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

1 Answer
Nov 24, 2016

Answer:

Inserting graph to give visual effect for the answer by Steffany R.. The graph reveals that as #x to 2^(1/3), y to +-oo#. 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^(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^(1/3)=1.71#, giving a local minimum 2.85, nearly.

nearly.'