# difficult to solve the turning point

• Oct 14th 2009, 12:59 AM
mastermin346
difficult to solve the turning point
how can i find the coordinates of the turning point at 4x(x-1)(x-2)?

this is my work to get the answer:

4x(x-1)(x-2)
=4x^3-12x^2+8x (divide 4)
=x^3-3x^2+2x

dy/dx=3x^2-6x+2=0

and how can i find the value of x??
is that true my work to get the right answer?

can u teach me step by step?tq.
• Oct 14th 2009, 01:52 AM
Quote:

Originally Posted by mastermin346
how can i find the coordinates of the turning point at 4x(x-1)(x-2)?

this is my work to get the answer:

4x(x-1)(x-2)
=4x^3-12x^2+8x (divide 4)
=x^3-3x^2+2x

dy/dx=3x^2-6x+2=0

and how can i find the value of x??
is that true my work to get the right answer?

can u teach me step by step?tq.

cant you solve the quadratic equation ?

3x^2-6x+2=0

Then substitute the values of x back into the original equation for y.
• Oct 14th 2009, 02:13 AM
mastermin346
Quote:

cant you solve the quadratic equation ?

3x^2-6x+2=0

Then substitute the values of x back into the original equation for y.

but how can i find the value of x??
• Oct 14th 2009, 03:10 AM
pacman
quadratic formula? 3x^2 - 6x + 2 = 0.

general form: ax^2 + bx + c = 0,

where a = 3, b = -6 and c = 2.