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

##### 1 Answer

Apr 10, 2016

vertical asymptotes

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]}