# How do you find the asymptotes for y=(x^2-5)/(x^2-3)?

Apr 10, 2016

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

#### Explanation:

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

solve:  x^2 - 3 = 0 → x^2 = 3 → x = +- sqrt3

$\Rightarrow x = \pm \sqrt{3} \text{ are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide terms on numerator/denominator by ${x}^{2}$

$\frac{{x}^{2} / {x}^{2} - \frac{5}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{3}{x} ^ 2} = \frac{1 - \frac{5}{x} ^ 2}{1 - \frac{3}{x} ^ 2}$

as $x \to \pm \infty , \frac{5}{x} ^ 2 \text{ and } \frac{3}{x} ^ 2 \to 0$

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

Here is the graph of the function.
graph{(x^2-5)/(x^2-3) [-10, 10, -5, 5]}