Maxima and Minima of Quadratic Function

Question

What’s the maxima of the quadratic function y=-x^2+3x-7? I believe that the answer is a value of -19/4 at at x=3/2. Is this correct?

Answer 1

Well, let's check.

Finding the critical points:

f'(x) = -2x+3 -> x = 3/2

Using the first derivative test

f'(1) = 1 thus when x < 3/2, f'(x) > 0

f'(2) = -1 thus when x > 3/2, f'(x) < 0

This means our maximum is at x = 3/2

We get the y coordinate by plugging back in our x into f(x).

f(3/2) = -(3/2)²+3(3/2)-7 = -19/4

thus, our maximum is

Maximum(3/2, -19/4)

You are correct!