calculus

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Find the cubic function of the form y=ax^3 + bx^2 + cx + d which has a relative maximum point at (0,2) and a point of inflection at (-1,-2). How do I even start this one?

  • calculus -

    First of all, both given points are on the curve, so
    for
    (0,2) ---> 2 = 0+0+0+d
    so we know d=2
    (-1,-2) ---> -2 = a(-1) + b(1) + c(1) + d
    -a + b + c + 2 = -2
    a - b - c = 4

    y' = 3ax^2 + 2bx + c
    since (0,2) is a max, we know y' = 0 when x = 0
    0 = 3a(0) + 2b(0) + c
    c = 0

    y'' = 6ax + 2b
    we know y'' = 0 when x = -1
    0 = -6a + 2b
    2b = 6a
    b = 3a

    so in a-b-c=4
    a - 3a - 0 = 4
    -2a = 4
    a = -2 and b=-6 , c=0 , d=2

    then the function is
    y = -2x^3 - 6x^2 + 2



    check my arithmetic.

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