# How do you graph y=(3x^3+1)/(4x^2-32) using asymptotes, intercepts, end behavior?

Nov 23, 2016

The vertical asymptotes are $x = \sqrt{8}$ and $x = - \sqrt{8}$
The slant asymptote is $y = \frac{3}{4} x$
No horozontal asymptote.

#### Explanation:

Let's factorise the denominator $\left(4 {x}^{2} - 32\right)$

$= 4 \left({x}^{2} - 8\right) = 4 \left(x + \sqrt{8}\right) \left(x - \sqrt{8}\right)$

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{\sqrt{8} , - \sqrt{8}\right\}$

As we cannot divide by $0$,

So $x \ne \sqrt{8}$ and $x \ne - \sqrt{8}$

The vertical asymptotes are $x = \sqrt{8}$ and $x = - \sqrt{8}$

As the degree of the numerator is $>$ the degree of the denominator, we expect a slant asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} + 1$$\textcolor{w h i t e}{a a a a}$∣$4 {x}^{2} - 32$

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} - 24 x$$\textcolor{w h i t e}{a a}$∣$\frac{3 x}{4}$

$\textcolor{w h i t e}{a a a a a}$$0 - 24 x + 1$

So, $\frac{3 {x}^{3} + 1}{4 {x}^{2} - 32} = \frac{3 x}{4} - \frac{24 x + 1}{4 {x}^{2} + 32}$

The slant asymptote is $y = \frac{3}{4} x$

To calculate the limits, we use the terms of highest degree.

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \frac{3 {x}^{3}}{4 {x}^{2}} = {\lim}_{x \to + \infty} \frac{3 x}{4} = + \infty$

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} \frac{3 {x}^{3}}{4 {x}^{2}} = {\lim}_{x \to - \infty} \frac{3 x}{4} = - \infty$

There are no horizontal asymptote

${\lim}_{x \to - {\sqrt{8}}^{-}} = - \infty$

${\lim}_{x \to - {\sqrt{8}}^{+}} = + \infty$

${\lim}_{x \to {\sqrt{8}}^{-}} = - \infty$

${\lim}_{x \to {\sqrt{8}}^{+}} = + \infty$

When $x = 0$, $\implies$, $y = - \frac{1}{32}$

When $y = 0$, $0 >$, $x = {\left(- \frac{1}{3}\right)}^{\frac{1}{3}}$

graph{(y-(3x^3+1)/(4x^2-32))(y-x3/4)=0 [-28.86, 28.9, -14.43, 14.43]}

Nov 23, 2016

Asymptotes: slant $y = \frac{3}{4} x$ and vertical $x = \pm \sqrt{8}$
y-intercept $= - \frac{1}{32}$ and x-intercept =$- \frac{1}{3} ^ \left(\frac{1}{3}\right) = - 6934$, nearly

#### Explanation:

Resolving into partial fractions,

$y = \frac{3}{4} x + \frac{6 x + \frac{1}{4}}{{x}^{2} - 8}$

Rearranging.

y-3/4x = (6x+1/4)/((x-sqrt8)(x+sqrt8)

The form reveals that the asymptotes are given by

the slant $y = \frac{3}{4} x$ and the vertical ones$x = \pm \sqrt{8}$.

Easily from the given equation, the intercepts can be obtained, as