# Stationary points - find horizontal line

• Jan 8th 2012, 05:03 AM
watp
Stationary points - find horizontal line
(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 $y=3x^2-x^3$ is shown below.

Attachment 23205
(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:

$y=3x^2-x^3$
$\frac{dy}{dx}=6x-3x^2$
$-3(x-2)=0$
$-3x=0$ or $x-2=0$
$x=0$ or $x=2$

When x=0, y=0, point(0,0)
When x=2, $y=3(2)^2-3^3=4$ point(2,4) (which is P).

• Jan 8th 2012, 05:21 AM
Plato
Re: Stationary points - find horizontal line
Quote:

Originally Posted by watp
b) The horizontal line through P meets the curve again at Q. Find the co-ordinates of Q.
Solve $4=3x^2-x^3$. That will give $(x,4)$
Solve $4=3x^2-x^3$. That will give $(x,4)$