How do you find the asymptotes for #f(x) =(x^2-1)/(x^2+4)#?

1 Answer
Sep 9, 2016

Answer:

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+4=0rArrx^2=-4#

This has no real solutions hence there are no vertical 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+4/x^2)=(1-1/x^2)/(1+4/x^2)#

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

#rArry=1" is the asymptote"#
graph{(x^2-1)/(x^2+4) [-10, 10, -5, 5]}