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

Nov 5, 2016

The vertical asymptote is $x = - 2$
The oblique asymptote is $y = 2 x + 4$

#### Explanation:

The vertical asymptote is $x = - 2$
As the degree of numerator is $>$ the degree of denominator, we expect an oblique asymptote.
Therefore, we do a long division
$2 {x}^{2}$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$∣x+2
$2 {x}^{2} + 4 x$$\textcolor{w h i t e}{a a}$color(white)(aaaa)∣2x-4
$\textcolor{w h i t e}{a}$$0 - 4 x - 3$
$\textcolor{w h i t e}{a}$$0 - 4 x - 8$
$\textcolor{w h i t e}{a a a a}$$- 0 + 5$
So, $\frac{2 {x}^{3} - 3}{x + 2} = 2 x - 4 + \frac{5}{x + 2}$
So the oblique asymptote is $y = 2 x - 4$
Limit $y = 2 {x}^{2} / x = 2 x = \pm \infty$
$x \to \pm \infty$
graph{(y-(2x^2-3)/(x+2))(y-2x+4)=0 [-33.56, 31.36, -22.34, 10.15]}