# How do you find the vertical, horizontal and slant asymptotes of: f(x)= ( x^3-2x-3) /( x^2+2)?

Oct 22, 2016

There is only a slant asymptote $y = x$

#### Explanation:

The domain of the function is $\mathbb{R}$ as the denominator is always $> 0$
So there is no vertical asymptote

The degree of the numerator is $>$ degree of denominator, so we expect a slant asymptote

A long division gives

$f \left(x\right) = \frac{{x}^{3} - 2 x - 3}{{x}^{2} + 2} = x - \frac{4 x + 3}{{x}^{2} + 2}$

So the slant asymptote is $y = x$
For horizontal asymptotes, we look at the limit of $f \left(x\right)$ as $x \to \pm \infty$

limit $f \left(x\right)$ as $x \to + \infty$ is $+ \infty$

limit $f \left(x\right)$ as $x \to - \infty$ is $- \infty$
So there is no horizontal asymptote

graph{(x^3-2*x-3)/(x^2+2) [-10, 10, -5, 5]}