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Math

posted by on .

If f(x)=ax^2+b
f(-3)=-3 and f'(-3)=2
find the coefficients a and b.

  • Math - ,

    f(x) = ax^2 + b
    f ' (x) = 2ax

    given :
    f ' (-3) = 2 ----> 2a(-3) = 2
    -3a = 1
    a = -1/3 , so f(x) = (-1/3)x^2 + b

    f(-3) = -3
    ---> (-1/3)(-3)^2 + b = -3
    -3 + b = -3
    b = 0

    so a = -1/3, b = 0 and f(x) = (-1/3)x^2

    check:
    f(-3) = -(1/3)(9) = -3
    f ' (x) = -(2/3)x
    f '(-3) = -(2/3)(-3) = 2
    all looks good.

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