The solution depends on whether you are in Calculus or Algebra. I will give the Calculus solution first, because it is much shorter.
This is a parabola that opens down, so its maximum point is at its vertex. The derivative will be zero there, so we find the derivative and set it equal to zero.
f'(x) = -8x + 2 = 0
-8x = -2
x = 1/4
f(1/4) = -4(1/4)2 +2(1/4) - 3 = -4/16 + 2/4 - 3 = -11/4
Here is how to do it with Algebra:
Since this is a parabola that opens down, its maximum point is at its vertex. The best way to find the vertex algebraically is by completing the square. This can only be done if the coefficient of x2 is 1. In this function, it is -4, so we need to factor a -4 out of the x terms:
f(x) = -4(x2 - ½x ) - 3 We just leave the -3 outside, because it doesn't contribute to this part.
I left a space, because that is where we will complete the square.
Step 1: Take the coefficient of x, which is -1/2, and divide it by 2 (which is the same as multiplying it by 1/2)
-1/2·1/2 = -1/4
Step 2: Square this result (-1/4)2 = 1/16
Step 3: Inside the parentheses, add and subtract the number obtained in step 2
f(x) = -4(x2 - (1/2)x + 1/16 - 1/16) -3
Step 4: Inside the parentheses, factor the trinomial formed by the first three terms
f(x) = -4[(x-1/2)2 - 1/16] - 3
Step 5: Distribute to remove the parentheses (or brackets, in this case)
f(x) = -4(x-1/2)2 + 1/4 - 3 Then combine the two number at the end
f(x) = -4(x - 1/2)2 - 11/4 So, the vertex is at (1/2, -11/4)
The maximum value of the function is -11/4 or -2.75