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Math Help - Peak of graph

  1. #1
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    [Solved] Peak of graph

    I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

    y=(x+p)^2+q

    is this correct? and how would you do for

    y=2x^2-4x+a^2-a

    ? Thanks for any advice.
    Last edited by davidman; January 23rd 2010 at 06:13 PM.
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  2. #2
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    Quote Originally Posted by davidman View Post
    I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

    y=(x+p)^2+q

    is this correct? and how would you do for

    y=2x^2-4x+a^2-a

    ? Thanks for any advice.
    If you write the quadratic equation as

    y = a(x - h)^2 + k

    then the turning point is (x, y) = (h, k).


    So for y = 2x^2 - 4x + a^2 - a, you complete the square

    y = 2\left(x^2 - 2x + \frac{1}{2}a^2 - \frac{1}{2}a\right)

    y = 2\left[x^2 - 2x + (-1)^2 - (-1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a\right]

    y = 2\left[(x - 1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a - 1\right]

    y = 2(x - 1)^2 + a^2 - a - 2.


    So the turning point is:

    (x, y) = (1, a^2 - a - 2).
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    wooah, you made that so easy to understand.

    thanks!
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  4. #4
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    Quote Originally Posted by davidman View Post
    I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

    y=(x+p)^2+q

    is this correct? and how would you do for

    y=2x^2-4x+a^2-a

    ? Thanks for any advice.
    If the vertex of the parabola is V(p,q) then the equation of the parabola whose axis is parallel to the y-axis is written as:

    y = a(x-p)^2+q

    With your example:

    y = 2x^2-4x+a^2-a = 2(x^2-2x+\tfrac12(a^2-a))

    y=2(x^2-2x\bold{\color{red}+1-1}+\tfrac12(a^2-a)) = 2(x-1)^2+a^2-a-2

    So the coordinates of the vertex are V\left(1, a^2-a-2\right)
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  5. #5
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    Thanks so much for the help!

    I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

    the second part is to get the minima of f(a)=a^2-a-2

    and I thought the answer would reveal itself if I factorised and solved for a, but the answer is apparently -\frac{9}{4}... I was not even close... what exactly are you supposed to do to get to that answer?
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  6. #6
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    Quote Originally Posted by davidman View Post
    Thanks so much for the help!

    I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

    the second part is to get the minima of f(a)=a^2-a-2

    and I thought the answer would reveal itself if I factorised and solved for a, but the answer is apparently -\frac{9}{4}... I was not even close... what exactly are you supposed to do to get to that answer?
    Probably you are trapped by the text of the question(?) It reads " ... get the minima of f(a) ..."

    I assume that you have calculated:

    f(a)=a^2-a-2

    f(a)=a^2-a\bold{\color{blue}+\frac14-\frac14}-2

    f(a)=\left(a-\frac12 \right)^2-\frac14-2

    f(a)=\left(a-\frac12 \right)^2-\frac94

    Therefore the vertex has the coordinates V\left(\tfrac12 ,~ -\tfrac94 \right)

    With a parabola opening up the f(a)-value of the vertex determines the minimum. The vertex is the lowest point of all points of the graph of f. Therefore

    \min(f(a)) = -\frac94
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  7. #7
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    Thanks a lot guys. earboth's post brings that question to perfection!

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