Reciprocal and linear rational functions

\(\mathrm{f}(x)=\dfrac1x\) has domain and range \(\mathbb R\setminus\{0\}\), asymptotes \(x=0\), \(y=0\), and \(\mathrm{f}^{-1}=\mathrm{f}\).

For \(\mathrm{f}(x)=\dfrac{ax+b}{cx+d}\), \(c\ne0\) and \(ad-bc\ne0\), the vertical asymptote is \(x=-\dfrac dc\) and horizontal asymptote is \(y=\dfrac ac\). Find intercepts by setting the numerator or \(x\) to zero, provided the original denominator is nonzero.

Example

\(\dfrac{2x+3}{x+1}=2+\dfrac1{x+1}\): asymptotes \(x=-1\), \(y=2\); intercepts \(\left(-\dfrac32,0\right)\) and \((0,3)\).

Check cancellation before asserting a vertical asymptote. \(\dfrac{2x+2}{x+1}=2\), \(x\ne-1\), has a missing point at \((-1,2)\), not a vertical asymptote.

From the formula to the branches

Write a nonconstant linear rational function as \(\mathrm f(x)=h+\dfrac{k}{x-v}\), where \(k\ne0\). Its centre is the intersection \((v,h)\) of the asymptotes. If \(k>0\), the right branch lies above \(y=h\) and the left below; if \(k<0\), these positions reverse.

As \(x\) approaches \(v\), inspect the sign of \(k/(x-v)\) on each side to choose the infinite direction. As \(|x|\) becomes large, that fraction approaches \(0\). Mark the intercepts before drawing the branches; do not connect across \(x=v\).

On its unrestricted real domain, \(h+\dfrac{k}{x-v}\) has range \(\mathbb R\setminus\{h\}\): solving \(y=h+\dfrac{k}{x-v}\) gives \(x=v+\dfrac{k}{y-h}\) for every \(y\ne h\), while \(y=h\) is impossible. Recheck this range if the input domain is restricted.