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Math Help - Proof of -b/2a

  1. #1
    Senior Member Paze's Avatar
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    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?
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    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}.
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    Senior Member Paze's Avatar
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    Re: Proof of -b/2a

    Thanks, Prove It!
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    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}
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    Member Henderson's Avatar
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    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.
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    Re: Proof of -b/2a

    Quote Originally Posted by Henderson View Post
    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...
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