# How do you identify all asymptotes or holes for g(x)=(x^3+3x^2-16x+12)/(x-2)?

Oct 22, 2016

The numerator is divisible by the denominator
$g \left(x\right) = \left(x + 6\right) \left(x - 1\right)$
There are no asymptotes or holes

#### Explanation:

Let's do a long division

${x}^{3} + 3 {x}^{2} - 16 x + 12$$\textcolor{w h i t e}{a a a}$∣$x - 2$
${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣ ${x}^{2} + 5 x - 6$
$0 + 5 {x}^{2} - 16 x$
$\textcolor{w h i t e}{a a a a a}$$0 - 10 x$
$\textcolor{w h i t e}{a a a a a a a a}$$- 6 x + 12$
$\textcolor{w h i t e}{a a a a a a a a a a a}$$0 + 0$

So $g \left(x\right) = \frac{{x}^{3} + 3 {x}^{2} - 16 x + 12}{x - 2} = {x}^{2} + 5 x - 6$

We can factorise so $g \left(x\right) = g \left(x\right) = \left(x + 6\right) \left(x - 1\right)$

This is a parabola
there are no asymptotes or holes

$g \left(x\right)$ is defined on $\mathbb{R}$