# How do you find vertical, horizontal and oblique asymptotes for 4/(1 -x^2)?

Sep 21, 2016

vertical asymptotes at x = ± 1
horizontal asymptote at y = 0

#### Explanation:

The denominator of the function cannot be zero as this would make the function 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 : $1 - {x}^{2} = 0 \Rightarrow \left(1 - x\right) \left(1 + x\right) = 0$

$\Rightarrow x = - 1 \text{ and " x=1" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$f \left(x\right) = \frac{\frac{4}{x} ^ 2}{\frac{1}{x} ^ 2 - {x}^{2} / {x}^{2}} = \frac{\frac{4}{x} ^ 2}{\frac{1}{x} ^ 2 - 1}$

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

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

Oblique 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 oblique asymptotes.
graph{(4)/(1-x^2) [-10, 10, -5, 5]}