# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (x-4)/(x^2-8x+16)?

Nov 2, 2016

$x = 4$ is a Vertical Asymptote

$y = 0$ is a Horizontal Asymptote

There is no Slant Asymptote.

#### Explanation:

This function is defined for all $x \in \mathbb{R}$ except $x = 4$

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

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

Domain of $f$ is:$\left(- \infty , 4\right) U \left(4 , + \infty\right)$

The vertical Asymptote is computed by setting the denominator to zero

$x - 4 = 0 \Rightarrow \textcolor{b l u e}{x = 4}$ is a $\textcolor{b l u e}{V . A}$

Let us start computing the asymptote:

The degree of numerator is greater than that of the denominator

Therefore ,$\textcolor{b l u e}{y = 0}$ is a $\textcolor{b l u e}{H . A}$

There is no Slant asymptote.