assdadsdasd

 y = ax^2 + bx + c with x as time and y as vertical displacement

    one root is (0,0), other is (duration,0)

    y = ax^2 + bx is valid bc there will never be a y intercept

    rules for simplifying quadratics:
    if (x+q)(x+g):

        b = q + g
        c = q * g

    for our case:

        b = q + g
        0 = q * g <-- either q or g has to be zero, it will be q in this case bc we want leftmost root to be origin.

    (x+0) (x+duration)

    so the vertex of the point is given by: (duration/2, maxVert)

    however, vertex is found by:

       vertex = -b/2a

    therefore:
        duration = d for simplification.

        d/2 = -b/2a

    factor in previous b:

        d/2 = -(q + g)/2a

    we know q = 0 and g = duration:

        d/2 = -d/2a

    solve for a:

        multiply by 2a:

        2ad/2 = -d

        reduce:

        ad = -d

        divide by d:

        a = -d/d

        a = -1


    factor our newly discovered variables into our original function:

    y = -1x^2 + (q + g)x + c

    reduce:

    y = -x^2 + (d)x