An algebra problem by Aly Ahmed

• The vertex gives the maximum point of the quadratic equation $-x^2 + 8x + 0$ . $x$ -coordinate $= \dfrac{-b}{2a} \implies \dfrac{-8}{-2} = 4$
• Substituting $x=4$ , the output we get is $-(4)^2 + 8(4) = \boxed{16}$

$\begin{array}{lll} 8x -x^2 &= &-x^2 +8x -16 + 16\\ &=&-(x-4)^2 +16\\ \end{array}$

Note that $-(x-4)^2$ has a maximum of $0$ . Therefore, the maximum value for the expression is $\boxed{16}$ .