# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= -1/(x+1)^2?

Apr 28, 2016

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

#### Explanation:

Finding the Vertical Asymptote

Given,

$f \left(x\right) = - \frac{1}{x + 1} ^ 2$

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

${\left(x + 1\right)}^{2} = 0$

$x + 1 = 0$

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

Finding the Horizontal Asymptote

Given,

$f \left(x\right) = - \frac{1}{x + 1} ^ 2$

If you foil the denominator, you will notice that the degree of the denominator is $2$.

$f \left(x\right) = - \frac{1}{{x}^{2} + 2 x + 1}$

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $f \left(x\right) = 0$.

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

Finding the Oblique Asymptote

Given,

$f \left(x\right) = - \frac{1}{x + 1} ^ 2$

There would be a slant asymptote if the degree of the leading term in the numerator is $1$ value larger than the degree of the leading term in the denominator. In your case, we see the opposite — ie. the degree in the denominator is greater than the degree in the numerator.

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