How to plot this graph $y^3=x^2$ I was solving a problem related to area under the integral. When I got a question with the curve $y^3=x^2$. Now this might seem trivial with plotting calculator and for some without plotting calculator. Now, since this question was the in booklet of an entrance exam with restriction to calculator usage.  I would like to understand the process which can be used and also some ideas to deal with such graphs in future, with the above stated exam environment. 
Thanks in advance!
 A: It helps to investigate the range of the function. Calculating on $\mathbb{R}$, you know that $x^2$ will be nonnegative: $x^2 \geq 0$. Since $y^3 = x^2$, it must hold that $y^3$ is nonnegative. By the nature of the cube-root, it follows that $y\geq 0$.
From this you can conclude there are no points below the $x$-axis.

You know there is symmetry in the line $x=0$, because if you fill in $-x$ for $x$ you will find the same function. So the only things you need to do is plotting the first quadrant ($x>0, y>0$) and mirroring it in $x=0$.
Now you can rewrite the problem to $x = \sqrt{y^3}$ for the first quadrant. This will be a continuous function for $x>0$ so you can plot that (compute some points, draw a curve through it). 

The only thing that may cause problems is the origin, because this is the place where mirroring occurs. Therefore the origin may cause discontinuities.
Use the representation to find that $\frac{d}{dy} \sqrt{y^3} = \frac{3}{2}\sqrt{y}$, so at $(0,0)$ the slope will be 0 in the y-direction, so this will resemble a square-root in the origin.
A: To know the nature of the graph you will have to have some ideas about graphs of $y=x^2$ and $y=x^3$ individually.
Look closely...It's given $x^2=y^3$.
So,$x$ must be greater that $y$ for all integer values.
SO, for any increase $\delta x$ in x-axis the increase $\delta y$ in y-axis must be smaller than it as you have a power of $3$ in y but $2$ in x.So,a small increment in $y$ will make up for a greater increment in $x$.
So,the graph must be inclined more towards the $x$ axis and it has a power greater than $1$.So,it is a curve more bent towards the $x$ axis.
The graph must pass through origin as $(0,0)$ are trivial solutions and there is no part of the graph below positive x axis as square of a number cannot give a negative value.
Can you reason why the left side of it is the mirror image of the first one?
