

The vertical asymptotes are calculated by performing the limits

${\lim}_{x \to - {5}^{-}} f \left(x\right) = {\lim}_{x \to - {5}^{-}} {x}^{2} / \left(x + 5\right) = \frac{25}{{0}^{-}} = - \infty$

${\lim}_{x \to - {5}^{+}} f \left(x\right) = {\lim}_{x \to - {5}^{+}} {x}^{2} / \left(x + 5\right) = \frac{25}{{0}^{+}} = + \infty$

The vertical asymptote is $x = - 5$

We perform a long division to calculate the slant asymptote

$\textcolor{w h i t e}{a a a a}$$x + 5$$\textcolor{w h i t e}{a a a a} |$${x}^{2} + 0 x + 0$$|$$\textcolor{w h i t e}{a a a a}$$x - 5$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$${x}^{2} + 5 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$0 - 5 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$- 5 x - 25$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$0 + 25$

Therefore,

$f \left(x\right) = x - 5 + \frac{25}{x + 5}$

${\lim}_{x \to - \infty} f \left(x\right) - \left(x - 5\right) = {\lim}_{x \to - \infty} \frac{25}{x + 5} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(x - 5\right) = {\lim}_{x \to + \infty} \frac{25}{x + 5} = {0}^{+}$

The slant asymptote is $y = x - 5$

To determine the horizontal asymptote, we calculate

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} {x}^{2} / \left(x + 5\right) = - \infty$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} {x}^{2} / \left(x + 5\right) = + \infty$

There is no horizontal asymptote

graph{(y-x^2/(x+5))(y-x+5)=0 [-58.5, 58.53, -29.27, 29.28]}