The coordinates of the vertex point are $(\frac{-b}{2a}, \frac{-b^2+4ac}{4a})$.
The x-coordinate of the vertex point, $\frac{-b}{2a}$, is the midpoint of the x-intercepts. Therefore, we can derive $\frac{-b}{2a}$ from the average of the two zeros. The two zeros of the quadratic are $\frac{-b\pm\sqrt {b^2-4ac}}{2a}$ and the sum is $\frac{-b}{a}$ and divide by two to get the average: $\frac{-b}{2a}$.
Now for the y-coordinate of the vertex point. I know that I can get to $\frac{-b^2+4ac}{4a}$ by completing the square for $y = ax^2 + bx + c$ and solve for $y$, but I don't know what it means in relation to the parabola, like the way $\frac{-b}{2a}$ does, nor do I really see why completing the square would give the coordinate points for the vertex.
I notice that $\frac{-b^2+4ac}{4a}$ kind of resembles the discriminant in the quadratic formula (the stuff under the radical, $\sqrt {b^2-4ac}$), but not sure if that leads to something or not.
Best Answer
It is just the $y$ coordinate of the vertex.
From $$ y = a x^2 + b x + c $$ with $a \neq 0,$ we get $$ y = \frac{(2ax+b)^2 + (4ac-b^2)}{4a}. $$ You should confirm that yourself.
When $x = -b/(2a),$ the term $(2ax+b)$ that is squared becomes zero. WHEN $a > 0,$ that gives the lowest possible point on the parabola.