(I apologise if this is in the incorrect section)

I have a stationary points question that I really do not get. I have worked out that the SP's are (0,0) and (2,-4), but I do not know how to find Q.

Question:

Part of the graph of the curve with equation $\displaystyle y=3x^2-x^3$ is shown below.

(I used a graphic calculator online to draw the graph)

a) Establish the coordinates of the stationary point P.

b) The horizontal line through P meets the curve again at Q. Find the co-ordinates of Q.

So far, I have worked out the SPs:

$\displaystyle y=3x^2-x^3$

$\displaystyle \frac{dy}{dx}=6x-3x^2$

$\displaystyle -3(x-2)=0$

$\displaystyle -3x=0$ or $\displaystyle x-2=0$

$\displaystyle x=0$ or $\displaystyle x=2$

When x=0, y=0, point(0,0)

When x=2, $\displaystyle y=3(2)^2-3^3=4$ point(2,4) (which is P).

But I don't know how I'm supposed to find Q. Can anyone please help?

Thanks in advance