What are the asymptotes of f(x)=(x^2+1)/(x^2-9)?

Sep 2, 2017

$\text{vertical asymptotes at } x = \pm 3$
$\text{horizontal asymptote at } y = 1$

Explanation:

The denominator of f(x) cannot be zero as this would make f(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.

$\text{solve } {x}^{2} - 9 = 0 \Rightarrow \left(x - 3\right) \left(x + 3\right) = 0$

$\Rightarrow x = - 3 \text{ and "x=3" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2} = \frac{1 + \frac{1}{x} ^ 2}{1 - \frac{9}{x} ^ 2}$

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

$\Rightarrow y = 1 \text{ is the asymptote}$
graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}