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Math Help - Another Graph of function

  1. #1
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    Another Graph of function

    Sketch the following curves :

    (1) y^2=x(1-x^2)

    Not sure how to do this when y is squared . Thanks .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Sketch the following curves :

    (1) y^2=x(1-x^2)

    Not sure how to do this when y is squared . Thanks .
    If you multiply the x bit out you can then square root both sides, to get y=\sqrt{x-x^3}
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  3. #3
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    Yes, you can do that but remember that y can be negative also:
    y= \pm\sqrt{x- x^3}.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Yes, you can do that but remember that y can be negative also:
    y= \pm\sqrt{x- x^3}.

    Yes i know i can do that . But i am unsure of how to find the horizontal and vertical asymtotes and other informations to sketch .
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  5. #5
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    What makes you think there are any asymptotes? You should do simple things like finding the x and y intercepts and turning points if there are any.
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  6. #6
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    Exactly , i am confusing myself .

    Well , let me try here

     y= \pm\sqrt{x- x^3}

    When y=0 ,x=0,1,-1

    When i differentiate to find the turning points , i am stucked here .

     \frac{dy}{dx}=\frac{(x-x^3)^{\frac{1}{2}}}{2-6x^2}

    Equate it to 0 , i got (x-x^3)^{\frac{1}{2}} , isn't it the same when i put y=0 and solve for x earlier .
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