# How do you find the Vertical, Horizontal, and Oblique Asymptote given (x^2+1)/(x^2-1)?

Sep 24, 2016

#### Answer:

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

#### 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: ${x}^{2} - 1 = 0 \Rightarrow \left(x - 1\right) \left(x + 1\right) = 0 \Rightarrow x = \pm 1$

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

Horizontal asymptotes occur as

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

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

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

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

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

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+1)/(x^2-1) [-10, 10, -5, 5]}