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Math Help - difficult to solve the turning point

  1. #1
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    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.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mastermin346 View Post
    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.
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    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??
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  4. #4
    Senior Member pacman's Avatar
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    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.

    ++++++++++++++++++++++++++++++++++++++++++++
    Attached Thumbnails Attached Thumbnails difficult to solve the turning point-x-squared.gif   difficult to solve the turning point-x-cubed.gif  
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