If you know the zeros of a quadratic function, how do you locate the vertex?

You calculate the average value of the roots.

Xm = (x'+x")/2

We have too:

y = ax^2 +bx + c

Or

y = a(x-x')(x-x")

You can substitute x' and x"

An exemple with, x' = 1 and x' = 2

We could stay:

y = a(x-1)(x-2)

y = a(x.x-2x -x+2)

y = a(x^2-3x+2)

y = ax^2-3ax+2a

We can to make a = 1

y = x^2 -3x +2

When we substitute Xm here,

We will have the value from Ym

As Xm = (x'+x")/2

Then,

Xm = (1+2)/2

Xm = 3/2

Then,

Ym = (xm)^2 -3(xm)+2

Ym = (3/2)^2 -3(3/2)+2

Ym = 9/4 -9/2 +2

Souvind this:

Ym = -1/4

This mean,

(Xm, ym) = (3/2, -1/4)

What am i doing here. add?