What are the asymptotes of #f(x)=(x^2+1)/(x^2-9)#?

1 Answer
Sep 2, 2017

Answer:

#"vertical asymptotes at "x=+-3#
#"horizontal asymptote at "y=1#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

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

#rArrx=-3" and "x=3" are the asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

Divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2+1/x^2)/(x^2/x^2-9/x^2)=(1+1/x^2)/(1-9/x^2)#

as #xto+-oo,f(x)to(1+0)/(1-0)#

#rArry=1" is the asymptote"#
graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}