# How do you find the vertical, horizontal or slant asymptotes for C(t) = t /(9t^2 +8)?

Jan 24, 2018

$\text{see explanation}$

#### Explanation:

The denominator of C(t) cannot be zero as this would make C(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.

$\text{solve } 9 {t}^{2} + 8 = 0 \Rightarrow {t}^{2} = - \frac{8}{9}$

$\text{this has no real solutions hence there are no vertical}$
$\text{asymptotes}$

$\text{horizontal asymptotes occur as }$

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

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

$C \left(t\right) = \frac{\frac{t}{t} ^ 2}{\frac{9 {t}^{2}}{t} ^ 2 + \frac{8}{t} ^ 2} = \frac{\frac{1}{t} ^ 2}{9 + \frac{8}{t} ^ 2}$

$\text{as } t \to \pm \infty , C \left(t\right) \to \frac{0}{9 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

$\text{Slant asymptotes occur if the degree of the numerator is }$
$> \text{degree of the denominator. This is not the case here }$
$\text{hence there are no slant asymptotes}$
graph{x/(9x^2+8) [-10, 10, -5, 5]}