A1.7 Determine algebraic representations of inverse linear and quadratic relations

Ontario Grade 11 Mathematics

Characteristics of Functions

MCR3U – A1.7 | Swapping Inputs and Outputs

Every relation takes an input and produces an output. The inverse of a relation simply reverses that process — it takes the original output and gives back the original input. In this lesson you will learn how to find the algebraic rule for the inverse of a linear relation and a quadratic relation. You will also see why restricting the domain of a quadratic relation sometimes becomes necessary. By the end, you will be able to write the equation of an inverse relation from scratch and verify your answer.

What you will learn

Find the algebraic inverse of a linear relation by swapping variables and solving for y.
Find the algebraic inverse of a quadratic relation by swapping variables and solving for y.
Describe how the domain of a quadratic relation can be restricted so that its inverse is also a function.
Verify an inverse relation by confirming that it reverses the output back to the original input.

Prerequisite Bridge: Relations, Functions, and Notation

A relation is any pairing of input values with output values. A function is a special relation where every input has exactly one output. We write

y = f(x)

to say that

y

is the output produced by the function

f

when the input is

x

.

Two skills from Grade 10 are needed here. First, solving a linear equation for one variable in terms of another — for example, rearranging

2x + y = 5

to get

y = 5 - 2x

. Second, solving a quadratic equation using the quadratic formula or by isolating a squared term. Both skills are used directly when finding an inverse.

A relation pairs inputs with outputs.
A function gives exactly one output for every input.
Solving for a variable means isolating it on one side of an equation.

What an Inverse Relation Does — and How to Find It

The inverse of a relation swaps every pair of coordinates. If a point

(a, b)

is on the original relation, then the point

(b, a)

is on the inverse. In terms of equations, the input and output simply trade places.

This swap leads directly to an algebraic method. To find the inverse of any relation written as an equation in

x

and

y

, follow two steps: swap

x

and

y

in the equation, then solve the new equation for

y

. The result is the algebraic representation of the inverse relation.

A useful way to picture this: the original relation asks 'given

x

, what is

y

?' The inverse asks 'given

y

, what is

x

?' Swapping the letters in the equation is exactly what converts the first question into the second.

x \leftrightarrow y

Swapping every $(x, y)$ pair to $(y, x)$ produces the inverse relation.
Algebraic method: replace $x$ with $y$ and $y$ with $x$ , then solve for $y$ .
The inverse reverses the original input-output relationship.

Inverse of a Linear Relation

A linear relation has the form

y = mx + b

, where

m \neq 0

is the slope and

b

is the

y

-intercept. Because the equation is already solved for

y

, swapping

x

and

y

gives

x = my + b

. Solving for

y

then produces the inverse.

Start from

x = my + b

. Subtract

b

from both sides to get

x - b = my

. Divide both sides by

m

to isolate

y

, giving

y = \frac{x - b}{m}

. This is also a linear equation, so the inverse of a linear relation is always a linear relation.

Notice that the slope of the inverse is

\frac{1}{m}

and the

y

-intercept is

-\frac{b}{m}

. You do not need to memorize those formulas — they follow automatically whenever you carry out the two steps.

y = \frac{x - b}{m}

Swap $x$ and $y$ , then solve for $y$ .
The inverse of a linear relation is itself linear.
The slope of the inverse is the reciprocal of the original slope.

Inverse of a Quadratic Relation — and Domain Restriction

A quadratic relation has the form

y = a(x - h)^2 + k

, written in vertex form. When you swap

x

and

y

, you get

x = a(y - h)^2 + k

. Solving for

y

requires isolating the squared term and then taking a square root.

Subtract

k

:

x - k = a(y - h)^2

. Divide by

a

:

\frac{x-k}{a} = (y-h)^2

. Take the square root of both sides:

y - h = \pm\sqrt{\frac{x-k}{a}}

, so

y = h \pm\sqrt{\frac{x-k}{a}}

. The

\pm

sign means two

y

-values come from one

x

-value, which means the inverse is not a function — it is a relation with two branches.

To make the inverse a function, restrict the domain of the original relation to one side of the vertex. If you keep only

x \geq h

(the right branch of the parabola), the inverse uses only the positive square root:

y = h + \sqrt{\frac{x-k}{a}}

. If you keep only

x \leq h

(the left branch), the inverse uses only the negative square root:

y = h - \sqrt{\frac{x-k}{a}}

.

This is why domain restriction matters: without it, the inverse of a quadratic relation is not a function, and the

\pm

in the answer is the algebraic signal telling you so.

y = h ±\sqrt{\frac{x - k}{a}}

Swap $x$ and $y$ , isolate the squared term, then take the square root.
The $\pm$ sign shows that the inverse of a full parabola is not a function.
Restricting the domain of the original relation to one side of the vertex makes the inverse a function.
Choose $+$ for the right branch ( $x \geq h$ ) and $-$ for the left branch ( $x \leq h$ ).

Verifying an Inverse

After finding an inverse, you can check your work by picking a specific point. Choose any

x

-value, compute the

y

-value using the original relation, then use that

y

-value as the input to the inverse. If the inverse returns your starting

x

-value, the inverse is correct.

For example, if the original relation gives the output

7

when the input is

3

, then the inverse must give the output

3

when the input is

7

. If it does not, an algebraic error was made somewhere in the swap-and-solve process.

Pick a point on the original relation and confirm the inverse maps it back.
A correct inverse always reverses the original input-output pairing.

Comparing a Relation and Its Inverse: Input-Output Table for y = 3x − 6

Original Input (x)	Original Output (y)	Inverse Input (x)	Inverse Output (y)
0	−6	−6	0
2	0	0	2
4	6	6	4
6	12	12	6

Worked example

Inverse of a Linear Relation

Find the algebraic representation of the inverse of the relation

y = 3x - 6

. Then verify your answer using the point where

x = 4

.

Identify the original relation
Write down the relation as given: $y = 3x - 6$ . The goal is to find the inverse by swapping inputs and outputs.
$y = 3x - 6$
Swap x and y
Replace every $x$ with $y$ and every $y$ with $x$ . This one substitution encodes the idea that the inverse trades inputs for outputs.
$x = 3y - 6$
Isolate the y-term
Add $6$ to both sides to move the constant away from the $y$ -term.
$x + 6 = 3y$
Solve for y
Divide both sides by $3$ to get $y$ alone. This is the algebraic representation of the inverse relation.
$y = \frac{x + 6}{3}$
Verify with x = 4
Substitute $x = 4$ into the original relation: $y = 3(4) - 6 = 12 - 6 = 6$ . So the original maps 4 \to 6. Now substitute $x = 6$ into the inverse: $y = \frac{6 + 6}{3} = \frac{12}{3} = 4$ . The inverse maps 6 \to 4, which confirms the answer is correct.

Answer: The inverse relation is

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

.

Check: Original:

x=4 \Rightarrow y=6

. Inverse:

x=6 \Rightarrow y=4

. ✓ The inverse correctly reverses the pairing.

Worked example

Inverse of a Quadratic Relation with Domain Restriction

The relation

y = (x - 2)^2 + 1

has its domain restricted to

x \geq 2

. Find the algebraic representation of its inverse and state its domain.

Write the original relation
The relation is in vertex form with vertex at $(2, 1)$ . The domain is restricted to $x \geq 2$ , which means only the right branch of the parabola is used.
$y = (x-2)^2 + 1, x ≥ 2$
Swap x and y
Replace every $x$ with $y$ and every $y$ with $x$ in the equation.
$x = (y - 2)^2 + 1$
Isolate the squared term
Subtract $1$ from both sides so that the squared expression is alone on the right.
$x - 1 = (y - 2)^2$
Take the square root of both sides
Taking the square root introduces a $\pm$ sign in general. However, because the domain of the original relation was restricted to $x \geq 2$ , the corresponding outputs $y$ of the inverse satisfy $y \geq 2$ . This means $y - 2 \geq 0$ , so only the positive square root applies.
$y - 2 = +\sqrt{x - 1}$
Solve for y
Add $2$ to both sides. The expression under the square root requires $x - 1 \geq 0$ , which means $x \geq 1$ . Since the vertex of the original parabola was at $y = 1$ , the smallest output value was $1$ , which becomes the smallest input value of the inverse.
$y = 2 + \sqrt{x - 1}$
Verify with a specific point
Choose $x = 5$ in the original: $y = (5-2)^2 + 1 = 9 + 1 = 10$ . Now substitute $x = 10$ into the inverse: $y = 2 + \sqrt{10 - 1} = 2 + \sqrt{9} = 2 + 3 = 5$ . The inverse maps 10 \to 5, which correctly reverses the original mapping 5 \to 10.

Answer: The inverse relation is

y = 2 + \sqrt{x-1}

, with domain

x \geq 1

.

Check: Original:

x=5 \Rightarrow y=10

. Inverse:

x=10 \Rightarrow y=5

. ✓

Common mistakes and how to avoid them

Forgetting to swap both x and y — for example, only replacing x with y but leaving the other variable unchanged.

Correction: Every x in the equation becomes y, and every y becomes x. Check both sides of the equation before solving.

Writing

\pm

in the final answer for a quadratic inverse without checking whether a domain restriction applies.

Correction: If the domain of the original quadratic has been restricted to one side of the vertex, use only the matching square root sign — positive for the right branch, negative for the left branch.

Dividing by the coefficient before subtracting the constant, which produces an incorrect result when isolating the squared term.

Correction: Always subtract the constant term first to isolate the squared expression, then divide, then take the square root.

Confusing the domain of the inverse with the domain of the original relation.

Correction: The domain of the inverse equals the range of the original relation, and the range of the inverse equals the domain of the original. Check by identifying what output values the original can produce — those become the allowed inputs of the inverse.

Claiming the inverse of every relation is a function.

Correction: The inverse of a quadratic relation without domain restriction is not a function because one input maps to two outputs. State the domain restriction clearly when it is needed.

Lesson summary

To find the inverse of any relation, swap x and y in the equation, then solve the new equation for y.
The inverse of a linear relation is always linear; its slope is the reciprocal of the original slope.
The inverse of a quadratic relation in vertex form involves a square root, and the ± sign shows that the inverse is generally not a function.
Restricting the domain of the original quadratic to one side of the vertex selects one square root branch and makes the inverse a function.
The domain of the inverse equals the range of the original relation.
Always verify an inverse by checking that it maps a known output of the original back to the corresponding input.

Check your understanding

Question 1

What is the correct first step when finding the algebraic inverse of the relation

y = 5x + 2

?

Subtract 2 from both sides to get $y - 2 = 5x$ .
Replace x with y and y with x to get $x = 5y + 2$ .
Divide both sides by 5 to get $\frac{y}{5} = x + \frac{2}{5}$ .
Set $y = 0$ and solve for x.

Show answer and explanation

Replace x with y and y with x to get

x = 5y + 2

.

The defining first step of the inverse process is swapping x and y. Any algebraic rearrangement done before the swap is just simplifying the original relation, not starting the inverse.

Question 2

What is the inverse of the relation

y = 4x - 8

?

$y = 4x + 8$
$y = \frac{x + 8}{4}$
$y = \frac{x - 8}{4}$
$y = -4x + 8$

Show answer and explanation

y = \frac{x + 8}{4}

Swap to get

x = 4y - 8

. Add 8:

x + 8 = 4y

. Divide by 4:

y = \frac{x+8}{4}

. Subtracting 8 (option C) instead of adding is the most common error here.

Question 3

The relation

y = (x + 1)^2 - 3

has domain

x \geq -1

. Which equation represents its inverse?

$y = -1 + \sqrt{x + 3}$
$y = -1 - \sqrt{x + 3}$
$y = 1 + \sqrt{x - 3}$
$y = \pm\sqrt{x + 3} - 1$

Show answer and explanation

y = -1 + \sqrt{x + 3}

Swap to get

x = (y+1)^2 - 3

. Add 3:

x + 3 = (y+1)^2

. Since the domain is

x \geq -1

, we have

y + 1 \geq 0

, so take the positive root:

y + 1 = \sqrt{x+3}

, giving

y = -1 + \sqrt{x+3}

.

Question 4

A student finds the inverse of

y = (x - 3)^2 + 5

(with no domain restriction) and writes

y = 3 + \sqrt{x - 5}

as the final answer. What is wrong with this answer?

The student should have added 5 instead of subtracting 5 under the square root.
Without a domain restriction on the original relation, the inverse requires a ± sign: $y = 3 \pm \sqrt{x-5}$ .
The student forgot to swap x and y before solving.
The inverse of a quadratic is always linear, not a square root.

Show answer and explanation

Without a domain restriction on the original relation, the inverse requires a ± sign:

y = 3 \pm \sqrt{x-5}

.

Because no domain restriction was stated, both branches of the parabola are included. Taking the square root requires ± to represent both branches. Writing only the positive root discards the left branch of the inverse.

Key terms

Relation: Any set of pairings between input values and output values, written as an equation, a table, or a set of ordered pairs.
Inverse relation: The relation formed by swapping every input-output pair of the original relation, so that $(a, b)$ becomes $(b, a)$ .
Domain: The set of all allowed input values (x-values) for a relation.
Range: The set of all output values (y-values) produced by a relation.
Vertex form: A way of writing a quadratic relation as $y = a(x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Domain restriction: Limiting the domain of a relation to a specific interval so that, for example, its inverse becomes a function.
Square root: The operation that reverses squaring; $\sqrt{n}$ is the non-negative number whose square equals $n$ .
± (plus or minus): A symbol indicating two possible values — one with addition and one with subtraction — arising here when taking the square root of both sides of an equation.

Continue through MCR3U

View the complete Ontario Grade 11 Mathematics learning path

About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation A1.7. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

A1.7 · Determine algebraic representations of inverse linear and quadratic relations

Ontario Grade 11 Mathematics

MCR3U – A1.7 | Swapping Inputs and Outputs

What you will learn

Prerequisite Bridge: Relations, Functions, and Notation

What an Inverse Relation Does — and How to Find It

Inverse of a Linear Relation

Inverse of a Quadratic Relation — and Domain Restriction

Verifying an Inverse

Comparing a Relation and Its Inverse: Input-Output Table for y = 3x − 6

Inverse of a Linear Relation

Inverse of a Quadratic Relation with Domain Restriction

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson