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.

Answer 1

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