2.8 Reciprocal and simple rational functionsIB Maths: Analysis and Approaches SL: Revision notes

The reciprocal function is $f(x) = \frac{1}{x}$, $x \ne 0$. Its graph is a rectangular hyperbola with two branches, in the first and third quadrants.

• Vertical asymptote $x = 0$: as $x \to 0$, $|f(x)|$ grows without bound.
• Horizontal asymptote $y = 0$: as $x \to \pm\infty$, $f(x) \to 0$.
• No intercepts with either axis. Domain and range are both $\mathbb{R}$ excluding 0.
• Symmetric in the lines $y = x$ and $y = -x$.

Put $y = \frac{1}{x}$ and interchange: $x = \frac{1}{y}$, so $y = \frac{1}{x}$ again. So $f^{-1}(x) = f(x)$: the function is self-inverse, and $(f \circ f)(x) = x$.

Graphically, the inverse is the reflection in $y = x$; since the graph of $\frac{1}{x}$ is symmetric in $y = x$, reflecting it gives the same graph. If $(a, b)$ is on the graph then so is $(b, a)$: $\left(\frac{1}{4}, 4\right)$ and $\left(4, \frac{1}{4}\right)$.

For $f(x) = \frac{ax + b}{cx + d}$, $c \ne 0$:

• Vertical asymptote where the denominator is zero: $x = -\frac{d}{c}$.
• Horizontal asymptote $y = \frac{a}{c}$, because for large $|x|$ the constants $b$ and $d$ become negligible and $f(x) \approx \frac{ax}{cx}$.

Example: $f(x) = \frac{3x - 6}{x + 2}$ has asymptotes $x = -2$ and $y = 3$. Domain $x \ne -2$, range $y \ne 3$.

Rewriting helps: $\frac{60t + 20}{t + 2} = 60 - \frac{100}{t + 2}$, which is $\frac{1}{x}$ stretched, reflected and translated — so the graph has the same hyperbola shape.

• $y$-intercept: put $x = 0$, giving $\left(0, \frac{b}{d}\right)$ (if $d \ne 0$).
• $x$-intercept: the numerator is zero, $ax + b = 0$, giving $\left(-\frac{b}{a}, 0\right)$ (if $a \ne 0$).

When a question describes a graph without a picture, list the asymptotes, the intercepts, the domain and the range. For $\frac{-4x + 1}{2x - 6}$: asymptotes $x = 3$, $y = -2$; intercepts $\left(\frac{1}{4}, 0\right)$ and $\left(0, -\frac{1}{6}\right)$; domain $x \ne 3$; range $y \ne -2$.

To find the inverse of $\frac{ax + b}{cx + d}$, interchange $x$ and $y$, multiply out the fraction and collect the $y$ terms: $x(cy + d) = ay + b \Rightarrow y(cx - a) = b - dx$, so $f^{-1}(x) = \frac{-dx + b}{cx - a}$.

Because the inverse is a reflection in $y = x$, the asymptotes swap: $x = -\frac{d}{c}$ becomes $y = -\frac{d}{c}$, and $y = \frac{a}{c}$ becomes $x = \frac{a}{c}$. The domain of $f^{-1}$ is the range of $f$.

In a model, a horizontal asymptote is a limiting value. If $W(t) = \frac{60t + 20}{t + 2}$ is a typing speed after $t$ hours of practice, then $W \to 60$ as $t \to \infty$: the model predicts the speed approaches, but never reaches, 60 words per minute. Also check the domain makes sense: $t \ge 0$, so the vertical asymptote $t = -2$ lies outside the model and has no meaning in context.