# Proof of -b/2a

• Apr 4th 2013, 12:55 AM
Paze
Proof of -b/2a
Can someone provide me with proof on the little formula $\frac{-b}{2a}$ used to find the mid-point on a second degree polynomial graph?
• Apr 4th 2013, 01:10 AM
Prove It
Re: Proof of -b/2a
Every quadratic function of the form $\displaystyle y=a\,x^2 + b\,x + c$ 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 $\displaystyle -\frac{b}{2a} = 0$ 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 \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 \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 $\displaystyle y = a\,x^2 + b\,x + c$ using $\displaystyle x = -\frac{b}{2a}$.
• Apr 4th 2013, 01:13 AM
Paze
Re: Proof of -b/2a
Thanks, Prove It!
• Apr 4th 2013, 01:24 AM
SworD
Re: Proof of -b/2a
Alternately, the turning point is where the derivative is zero;

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

$2ax + b = 0$ leads to

$x = -\frac{b}{2a}$
• Apr 4th 2013, 06:50 AM
Henderson
Re: 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:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,

then it is easier to see that the symmetric line would be at $x = \frac{-b}{2a}$, since from there you either add or subtract the same amount $(\frac{\sqrt{b^2 - 4ac}}{2a})$ to get to the roots.
• Apr 5th 2013, 07:49 AM
Prove It
Re: Proof of -b/2a
Quote:

Originally Posted by Henderson
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:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,

then it is easier to see that the symmetric line would be at $x = \frac{-b}{2a}$, since from there you either add or subtract the same amount $(\frac{\sqrt{b^2 - 4ac}}{2a})$ to get to the roots.

Assuming of course that your equation HAS roots...