# How do you find vertical, horizontal and oblique asymptotes for f(x)= (5x^5)/(x^3-2)?

Nov 24, 2016

Inserting graph to give visual effect for the answer by Steffany R.. The graph reveals that as $x \to {2}^{\frac{1}{3}} , y \to \pm \infty$. See explanation for this aspect of the graph.

#### Explanation:

graph{10y(x^3-2)-5x^5=0 [-10, 10, -5, 5]}

graph{y(x^3-2)-5x^5=0 [-5, 5, -2.5, 2.5]} graph{10y(x^3-2)-5x^5=0 [-10, 10, -5, 5]}

The second graph reveals only the portion of the graph for

$x < {2}^{\frac{1}{3}} = 1.26$, nearly.

I have introduced scale factor 1 for 10 to bring to our page the

other invisible part.

Observe that the tangent is horizontal at x = 0 ( a point of inflexion ) a
nd $x = {5}^{\frac{1}{3}} = 1.71$, giving a local minimum 2.85, nearly.

nearly.'