Sketch the following curves :
(1) y^2=x(1-x^2)
Not sure how to do this when y is squared . Thanks .
Exactly , i am confusing myself .
Well , let me try here
$\displaystyle y= \pm\sqrt{x- x^3}$
When y=0 ,x=0,1,-1
When i differentiate to find the turning points , i am stucked here .
$\displaystyle \frac{dy}{dx}=\frac{(x-x^3)^{\frac{1}{2}}}{2-6x^2}$
Equate it to 0 , i got $\displaystyle (x-x^3)^{\frac{1}{2}}$ , isn't it the same when i put y=0 and solve for x earlier .