Represent and evaluate linear and quadratic functions using function notation

Ontario Grade 11 Mathematics

Characteristics of Functions

MCR3U – A1.2: Linear and Quadratic Functions in Function Notation

You already know how to write equations like

y = 2x + 1

or

y = x^2 - 3x + 4

from Grade 10. In this lesson, you will learn a more powerful way to write the same relationships — using function notation. Function notation, such as

f(x) = 2x + 1

, does everything

y

does, plus it lets you name the function, track inputs and outputs clearly, and communicate results efficiently. By the end of this lesson, you will be comfortable reading, writing, and evaluating linear and quadratic functions expressed in function notation.

What you will learn

Explain what function notation means and why it is used instead of y.
Write the equation of a linear or quadratic function using function notation.
Evaluate a function by substituting a value or expression into it.
Interpret the meaning of f(a) = b in context.

Prerequisite Bridge: Relations, Functions, and y = mx + b

A relation is any rule that connects input values to output values. A function is a special relation where every input has exactly one output. For example, if you put 3 into the rule 'double and add 1,' you always get 7 — never two different answers. That consistency is what makes something a function.

In Grade 10 you represented linear functions as

y = mx + b

, where

m

is the slope and

b

is the y-intercept, and quadratic functions as

y = ax^2 + bx + c

, where

a \neq 0

. The variable

y

told you the output, and

x

was the input. Function notation keeps that same idea but adds a label so you can refer to specific functions by name.

A function has exactly one output for every input.
Linear functions have the form $y = mx + b$ ; quadratic functions have the form $y = ax^2 + bx + c$ .
The input variable is usually $x$ ; the output is usually $y$ .

What Is Function Notation?

Function notation replaces

y

with a name followed by

(x)

. The most common name is

f

, read as 'f of x,' but any letter works —

g

,

h

,

C

,

A

, and so on. The letter inside the parentheses is the input variable.

So

y = 3x - 5

becomes

f(x) = 3x - 5

. Both equations describe exactly the same line. The difference is that

f(x) = 3x - 5

has a built-in name (

f

) that you can use when communicating results. For instance,

f(2)

means 'evaluate function

f

when the input is 2.' It does NOT mean

f

multiplied by

x

— the parentheses here signal substitution, not multiplication.

You can have more than one function in the same problem. Calling them

f

and

g

keeps them separate without confusion. For a quadratic,

y = -2x^2 + x + 6

becomes

g(x) = -2x^2 + x + 6

. The notation works the same way regardless of whether the function is linear or quadratic.

Either form can be used depending on the type of relationship you are representing.

f(x) = mx + b

$f(x)$ is read 'f of x' and means the output of function $f$ when the input is $x$ .
The parentheses in $f(x)$ signal substitution, not multiplication.
Any letter can name a function: $f$ , $g$ , $h$ , $C$ , $A$ , etc.
Function notation works identically for linear and quadratic functions.

Evaluating a Function: Substituting a Number

To evaluate

f(x)

at a specific number, replace every

x

in the rule with that number and simplify. This is called substitution. The result is a single number — the output.

For example, given

f(x) = 4x - 7

, evaluating at

x = 3

means computing

f(3) = 4(3) - 7 = 12 - 7 = 5

. You write this as

f(3) = 5

, which tells anyone reading your work exactly which function you used and what input produced the output 5.

The same process applies to quadratic functions. Given

g(x) = x^2 - 2x + 3

, evaluating at

x = -1

gives

g(-1) = (-1)^2 - 2(-1) + 3 = 1 + 2 + 3 = 6

. Always use brackets around a negative substitution value to avoid sign errors.

f(a) = 4(a) - 7

Replace every $x$ with the given value, then simplify step by step.
Use brackets around negative inputs: $(-1)^2 = 1$ , not $-1$ .
The answer $f(a) = b$ means input $a$ produces output $b$ .

Evaluating a Function at an Expression

Function notation is especially useful when the input is an expression rather than a single number. For example,

f(x + 1)

means replace every

x

in the rule with the entire expression

(x + 1)

.

Given

f(x) = 3x + 2

, finding

f(x + 1)

works like this: replace

x

with

(x + 1)

to get

f(x+1) = 3(x+1) + 2 = 3x + 3 + 2 = 3x + 5

. Notice the result is still a linear expression — you expanded and simplified using the distributive property from Grade 9.

For a quadratic,

h(x) = x^2 - x

, evaluating

h(2t)

means replace

x

with

2t

:

h(2t) = (2t)^2 - (2t) = 4t^2 - 2t

. This kind of evaluation appears when you investigate how a function behaves for a scaled input, and it requires careful use of exponent rules.

f(x+1) = 3(x+1) + 2

Treat the entire input expression as a single unit and substitute it everywhere $x$ appears.
Use brackets to avoid missing terms or sign errors during expansion.
Simplify completely using the distributive property and exponent rules.

Interpreting Function Notation in Context

Function notation is not just symbolic — it carries meaning. If

C(n) = 5n + 20

represents the cost in dollars of printing

n

posters, then

C(0) = 20

means it costs CAD 20 even before any posters are printed (a fixed setup fee).

C(10) = 5(10) + 20 = 70

means 10 posters cost CAD 70.

A statement like

C(n) = 95

is an equation you can solve:

5n + 20 = 95 \Rightarrow 5n = 75 \Rightarrow n = 15

. So 15 posters cost exactly CAD 95. Reading and writing statements in function notation this way makes the relationship between input and output explicit and easy to follow.

For a quadratic in context, if

h(t) = -5t^2 + 20t

models the height in metres of a ball

t

seconds after it is thrown, then

h(0) = 0

(ball starts at ground level),

h(2) = -5(4) + 20(2) = -20 + 40 = 20

(ball is 20 m high at 2 s), and

h(4) = -5(16) + 80 = 0

(ball returns to ground at 4 s).

$f(a) = b$ means the output is $b$ when the input is $a$ .
$f(a) = b$ can also be written as an equation to solve for an unknown input.
The function's name and variable should reflect what they represent in a real situation.

Function Notation at a Glance: Linear vs. Quadratic

Feature	Linear Example	Quadratic Example
Standard form	$f(x) = mx + b$	$g(x) = ax^2 + bx + c,\ a \neq 0$
Sample rule	$f(x) = 3x - 1$	$g(x) = 2x^2 - x + 4$
Evaluate at $x = 2$	$f(2) = 3(2)-1 = 5$	$g(2) = 2(4)-2+4 = 10$
Evaluate at $x = -3$	$f(-3) = 3(-3)-1 = -10$	$g(-3) = 2(9)-(-3)+4 = 25$
Evaluate at expression	$f(x+1) = 3x+2$	$g(2x) = 8x^2-2x+4$

Worked example

Evaluating a Linear Function at a Number and an Expression

Let

f(x) = 6x - 4

. Find: (a)

f(-2)

, (b)

f(0)

, (c)

f(x + 3)

.

Write the function rule
Start by clearly restating the rule so every substitution is visible.
$f(x) = 6x - 4$
Evaluate part (a): substitute $x = -2$
Replace every $x$ with $(-2)$ . Use brackets to protect the negative sign, especially in front of the coefficient 6.
$f(-2) = 6(-2) - 4 = -12 - 4 = -16$
Evaluate part (b): substitute $x = 0$
Replace $x$ with $0$ . Multiplying by 0 removes the $x$ -term, leaving only the constant.
$f(0) = 6(0) - 4 = 0 - 4 = -4$
Evaluate part (c): substitute the expression $(x + 3)$ for every $x$
Replace $x$ with the entire expression $(x + 3)$ . Distribute 6 across the bracket using the distributive property, then collect like terms.
$f(x+3) = 6(x+3) - 4 = 6x + 18 - 4 = 6x + 14$

Answer:

f(-2) = -16

,

f(0) = -4

,

f(x+3) = 6x + 14

.

Check: Verify

f(-2)

:

6(-2) - 4 = -12 - 4 = -16

, correct. Verify

f(x+3)

by testing

x = -2

:

f(-2+3) = f(1) = 6(1)-4 = 2

. Using the simplified expression:

6(-2)+14 = -12+14 = 2

, matches.

Worked example

Evaluating a Quadratic Function and Solving f(x) = k

Let

g(x) = -x^2 + 5x - 2

. Find: (a)

g(3)

, (b)

g(-1)

, (c) the value(s) of

x

such that

g(x) = 4

.

Write the function rule
Restate the quadratic rule to keep track of the negative leading coefficient.
$g(x) = -x^2 + 5x - 2$
Evaluate part (a): substitute $x = 3$
Replace $x$ with $3$ . Square first, then multiply by the coefficient $-1$ , then add the remaining terms.
$g(3) = -(3)^2 + 5(3) - 2 = -9 + 15 - 2 = 4$
Evaluate part (b): substitute $x = -1$
Replace $x$ with $(-1)$ . Squaring a negative gives a positive: $(-1)^2 = 1$ . Multiply by the leading coefficient $-1$ to get $-1$ , then continue.
$g(-1) = -(-1)^2 + 5(-1) - 2 = -1 - 5 - 2 = -8$
Set up the equation for part (c)
Writing $g(x) = 4$ means you need the input(s) that produce an output of 4. Replace $g(x)$ with the rule and set it equal to 4.
$-x^2 + 5x - 2 = 4$
Rearrange into standard form
Move all terms to one side so the equation equals 0. Add $x^2$ , subtract $5x$ , and add $2$ to both sides.
$x^2 - 5x + 6 = 0$
Factor the quadratic
Find two numbers that multiply to $6$ and add to $-5$ . Those numbers are $-2$ and $-3$ , so the factored form is $(x-2)(x-3)$ .
$(x - 2)(x - 3) = 0$
Apply the zero-product property and solve
If a product equals zero, at least one factor must equal zero. Set each factor equal to zero separately to find the two possible input values.
$x = 2, \ x = 3$

Answer:

g(3) = 4

,

g(-1) = -8

, and

g(x) = 4

when

x = 2

or

x = 3

.

Check: Verify

g(2) = -(4) + 10 - 2 = 4

, correct. Verify

g(3) = -(9) + 15 - 2 = 4

, correct. Notice that the answer to part (a) already told us

g(3) = 4

, which is consistent.

Common mistakes and how to avoid them

Reading

f(x)

as multiplication: '

f

times

x

.'

Correction:

f(x)

means 'function

f

evaluated at input

x

.' The parentheses signal substitution, not multiplication.

Forgetting to square the entire substituted value, e.g. writing

-x^2

at

x = -3

as

-(-3)^2 = 9

instead of

-9

.

Correction:

(-3)^2 = 9

, so

-(-3)^2 = -(9) = -9

. Always square first, then apply the negative coefficient.

When evaluating

f(x + 1)

, only replacing one occurrence of

x

and missing others.

Correction: Every single

x

in the rule must be replaced by

(x+1)

. Rewrite the full rule, then substitute.

Dropping brackets around a negative input, e.g. writing

3 \cdot -2

instead of

3(-2)

, which can cause sign errors.

Correction: Always use brackets:

3(-2) = -6

. This habit prevents errors in longer expressions.

Confusing

f(x) = k

(find the input) with

f(k)

(find the output).

Correction:

f(k)

asks you to substitute

k

and calculate the output.

f(x) = k

asks you to set up and solve an equation to find the input(s) that give output

k

.

Lesson summary

Function notation replaces $y$ with a named expression like $f(x)$ , making it easier to name functions, identify inputs and outputs, and communicate results.
A linear function in function notation looks like $f(x) = mx + b$ ; a quadratic looks like $g(x) = ax^2 + bx + c$ with $a \neq 0$ .
To evaluate $f(a)$ , replace every $x$ in the rule with $a$ (using brackets) and simplify step by step.
To evaluate $f(\text{expression})$ , replace every $x$ with the entire expression in brackets, then expand and simplify.
$f(a) = b$ states that input $a$ produces output $b$ ; setting $f(x) = k$ creates an equation you can solve for the input.
Always use brackets around negative or multi-term substitutions to avoid sign and expansion errors.

Check your understanding

Question 1

Given

f(x) = 5x - 3

, what is

f(-2)

?

-13
7
-7
13

Show answer and explanation

-13

f(-2) = 5(-2) - 3 = -10 - 3 = -13

. The key is to use brackets:

5 \times (-2) = -10

, then subtract 3.

Question 2

Which expression correctly represents

f(x + 2)

for

f(x) = 4x - 1

?

$4x + 1$
$4x + 7$
$4x - 1 + 2$
$4x^2 + 7$

Show answer and explanation

4x + 7

Replace every

x

with

(x+2)

:

f(x+2) = 4(x+2) - 1 = 4x + 8 - 1 = 4x + 7

. Option C forgets to distribute the 4; options A and D are expansion errors.

Question 3

Let

g(x) = -x^2 + 3x

. What is

g(-2)

?

$-10$
$10$
$-2$
$2$

Show answer and explanation

-10

g(-2) = -(-2)^2 + 3(-2) = -(4) + (-6) = -4 - 6 = -10

. A common error is computing

-(-2)^2

as

+4

instead of

-4

.

Question 4

For

h(x) = x^2 - 4x + 3

, which values of

x

satisfy

h(x) = 0

?

$x = 1$ and $x = 3$
$x = -1$ and $x = -3$
$x = 2$ only
$x = 4$ and $x = 0$

Show answer and explanation

x = 1

and

x = 3

Set

x^2 - 4x + 3 = 0

and factor:

(x-1)(x-3) = 0

, giving

x = 1

or

x = 3

. You can verify:

h(1) = 1 - 4 + 3 = 0

and

h(3) = 9 - 12 + 3 = 0

, both correct.

Key terms

Function: A relation where every input value produces exactly one output value.
Function notation: A way of writing a function using a name and an input variable, such as $f(x)$ , instead of just $y$ .
Input: The value you substitute into a function; also called the independent variable, usually represented by $x$ .
Output: The result you get after substituting the input into the function rule; also called the dependent variable, represented by $f(x)$ or $y$ .
Evaluate: To find the output of a function by substituting a specific value or expression for the input variable.
Linear function: A function whose rule has degree 1, producing a straight-line graph. In function notation: $f(x) = mx + b$ .
Quadratic function: A function whose rule has degree 2, producing a parabolic graph. In function notation: $g(x) = ax^2 + bx + c$ , where $a \neq 0$ .
Substitution: The process of replacing a variable with a specific number or expression and then simplifying the result.

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