Find the minimum or maximum of the function y=−x^2+4x−5.

maximum value :  y = - 1  at  (2, -1)

No local minimum

Step-by-step explanation:

The   function y=−x^2+4x−5.  is a downward opening parabola.

The local maximum is at the vertex

y = - (xx - 4x )  - 5

y = - (xx - 4x +4  - 4)  - 5

y = - (xx - 4x + 4) + 4 - 5

y = -(x - 2)^2  - 1

vertex at (2, -1)

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong