# Thread: Another Graph of function

1. ## 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 .

2. Originally Posted by thereddevils
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 $\displaystyle y=\sqrt{x-x^3}$

3. Yes, you can do that but remember that y can be negative also:
$\displaystyle y= \pm\sqrt{x- x^3}$.

4. Originally Posted by HallsofIvy
Yes, you can do that but remember that y can be negative also:
$\displaystyle y= \pm\sqrt{x- x^3}$.

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 .

5. 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.

6. 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 .