Thread: difficult to solve the turning point

1. 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.

2. 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.

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??

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

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

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