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Math Help - Determining an expression for a function

  1. #1
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    Determining an expression for a function

    I can't figure this out for the life of me. Help is much appreciated.

    The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.

    Thanks!
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  2. #2
    Newbie sir nerdalot's Avatar
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    Quote Originally Posted by ocmisssunshine View Post
    The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.
    Here we want to make a quadratic equation.

    So x must = either -1 or 3 for y to be 0 and to intercept the x-axis

    Therefore (x+1)(x-3) = 0
    expanding gives you... x^2 - 3x + x - 3 = 0 where x <= 4

    the graph is as seen below
    Attached Thumbnails Attached Thumbnails Determining an expression for a function-graph.jpg  
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    Oh thanks!!! That helps alot!!
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  4. #4
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    Quote Originally Posted by ocmisssunshine View Post
    I can't figure this out for the life of me. Help is much appreciated.

    The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.

    Thanks!
    (We were sent home due to heavy rain, so I can help you now.)

    The parabola has two x-intercepts. That means it is a "vertical" parabola, or a parabola whose axis of symmetry is vertical, or a parabola that opens upward or downward.

    The range is all numbers less than or equal to 4. That means the parabola is below, or on, the y=4 horizontal line. That means further that the vertical parabola opens downward.
    The "equal to 4" means the parabola touches the y=4 at one point only. That means further that the the vertex is at y=4.

    A standard form of a vertical parabola that opens downward is
    (y-k) = -a(x-h)^2 ---------(i)
    where
    (h,k) is the vertex
    "a" is any rational number.

    The two x-intercepts -1 and 3 are the points (-1,0) and (3,0) respectively.

    For point (-1,0),
    (y-k) = -a(x-h)^2 ---------(i)
    (0-k) = -a(-1-h)^2
    -k = -a(1+h)^2
    k = a(1+h)^2 ---------------(ii)

    For point (3,0),
    (y-k) = -a(x-h)^2 ---------(i)
    (0-k) = -a(3-h)^2
    -k = -a(3-h)^2
    k = a(3-h)^2 ---------------(iii)

    The k in (ii) equals the k in (iii),
    a(1+h)^2 = a(3-h)^2
    (1+h)^2 =(3-h)^2
    1+h = 3-h
    h+h = 3-1
    2h = 2
    h = 1 ---------------**

    We already know that k = 4, so (h,k) = (1,4).
    Plug those into (i), using the x-intercept point (-1,0)
    (y-k) = -a(x-h)^2 ---------(i)
    (0-4) = -a(-1-1)^2
    -4 = -4a
    a = 1 ------------**

    Therefore, the equation of the parabola is
    (y-k) = -a(x-h)^2 ---------(i)
    y -4 = -1(x-1)^2
    y -4 = -(x^2 -2x +1)
    y = -x^2 +2x -1 +4
    y = -x^2 +2x +3 -------------------answer.
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