calculus

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find the point on the graph of the given function at which the slope of the tangent line is the given slope

x^3+9x^2+33x+15 slope= 6

  • calculus -

    You should have properly written the function as
    y = x^3 + 9x^2 + 33x + 15
    or
    f(x) = x^3 + 9x^2 + 33x + 15

    then dy/dx = 3x^2 + 18x + 33
    but you want that to be 6
    3x^2 + 18x + 33 = 6
    x^2 + 6x + 9 = 0
    (x+3)^2 = 0
    x+3 = 0
    x = -3
    then f(3) = -27 + 81 - 99 + 15 = -30
    The point is (-3,-30)

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