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

Answer:

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#.