# How do you find the vertical, horizontal and slant asymptotes of: (x^2-25)/(x^2+5x)?

Apr 30, 2016

vertical asymptote: $x = 0$
horizontal asymptote: $f \left(x\right) = 1$
slant asymptote: does not exist

#### Explanation:

Finding the Vertical Asymptote

Given,

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

Factor the numerator.

f(x)=((x+5)(x-5))/(x(x+5)

Cancel out any factors that appear in the numerator and denominator.

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

Set the denominator equal to $0$ and solve for $x$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = 0} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finding the Horizontal Asymptote

Given,

$f \left(x\right) = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{1} {x}^{2} - 25}{\textcolor{p u r p \le}{1} {x}^{2} + 5 x}$

Divide the $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{leading coefficient}}$ of the leading term in the numerator by the $\textcolor{p u r p \le}{\text{leading coefficient}}$ of the leading term in the denominator.

$f \left(x\right) = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{1}}{\textcolor{p u r p \le}{1}}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finding the Slant Asymptote

Given,

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

There would be a slant asymptote if the degree of the leading term in the numerator was $1$ value larger than the degree of the leading term in the denominator. In your case, we see that the degree in both the numerator and denominator are equal.

$\therefore$, the slant asymptote does not exist.