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

##### 1 Answer
Oct 25, 2016

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: ${t}^{2} + 36 = 0 \Rightarrow {t}^{2} = - 36$

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

Horizontal asymptotes occur as

${\lim}_{t \to \pm \infty} , g \left(t\right) \to c \text{ ( a constant)}$

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

$g \left(t\right) = \frac{\frac{t}{t} ^ 2 - \frac{6}{t} ^ 2}{{t}^{2} / {t}^{2} + \frac{36}{t} ^ 2} = \frac{\frac{1}{t} - \frac{6}{t} ^ 2}{1 + \frac{36}{t} ^ 2}$

as $t \to \pm \infty , g \left(t\right) \to \frac{0 - 0}{1 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(x-6)/(x^2+36) [-10, 10, -5, 5]}