Answer
$ y=\frac{137}{16}$ is the minimum, when $ x=\frac{8}{3}$
Work Step by Step
Given $$ y= -4 x^{2}+3 x+8 $$
So, we have
\begin{aligned} y&-4 x^{2}+3 x+8 \\
&=-4\left(x^{2}-\frac{3}{4} x\right)+8\\
& =-4\left(x^{2}-\frac{3}{4} x+\frac{9}{64}-\frac{9}{64}\right)+8\\
& =-4\left(x^{2}-\frac{3}{4} x+\frac{9}{64}\right)+8+\frac{9}{16}\\
& =-4\left(x-\frac{3}{8}\right)^{2}+\frac{137}{16}\\
\end{aligned}
The lowest value for the squared term is $0$. So, we see that $ y=\frac{137}{16}$ is the minimum, when $ x=\frac{8}{3}$.
