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Thread: Turning Points

  1. #1
    Newbie Ryannnn's Avatar
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    Turning Points

    Sketch this curve on a graph

    $\displaystyle y=4x^2-x^3$

    I differentiated and got

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

    At TP dy by dx = o

    So $\displaystyle 8x-3x^2=0$

    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

    $\displaystyle y=4x^2-x^3$

    I differentiated and got

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

    At TP dy by dx = o

    So $\displaystyle 8x-3x^2=0$

    Now where do I go?
    Factorise.

    $\displaystyle 8x - 3x^2 = 0$

    $\displaystyle x(8 - 3x) = 0$

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

    $\displaystyle 8x - 3x^2 = 0$

    $\displaystyle x(8 - 3x) = 0$

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

    $\displaystyle y=0$

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

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

    $\displaystyle y = 4x^2 - x^3$.

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

    $\displaystyle y = 4x^2 - x^3$.

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

    When $\displaystyle x=\frac{8}{3}$

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

    $\displaystyle y=\frac{256}{27}$

    $\displaystyle y=9\frac{13}{27}$

    Sorry for going on abit.
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