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)