Learn9MA0Paper 2 · Pure Mathematics 2Algebra and functionsGraphs of a function and its inverse

Algebra and functions · Subtopic 15 of 16

Graphs of a Function and its Inverse

Why the graph of $f^{-1}$ is the reflection of $f$ in the line $y=x$ , how points, intercepts and asymptotes swap over, and where a function meets its own inverse.

Cambridge 9709 Paper 1Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2≈ 30 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Sketch $y=f^{-1}(x)$ by reflecting $y=f(x)$ in the line $y=x$ .
Swap the coordinates of key points, intercepts and asymptotes.
Find where the graphs of $f$ and $f^{-1}$ meet.
Recognise self-inverse functions from their symmetry.

Pause and think

Warm-up: $f(2)=7$ . What is $f^{-1}(7)$ ? Which point is on $y=f(x)$ , and which is on $y=f^{-1}(x)$ ?

Section 2 of 14

Core idea

Start hereA mirror on the diagonal

If $f(2)=5$ , then $f^{-1}(5)=2$ . So the point $(2,5)$ is on $y=f(x)$ and the point $(5,2)$ is on $y=f^{-1}(x)$ : the coordinates are swapped. Swapping $x$ and $y$ for every point is exactly what a mirror placed along the line $y=x$ does. So you can draw an inverse without any algebra: just reflect.

The big idea

The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$ . Where $f$ and $f^{-1}$ meet (if at all) is usually on $y=x$ , so solving $f(x)=x$ finds those points.

1. Why reflection swaps coordinates

A point and its reflection

x on f = 0.0

The joining line always crosses

y=x

at right angles, halfway between the two points.

Reflecting

(a,b)

y=x

gives

(b,a)

$\text{midpoint of }(a,b),(b,a)=\left(\tfrac{a+b}2,\tfrac{a+b}2\right)$
Its coordinates are equal, so it lies on $y=x$ .
$\text{gradient}=\frac{a-b}{b-a}=-1$
The segment joining them has gradient $-1$ .
$(-1)\times1=-1$
$y=x$ has gradient $1$ : the segment is perpendicular to it.

y=x

is the perpendicular bisector of the segment, so

(b,a)

is the mirror image of

(a,b)

2. Sketching an inverse

f(x) = x²/4 + 1 (x ≥ 0) and its inverse

(4,5)

f

becomes

(5,4)

f^{-1}

. The dashed segment joining them crosses

y=x

at right angles: that is what a reflection looks like.

Worked example $f(x)=\tfrac14x^2+1$ , $x\ge0$

Sketch $f$ and $f^{-1}$ on the same axes; find $f^{-1}$ and any intersection.

Goal: Graphs, formula and meeting point.
Strategy: Reflect; for the meeting point solve $f(x)=x$ .

$f$ starts at $(0,1)$ ; $f^{-1}$ starts at $(1,0)$ (swap).
$y=\tfrac14x^2+1\Rightarrow x^2=4(y-1)\Rightarrow f^{-1}(x)=2\sqrt{x-1},\ x\ge1$
Positive root since $x\ge0$ .
$\tfrac14x^2+1=x\Rightarrow x^2-4x+4=0\Rightarrow(x-2)^2=0$
Meeting on $y=x$ .

Answer

f^{-1}(x)=2\sqrt{x-1}

; the graphs touch at

(2,2)

✓ Check

f(2)=1+1=2

and

f^{-1}(2)=2\sqrt1=2

✓.

Edge caseUse equal scales The reflection only looks right if both axes use the same scale; otherwise

y=x

isn't at

45^\circ

Pause and think

$f$ passes through $(0,3)$ and $(4,7)$ . Name two points on $f^{-1}$ .

Summary

$f^{-1}$ is $f$ reflected in $y=x$ .
Points swap: $(a,b)\to(b,a)$ .
Solve $f(x)=x$ for meeting points on $y=x$ .

Reflection in y = x: Swaps x- and y-coordinates.

Every point swaps over

Loading the interactive graph…

Try this

Drag $p$ . The point on $f$ and the point on $f^{-1}$ always have their coordinates swapped.
The two points are mirror images in $y=x$ . Fold the screen along that line in your head: the curves land on each other.

A $y$ -intercept $(0,c)$ of $f$ becomes an $x$ -intercept $(c,0)$ of $f^{-1}$ , and the other way round.
A horizontal asymptote $y=k$ of $f$ becomes a vertical asymptote $x=k$ of $f^{-1}$ .
The domain and range swap, so a curve that starts at $(2,1)$ has an inverse that starts at $(1,2)$ .

HistoryGraphs are younger than you might think. Drawing equations as curves on two axes became standard after René Descartes published La Géométrie in 1637. Before that, algebra and geometry were separate subjects. Reflecting a graph in

y=x

is a purely Cartesian idea: it simply swaps the two coordinates.

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