How do you find the vertical, horizontal and slant asymptotes of: g(x)=2 /(x^2-16)?

Sep 5, 2016

vertical asymptotes at x = ± 4
horizontal asymptote at y = 0

Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: ${x}^{2} - 16 = 0 \Rightarrow \left(x - 4\right) \left(x + 4\right) = 0$

$\Rightarrow x = - 4 \text{ and " x=4" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , g \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$

$g \left(x\right) = \frac{\frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{16}{x} ^ 2} = \frac{\frac{2}{x} ^ 2}{1 - \frac{16}{x} ^ 2}$

as $x \to \pm \infty , g \left(x\right) \to \frac{0}{1 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2) Hence there are no slant asymptotes.
graph{2/(x^2-16) [-10, 10, -5, 5]}