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Thread: Stationary points - find horizontal line

  1. #1
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    Stationary points - find horizontal line

    (I apologise if this is in the incorrect section)

    I have a stationary points question that I really do not get. I have worked out that the SP's are (0,0) and (2,-4), but I do not know how to find Q.

    Question:
    Part of the graph of the curve with equation $\displaystyle y=3x^2-x^3$ is shown below.

    Stationary points - find horizontal line-diagram.png
    (I used a graphic calculator online to draw the graph)

    a) Establish the coordinates of the stationary point P.
    b) The horizontal line through P meets the curve again at Q. Find the co-ordinates of Q.

    So far, I have worked out the SPs:

    $\displaystyle y=3x^2-x^3$
    $\displaystyle \frac{dy}{dx}=6x-3x^2$
    $\displaystyle -3(x-2)=0$
    $\displaystyle -3x=0$ or $\displaystyle x-2=0$
    $\displaystyle x=0$ or $\displaystyle x=2$

    When x=0, y=0, point(0,0)
    When x=2, $\displaystyle y=3(2)^2-3^3=4$ point(2,4) (which is P).

    But I don't know how I'm supposed to find Q. Can anyone please help?

    Thanks in advance
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  2. #2
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    Re: Stationary points - find horizontal line

    Quote Originally Posted by watp View Post
    b) The horizontal line through P meets the curve again at Q. Find the co-ordinates of Q.
    But I don't know how I'm supposed to find Q. Can anyone please help?
    Solve $\displaystyle 4=3x^2-x^3$. That will give $\displaystyle (x,4)$
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  3. #3
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    Re: Stationary points - find horizontal line

    Quote Originally Posted by Plato View Post
    Solve $\displaystyle 4=3x^2-x^3$. That will give $\displaystyle (x,4)$
    Thanks! I never knew it would be that simple
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