# How do you find the asymptotes for y=(x^3-2x^2-x+2)/(x^2-4)?

Jan 7, 2017

The vertical asymptote is $x = - 2$
A hole at $x = 2$
The slant asymptote is $y = x - 2$
No horizontal asymptote.

#### Explanation:

The denominator is ${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2} - x + 2$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} - 4$

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a a}$$- 4 x$$\textcolor{w h i t e}{a a a a a a a}$∣$x - 2$

$\textcolor{w h i t e}{a a a a}$$0 - 2 {x}^{2}$$\textcolor{w h i t e}{a a a}$$3 x + 2$

$\textcolor{w h i t e}{a a a a a a}$$- 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a}$$+ 8$

$\textcolor{w h i t e}{a a a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a}$$3 x - 6$

Therefore,

$\frac{{x}^{3} - 2 {x}^{2} - x + 2}{{x}^{2} - 4} = \left(x - 2\right) + \frac{3 \left(x - 2\right)}{{x}^{2} - 4}$

=(x-2)+(3cancel((x-2)))/((x+2)cancel((x-2))

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

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

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

The vertical asymptote is $x = - 2$

${\lim}_{x \to - \infty} f \left(x\right) - \left(x - 2\right) = {\lim}_{x \to - \infty} \frac{3}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(x - 2\right) = {\lim}_{x \to + \infty} \frac{3}{x} = {0}^{+}$

The slant asymptote is $y = x - 2$

graph{(y-(x^3-2x^2-x+2)/(x^2-4))(y-x+2)(y-50x-100)=0 [-23.33, 22.3, -16.34, 6.48]}

Jan 7, 2017

Alternative format for long division + graph

#### Explanation:

Note that I have used place keepers where there is no value solely so that things line up in formatting. Example $0 {x}^{2}$

$\text{ } {x}^{3} - 2 {x}^{2} - x + 2$
$\textcolor{red}{x} \left({x}^{2} - 4\right) \to \underline{{x}^{3} + 0 {x}^{2} - 4 x} \leftarrow \text{ subtract}$
$\text{ } 0 \textcolor{w h i t e}{.} - 2 {x}^{2} + 3 x + 2$
$\textcolor{red}{- 2} \left({x}^{2} - 4\right) \to \underline{\textcolor{w h i t e}{.} - 2 {x}^{2} + 0 x + 8} \leftarrow \text{ subtract}$
color(red)("Remainder") ->" "0color(white)(.)+3x-6

$y \text{ "=" "(x^3-2x-x+2)/(x^2-4)" " =" } \left(x - 2\right) + \frac{3 x - 6}{{x}^{2} - 4}$