Construct function with known asymptotes Suppose I have two lines, described by
$$ f_1(x)=r_1(x-x_0)+y_0$$
$$ f_2(x)=r_2(x-x_0)+y_0$$
E.g, for $(x_0,y_0)=(1,\frac{1}{2})$, $r_1=\frac{1}{2}$ and $r_2=4$ it looks like:

I am looking for a function $f(x)$, with

*

*$f(x) \geq f_1(x) \forall x$

*$f(x) \geq f_2(x) \forall x$

*$\displaystyle\lim_{x\to-\infty} f(x)-f_1(x)=0$

*$\displaystyle\lim_{x\to\infty}f(x)-f_2(x)=0$

*$r_2>r_1>0$

*$f$ and $f_x$ are continuous

That is, far from $(x_0,y_0)$ it asymptotes to either $f_1$ or $f_2$, and it should always be above both function (otherwise there are four solution I guess).
I understand that this function would look a bit like the hyperbola $\frac{1}{x}$, if $f_1$ and $f_2$ are transformed such that they align with the $x$ and $y$ axis. It is not that obvious to me now how that can be achieved.
 A: You have one obvious answer:
$g(x)=f_1(x)$ for $x<\alpha$
$g(x)=f_2(x)$ for $x\geq\alpha$
with $\alpha$ being the point where $f_1$ and $f_2$ intersects.
EDIT: a hint for an answer closer to what you are looking for...
You can change the coordinates, in order to get $f_1$ and $f_2$ symmetrical with regard to the new $x'$ axis. Within this new set of coordinates, the equation of the hyperbola will take the form of $\frac{x²}{a²}-\frac{y²}{b²}=1$, with $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$ being the equation of the asymptotes in the new set of coordinates. 
You then revert to your old set of coordinates and get the equation. 
A: If you're looking for something "fancy", and continuous or something, pick a function $g(x)$ that goes from $0$ to $1$ as $x$ goes from $-\infty$ to $\infty$. Simple manipulations of the inverse tangent function $\arctan(x)$ or the error function $\operatorname{erf}(x) = \int_{-\infty} ^x e^{-t^2} dt$ might do the trick. Then what you do is the following:
$$
f(x) = f_1(x)(1-g(x)) + f_2(x)g(x)
$$
This will lie between the two lines, so it's not exactly what you want, but I hope it can put you on some kind of track.
