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

Mar 22, 2016

vertical asymptotes , x = ± 4
horizontal asymptote y = 2

#### Explanation:

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

solve:  x^2 - 16 = 0 → (x-4)(x+4) = 0 → x = ± 4

Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0
divide all terms on numerator and denominator by ${x}^{2}$

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

As x →∞  8/x^2" and " 16/x^2 → 0

$\Rightarrow y = \frac{2}{1} = 2 \text{ is the asymptote }$

Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator , which is not the case here , so there are no slant asymptotes.

Here is the graph of the function.
graph{(2x^2-8)/(x^2-16) [-10, 10, -5, 5]}