Find the equation of tangent

Let the touching point (a,b)

dy/dx = 3x^2.

At (a,b), the slope of tangent line that pass through the point (0,2)
= 3a^2

This value should be equal to (b-2)/a

So (b-2)/a = 3a^2 => b-2 = 3a^3

Sub. back into y = 3x^3, we have

3a^3+2=a^3

2a^3 = -2

a = -1, b = -1

The touching point is (-1,-1).

Equation of the tangent line is y - 2 = 3x or y - 3x - 2 = 0

2011-01-21 21:34:51 補充：
thery are both the slopes of the same line

y = x^3
dy/dx = 3x^2

Let (h,k) be a point on the curve such that k = h^3
dy/dx = 3h^2
Equation of tangent:
y - k = 3h^2 (x - h)

When the tangent passes through the point (0,2),
the equation of tangent becomes
2 - k = 3h^2 (0 - h)
= - 3h^3

Converting (h,k) into (x,y), we obtain
2 - y = - 3x^3
y = 3x^3 + 2