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

Dec 4, 2016

The oblique asymptote is $y = x$
No vertical and horizontal asymptotes

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

The denominator ${x}^{2} + 9 > 0$, $\forall x \in \mathbb{R}$

So there are no vertical asymptotes.

As the degree of the numerator is $>$ degree of the denominator, there is an oblique asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 8$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} + 9$

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 9 x$$\textcolor{w h i t e}{a a a}$∣$x$

$\textcolor{w h i t e}{a a a a}$$0 - 9 x + 8$

So,

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

So, the oblique asymptote is $y = x$

To calculate the limits as $x - \pm \infty$, we take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} {x}^{3} / {x}^{2} = {\lim}_{x \to \pm \infty} x = \pm \infty$

graph{(y-(x^3+8)/(x^2+9))(y-x)=0 [-5.546, 5.55, -2.773, 2.774]}