## Calculus (3rd Edition)

$y=\frac{137}{16}$ is the minimum, when $x=\frac{8}{3}$
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 $x=\frac{8}{3}$.