f(x)=-2x²+4x+4 ··· Eq.1
The graph of a quadratic equation y=ax²+bx+c (a≠0) is a parabola. If a>0 (y">0), the curve is concave downwards*, and has the minimum at its critical point (y'=0). If a<0 (y"<0), the curve is convex upwards*, and has the maximum at its critical point (y'=0).
Therefore, the given equation doesn't have the minimum, but the maximum.
Here, we examine the maximum. From Eq.1, we have:
f'(x)=-4x+4=-4(x-1), so at x=1, f'(1)=0
f"(x)=(f'(x))'=(-4x+4)'=-4 (<0)**
Therefore, the given function has the maximum at x=1 and the maximum is: f(1)=-2(1)²+4(1)+4=6
* f(x)=ax²+bx+c=ax²(1 + b/(ax) + c/(ax²)). If a>0, and x approaches (+) or (-) infinity, f(x) approaches (+) infinity: the graph is concave downwards. If a<0, f(x) approaches (-) infinity: convex upwards.
** If f'(x)=0 (the tangent is parallel to x-axis), f(x)">0 at the minimum point. If f'(x)=0, f(x)"<0 at the maximum point.