# How do you graph y=(x^3+27)/(3x^2+x) using asymptotes, intercepts, end behavior?

Nov 29, 2017

See below.

#### Explanation:

$y$ axis intercepts occur when $x = 0$

$y = \frac{{\left(0\right)}^{3} + 27}{x {\left(0\right)}^{2} + \left(0\right)} = 0$

(This is undefined, so no y axis intercept)

$x$ axis intercepts occur when $y = 0$ i.e. when ${x}^{3} + 27 = 0$

${x}^{3} + 27 = 0 \implies x = \sqrt[3]{- 27} = - 3$

(This will also have complex roots, but these will be of no help in graphing).

Point: $\left(- 3 , 0\right)$

Vertical asymptotes occur where the function is undefined. i.e. $3 {x}^{2} + x = 0$

$3 {x}^{2} + x = x \left(3 x + 1\right) = 0 \implies x = 0 \mathmr{and} x = - \frac{1}{3}$

Asymptotes are the lines:

$\textcolor{b l u e}{x = 0} \mathmr{and} \textcolor{b l u e}{x = - \frac{1}{3}}$

Because the rational function is an improper fraction ( degree of numerator is higher than degree of denominator) an oblique asymptote occurs. The can be found by dividing numerator by denominator.

$\frac{{x}^{3} + 27}{3 {x}^{2} + x} = \frac{1}{3} x - \frac{1}{9}$

( we only need to divide enough to give us the terms of a line equation)

So the line:

$\textcolor{b l u e}{y = \frac{1}{3} x - \frac{1}{9}}$ is an oblique asymptote:

End behaviour:

${x}^{3} / \left(3 {x}^{2}\right) = \frac{1}{3} x$

as $x \to \infty$,$\textcolor{w h i t e}{88} \frac{1}{3} x \to \infty$

as $x \to - \infty$,$\textcolor{w h i t e}{88} \frac{1}{3} x \to - \infty$

Graph: