The relation $y^2=f(x)$ Given a sketch of a function $f(x)$ , how to sketch the graph of $y^2=f(x)$? 
Is there any relationship between the features of two graphs? 
Like vertical asymptotes or the zeros.
 A: As far as I'm aware there is no answer that fits all here as different functions will be behave in different ways under the square-root but one thing you must do is (assuming you are trying to graph a real function) take the modulus of $f(x)$ first since if $f(x) < 0$, then $\sqrt{f(x)}\notin\mathbb{R}$. Now you are trying to graph $y=\pm\sqrt{\mid f(x)\mid}$ so there are a few different forms this can take depending on $f(x)$. 
If $f(x) = ax^{2}$, for some $a\in\mathbb{R}$, then the graph of $y$ will be a straight line "V" shape. But then don't forget the $\pm$ in the expression for $y$ which gives you an overall "X" shape to the graph. 
If $f(x) = e^{x}$, then $y=e^{\frac{x}{2}}$ will be a similar shape to $f(x)$. Again the $\pm$ part of the expression will give you another curve passing through $(0,-1)$ and tending off to $-\infty$.
Another interesting example is $f(x)=x^{3}$ which I will leave for you to think about. 
So my point is given a sketch of $f(x)$ you can see what form the function approximately takes and, given the above examples, you can estimate what $y$ should look like. 
A: If you are sketching over $\Bbb R$, the graph of $y$ is discontinuous whenever $f(x)<0$. $y$ will have the same zeroes and poles as $f(x)$ as well as a vertical tangent line at $f(x)=0$. This is because $$y=\pm\sqrt {f(x)}$$ so $$\frac{dy}{dx}=\pm \frac{f'(x)}{2\sqrt{f(x)}}$$
Only if $f'(x)=f(x)=0$ will there not necessarily be a vertical tangent, since $\frac{dy}{dx}$ will be indeterminate. For each value of $f(x)$ there will be two values of $y$, due to $y^2$. 
