calculus

A curve passes through the point (1,-11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,-16) is parallel to the x-axis. Find
i) the values of a and b
ii) the equation of the curve

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  1. since dy/dx = ax^2 + b
    the equation must have been

    y = (1/3)a x^3 + bx + c

    at (1,-11) ---> -11 = (1/3) a + b + c
    or
    a + 3b + 3c = -33 (#1)

    at (2, -16) ---> -16 = (8/3)a + 2b + c
    or
    8a + 6b + 3c = -48 (#2)

    #2 - #1 :
    7a + 3b = -15 (#3)

    We also know that at (2,-16), the slope is zero
    ax^2 + b = 0
    4a + b = 0
    b = -4a

    sub into #3
    7a - 12a = -15
    -5a=-15
    a=3
    then b= -12

    and in #1
    3 -36+ 3c = -33
    3c = 0
    c = 0

    function is y = (1/3)(3)x^3 - 12x

    y = x^3 - 12x

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  2. I like the way you solve it

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