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Thread: 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)

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

    -----------

    For b)

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

    $\displaystyle f(x) = (x^2 - 6x - 16) $

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

    $\displaystyle x = 3 $

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

    Then $\displaystyle 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?

    $\displaystyle k^2 - 6k - 16 = 0 $

    Of course you would!

    But look at what i did:

    I simply treated $\displaystyle (x + 3) $ as $\displaystyle k $

    So let's solve this one:

    $\displaystyle k^2 - 6k - 16 = 0 $

    $\displaystyle (k - 8)(k + 2) = 0 $

    $\displaystyle k = 8 \ and \ k = -2 $

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

    So $\displaystyle (x + 3) = 8 $ and $\displaystyle (x + 3) = -2 $

    $\displaystyle x = 5 $ or $\displaystyle x = -5 $
    Last edited by janvdl; Jul 8th 2007 at 07:41 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.

    $\displaystyle y = x^2 - 6x - 16$

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

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

    when $\displaystyle x = 3$

    $\displaystyle y = 3^2 - 6(3) - 16 = -25$

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