Can someone provide me with proof on the little formula $\displaystyle \frac{-b}{2a}$ used to find the mid-point on a second degree polynomial graph?
Every quadratic function of the form $\displaystyle \displaystyle y=a\,x^2 + b\,x + c $ will have a y-intercept where x = 0. There are two possibilities.
The first is that the y-intercept is the turning point. If this is the case, the turning point is at (0, c). It is also well known that if the turning point is on the y-axis, that b = 0. So that would mean $\displaystyle \displaystyle -\frac{b}{2a} = 0$ as well and therefore works as a formula for finding the turning point when the turning point is on the y-axis.
The second case is where the y-intercept is not the turning point. We know that the graph of a quadratic is symmetric. Since the turning point is not on the y-axis, that means there will be a second point which has the same y-value of c. If we average these x-values, we will get the axis of symmetry / x-coordinate of the turning point.
$\displaystyle \displaystyle \begin{align*} c &= a\,x^2 + b\,x + c \\ 0 &= a\,x^2 + b\,x \\ 0 &= x\left( a\,x + b \right) \\ x = 0 \textrm{ or } x &= -\frac{b}{a} \end{align*}$
So the axis of symmetry will be the average of these two points due to the symmetry of the graph. Averaging these we find
$\displaystyle \displaystyle \begin{align*} x &= \frac{-\frac{b}{a} + 0}{2} \\ &= -\frac{b}{2a} \end{align*}$
So this means that we can always find the axis of symmetry of a quadratic of the form $\displaystyle \displaystyle y = a\,x^2 + b\,x + c $ using $\displaystyle \displaystyle x = -\frac{b}{2a}$.
Knowing that the parabola is symmetric about the vertical line through that turning point, you could say that the x-value must be the average of the two roots of the equation. If you know the quadratic formula for finding those roots:
$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $,
then it is easier to see that the symmetric line would be at $\displaystyle x = \frac{-b}{2a} $, since from there you either add or subtract the same amount $\displaystyle (\frac{\sqrt{b^2 - 4ac}}{2a}) $ to get to the roots.