# How do you find the asymptotes for f(x) = (3 - x) / x^2?

Nov 7, 2016

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

#### Explanation:

As you cannot divide by 0, the vertical asymptote is $x = 0$

There is no slant asymptote as the degree of the numerator is less than the degree of the denominator:

${\lim}_{n \rightarrow - \infty} \frac{3 - x}{x} ^ 2 = {\lim}_{n \rightarrow - \infty} - \frac{x}{x} ^ 2 = {\lim}_{n \rightarrow - \infty} - \frac{1}{x} = {0}^{+}$

${\lim}_{n \rightarrow + \infty} \frac{3 - x}{x} ^ 2 = {\lim}_{n \rightarrow + \infty} - \frac{x}{x} ^ 2 = {\lim}_{n \rightarrow + \infty} - \frac{1}{x} = {0}^{-}$

The horizontal asymptote is $y = 0$
graph{(3-x)/x^2 [-10, 10, -5, 5]}