# How do you find the slant asymptote of (x^2+3x+2)/(x-2)?

Dec 27, 2015

Divide to find the quotient polynomial $y = x + 5$, which is the slant asymptote (otherwise known as the oblique asymptote).

#### Explanation:

Long divide or do something like this to separate out the quotient $\left(x + 5\right)$ and remainder $12$:

$\frac{{x}^{2} + 3 x + 2}{x - 2}$

$= \frac{{x}^{2} - 2 x + 5 x - 10 + 12}{x - 2}$

$= \frac{\left({x}^{2} - 2 x\right) + \left(5 x - 10\right) + 12}{x - 2}$

$= \frac{x \left(x - 2\right) + 5 \left(x - 2\right) + 12}{x - 2}$

$= \frac{\left(x + 5\right) \left(x - 2\right) + 12}{x - 2}$

$= x + 5 + \frac{12}{x - 2}$

Then as $x \to \pm \infty$ the term $\frac{12}{x - 2} \to 0$

So the slant asymptote is $y = x + 5$