find the equation of tangent

Let the touching point (a,b)

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

At (a,b), the slope of tangent line that pass through the point (0,-1)
= 3a^2 - 2a - 2

This value should be equal to (b + 1)/a

So (b + 1)/a = 3a^2 - 2a - 2 => b + 1 = 3a^3 - 2a^2 - 2a

Sub. back into y = x^3 - x^2 - 2x, we have

3a^3 - 2a^2 - 2a - 1 = a^3 - a^2 - 2a

2a^3 - a^2 = 1

a = 1, b = -2

The touching point is (1,-2).

Equation of the tangent line is y + 2 = -(x - 1) or x + y + 1 = 0

Let P(a,b) be a point on the curve such that
b = a^3 - a^2 - 2a ... (1)

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

At P, dy/dx = 3a^2 - 2a - 2

When P lies on a tangent that passes through (0,-1),
slope of tangent = [b - (-1)] / (a - 0)
= (b + 1) / a
3a^2 - 2a - 2 = (b + 1) / a
b + 1 = 3a^3 - 2a^2 - 2a
b = 3a^3 - 2a^2 - 2a - 1 ... (2)

From (1) and (2),
a^3 - a^2 - 2a = 3a^3 - 2a^2 - 2a - 1
2a^3 - a^2 - 1 = 0
(a - 1)(2a^2 + a + 1) = 0
a - 1 = 0 or 2a^2 + a + 1 = 0
a = 1

For 2a^2 + a + 1 = 0,
As delta = 1^2 - 4(2)(1) = -7 < 0,
there are no real solutions.

So, a = 1
From (1), b = 1^3 - 1^2 - 2(1) = -2
The point on the curve is (1,-2)

Slope of tangent = [(-2) - (-1)] / (1 - 0) = -1

The equation of the tangent is
y - (-1) = -1(x - 0)
y + 1 = -x
y = -x - 1