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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.

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closed as too broad by Jack, Nosrati, Alex Provost, Daniel W. Farlow, user354271 Nov 1 '17 at 20:33

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Without assuming $f$ is nonnegative, the question is not well-defined. $\endgroup$ – Jack Nov 1 '17 at 14:52
  • $\begingroup$ @Jack It is assumed $\endgroup$ – Siwei Nov 1 '17 at 14:54
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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.

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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 $y^2$.

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  • $\begingroup$ I think you meant to say "Only if $f(x)=0$," not $f'(x)=0$. $\endgroup$ – Barry Cipra Nov 1 '17 at 15:05
  • $\begingroup$ I meant to say if $f(x)=f'(x)=0$. $\endgroup$ – aleden Nov 1 '17 at 15:08
  • $\begingroup$ If $f(x)=f'(x)=0$, you can L'Hopitate $(f'(x))^2/f(x)$ to $2f'(x)f''(x)/f'(x)=2f''(x)$. (You need $f''$ to be non-negative near $x$ in order for $f$ to be non-negative near $x$.) $\endgroup$ – Barry Cipra Nov 1 '17 at 15:23

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