# How do you find vertical, horizontal and oblique asymptotes for y = (x-4)^2/(x^2-4)?

Apr 1, 2016

vertical asymptotes x = ± 2
horizontal asymptote y = 1

#### Explanation:

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

solve :  x^2 - 4 = 0 → (x-2)(x+2) = 0

rArr x = ± 2 " are the asymptotes "

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

now numerator = ${\left(x - 4\right)}^{2} = {x}^{2} - 8 x + 16$

and y = $\frac{{x}^{2} - 8 x + 16}{{x}^{2} - 4}$

divide all terms on numerator/denominator by ${x}^{2}$

$\frac{{x}^{2} / {x}^{2} - \frac{8 x}{x} ^ 2 + \frac{16}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{1 - \frac{8}{x} + \frac{16}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

As $x \to \infty , \frac{8}{x} , \frac{16}{x} ^ 2 \text{ and } \frac{4}{x} ^ 2 \to 0$

$\Rightarrow y = \frac{1}{1} = 1 \text{ is the asymptote }$

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(x-4)^2/(x^2-4) [-10, 10, -5, 5]}