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Math Help - Determine eq of quad

  1. #1
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    Determine eq of quad

    Q askes: Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are 2 +\sqrt{3} and  2 - \sqrt{3}

    To get started, I'm thinking of factored form f(x) = a(0 - \sqrt{3}) (0  +\sqrt{3})
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  2. #2
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    You have f(x)=ax^2+bx+c and know that (-4, 5) is on graph of f, then: 16a-4b+c=5

    Indication:
    Use Vičte's formulas.
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  3. #3
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    \frac{5}{33}(x^2 -4x +1) is suppost to be the answer.

    However, I cannot seem to get to the answer. I'm using:

     f(x) = a(x -2 \sqrt3)(x -2+\sqrt3)
     f(x)= a(x+1)(x-5)
     f(x)=a(x^2 -4x -5)

    ... this does not seem to work.
    Last edited by reallylongnickname; April 2nd 2011 at 11:22 AM.
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  4. #4
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    Well, probably because  f(x) = a[x -(2 -\sqrt3)][x -(2+\sqrt3)], not that expression you wrote.

    Now use what you know: (-4, 5) is on the graph of f.
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  5. #5
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    Question askes word for word:
    Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are 2 +\sqrt{3} and  2 - \sqrt{3}

    Wheather wrong or right, the formula below is what I came up with to use.
    f(x) = a[x -(2-\sqrt3)] [(x -(2+\sqrt3)]

    How does it look? It lead me to this:  f(5) = a[-4 -(2 -\sqrt3)] [-4 -(2+\sqrt3)]
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  6. #6
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    If (-4, 5) is on the graph of f, then f(-4)=5. After that calculate.
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  7. #7
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    Quote Originally Posted by reallylongnickname View Post
    Question askes word for word:
    Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are 2 +\sqrt{3} and  2 - \sqrt{3}

    Wheather wrong or right, the formula below is what I came up with to use.
    f(x) = a[x -(2-\sqrt3)] [(x -(2+\sqrt3)]

    How does it look? It lead me to this:  f(5) = a[-4 -(2 -\sqrt3)] [-4 -(2+\sqrt3)]
    It's simplest to think of that as a "product of sum and difference": (a+ b)(a- b)= a^2- b^2
    f(x= a((x-2)+\sqrt{3})((x-2)- \sqrt{3}= a((x- 2)^2- 3)= a(x^2- 4x+ 4-3)= a(x^2- 4x+ 1)

    But you don't calculate f(5): f(-4)= 5
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    It's simplest to think of that as a "product of sum and difference": (a+ b)(a- b)= a^2- b^2
    f(x= a((x-2)+\sqrt{3})((x-2)- \sqrt{3}= a((x- 2)^2- 3)= a(x^2- 4x+ 4-3)= a(x^2- 4x+ 1)

    But you don't calculate f(5): f(-4)= 5
    Ahhh, i see. thx. I'm not sure what u mean about not calculating f(5).
    When I plug f(5) and x= -4 into your final equation, the result is the same as the answer that was given.

    f(5) =   a((-x)^2 -4(-x) +1)
    f(5) = a((-4)^2 -4(-4) +1)
    f(5) = a33
    \frac{5}{33} = a
    =\frac{5}{33} ((x)^2 -4(x) +1)
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