Is $y=x+\frac{1}{x}$ a hyperbola? How do we confirm or disprove that? And is there a name for this kind of function? $$f(x)=c(x-a)+\frac{d}{x-a}+b$$
If we restrict that $x-a>0$ and $c,d>0$, an observation is that the minimum $2\sqrt{cd}+b$ is reached when $$x=\frac{d}{\sqrt{cd}}+a.$$ This can be confirmed by differentiation. However, noticing that's also exactly when $$c(x-a)=\frac{d}{x-a},$$ I'd like to ask if there's a simpler explanation why this function reaches its minimum when two of its components are equal?
 A: Note that $$y = x + \dfrac1x \implies xy = x^2 + 1 \implies x \underbrace{(y-x)}_{z} = 1$$
Hence, we have $xz = 1$, where $z=y-x$. This is the (rectangular) hyperbola in the $XZ$ plane with the lines $x=0$ i.e. $y$ axis (or $z$ axis) and the line $z=0$ i.e. the line $y=x$ as asymptotes.
A: The graph of this function is a true hyperbola, but let us say that a function $f(x)$ is "hyperbol-ish" if 


*

*There is a line $L: y=ax+b$ such that as $x\to \infty$ or $\to -\infty$, the function $f(x)$ is asymptotic to $L$, i.e.,
$$\lim_{x\to \pm \infty} f(x) - (ax+b)=0,$$
and

*There is a vertical line $L':x=c$ such that as $x\to c^+$ or as $x\to c^-$, the function $f(x)$ is asymptotic to $L'$, i.e.,
$$\lim_{x\to c^\pm} f(x)=\pm\infty.$$


Let us now show that $f(x)=x+\frac{1}{x}$ is hyperbol-ish. Let $L: y=x$. Then:
$$\lim_{x\to \pm\infty} f(x) - x=\lim_{x\to \pm\infty} x+\frac{1}{x}-x=\lim_{x\to \pm\infty} \frac{1}{x}=0.$$
Now, let $L': x=0$. Then:
$$\lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} x+\frac{1}{x}=\infty$$
and
$$\lim_{x\to 0^-} f(x)=\lim_{x\to 0^-} x+\frac{1}{x}=-\infty,$$
as desired.
