the max is obtained at the vertex.
the x of the vertex is -b/-2 = b/2
so f(b/2) = 25
-b^2/4 + b^2/2 - 75 = 25
- b^2/4 + 2b^2/4 = 100
b^2/4 = 100
b^2 = 400
b = ± 20
Find B so that F(x)=-x^2+bx-75 has a maximum value of 25.
1 answer
1 answer