I’m attempting to verify a limited relationship one of my high school students observed.

Question

When graphing a quadratic function $f(x)=ax^2+bx+c$, where $a=1$ with two roots, $x_1$ and $x_2$, it appears that the y-coordinate of the vertex is equal to $(\frac{x_2-x_1}{2})^2$

I’ve tested out several cases and it does appear to consistently hold true. However, I haven’t been able to verify it nor make sense as why it is true. I was hoping someone here would have insight to share or possibly generalize it for cases where $a\neq 1$.

Answer 1

First, your function is $$ f(x) = a(x-x_1)(x-x_2). $$ There are several ways to see this, but note that the polynomial $a(x-x_1)(x-x_2)$ has roots $x_1$ and $x_2$ and leading coefficient $a$, and you probably know that these three things determine a quadratic polynomial.

Next, the vertex is at $\frac{x_1+x_2}2$. That should be geometrically obvious, and if this is good enough for you, then you can just plug that in to get $$f\left(\frac{x_1+x_2}2\right)=a\left(\frac{x_1+x_2}2-x_1\right)\left(\frac{x_1+x_2}2-x_2\right)=a\cdot\frac{-x_1+x_2}2\cdot\frac{x_1-x_2}2=-a\left(\frac{x_1-x_2}2\right)^2. $$

Otherwise you can see that $$ f(x) = f\left(x - \frac{x_1+x_2}2 + \frac{x_1+x_2}2\right) =a\left(x - \frac{x_1+x_2}2 - \frac{x_1-x_2}2\right)\left(x - \frac{x_1+x_2}2 + \frac{x_1-x_2}2\right) =a\left(x - \frac{x_1+x_2}2 \right)^2 - a\left(\frac{x_1-x_2}2\right)^2, $$ with a minimum at $\frac{x_1+x_2}2$ that has the value $-a\left(\frac{x_1-x_2}2\right)^2$.

Answer 2

$(x_1-x_2)^2/4=(x_1+x_2)^2/4-x_1x_2 = (-b/2a)^2 -c/a = (b^2-4ac)/(4a^2)$

At the vertex, $x=b/2a$ and the y-coordinate is $(-b^2+4ac)/(4a)$

So the second is $-a$ times the first.

Answer 3

If you are familiar with differention, you may consider the function $f(x) =a(x-x_1)(x-x_2)$. Then the vertex of the parabola comes at the point where $f'(x)=0$ and that is $t=\frac{x_1+x_2}{2}$. So $f(t) =-a\frac{(x_1-x_2)^2}{4}$