Calculus

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Find an equation of the tangent line to the curve of f (x) = x^3 − 8, at the point where the curve crosses the x-axis.

  • Calculus -

    f ' (x) = 3x^2

    when the curve crossess the x-axis, y = 0
    x^3 - 8 = -
    x^3 = 8
    x = 2 , so we have the point (2,0)

    when x = 2, slope = 3(4) = 12

    y - 0 = 12(x-2)
    y = 12x - 24

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