# How do you find vertical, horizontal and oblique asymptotes for ( 4x^5)/(x^3-1)?

Aug 8, 2018

This function has a vertical asymptote $x = 1$ and no horizontal or oblique asymptotes.

#### Explanation:

Given:

$f \left(x\right) = \frac{4 {x}^{5}}{{x}^{3} - 1}$

Note that:

$\frac{4 {x}^{5}}{{x}^{3} - 1} = \frac{4 {x}^{5} - 4 {x}^{2} + 4 {x}^{2}}{{x}^{3} - 1}$

$\textcolor{w h i t e}{\frac{4 {x}^{5}}{{x}^{3} - 1}} = \frac{4 {x}^{2} \left({x}^{3} - 1\right) + 4 {x}^{2}}{{x}^{3} - 1}$

$\textcolor{w h i t e}{\frac{4 {x}^{5}}{{x}^{3} - 1}} = 4 {x}^{2} + \frac{4 {x}^{2}}{{x}^{3} - 1}$

$\textcolor{w h i t e}{\frac{4 {x}^{5}}{{x}^{3} - 1}} = 4 {x}^{2} + \frac{4 {x}^{2}}{\left(x - 1\right) \left({x}^{2} + x + 1\right)}$

The only real zero of the denominator is $x = 1$, which is not a zero of the numerator.

Hence, $f \left(x\right)$ has a vertical asymptote at $x = 1$

Note also that:

${\lim}_{x \to \pm \infty} \frac{4 {x}^{2}}{{x}^{3} - 1} = {\lim}_{x \to \pm \infty} \frac{\frac{4}{x}}{1 - \frac{1}{x} ^ 3} = \frac{0}{1 - 0} = 0$

Hence $f \left(x\right)$ is asymptotic to the parabola $4 {x}^{2}$ and has no horizontal or oblique asymptotes.

graph{(4x^5)/(x^3-1) [-5, 5, -35, 75]}