1.) f'(x) = -4x + 8

0 = -4x + 8;

Solve for x;

x = 2

Thus, the critical point is at x = 2, and that's either a min or a max.

-2(2)^2 + 8(2) + 3

-2*4 + 16 + 3

-8 + 16 + 3

8 + 3 = 11

And thus, it'll be at (2, 11).

We can check by seeing whether the second deriv is pos or neg, and thus determine concavity.

f''(x) = -4, and thus it is a max, since it's concave down.

Therefore, we have a maximum at (2, 11).