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Math Help - Parabola Problem

  1. #1
    Bar0n janvdl's Avatar
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    Parabola Problem

    A parabola has x-axis values of -3 and 5. And it symmetry axis is: x = 1

    The a value is negative, but i cannot say what its exact value is.

    It says to prove that  f(x) = -x^2 + 2x + 15

    Here's what i did:

     f(x) = a(x + 3)(x - 5)

     f(x) = a(x^2 - 2x - 15)

    And now i KNOW that  a must be equal to  -1 but i have no idea how to prove it

    Please help guys.

    EDIT: Sorry, i forgot to mention that there is also a graph  g(x) = |x| - 3
    Last edited by janvdl; July 23rd 2007 at 08:47 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    A parabola has x-axis values of -3 and 5. And it symmetry axis is: x = 1

    The a value is negative, but i cannot say what its exact value is.

    It says to prove that  f(x) = -x^2 + 2x + 15

    Here's what i did:

     f(x) = a(x + 3)(x - 5)

     f(x) = a(x^2 - 2x - 15)

    And now i KNOW that  a must be equal to  -1 but i have no idea how to prove it

    Please help guys.

    EDIT: Sorry, i forgot to mention that there is also a graph  g(x) = |x| - 3
    I presume that "x-axis values of -3 and 5" means these are the x intercepts (-3, 0) and (5, 0)?

    Then
    f(x) = a(x^2 - 2x - 15)
    is as good as you are going to get with the current information. The only thing you can do with it now is to find the axis of symmetry, which is supposed to be the line x = 1. But:
    f(x) = a(x^2 - 2x - 15) = ax^2 + -2ax - 15a
    has an axis of symmtry at
    x = -\frac{-2a}{2 \cdot a} = 1
    and the "a" cancels out naturally. So we have no new information here.

    What is the significance of the graph of  g(x) = |x| - 3 ?

    -Dan
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    I presume that "x-axis values of -3 and 5" means these are the x intercepts (-3, 0) and (5, 0)?

    Then
    f(x) = a(x^2 - 2x - 15)
    is as good as you are going to get with the current information. The only thing you can do with it now is to find the axis of symmetry, which is supposed to be the line x = 1. But:
    f(x) = a(x^2 - 2x - 15) = ax^2 + -2ax - 15a
    has an axis of symmtry at
    x = -\frac{-2a}{2 \cdot a} = 1
    and the "a" cancels out naturally. So we have no new information here.

    What is the significance of the graph of  g(x) = |x| - 3 ?

    -Dan
    Apparently i'm supposed to use  f(x) = a(x - p)^2 + q

    I'm still busy figuring it out, i'm quite tired

    Thanks anyway Topsquark.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    Apparently i'm supposed to use  f(x) = a(x - p)^2 + q

    I'm still busy figuring it out, i'm quite tired

    Thanks anyway Topsquark.
    That form would be helpful if you
    1) Had three points on the parabola
    or
    2) Knew the vertex of the parabola

    Since neither of these conditions are met, you can't get any further.

    -Dan
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  5. #5
    Bar0n janvdl's Avatar
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    Download and zoom out to see better...
    Attached Thumbnails Attached Thumbnails Parabola Problem-photo-0535.jpg  
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  6. #6
    Forum Admin topsquark's Avatar
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    The problem is that
    f(x) = a(x^2 - 2x - 15)

    f(x) = a(x^2 - 2x + 1) - a(1) - 15a

    f(x) = a(x - 1)^2 - 16a
    has multiple solutions in terms of what you are being asked for. Any value of "a" will give you a parabola with the given properties.

    -Dan
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    The problem is that
    f(x) = a(x^2 - 2x - 15)

    f(x) = a(x^2 - 2x + 1) - a(1) - 15a

    f(x) = a(x - 1)^2 - 16a
    has multiple solutions in terms of what you are being asked for. Any value of "a" will give you a parabola with the given properties.

    -Dan
    So the memo is wrong?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    So the memo is wrong?
    I couldn't read it that well, but it appeared to give a = -1 explicitly and I can construct the parabola for a = -1/2, a = -1/sqrt(3), etc. with intercepts at x = -3 and 5 with an axis of symmetry x = 1 (as well as for a = -1.)

    -Dan
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  9. #9
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    I couldn't read it that well, but it appeared to give a = -1 explicitly and I can construct the parabola for a = -1/2, a = -1/sqrt(3), etc. with intercepts at x = -3 and 5 with an axis of symmetry x = 1 (as well as for a = -1.)

    -Dan
    But then a shouldn't necessarily only be -1 ?

    Then a could be any value? So it could also have an equation of  f(x) = -2x^2 ... \ -3x^2 ... \ -4x^2... ?

    This is the weirdest parabola prob in all my high school years...
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    But then a shouldn't necessarily only be -1 ?

    Then a could be any value? So it could also have an equation of  f(x) = -2x^2 ... \ -3x^2 ... \ -4x^2... ?

    This is the weirdest parabola prob in all my high school years...
    Yup. Not only that, "a" could also be positive and give a "correct" parabola. The only reason it can't be is that problem told you to assume it wasn't.

    It's only this weird because they didn't specify enough information. If another condition on the parabola were given then we could set up an equation to solve for "a."

    -Dan
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