What is the vertex of #y= -x^2+40x-16#?

1 Answer
Aug 1, 2018

The vertex is at #(20, 384)#.

Explanation:

Given: #y = -x^2 + 40x - 16#

This equation is in standard quadratic form #(y = ax^2 + bx + c)#, meaning we can find the #x#-value of the vertex using the formula #(-b)/(2a)#.

We know that #a = -1#, #b = 4#, and #c = -16#, so let's plug them into the formula:
#x = (-40)/(2(-1)) = 20#

Therefore, the #x#-coordinate is #20#.

To find the #y#-coordinate of the vertex, plug in the #x#-coordinate and find #y#:
#y = -x^2 + 40x - 16#

#y = -(20)^2 + 40(20) - 16#

#y = -400 + 800 - 16#

#y = 384#

Therefore, the vertex is at #(20, 384)#.

Hope this helps!