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Math Help - Equation Graph

  1. #1
    Member GAdams's Avatar
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    Equation Graph

    For (a); I tried this:

    y = x^2 + qx + r
    y = o^2 + 6 * 0 + 16
    y = -16


    y = x^2 -6x - 16
    y = 8^2 - 6*8 - 16
    y = 0

    I am lost with (b) and not sure about (c)

    Thanks.
    Attached Thumbnails Attached Thumbnails Equation Graph-equation-graph-i.jpg   Equation Graph-equation-graph-ii.jpg  
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  2. #2
    Bar0n janvdl's Avatar
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    For a)

     f(x) = a(x - \alpha)(x - \beta)
     f(x) = a(x - 8)(x + 2)
     f(x) = a(x^2 - 6x - 16)

    -----------

    For b)

    I think we can get the minimum point by using Calculus.

     f(x) = (x^2 - 6x - 16)

     \frac{dy}{dx} = 2x - 6 = 0

     x = 3

    Set  x = 3 into  f(x) = (x^2 - 6x - 16)

    Then  f(3) = -25

    So point P's coordinates is: (3 ; - 25)

    -------------------

    For c)

    Now this one really isn't hard.
    Look at this:

    Would you be able to solve this?

     k^2 - 6k - 16 = 0

    Of course you would!

    But look at what i did:

    I simply treated  (x + 3) as  k

    So let's solve this one:

     k^2 - 6k - 16 = 0

     (k - 8)(k + 2) = 0

     k = 8 \ and \ k = -2

    But remember that  k is actually equal to  (x + 3)

    So  (x + 3) = 8 and  (x + 3) = -2

     x = 5 or  x = -5
    Last edited by janvdl; July 8th 2007 at 08:41 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    To (b)

    I don't think GAdams has taken a calculus class as yet. We can use the vertex formula to get the lowest Point.

    y = x^2 - 6x - 16

    This is an upward opening parabola, so the vertex occurs at the minimum point.

    For vertex, x = \frac {-b}{2a} = \frac {6}{2} = 3

    when x = 3

    y = 3^2 - 6(3) - 16 = -25

    So the coordinates for the lowest point is (3,-25)
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