# How do you find vertical, horizontal and oblique asymptotes for (x^3+5x^2)/(x^2-25)?

Feb 7, 2017

The vertical asymptote is $x = 5$
The oblique asymptote is $y = x + 5$
No horizontal asymptote

#### Explanation:

The denominator is

${x}^{2} - 25 = \left(x + 5\right) \left(x - 5\right)$

and the numerator is

${x}^{3} + 5 {x}^{2} = {x}^{2} \left(x + 5\right)$

Therefore,

$\frac{{x}^{3} + 5 {x}^{2}}{{x}^{2} - 25} = \frac{{x}^{2} \cancel{x + 5}}{\cancel{x + 5} \left(x - 5\right)}$

$= {x}^{2} / \left(x - 5\right)$

Let $f \left(x\right) = {x}^{2} / \left(x - 5\right)$

As we cannot divide by $0$, $x \ne 5$

The vertical asymptote is $x = 5$

The degree of the numerator is $>$ than the degree of the denominator, there is an oblique asymptote.

Let 's do a long division

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

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

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

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

Therefore,

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

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

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

The oblique asymptote is $y = x + 5$

graph{(y-(x^2)/(x-5))(y-x-5)(y+50x-250)=0 [-65.86, 65.94, -32.9, 32.9]}