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Math Help - quadratic curve

  1. #1
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    quadratic curve

    ImageShack - Hosting :: mathdo0.jpg

    sorry for the scribbling.
    could someone please show me how to do this.
    thanks
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  2. #2
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    Here's a graph of two such parabolas.

    Since they cross the y-axis at 0, then c=0.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
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    Quote Originally Posted by poniekid View Post
    ImageShack - Hosting :: mathdo0.jpg

    sorry for the scribbling.
    could someone please show me how to do this.
    thanks
    Use the given equation to calculate the missing coefficients:

    Parabola 1:
    y=-2x^2+bx+c

    P(-2,0) and Q(0, 0)

    Plug in the coefficients into the equation:

    Q: 0=0+0+c~\implies~ c=0

    P: 0=-2\cdot (-2)^2+b\cdot(-2)+0~\implies~ b= -4 and therefore the equation of parabola1 becomes:

    p_1:y=-2x^2-4x

    Do just the same to calculate the equation of parabola2. You should come out with

    p_2:y=-2x^2+4x

    The domain is [-2, 0] or [0, 2] respectively.

    By the way: You can get one function whose graph is the required "M":

    y = -2x^2+4\cdot |x|~,~x \in [-2, 2]

    The height of the letter is the y-coordinate of the vertex of one parabola. I don't know which method you prefer to calculate the vertex but you should get: V_1(-1, 2)

    To answer #5 I've sketched a family of curves:

    y_a=-ax^2+2a \cdot |x| for a running from 2 to 6 in steps of 0.5. Now make your choice!
    Attached Thumbnails Attached Thumbnails quadratic curve-m_logo.gif  
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