# How do you find the vertical, horizontal or slant asymptotes for y= (x-4)/(x^2-8x+16)?

##### 1 Answer
May 23, 2016

vertical asymptote x = 4
horizontal asymptote y = 0

#### Explanation:

The first step here is to factorise and simplify the function.

$\Rightarrow \frac{\cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)} \left(x - 4\right)} = \frac{1}{x - 4}$

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 4 = 0 → x = 4 is the asymptote.

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , y \to 0$

divide terms on numerator/denominator by x

$\frac{\frac{1}{x}}{\frac{x}{x} - \frac{4}{x}} = \frac{\frac{1}{x}}{1 - \frac{4}{x}}$

as $x \to \pm \infty , y \to \frac{0}{1 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0,denominator-degree 1).Hence there are no slant asymptotes.
graph{1/(x-4) [-10, 10, -5, 5]}