The curve $\displaystyle y=-2x^2+8x-3$ has a maximum point at $\displaystyle x=p$,where $\displaystyle p$ is constant.Find the value of $\displaystyle p$
A parabola opening downward has its maximum value at the vertex- and you can get that by completing the square:
$\displaystyle -2x^2+ 8x- 3= -2(x^2- 4x)- 3= -2(x^2- 4x+ 4- 4)- 3$
$\displaystyle = -2(x^2- 4x+ 4)+ 8- 3= -2(x- 2)^2+ 5$
Since a square is never negative, that will be largest when x-2= 0.