Given
[tex]\begin{gathered} y=-2x^2+8x-4 \\ \frac{dy}{dx}=-4x+8 \\ \text{put it equal to zero.} \\ -4x+8=0 \\ -4x=-8 \\ x=\frac{8}{4} \\ x=2 \end{gathered}[/tex][tex]\frac{d^2y}{dx^2}=-4[/tex]
Since, double derivative is negative. So, x=2 gives the point of maxima.
And, the maxima is given as,
[tex]\begin{gathered} y=-2x^2+8x-4 \\ y_{\max }=-2(2)^2+8(2)-4 \\ y_{\max }=-2(4)+16-4 \\ y_{\max }=-8+16-4 \\ y_{\max }=8-4 \\ y_{\max }=4 \end{gathered}[/tex]
