# Thread: Find the coordinates of the turning point

1. ## Find the coordinates of the turning point

how can i find the coordinates of the turning point 4x(x-1)(x-2)?anyone can help.

2. Originally Posted by mastermin346
how can i find the coordinates of the turning point 4x(x-1)(x-2)?anyone can help.
$y=4x(x-1)(x-2)=4(x^3-3x^2+2x)$

$\frac{dy}{dx}=4(3x^2-6x+2)$

At the turning point, $\frac{dy}{dx}=0$

$x=1\pm{\frac{1}{\sqrt{3}}}$

3. thank you