Answer
$h(t) = \frac{1}{3}\left(t+9\right)^2-19$
Work Step by Step
The conversion of the function to the standard vertex form is self explanatory. All mathematical steps are shown below. $$
\begin{aligned}
h(t)&=\frac{1}{3} t^2+6 t+8\\
& = \frac{1}{3}\left( t^2+3\cdot 6t \right) +8\\
&= \frac{1}{3}\left( t^2+18t\right) +8\\
& = \frac{1}{3}\left[ t^2+18t+\left( \frac{18}{2}\right)^2\right]+8-\frac{1}{3}\cdot \left( \frac{18}{2}\right)^2\\
& = \frac{1}{3}\left(t^2+18t+ 9^2 \right)+8-\frac{9^2}{3} \\
& =\frac{1}{3}\left(t+9\right)^2-19.
\end{aligned}
$$
