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.
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.
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