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

1 Answer
Aug 8, 2018

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

Explanation:

Given:

f(x) = (4x^5)/(x^3-1)

Note that:

(4x^5)/(x^3-1)=(4x^5-4x^2+4x^2)/(x^3-1)

color(white)((4x^5)/(x^3-1))=(4x^2(x^3-1)+4x^2)/(x^3-1)

color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/(x^3-1)

color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/((x-1)(x^2+x+1))

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

Hence, f(x) has a vertical asymptote at x=1

Note also that:

lim_(x->+-oo) (4x^2)/(x^3-1) = lim_(x->+-oo) (4/x)/(1-1/x^3) = 0/(1-0) = 0

Hence f(x) is asymptotic to the parabola 4x^2 and has no horizontal or oblique asymptotes.

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