Determine the equations of the vertical and horizontal asymptotes, if any, for g (x) = x^3 / (x-2) (x+1)

for a rational, we find the vertical asymptotes where its denominator is 0, thus

(x-2) (x+1) = 0, gives us two vertical asymptotes when that happens, x = 2 and x = - 1.

if we expand the denominator, we'll end up with a quadratic equation, namely a 2nd degree equation, whilst the numerator is of 3rd degree. Whenever the numerator has a higher degree than the denominator, the rational has no horizontal asymptotes, however when the numerator is exactly 1 degree higher like in this case, it has an oblique asymptote instead.