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Math Help - Another Parabola help

  1. #1
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    Another Parabola help

    Hello, I Got a problem with this question im tryin to solve
    and this is involved wth fractions which confuses me alot
    Problem:

    Axis VERTIcal;
    Passes Through (-1,-3),(1,-2), & (2,1)
    EDIT: Very Sorry For the problem!

    The Answer at back of book is:
    5x^2 + 3x - 6y - 20 = 0

    Thanks and very appreciated
    Last edited by ^_^Engineer_Adam^_^; January 23rd 2007 at 07:26 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hello, I Got a problem with this question im tryin to solve
    and this is involved wth fractions which confuses me alot
    Problem:

    Axis Horizontal;
    Passes Through (-1,-3),(1,-2), & (2,1)
    EDIT: Very Sorry For the problem!

    The Answer at back of book is:
    5x^2 + 3x - 6y - 20 = 0

    Thanks and very appreciated
    First observation: the answer in the back of the book is a parabola with
    its axis vertical.

    RonL
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hello, I Got a problem with this question im tryin to solve
    and this is involved wth fractions which confuses me alot
    Problem:

    Axis Horizontal;
    Passes Through (-1,-3),(1,-2), & (2,1)
    EDIT: Very Sorry For the problem!

    The Answer at back of book is:
    5x^2 + 3x - 6y - 20 = 0

    Thanks and very appreciated
    Axis, horizontal.
    Then the horizontal parabola is in the form
    x = A*y^2 +B*y +C -------------------------(i)
    But solving it that way, we get
    5y^2 +y +12x -30 = 0 --------------answer
    Which is not what the back of the book says.

    --------------------------------------------------------------
    So I tried Axis, vertical.
    The form then is
    y = A*x^2 +B*x +C -------------------------(ii)

    At point (-1,-3):
    -3 = A -B +C ---------------(1)

    At point (1,-2):
    -2 = A +B +C ---------------(2)

    At point (2,1):
    1 = 4A +2B +C --------------(3)

    Solve those 3 equations simultaneously you'd get
    A = 5/6
    B = 1/2
    C = -10/3
    So, plugging those into (ii),
    y = (5/6)x^2 +(1/2)x -10/3
    Clear the fractions, multiply both sides by 6,
    6y = 5x^2 +3x -20
    Therefore,
    0 = 5x^2 +3x -6y -20
    5x^2 +3x -6y -20 = 0 --------------answer.

    And the back of the book is happy.
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!

    Axis Horizontal
    Passes Through (-1,-3), (1,-2), and (2,1)


    The Answer at back of book is: 5x^2 + 3x - 6y - 20 = 0 . Impossible!

    Their equation has x^2 and y ... a parabola with a vertical axis.




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  5. #5
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    Sory for the problem!
    its given is vertical
    And Thanks a lot for answering!
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