Can someone provide me with proof on the little formula used to find the mid-point on a second degree polynomial graph?

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- Apr 4th 2013, 01:55 AMPazeProof of -b/2a
Can someone provide me with proof on the little formula used to find the mid-point on a second degree polynomial graph?

- Apr 4th 2013, 02:10 AMProve ItRe: Proof of -b/2a
Every quadratic function of the form 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 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.

So the axis of symmetry will be the average of these two points due to the symmetry of the graph. Averaging these we find

So this means that we can always find the axis of symmetry of a quadratic of the form using . - Apr 4th 2013, 02:13 AMPazeRe: Proof of -b/2a
Thanks, Prove It!

- Apr 4th 2013, 02:24 AMSworDRe: Proof of -b/2a
Alternately, the turning point is where the derivative is zero;

leads to

- Apr 4th 2013, 07:50 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:

,

then it is easier to see that the symmetric line would be at , since from there you either add or subtract the same amount to get to the roots. - Apr 5th 2013, 08:49 AMProve ItRe: Proof of -b/2a