# How do you find the vertical, horizontal or slant asymptotes for  f(x)= (x^2)/(x-2)^2?

Sep 17, 2016

The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 1$.

#### Explanation:

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

Vertical Asymptote :

Find the value of x that makes the denominator equal zero.
${\left(x - 2\right)}^{2} = 0$
$\left(x - 2\right) \left(x - 2\right) = 0$
$x - 2 = 0$
$x = 2$ is the VA

Horizontal Asymptote :

Compare the degree of the numerator and denominator. The degree is given by the greatest exponent.
If they are the same, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
If the degree of the denominator is larger, the HA is $y = 0$.
*If the degree of the numerator is larger, there is a slant asymptote.

In this problem, the degree of both the numerator and the denominator is $\textcolor{\mathmr{and} a n \ge}{2}$. The HA is y=LC/LC.

f(x)=(color(red)(1)x^color(orange)2)/(color(blue)(1)(x-2)^color(orange)2

$y = \frac{\textcolor{red}{1}}{\textcolor{b l u e}{1}}$ or $y = 1$ is the HA.

Slant Asymptote :

This function has no slant asymptote because the degree of the numerator is not greater than the degree of the denominator.