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?

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- Apr 4th 2013, 12:55 AMPazeProof of -b/2a
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?

- Apr 4th 2013, 01:10 AMProve ItRe: Proof of -b/2a
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 $\displaystyle \displaystyle x = -\frac{b}{2a}$. - Apr 4th 2013, 01:13 AMPazeRe: Proof of -b/2a
Thanks, Prove It!

- Apr 4th 2013, 01:24 AMSworDRe: Proof of -b/2a
Alternately, the turning point is where the derivative is zero;

$\displaystyle \frac{d}{dx} (ax^2+bx+c) = 2ax + b$

$\displaystyle 2ax + b = 0$ leads to

$\displaystyle x = -\frac{b}{2a}$ - Apr 4th 2013, 06:50 AMHendersonRe: Proof of -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. - Apr 5th 2013, 07:49 AMProve ItRe: Proof of -b/2a