Calculus
posted by Anonymous .
Find an equation of the tangent line to the curve of f (x) = x^3 − 8, at the point where the curve crosses the xaxis.

f ' (x) = 3x^2
when the curve crossess the xaxis, 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(x2)
y = 12x  24