Analyze reflections across the axes and the line y = x

Alberta Grade 12 Mathematics

Relations and Functions

Track how points, equations, domains, and ranges change

A reflection flips a graph across a line, like an image in a mirror. Each point moves to the other side of that line by the same distance. This lesson focuses on three mirror lines: the

x

-axis, the

y

-axis, and the line

y=x

. Coordinate rules make the changes predictable. They also help you write an equation for the reflected graph and check whether that graph is still a function.

What you will learn

Apply coordinate rules for reflections across the $x$ -axis, $y$ -axis, and line $y=x$ .
Write and check equations for reflected graphs.
Compare the domain and range of a graph with those of its reflection across $y=x$ .
Decide whether a reflected relation is a function.

Review: coordinates, functions, domain, and range

A point on a coordinate plane is written as an ordered pair, such as

(2,3)

. The first coordinate gives the horizontal position. The second gives the vertical position. A graph is a set of points that satisfy an equation.

A function assigns exactly one output to each allowed input. In the notation

y=f(x)

,

x

is the input and

y

is the output. For example, if

f(x)=2x-1

, then

f(2)=3

, so

(2,3)

is on the graph.

The domain is the set of

x

-values on a graph. The range is the set of

y

-values. A useful graph check is the vertical-line test: a graph is a function if every vertical line meets it no more than once. The horizontal-line test asks whether every horizontal line meets the graph no more than once. This second test helps when reflecting across

y=x

.

The coordinates of a point are written $(x,y)$ .
For $y=f(x)$ , the domain is the set of inputs and the range is the set of outputs.
A function passes the vertical-line test.

Coordinate rules for the three reflections

Start with a point

(x,y)

. Across the

x

-axis, its horizontal coordinate stays the same and its vertical coordinate changes sign. Across the

y

-axis, its vertical coordinate stays the same and its horizontal coordinate changes sign. Across

y=x

, the coordinates switch places.

For a graph written as

y=f(x)

, these point rules give equation rules. A reflection across the

x

-axis changes each output to its opposite, so the equation becomes

y=-f(x)

. A reflection across the

y

-axis changes each input to its opposite, so replace every input

x

with

-x

to get

y=f(-x)

.

For a reflection across

y=x

, switch the coordinates of every point. To find the reflected equation, interchange

x

and

y

in the original equation. Then solve for

y

if possible. The result may be a relation rather than a function.

To decide whether the reflection across

y=x

is a function, use the horizontal-line test on the original graph. If the original passes that test, its reflection passes the vertical-line test. If it fails the horizontal-line test, the reflected graph is not a function.

(x,y)\mapsto(x,-y),\quad (x,y)\mapsto(-x,y),\quad (x,y)\mapsto(y,x)

Across the $x$ -axis, change the sign of the $y$ -coordinate.
Across the $y$ -axis, change the sign of the $x$ -coordinate.
Across $y=x$ , switch the coordinates.
A reflection across $y=x$ is a function exactly when the original graph passes the horizontal-line test.

Reading the reflected graph

Consider the point

(2,3)

. Its reflection across the

x

-axis is

(2,-3)

. Its reflection across the

y

-axis is

(-2,3)

. Its reflection across

y=x

is

(3,2)

. These images show which coordinate changes and which stays fixed.

The equation rules match the point rules. In an

x

-axis reflection, the output changes sign. In a

y

-axis reflection, the input changes sign. In a reflection across

y=x

, the old output becomes the new input, and the old input becomes the new output.

A reflection across

y=x

swaps the domain and range. For example, if a graph has domain

[1,4]

and range

[-2,3]

, the reflected relation has domain

[-2,3]

and range

[1,4]

. The reflection does not guarantee that the new relation is a function.

Points on the mirror line do not move. A point on either axis stays fixed when reflected across that axis. A point on the line

y=x

has equal coordinates, so switching its coordinates leaves it unchanged. A reliable check is to choose a point on the original graph, apply the point rule, and test the image in the reflected equation.

A point check can confirm a reflected equation.
A reflection across $y=x$ exchanges domain and range.
Check the function condition before writing a reflected relation as $y=f(x)$ .

Worked example

Reflect a line across both axes

The line is

y=2x-1

. Find its reflections across the

x

-axis and the

y

-axis. Check each result using the point on the original line when

x=1

.

A line and its axis reflections

The graph shows only a viewing window of the three full lines.

Reflect across the x-axis
At $x=1$ , the original line has $y=1$ , so $(1,1)$ is on the line. The $x$ -axis rule keeps the first coordinate and changes the sign of the second. For the equation, change the sign of the whole output expression.
$y=-(2x-1)=-2x+1$
Reflect across the y-axis
The point $(1,1)$ becomes $(-1,1)$ across the $y$ -axis. For the equation, replace every input $x$ with $-x$ . This changes the horizontal position while leaving the output value associated with the reflected point unchanged.
$y=2(-x)-1=-2x-1$
Check both images
The first equation gives $y=-1$ at $x=1$ , matching the image $(1,-1)$ . The second gives $y=1$ at $x=-1$ , matching $(-1,1)$ . Both checks agree with the coordinate rules.
$-(2(1)-1)=-1,\qquad 2(-(-1))-1=1$

Answer: Across the

x

-axis, the line is

y=-2x+1

. Across the

y

-axis, it is

y=-2x-1

.

Check: The original point

(1,1)

becomes

(1,-1)

across the

x

-axis and

(-1,1)

across the

y

-axis.

Worked example

Reflect a line across $y=x$

Reflect the line

y=2x+3

across

y=x

. Write the reflected equation as

y=mx+b

, and state the domain and range of each line.

A line and its reflection across y = x

The graph shows only a viewing window of the two full lines.

Switch the variables
Reflection across $y=x$ switches the coordinates of each point. Apply that rule to the equation by interchanging $x$ and $y$ . The new equation describes the reflected relation.
$x=2y+3$
Solve for y
Subtract $3$ from both sides, then divide by $2$ . The resulting equation is in function form.
$y=\frac{x-3}{2}=\frac{1}{2}x-\frac{3}{2}$
Compare domain and range
The original line and the reflected line both extend without restriction in either coordinate, so both have all real numbers as their domain and range. Check a point: $(0,3)$ on the original becomes $(3,0)$ . The reflected equation gives $0$ when $x=3$ .
$\text{Domain}=\text{Range}=\mathbb{R}$

Answer: The reflected line is

y=\frac{1}{2}x-\frac{3}{2}

. Both lines have domain

\mathbb{R}

and range

\mathbb{R}

.

Check: The point

(0,3)

becomes

(3,0)

, and

(3,0)

satisfies the reflected equation.

Worked example

A reflection across $y=x$ that is not a function

Reflect

y=x^2

across

y=x

. Write the reflected relation and determine whether it is a function.

The parabola and its reflected relation

The graph shows only a viewing window of the full parabola and its reflection.

Switch the variables
Interchange $x$ and $y$ in the original equation. This gives an equation for the reflected relation. The equation can be left in this form even if it cannot be written as one output for every input.
$x=y^2$
Check whether the reflection is a function
The original graph fails the horizontal-line test: for example, the horizontal line $y=1$ meets it at two points. Therefore its reflection fails the vertical-line test. For the reflected relation, setting the input to $1$ gives two possible outputs. x=1\Longrightarrow y=1 or y=-1
Confirm with points
The original points $(1,1)$ and $(-1,1)$ reflect to $(1,1)$ and $(1,-1)$ . The reflected relation therefore has two outputs for the input $x=1$ , which confirms that it is not a function.
$(1,1),\;(-1,1)\mapsto(1,1),\;(1,-1)$

Answer: The reflected relation is

x=y^2

. It is not a function of

x

because the input

x=1

has two outputs,

y=1

and

y=-1

.

Check: The original graph fails the horizontal-line test, so the reflected graph fails the vertical-line test.

Common mistakes and how to avoid them

Using

-f(x)

for a reflection across the

y

-axis.

Correction: Use

f(-x)

for a

y

-axis reflection. The expression

-f(x)

describes a reflection across the

x

-axis.

Changing signs instead of switching coordinates for a reflection across

y=x

.

Correction: A point

(x,y)

becomes

(y,x)

. Neither coordinate must change sign.

Assuming every reflection across

y=x

is a function.

Correction: Test the original graph with the horizontal-line test. If it fails, the reflected relation fails the vertical-line test.

Switching variables in only part of an equation.

Correction: For a reflection across

y=x

, interchange

x

and

y

throughout the full equation.

Lesson summary

Across the $x$ -axis, $(x,y)$ becomes $(x,-y)$ ; for $y=f(x)$ , use $y=-f(x)$ .
Across the $y$ -axis, $(x,y)$ becomes $(-x,y)$ ; for $y=f(x)$ , use $y=f(-x)$ .
Across $y=x$ , $(x,y)$ becomes $(y,x)$ ; interchange variables in the equation.
A reflection across $y=x$ swaps domain and range.
The reflected relation is a function exactly when the original graph passes the horizontal-line test.

Check your understanding

Question 1

The point

(4,-2)

is reflected across the

x

-axis. What is its image?

$(4,2)$
$(-4,-2)$
$(-2,4)$
$(4,-2)$

Show answer and explanation

(4,2)

An

x

-axis reflection keeps the

x

-coordinate and changes the sign of the

y

-coordinate.

Question 2

What is the reflection of

y=x^2+1

across the

y

-axis?

$y=-x^2-1$
$y=(-x)^2+1$
$y=x^2-1$
$y=-(x^2+1)$

Show answer and explanation

y=(-x)^2+1

Replace the input

x

with

-x

. Since

(-x)^2=x^2

, the equation is unchanged after simplification.

Question 3

Reflect

y=3x-6

across

y=x

. Which equation is the reflection?

$y=3x+6$
$y=\frac{x+6}{3}$
$y=\frac{x-6}{3}$
$y=-3x+6$

Show answer and explanation

y=\frac{x+6}{3}

Switch the variables to get

x=3y-6

. Solving for

y

gives

y=\frac{x+6}{3}

.

Question 4

The graph

y=x^2

is reflected across

y=x

. Is the reflected relation a function of

x

?

Yes, because the original graph is a function.
Yes, because switching coordinates always gives a function.
No, because the original graph fails the horizontal-line test.
No, because the original graph fails the vertical-line test.

Show answer and explanation

No, because the original graph fails the horizontal-line test.

The original parabola fails the horizontal-line test. Its reflection therefore fails the vertical-line test and is not a function of

x

.

Key terms

Reflection: A flip across a line. Each point and its image are the same distance from the mirror line.
Domain: The set of $x$ -values on a graph.
Range: The set of $y$ -values on a graph.
Vertical-line test: A graph is a function if no vertical line meets it more than once.
Horizontal-line test: A graph passes this test if no horizontal line meets it more than once.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation RF5. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

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RF5 · Analyze reflections across the axes and the line y = x

Alberta Grade 12 Mathematics

Track how points, equations, domains, and ranges change

What you will learn

Review: coordinates, functions, domain, and range

Coordinate rules for the three reflections

Reading the reflected graph

Reflect a line across both axes

Reflect a line across $y=x$

A reflection across $y=x$ that is not a function

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through Math 30-1

About this lesson and its review