# How do you find the Vertical, Horizontal, and Oblique Asymptote given  f(x) = (x^2) / (x-2)?

Dec 29, 2016

The vertical asymptote is $x = 2$
The oblique asymptote is $y = x + 2$
No horizontal asymptote

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{2\right\}$

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

Therefore,

The vertical asymptote is $x = 2$

As the degree of the numerator is $>$ than the degree of the denominator, we have an oblique asymptote.

Let's do a long division

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

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

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

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

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 + 4$

So,
$f \left(x\right) = x + 2 + \frac{4}{x - 2}$

Now we calculate the limits

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(x + 2\right)\right) = {\lim}_{x \to - \infty} \frac{4}{x - 2} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(x + 2\right)\right) = {\lim}_{x \to + \infty} \frac{4}{x - 2} = {0}^{+}$

So,

The oblique asymptote is $y = x + 2$

graph{(y-(x^2)/(x-2))(y-x-2)=0 [-28.86, 28.86, -14.44, 14.44]}