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

Aug 3, 2017

The vertical asymptote is $x = 2$
The horizontal asymptote is $y = 1$
No oblique asymptote

#### Explanation:

The vertical asymptotes are calculated by performing the limits

${\lim}_{x \to - {2}^{-}} f \left(x\right) = {\lim}_{x \to - {2}^{-}} {x}^{2} / {\left(x - 2\right)}^{2} = \frac{4}{{0}^{+}} = + \infty$

${\lim}_{x \to - {2}^{+}} f \left(x\right) = {\lim}_{x \to - {2}^{+}} {x}^{2} / {\left(x - 2\right)}^{2} = \frac{4}{{0}^{+}} = + \infty$

The vertical asymptote is $x = 2$

To determine the horizontal asymptote, we calculate

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} {x}^{2} / \left({x}^{2}\right) = 1$

The horizontal asymptote is $y = 1$

As the degree of the numerator is $=$ to the degree of the denominator, there is no oblique asymptote

graph{x^2/(x-2)^2 [-20.27, 20.28, -10.14, 10.14]}