What is the vertex of y= -x^2+40x-16y=x2+40x16?

1 Answer
Aug 1, 2018

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

Explanation:

Given: y = -x^2 + 40x - 16y=x2+40x16

This equation is in standard quadratic form (y = ax^2 + bx + c)(y=ax2+bx+c), meaning we can find the xx-value of the vertex using the formula (-b)/(2a)b2a.

We know that a = -1a=1, b = 4b=4, and c = -16c=16, so let's plug them into the formula:
x = (-40)/(2(-1)) = 20x=402(1)=20

Therefore, the xx-coordinate is 2020.

To find the yy-coordinate of the vertex, plug in the xx-coordinate and find yy:
y = -x^2 + 40x - 16y=x2+40x16

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

y = -400 + 800 - 16y=400+80016

y = 384y=384

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

Hope this helps!