How do you find vertical, horizontal and oblique asymptotes for (x^3+8) /(x^2+9)?

May 19, 2016

$\frac{{x}^{3} + 8}{{x}^{2} + 9}$ has an oblique asymptote $y = x$ and no other asymptotes.

Explanation:

For all Real values of $x$, the denominator ${x}^{2} + 9 \ge 9 > 0$.

So there are no vertical asymptotes.

$\frac{{x}^{3} + 8}{{x}^{2} + 9} = \frac{{x}^{3} + 9 x - 9 x + 8}{{x}^{2} + 9} = \frac{x \left({x}^{2} + 9\right) - 9 x + 8}{{x}^{2} + 9} = x + \frac{- 9 x + 8}{{x}^{2} + 9}$

Then since the denominator has higher degree than the numerator:

$\frac{- 9 x + 8}{{x}^{2} + 9} \to 0$ as $x \to \pm \infty$

Hence:

$\frac{{x}^{3} + 8}{{x}^{2} + 9} = x + \frac{- 9 x + 8}{{x}^{2} + 9}$

is asymptotic to $x$ as $x \to \pm \infty$

So $f \left(x\right)$ has no vertical or horizontal asymptotes, but has one oblique asymptote:

$y = x$

graph{(y-(x^3+8)/(x^2+9))(y-x) = 0 [-20, 20, -10, 10]}