# How do you identify all asymptotes or holes for g(x)=(x^2-6x+8)/(x+2)?

Sep 16, 2017

$g \left(x\right)$ has a vertical asymptote $x = - 2$ and an oblique (slant) asymptote $y = x - 8$

#### Explanation:

Given:

$g \left(x\right) = \frac{{x}^{2} - 6 x + 8}{x + 2}$

Divide the numerator by the denominator by grouping:

$g \left(x\right) = \frac{{x}^{2} - 6 x + 8}{x + 2}$

$\textcolor{w h i t e}{g \left(x\right)} = \frac{{x}^{2} + 2 x - 8 x - 16 + 24}{x + 2}$

$\textcolor{w h i t e}{g \left(x\right)} = \frac{x \left(x + 2\right) - 8 \left(x + 2\right) + 24}{x + 2}$

$\textcolor{w h i t e}{g \left(x\right)} = \frac{\left(x - 8\right) \left(x + 2\right) + 24}{x + 2}$

$\textcolor{w h i t e}{g \left(x\right)} = x - 8 + \frac{24}{x + 2}$

From this alternative expression we can see that $g \left(x\right)$ has a vertical asymptote at $x = - 2$ and an oblique (a.k.a. slant) asymptote $y = x - 8$.

graph{(y-(x^2-6x+8)/(x+2))(y-x+8) = 0 [-79.6, 80.4, -45.64, 34.36]}