Question

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

Answer 1

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]}