Reciprocal functions and inverse functions (QCE Math Methods Unit 1)

Syllabus dot point

“Sketch reciprocal functions of the form $y = \dfrac{a}{x - h} + k$ , identify asymptotes, domain and range, and find the inverse of a one-to-one function”

7 min readUpdated 7 Oct 2026

Quick answer

A reciprocal function $y=\dfrac{a}{x-h}+k$ has a vertical asymptote at $x=h$ and a horizontal asymptote at $y=k$ . A function has an inverse function only if it is one-to-one, and the inverse is found by swapping $x$ and $y$ and solving. The graph of the inverse is the reflection in $y=x$ , and the domain and range swap.

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What this dot point is asking
Reciprocal functions
One-to-one functions and inverses
Practice questions

What this dot point is asking

In the 2025 QCAA Mathematical Methods syllabus, Unit 1 Topic 3 is titled Functions and relations. Two skills from this topic are working with graphs of reciprocal functions and finding inverse functions. The official syllabus is at qcaa.qld.edu.au.

Reciprocal functions

The basic graph $y=\dfrac{1}{x}$ has two branches, with $x=0$ and $y=0$ as asymptotes. The domain is $x\ne 0$ and the range is $y\ne 0$ .

Transformations move the asymptotes. For $y=\dfrac{a}{x-h}+k$ :

the vertical asymptote is $x=h$
the horizontal asymptote is $y=k$
the domain is $x\ne h$ and the range is $y\ne k$
a positive $a$ puts the branches in the top right and bottom left of the asymptote cross, and a negative $a$ reflects them

Sketch

y=\dfrac{1}{x-2}+1

The vertical asymptote is $x=2$ and the horizontal asymptote is $y=1$ .

For the $y$ -intercept set $x=0$ : $y=\dfrac{1}{-2}+1=\dfrac12$ .

For the $x$ -intercept set $y=0$ : $\dfrac{1}{x-2}=-1$ , so $x-2=-1$ and $x=1$ .

The graph passes through $(1,0)$ and $\left(0,\tfrac12\right)$ , and both points lie on the left branch. The right branch lies above $y=1$ for $x>2$ . The domain is $x\ne 2$ and the range is $y\ne 1$ .

One-to-one functions and inverses

A function is one-to-one if each output comes from exactly one input. The horizontal line test is a quick check. If any horizontal line cuts the graph more than once, the function is not one-to-one and has no inverse on that domain.

For example, $y=x^2$ fails the test on all real numbers, but it passes if the domain is restricted to $x\ge 0$ .

To find the inverse $f^{-1}$ :

Write $y=f(x)$ .
Swap $x$ and $y$ .
Solve for $y$ .
State the domain of $f^{-1}$ , which is the range of $f$ , and its range, which is the domain of $f$ .

The graphs of $f$ and $f^{-1}$ are reflections of each other in the line $y=x$ , and $f(f^{-1}(x))=x$ .

Inverse of

f(x)=x^2-4

for

x\ge 0

Write $y=x^2-4$ with $x\ge 0$ . The range of $f$ is $y\ge -4$ .

Swap to get $x=y^2-4$ , so $y^2=x+4$ . Because the original domain was $x\ge 0$ , the inverse takes the positive root.

f^{-1}(x)=\sqrt{x+4},\quad x\ge -4.

Check: $f(3)=5$ and $f^{-1}(5)=\sqrt{9}=3$ .

Inverse of

f(x)=\dfrac{1}{x-2}+1

Write $y=\dfrac{1}{x-2}+1$ and swap: $x=\dfrac{1}{y-2}+1$ . Then $x-1=\dfrac{1}{y-2}$ , so $y-2=\dfrac{1}{x-1}$ .

f^{-1}(x)=2+\frac{1}{x-1}=\frac{2x-1}{x-1},\quad x\ne 1.

The domain of $f^{-1}$ is $x\ne 1$ , which is the range of $f$ . Check: $f(3)=2$ and $f^{-1}(2)=3$ .

Common trap

Do not write $f^{-1}(x)$ as $\dfrac{1}{f(x)}$ . The inverse function is not the reciprocal. Also, remember the domain restriction when you find an inverse from a quadratic, and state the domain of the inverse, because an answer without a domain is incomplete.

Practice questions

1. State the asymptotes, domain and range of $y=\dfrac{3}{x+1}-2$ .

Model answer: The vertical asymptote is $x=-1$ and the horizontal asymptote is $y=-2$ . The domain is $x\ne -1$ and the range is $y\ne -2$ .

2. Find the inverse of $f(x)=2x+3$ and state its domain.

Model answer: Swap to get $x=2y+3$ , so $y=\dfrac{x-3}{2}$ . The domain of $f^{-1}$ is all real numbers.

3. Explain why $f(x)=x^2$ has no inverse function on $\mathbb{R}$ , and give a domain on which it does.

Model answer: A horizontal line such as $y=4$ cuts the graph at $x=2$ and $x=-2$ , so the function is not one-to-one. Restricting the domain to $x\ge 0$ makes it one-to-one, and the inverse is $f^{-1}(x)=\sqrt{x}$ for $x\ge 0$ .

4. The graph of $f$ passes through $(3,5)$ . State one point on the graph of $f^{-1}$ .

Model answer: Reflection in $y=x$ swaps coordinates, so $(5,3)$ lies on the graph of $f^{-1}$ .

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA