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

Nov 27, 2016

The vertical asymptote is $x = - 2$
The oblique asymptote is $y = - 3 x + 6$
No horizontal asymptote

#### Explanation:

Let $f \left(x\right) = \frac{- 3 {x}^{2} + 2}{x + 2}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2\right\}$

As we cannot divide by $0$, $x \ne - 2$

So, the vertical asymptote is $x = - 2$

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

$\textcolor{w h i t e}{a a a a}$$- 3 {x}^{2} - 6 x$$\textcolor{w h i t e}{a a a}$∣$- 3 x + 6$

$\textcolor{w h i t e}{a a a a}$$- 0 + 6 x + 2$

$\textcolor{w h i t e}{a a a a a a a}$$+ 6 x + 12$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 - 10$

Therefore,

$\frac{- 3 {x}^{2} + 2}{x + 2} = \left(- 3 x + 6\right) - \frac{10}{x + 2}$

So, the oblique asymptote is $y = - 3 x + 6$

To calculate the limits to $\infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} - \frac{3 {x}^{2}}{x} = {\lim}_{x \to - \infty} - 3 x = + \infty$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} - \frac{3 {x}^{2}}{x} = {\lim}_{x \to + \infty} - 3 x = - \infty$

graph{(y-(-3x^2+2)/(x+2))(y+3x-6)=0 [-74.04, 74.05, -37, 37]}