Math

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I am trying to complete this input/output table. Input 0 output blank, input 3/2 output 17/6, input 11/4 output blank, input 3 output 13/3, input blank output 13/2

  • Math -

    Your relation could be linear:
    using (3/2 , 17,6) and (3, 13/3)
    slope = (17/6 - 13/3)/(3/2 - 3) = (-3/2) / (-3/2) = 1
    y - 13/3 = 1(x-3)

    3y - 13 = 3x-9
    3x - 3y = -4

    now plug in the given input as x or the given output as y
    e.g input = x = 11/4
    3(11/4) - 3y = -4
    times 4
    33 - 12y = -16
    -12y = -49
    y = -49/-12 = 49/12

    do the remaining one in the same way

    your relation could have been quadratic:
    let's make (3,13/3) the vertex
    then
    y = a(x-3)^2 + 13/3
    but (3/2 , 17/6) lies on it
    17/6 = a(3/2 - 3)^2 + 13/3
    17/6 = a(9/4) + 13/3
    times 12
    34 = 27a + 52
    27a = -18
    a = -18/27 = 2/3

    so it could have been
    y = (2/3)(x-3)^2 + 13/3

    There could have been an infinite number of possible quadratics, not to mention, we could have made it into a cubic, etc.

  • Math -

    Hmmm. In general, you cannot fit a quadratic to 4 points. Only a single cubic, but many of higher degree.

  • Math -

    Steve, only 2 points were actually given.
    (3/2, 17/6)
    (3, 13/3)

    The others were
    (0, ??)
    (11/4, ??)

    So I could form a unique linear, but an infinite number of quadratics, cubics, etc that pass through those two points.
    Once I have an equation, the missing y values of the other 2 points are then found

  • Reiny - Oww - my Bad! -

    Oww. My Bad!

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