Thread: 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??
please help me...
is that true my work to get the right answer?

can u teach me step by step?tq.

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??
please help me...
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.

Originally Posted by mathaddict

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??

quadratic formula? 3x^2 - 6x + 2 = 0.

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

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

Substitute in the quadratic formula, answers are

x = 0.423 and x = 1.577.

i included a graphs, see for yourself. Double click the graph, for a better view

A) 3x^2 - 6x + 2 = 0.

B) x^3-3x^2+2x = x(x^2 - 3x + 2) = 0.

++++++++++++++++++++++++++++++++++++++++++++