What's the vertex of the following equation X^2 - 10x + 25

(5,0)

Step-by-step explanation:

X^2 - 10x + 25

The vertex is at

h = - b/2a where ax^2 + bx + c

b = - 10 and a = 1

h = - (-10) / 2 (1)

= 10/2 = 5

The vertex is at x=5

To find the y value stick x=5 into the function

5^2 - 10 (5) + 25

25-50+25 = 0

The vertex is at 5,0

It would be (5,0) !