# How do you find the asymptotes for f(x) =(2x^2-5x+6)/(x+2)?

Dec 16, 2016

The vertical asymptote is $x = - 2$
The slant asymptote is $y = 2 x - 9$
No horizontal asymptote

#### Explanation:

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

The degree of the numerator is $>$ than 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}$$2 {x}^{2} - 5 x + 6$∣$x + 2$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a}$∣$2 x - 9$

$\textcolor{w h i t e}{a a a a a}$$0 - 9 x + 6$

$\textcolor{w h i t e}{a a a a a a a}$$- 9 x - 18$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$+ 24$

So,

$\frac{2 {x}^{2} - 5 x + 6}{x + 2} = 2 x - 9 + \frac{24}{x + 2}$

The slant asymptote is $y = 2 x - 9$

For the horizontal asymptotes, we calculate the limita as ${x}_{>} \infty$

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} 2 {x}^{2} / x = {\lim}_{x \to \pm \infty} 2 x = \pm \infty$

There is no horizontal asymptote.

graph{(y-(2x^2-5x+6)/(x+2))(y-2x+9)=0 [-93.7, 93.9, -46.8, 46.9]}