How do you find the asymptotes for #y=(x^2+1)/(x^2-9)#?

1 Answer
Feb 13, 2016

Answer:

vertical asymptotes at x = ± 3
horizontal asymptote at y = 1

Explanation:

vertical asymptotes occur when the denominator of a rational function tends to zero. To find equation let denominator = 0

solve # x^2 - 9 = 0 rArr (x+3)(x-3)= 0 rArr x = ± 3#

horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0#

If the degree of the numerator and denominator are equal, the equation can be found by taking the ratio of leading coefficients.

in this case they are both of degree 2 and equation is

# y = 1/1 = 1 #
here is the graph as an illustration
graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}