# How do you identify all horizontal and slant asymptote for f(x)=4/(x-2)^3?

Nov 7, 2016

The vertical asymptote is $x = 2$
The horizontal asymptote is $y = 0$

#### Explanation:

As you cannot divide by $0$, you have a vertical asymptote $x = 2$
There is no slant asymptote as the degree of the numerator is smaller than the degree of the denominator.
${\lim}_{n \rightarrow - \infty} f \left(x\right) = {\lim}_{n a r r - \infty} \frac{4}{x} ^ 3 = {0}^{-}$

${\lim}_{n \rightarrow + \infty} f \left(x\right) = {\lim}_{n a r r + \infty} \frac{4}{x} ^ 3 = {0}^{+}$

So there is a horizontal asymptote $y = 0$
graph{4/(x-2)^3 [-16.02, 16.01, -8.01, 8.01]}