The tangent to the cubic function that is defined by y = x^3 - 6x^2 + 8x at point A(3,-3) intesects the curve at another point, B. Find the coordinates of point B. Illustrate with a sketch.

y' = 3x^2 - 12x + 8
y' = 3(3)^2 - 12(3) + 8
y = -1

Question

The tangent to the cubic function that is defined by y = x^3 - 6x^2 + 8x at point A(3,-3) intesects the curve at another point, B. Find the coordinates of point B. Illustrate with a sketch.

y' = 3x^2 - 12x + 8
y' = 3(3)^2 - 12(3) + 8
y = -1 <-- slope of the line of A

This line also has the point B.

y = mx + b
-3 = -1(3) + b 
0 = b
Therefore, line equation is y = -1x.

So, I equate:

-1x = x^3 - 6x^2 + 8x
0 = x^3 - 6x^2 + 9x
0 = x(x^2 - 6x + 9)
0 = x(x-3)(x-3)

x = 3, or x = 0.

X = 3 we already have, so x = 0 should be point B. I sub back in X = 0 into the main cubic function and I retrieve