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

Dec 6, 2016

Vertical+ $\downarrow x = 1 \uparrow$ The contracted graph illustrates the asymptoticity of x = 1.

#### Explanation:

By actual division $y = 4 {x}^{2} + \frac{4 {x}^{2}}{\left(x - 1\right) \left({x}^{2} + x + 1\right)}$.

As $x \to 1 , y \to 4 \left(1 \pm \infty \left(\frac{1}{3}\right)\right) = \pm \infty$.

So, x = 1 is the asymptote.

I surmise that the origin is not point of inflexion, despite that

seemingly it is so.

The curve is flat there, upon the x-axis, in infinitesimal

neighborhood. In other words, y and all its derivatives are 0, And so,

thereabout the origin , there is no series expansion.

graph{y(x^3-1)-4x^5=0 [-40, 40, -20, 20]} }