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Math Help - Turning Points

  1. #1
    Newbie Ryannnn's Avatar
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    Turning Points

    Sketch this curve on a graph

    y=4x^2-x^3

    I differentiated and got

    \frac{dy}{dx}=8x-3x^2

    At TP dy by dx = o

    So 8x-3x^2=0

    Now where do I go?
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  2. #2
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    Quote Originally Posted by Ryannnn View Post
    Sketch this curve on a graph

    y=4x^2-x^3

    I differentiated and got

    \frac{dy}{dx}=8x-3x^2

    At TP dy by dx = o

    So 8x-3x^2=0

    Now where do I go?
    Factorise.

    8x - 3x^2 = 0

    x(8 - 3x) = 0

    x = 0 or x = \frac{8}{3}.
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  3. #3
    Newbie Ryannnn's Avatar
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    Quote Originally Posted by Prove It View Post
    Factorise.

    8x - 3x^2 = 0

    x(8 - 3x) = 0

    x = 0 or x = \frac{8}{3}.
    So when x=0

    y=0

    But how do you find y when x=\frac{8}{3}

    Thanks for your help.
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  4. #4
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    You're told that

    y = 4x^2 - x^3.

    Now let x = \frac{8}{3} and find y.
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  5. #5
    Newbie Ryannnn's Avatar
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    Quote Originally Posted by Prove It View Post
    You're told that

    y = 4x^2 - x^3.

    Now let x = \frac{8}{3} and find y.
    So...

    When x=\frac{8}{3}

    y=\frac{256}{9}-\frac{512}{27}

    y=\frac{256}{27}

    y=9\frac{13}{27}

    Sorry for going on abit.
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