How do you find vertical, horizontal and oblique asymptotes for f(x)= (6x^4-x^3+6x+5)/(x^3+1)?

1 Answer
Jun 14, 2018

Vertical asypmtote: x=-1
Oblique asymptote: y=6x-1

Explanation:

The vertical asymptote occurs at the point where the denominator is equal to zero, and in this case, x=-1 satisfies x^3+1=0. Since the degree of the numerator is larger than the degree of the denominator, there is an oblique asymptote instead of a horizontal asymptote. Using polynomial long division, we find that f(x)=6x-1+6/(x^3+1), so the oblique asyptote is y=6x-1.