Question

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

Answer 1

vertical asymptotes `x = +-sqrt3`
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`

`rArr x = +- sqrt3 " are the asymptotes "`

Horizontal asymptotes occur as `lim_(xto+-oo) f(x) to 0`

divide terms on numerator/denominator by `x^2`

`(x^2/x^2 - 5/x^2)/(x^2/x^2 - 3/x^2) = (1 - 5/x^2)/(1-3/x^2)`

as `x to+-oo , 5/x^2" and " 3/x^2 to 0`

`rArr y = 1/1 = 1" is the asymptote "`

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