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I have the function $f(x)=-\frac{1}{x}$, plotted on standard 2D axes.

I take the vertical asymptote and rotate it $45$ degrees clockwise, so that it's now $y=x$ instead of the $y$-axis. The horizontal asymptote is still the $x$-axis ($y=0$).

What function do I have now?

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Using the formulae for the rotation of the coordinate frame $$x=x'\cos\alpha-y'\sin\alpha$$ $$y=x'\sin\alpha+y'\cos\alpha$$ we obtain $$\alpha=\frac{\pi}{4}\implies\cos\alpha=\sin\alpha=\frac{1}{\sqrt{2}}$$ hence $$x=\frac{1}{\sqrt{2}}\left(x'-y'\right)$$ $$x=\frac{1}{\sqrt{2}}\left(x'+y'\right)$$ so $$xy=-1$$ transforms into $$y'^2-x^2=2$$ EDIT: this is an indication of an alternative answer to the actual question. What is in fact required is a conformal mapping that halves angles. It is well-known that square root will do the trick: $$\sqrt{x+iy}=\frac{1}{2}\sqrt{2}\left[\sqrt{\sqrt{x^2+y^2}+x}+i\sqrt{\sqrt{x^2+y^2}-x} \right]$$ Which seems to give the correct answer on the RHS when $y=\pm 2$. To understand why it must be so, denote the original coordinates as $u$, $v$, then $$(u+iv)^2=x+iy$$ $$u^2-v^2+2uvi=x+iy$$ If the equation of your hyperbola in the original plane is $$uv=-1$$, then comparing the imaginary parts we deduce that $y=-2$

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Right idea, but if we are using the formulae you gave, we are rotating by $\alpha$ to go from the $(x',y')$ frame to the $(x,y)$ frame. In other words, we are starting with the curve $x'y'=-1$, and from there, we should obtain the curve $y^2-x^2=2$. –  Cameron Buie Jun 4 '12 at 23:08
    
Thanks for the try! This rotates both axes, not just the vertical one. I've edited the question to be clearer that the horizontal asymptotes should remain where they were. –  WBT Jun 5 '12 at 0:25
    
rotating target ;) –  Valentin Jun 5 '12 at 9:57
    
i did confuse direct and reverse transformations as i always do, so even that the focus of the question has shifted, i will change the sign of the angle –  Valentin Jun 6 '12 at 20:19
    
The focus of the question never shifted - it was always just the vertical asymptote that was rotating and the horizontal asymptote remaining the same, but the current problem statement is more emphatic about that than the original. Thanks for the answer and edit! –  WBT Jul 9 '12 at 22:02
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Here is a correct answer from someone who, as far as I can tell, is not on this site, but is welcome to credit if she wants it, and welcome to anonymity if she doesn't want an overwhelming number of help requests.

I've cleaned it up a little, added some formatting, and solved the second half.

In case this is a trick question, you don't have a function anymore.
For a given value of $x$, you will have two values of $y$, so $f(x)$ is not defined.

Anyway, let's pick the negative portion of $y = -\frac{1}{x}$ and rotate the asymptote for that.
Note that the function $-\frac{1}{x} + x$ looks like what we want but rotated by 45 degrees.
If I apply a rotation transformation to the $x$ and $y$ axes (ie rotate the coordinates by 45 degrees) we should get the correct function.
Let $x = x'cos(\theta) - y'sin(\theta)$
and $y = x'sin(\theta) + y'cos(\theta)$
where $x'$ and $y'$ are our new coordinates.
Plug that stuff into $y = -1/x + x$ and solve for $y'$.
If we assume $y'$ is positive (you can also assume $y'$ is negative if you want the other possible function), we get $y' = (\sqrt(x'^2 + 4) +x')/2$.
To check our solution, as $x \to -\infty, y = 0$. As $x \to +\infty, y = x$. Yay!

It's not meant to be a trick question, and this is an accurate solution.
$y' = (\sqrt(x'^2 + 4) \pm x')/2$.
Thanks to the math cat who answered it!

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