The given function is
[tex]f(x)=-16t^2+40[/tex]First, we find the vertex V(h,k), where
[tex]h=-\frac{b}{2a}[/tex]We know that a = -16 and b = 0, so
[tex]h=-\frac{0}{2\cdot(-16)}=0[/tex]Then, we find k evaluating the function when x = 0.
[tex]k=f(x)=-16(0)^2+40=40[/tex]The vertex is V(0, 40).
Then, we find the intercepts of the function when y = 0.
[tex]\begin{gathered} -16t^2+40=0 \\ -16t^2=-40 \\ t^2=-\frac{40}{-16} \\ t=\sqrt[]{2.5} \\ t\approx\pm1.58 \end{gathered}[/tex]The intercepts are (-1.58, 0) and (1.58, 0).
Now, we use all three points to graph the function.
The image below shows the graph of the function:
In the image, we used t = x.