# How do you identify all horizontal and slant asymptote for f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2)?

Nov 9, 2016

The vertical asymptote is $x = - 2$
The slant asymptote is $y = 2 x - 7$
There are no horizontal asymptote

#### Explanation:

The denominator can be factorised as $\left(x + 2\right) \left(x + 1\right)$
So the vertcal asymptote is $x = - 2$
Let's do a long division to simplify $f \left(x\right)$
$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - {x}^{2} - 2 x + 1$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} + 3 x + 2$
$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + 6 {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a a a a}$∣$2 x - 7$
$\textcolor{w h i t e}{a a a a a a}$$0 - 7 {x}^{2} - 6 x + 1$
$\textcolor{w h i t e}{a a a a a a a a}$$- 7 {x}^{2} - 21 x - 14$
$\textcolor{w h i t e}{a a a a a a a a a}$$- 0 + 15 x + 15$

$\frac{2 {x}^{3} - {x}^{2} - 2 x + 1}{{x}^{2} + 3 x + 2} = 2 x - 7 + \frac{15 \cancel{x + 1}}{\left(x + 2\right) \cancel{x + 1}}$

$\therefore y = 2 x - 7$ is a slant asymptote
Also, $f \left(x\right) = \frac{\left(2 x - 1\right) \left(x + 1\right) \left(x - 1\right)}{\left(x + 2\right) \left(x + 1\right)} = \frac{\left(2 x - 1\right) \left(x - 1\right)}{x + 2}$
${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} 2 x = + \infty$
${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} 2 x = - \infty$

graph{(y-((2x^3-x^2-2x+1)/(x^2+3x+2)))(y-2x+7)=0 [-83.3, 83.4, -41.7, 41.74]}