Answer
$f(t)=-16\left(t-\frac{47}{32}\right)^2+\frac{2401}{64}$
Vertex: $\left(\frac{47}{32}, \frac{2401}{64}\right)$
Work Step by Step
Complete the square to write the function in vertex form:
$$
\begin{aligned}
f(t) & =-16\left(t^2-\frac{47}{16} t-\frac{3}{16}\right) \\
& =-16\left(t^2-\frac{47}{16} t+\left(\frac{47}{32}\right)^2-\left(\frac{47}{32}\right)^2-\frac{3}{16}\right) \\
& =-16\left(\left(t-\frac{47}{32}\right)^2-\frac{2209}{1024}-\frac{192}{1024}\right) \\
& =-16\left(\left(t-\frac{47}{32}\right)^2-\frac{2401}{1024}\right) \\
& =-16\left(t-\frac{47}{32}\right)^2+\frac{2401}{64} .
\end{aligned}
$$
Vertex: $\left(\frac{47}{32}, \frac{2401}{64}\right)$
