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Math Help - Quadratic Question

  1. #1
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    Quadratic Question

    Find the quadratic function f such that f (2) = f (4) = 0 and 7 is the greatest value of f (x).

    I know that

     -k(x-2)(x-4) because the parabola is a negative one because it has a greatest value and because the x ints are 2 and 4

    I don't know what to do from here
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    k=7 and you done!
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  3. #3
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    A better solution:

    Your \displaystyle x intercepts are at \displaystyle x = 2 and \displaystyle x = 4, so the axis of symmetry is \displaystyle x = \frac{2+4}{2} = 3 So the turning point is \displaystyle (3, 7).


    Therefore the equation of the parabola is:

    \displaystyle y = a(x - 3)^2 + 7.


    Now substituting another of your points that lie on the curve, e.g. \displaystyle (2, 0), you find

    \displaystyle 0 = a(2 - 3)^2 + 7

    \displaystyle 0 = a + 7

    \displaystyle a = -7.


    So the equation of your parabola is:

    \displaystyle y = -7(x - 3)^2 + 7.
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  4. #4
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    Yet another way to do this if you don't happen to know off-hand that the vertex is always exactly midway between the x-intercepts: -k(x- 2)(x- 4)= -k(x^2- 6x+ 8). Now complete the square to find the highest point:
    -k(x^2- 6x+ 8)= -k(x^2- 6x+ 9- 9+ 8= -k((x- 3)^2- 1)

    Since a square is never negative (x- 3)^2- 1 will have its lowest point (and -k((x-3)^2- 1) its highest) when, just as Prove It said, when x= 3. At that point -k((3- 3)^2- 1)= k= 7.
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