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

1 Answer
Apr 10, 2016

Answer:

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]}