Finding the vertex of a quadratic equation

So I found the vertex formula -b/2(a) online because I couldn't figure out how to find the vertex of a quadratic equation based on the way my teacher taught us (she really is a horrible teacher). I did my math worksheet and I got all of the answers right when I used this formula. I asked my teacher today if I could use that formula to find the vertex instead of the one she taught us and she told me I had to find out where this formula came from and explain it to her, then I could use it.
Does anyone know where this formula comes from? I don't understand what she meant, cause formulas are just made..and we use them

y = ax^2 + bx + c
y' = 2ax + b = 0
2ax = -b
x = -b/2a

Edit: well the derivative of y => dy/dx = y' , but have you covered derivatives yet?

I was afraid of that. OK, the vertex is located between the x-intercepts hence:

{[-b + √(b² - 4ac) ] / 2a + [-b - √(b² - 4ac) ] / 2a}/2 => -b/2a

You are welcome & thanks for the BA. And about -2b: yes you are right but it simplifies as:
{[-b + √(b² - 4ac) ] / 2a + [-b - √(b² - 4ac) ] / 2a}/2 => common denominator = 2a then:
= [-b + √(b² - 4ac) - b - √(b² - 4ac)]/ 2a/2 => simplify:
= [(-2b)/(2a)]/2 = (-b/a)/2 = -b/2a

I hope this helps.

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