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

Vertical asypmtote: $x = - 1$
Oblique asymptote: $y = 6 x - 1$
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 $f \left(x\right) = 6 x - 1 + \frac{6}{{x}^{3} + 1}$, so the oblique asyptote is $y = 6 x - 1$.