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

Step 1:

Existence : The curve does not exist when $x >1,$ ie for $1 < x < \infty $

So exists in $ -\infty < x \leq 1$

Step 2:

(ii) Symmetry: The curve is symmetric about x-axis only

Step 3:

(iii) Asymptotes: The curve does not admit asymptotes since there are no further values of $ x(or\;y)$ for which $y \to \pm \infty(or\; x \to \pm \infty)$

Step 4:

Loops:The curve passes through (0,0) twice and (1,0) once

So a loop is formed between $(0,0)$ and $(1,0)$