How do you find the asymptotes for # g(t) = (t − 6) / (t^(2) + 36)#?

1 Answer
Oct 25, 2016

Answer:

horizontal asymptote at y = 0

Explanation:

The denominator of g(t) cannot be zero as this would make g(t) undefined. Equating the denominator to zero and solving gives the value that t cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #t^2+36=0rArrt^2=-36#

This has no real solutions, hence there are no vertical asymptotes.

Horizontal asymptotes occur as

#lim_(t to+-oo),g(t)toc" ( a constant)"#

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

#g(t)=(t/t^2-6/t^2)/(t^2/t^2+36/t^2)=(1/t-6/t^2)/(1+36/t^2)#

as #t to+-oo,g(t)to(0-0)/(1+0)#

#rArry=0" is the asymptote"#
graph{(x-6)/(x^2+36) [-10, 10, -5, 5]}