# How do you graph y=x^3/(x^2+7) using asymptotes, intercepts, end behavior?

Nov 20, 2016

See explanation below

#### Explanation:

The domain of y is ${D}_{y} = \mathbb{R}$

The denominator is $\left({x}^{2} + 7\right) > 0$, so we don't have a vertical asymptote.

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

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a a a a}$∣${x}^{2} + 7$

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 7 x$$\textcolor{w h i t e}{a a a a}$∣$x$

$\textcolor{w h i t e}{a a a a a}$$0 - 7 x$

So, ${x}^{3} / \left({x}^{2} + 7\right) = x - \frac{7 x}{{x}^{2} + 7}$

Therefore, $y = x$ is an oblique asymptote

For the limits, we take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} {x}^{3} / {x}^{2} = {\lim}_{x \to \pm \infty} x = \pm \infty$

For the intercepts, when $x = 0$, $\implies$ $y = 0$

graph{(y-x^3/(x^2+7))(y-x)=0 [-7.9, 7.904, -3.956, 3.946]}