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Module 1 · L2 of 15 ~40 min ⚡ +50 XP in Learn · +25 to complete

Function Notation & Evaluation

How does a taxi meter know what to charge? It follows a simple rule: a fixed cost plus a rate for every kilometre travelled. In mathematics, we write this rule using function notation — and it opens the door to everything from economics to engineering.

Today's hook — A taxi charges a $3 flag fall plus $2 for every kilometre travelled. How much would a 5 km trip cost? How much would a 10 km trip cost? Can you write a general rule using $C$ for cost and $d$ for distance?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

A taxi charges a $3 flag fall plus $2 for every kilometre travelled. How much would a 5 km trip cost? How much would a 10 km trip cost? Can you write a general rule using $C$ for cost and $d$ for distance?

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02
Formula reference · this lesson
core notation

$$f(a) \quad \text{replace every } x \text{ with } a$$

$f(a)$  = replace every $x$ in the rule with $a$, then simplify
$\dfrac{f(a+h) - f(a)}{h}$ = difference quotient — average rate of change from $a$ to $a+h$

Key insight: when substituting algebraic expressions, always use brackets. This prevents sign errors.

03
What you'll master
Know

Key facts

  • how to evaluate $f(a)$ for numerical and algebraic inputs
  • the meaning of the difference quotient
  • how to interpret function notation in real-world contexts
Understand

Concepts

  • that $f(x)$ describes a rule, not a multiplication
  • why brackets are essential when substituting negatives or algebraic terms
  • how functions model relationships like cost, distance, and temperature
Can do

Skills

  • evaluate functions for numerical inputs, negative inputs, and zero
  • evaluate functions for algebraic inputs such as $f(x+h)$ and $f(a)$
  • set up and interpret functions in real-world contexts
  • simplify and evaluate the difference quotient for linear and quadratic functions
04
Key terms
Evaluate
Find the output of a function for a given input by substituting and simplifying.
Brackets (when substituting)
Essential when substituting negatives or algebraic terms to prevent sign errors.
Function rule
The recipe: replace every instance of the independent variable with the given value, then simplify.
Difference quotient
$\dfrac{f(a+h)-f(a)}{h}$ — measures the average rate of change of $f$ from $a$ to $a+h$.
Context function
A function that models a real-world situation, e.g. $C(d) = 3 + 2d$ for taxi cost.
Negative inputs
A common source of errors; always write brackets around negative substituted values.
05
Evaluating functions
core concept · +3 XP at end

To evaluate a function means to find the output for a given input. The process is always the same: replace every instance of the independent variable with the given value, then simplify using the correct order of operations.

Think of it like following a recipe. The function rule is the recipe, and the input is the ingredient you're using. If the recipe says "double the amount and subtract 3," then with an input of 5 you get $2(5) - 3 = 7$. The recipe doesn't change — only the ingredient does.

Suppose $f(x) = x^2 - 3x + 2$. To find $f(4)$:

  • Replace $x$ with $4$: $f(4) = (4)^2 - 3(4) + 2$
  • Simplify: $16 - 12 + 2 = 6$

Negative inputs are a common source of errors. Always use brackets:

  • $f(-2) = (-2)^2 - 3(-2) + 2$
  • Simplify: $4 + 6 + 2 = 12$

Without brackets, $(-2)^2$ can easily become $-2^2 = -4$, which is incorrect. The bracket protects the sign.

Functions can also be evaluated for algebraic expressions. If $f(x) = 2x + 1$, then:

  • $f(a) = 2a + 1$
  • $f(x+h) = 2(x+h) + 1 = 2x + 2h + 1$

Again, brackets are your best defence against mistakes. Every $x$ in the original rule must be replaced by the entire expression in parentheses.

Evaluate = replace every $x$ with the given value, then simplify; Always put brackets around substituted values: $f(-2) = (-2)^2$, not $-2^2$

Pause — copy the substitution rule (replace every $x$ with the input in brackets) and the negative-input bracket warning into your book.

Did you get this? True or false: to find $f(-3)$ for $f(x) = x^2$, you write $-3^2 = -9$.

Quick check: If $f(x) = 4x - 5$, what is $f(3)$?

06
Functions in context & the difference quotient
core concept

We just saw that evaluating a function means substituting a value for $x$ and simplifying — even an algebraic expression like $x + h$ can be substituted. That raises a question: if we substitute $a + h$ and $a$ into the same function, what does the difference in outputs tell us? This card answers it → the difference quotient $\frac{f(a+h)-f(a)}{h}$ measures average rate of change over an interval.

Function notation becomes powerful when we use it to model real situations. A taxi fare might be written as $C(d) = 3 + 2d$, where $C$ is the cost in dollars and $d$ is the distance in kilometres. The notation tells us instantly what the variables represent and how they are related.

Context functions often use descriptive names. $C(t)$ might mean "cost after $t$ minutes." $T(h)$ might mean "temperature after $h$ hours." The letter inside the brackets is always the input, and the letter before the brackets names the function (or the output quantity).

The difference quotient measures the average rate of change of a function over an interval:

$$\frac{f(a+h) - f(a)}{h}$$

This expression appears throughout calculus. For now, think of it as the average slope of the function between the points $x = a$ and $x = a+h$. If the difference quotient is constant for all values of $a$ and $h$, the function is a straight line.

THE FUNCTION MACHINE x = 3 f(x) = 2x + 5 Rule machine f(3) = 11 One input → apply the rule → exactly one output

Context functions: the letter before brackets = output name; letter inside = input name (e.g. $C(d)$: cost depends on distance); $f(0)$ often represents a fixed fee, initial value, or starting point

Pause — copy the difference quotient formula $\dfrac{f(a+h)-f(a)}{h}$, its meaning (average rate of change), and the context-function naming convention into your book.

Fill the blanks: drag each token into the matching blank.

input output fixed rate

In $C(d) = 3 + 2d$, the distance $d$ is the ___. The $3$ is a ___ fee. The coefficient $2$ is the ___ of change. The cost $C$ is the ___.

Worked example 1 · numerical & algebraic evaluation +5 XP on full reveal

If $f(x) = 2x^2 - 3x + 1$, find $f(2)$, $f(-1)$, and $f(a)$.

1
$f(2) = 2(2)^2 - 3(2) + 1$
Substitute $x = 2$ — use brackets to avoid sign errors
2
$f(2) = 8 - 6 + 1 = \mathbf{3}$
Simplify
3
$f(-1) = 2(-1)^2 - 3(-1) + 1$
Substitute $x = -1$ using brackets — essential with negatives
4
$f(-1) = 2(1) + 3 + 1 = \mathbf{6}$
Simplify
5
$f(a) = 2a^2 - 3a + 1$ ✓
Substitute $x = a$ — replace every $x$ with $a$
Worked example 2 · function in context +5 XP on full reveal

A taxi charges a fare according to the rule $C(d) = 5 + 1.8d$, where $C$ is the cost in dollars and $d$ is the distance in kilometres. Find the cost of a 15 km trip and a 30 km trip. Interpret the meaning of $C(0)$.

1
$C(d) = 5 + 1.8d$
Cost depends on distance. $5$ is the fixed flag fall and $1.8d$ is the variable charge.
2
$C(15) = 5 + 1.8(15) = 5 + 27 = \mathbf{32}$
Substitute $d = 15$
3
$C(30) = 5 + 1.8(30) = 5 + 54 = \mathbf{59}$
Substitute $d = 30$
4
$C(0) = 5 + 1.8(0) = \mathbf{5}$
Substitute $d = 0$
5
$C(0) = \$5$ represents the flag fall ✓
The cost is $5 even if the taxi travels 0 km — this is the fixed booking fee.
Worked example 3 · the difference quotient +5 XP on full reveal

For $f(x) = x^2 + 2x$, find $\dfrac{f(x+h) - f(x)}{h}$ in simplified form.

1
$f(x+h) = (x+h)^2 + 2(x+h)$
Substitute $(x+h)$ for $x$ — use brackets around the whole expression
2
$f(x+h) = x^2 + 2xh + h^2 + 2x + 2h$
Expand $(x+h)^2 = x^2 + 2xh + h^2$ and distribute
3
$f(x+h) - f(x) = (x^2 + 2xh + h^2 + 2x + 2h) - (x^2 + 2x)$
Subtract $f(x) = x^2 + 2x$
4
$= 2xh + h^2 + 2h$
Simplify the numerator — the $x^2 + 2x$ terms cancel
5
$\dfrac{2xh + h^2 + 2h}{h} = 2x + h + 2$ ✓
Divide every term by $h$ (factor $h$ from the numerator)
1

Treating $f(x)$ as $f$ multiplied by $x$

This is the most common error in function notation. $f(x)$ means "the function $f$ evaluated at $x$" — it is not an algebraic product. Writing $f(3) = 3f$ or expanding $f(x+1) = f \cdot (x+1)$ shows a fundamental misunderstanding.

✓ Fix: Read $f(x)$ aloud as "$f$ of $x$." The parentheses contain the input, just like $g(2)$ or $h(a+b)$.

2

Forgetting brackets with negative or algebraic inputs

If $f(x) = x^2$ and you want $f(-2)$, writing $-2^2$ gives $-4$ because the exponent only applies to the $2$. The correct substitution is $(-2)^2 = 4$.

✓ Fix: Always write brackets around substituted values before simplifying: $f(-2) = (-2)^2$.

3

Only replacing the first occurrence of $x$

In $f(x) = x^2 - 3x + 1$, some students substitute correctly for the first $x$ but leave the second one unchanged, writing $f(a) = a^2 - 3x + 1$.

✓ Fix: Replace every instance of the variable in the rule. Count them before you simplify.

4

Errors when expanding $(x+h)^2$

A classic algebraic slip is writing $(x+h)^2 = x^2 + h^2$. The middle term $2xh$ is missing.

✓ Fix: Memorise the identity $(x+h)^2 = x^2 + 2xh + h^2$. It appears constantly in difference quotient problems.

Evaluate each function for the given input. Show your substitution step before simplifying.

1

If $f(x) = 4x - 5$, find $f(3)$ and $f(-2)$.

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2

If $g(x) = x^2 - 3x + 2$, find $g(0)$ and $g(1)$.

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3

If $h(x) = 2x^2 + x - 1$, find $h(a)$ and $h(x+h)$ in expanded form.

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Odd one out: Which evaluation below contains an error?

Work mode · how are you completing this lesson?
1

Given $f(x) = 3x - 7$, find $f(4)$.

2

Given $g(x) = x^2 + 2x$, find $g(-3)$.

3

For $C(d) = 5 + 1.8d$, what does $C(0) = 5$ represent?

4

For $f(x) = 2x + 3$, show that the difference quotient $\dfrac{f(a+h)-f(a)}{h}$ equals $2$.

5

For $f(x) = x^2 - 4x + 3$, find $f(a+1)$ in simplified form.

13
Revisit your thinking

Earlier you were asked: A taxi charges a $3 flag fall plus $2 for every kilometre. How much would a 5 km trip cost? How much would a 10 km trip cost? Can you write a general rule?

A 5 km trip costs $3 + 2(5) = \$13$, and a 10 km trip costs $3 + 2(10) = \$23$. The general rule is $C(d) = 3 + 2d$, where $C$ is the cost in dollars and $d$ is the distance in kilometres. This is a function because each distance input produces exactly one cost output. The $3$ is a constant (the flag fall) and the $2$ is the rate of change per kilometre — ideas that will follow you through calculus and beyond.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

02
Short answer
UnderstandBand 32 marks

Q8. Explain the difference between $f(x)$ and $f \times x$. Use a specific example with $f(x) = 2x + 3$ to illustrate your answer. (2 marks)

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ApplyBand 34 marks

Q9. Consider the function $f(x) = x^2 - 4x + 3$. (a) Find $f(2)$. (b) Find $f(-1)$. (c) Find $f(a+1)$ in simplified form. Show all working. (4 marks)

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AnalyseBand 54 marks

Q10. The difference quotient $\dfrac{f(a+h) - f(a)}{h}$ measures the average rate of change of a function. (a) For $f(x) = 2x + 3$, show that the difference quotient equals $2$. (b) Explain what this result tells you about the graph of $f(x) = 2x + 3$. (4 marks)

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📖 Comprehensive answers (click to reveal)

Multiple choice — drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Activity 1 — Evaluate and explain model answers

1. $f(3) = 4(3) - 5 = 12 - 5 = 7$; $f(-2) = 4(-2) - 5 = -8 - 5 = -13$

2. $g(0) = (0)^2 - 3(0) + 2 = 2$; $g(1) = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$

3. $h(a) = 2a^2 + a - 1$; $h(x+h) = 2(x+h)^2 + (x+h) - 1 = 2(x^2 + 2xh + h^2) + x + h - 1 = 2x^2 + 4xh + 2h^2 + x + h - 1$

Short answer model answers

Q8 (2 marks): $f(x)$ is function notation meaning "the output of function $f$ when the input is $x$" [1]. In contrast, $f \times x$ means the variable $f$ multiplied by $x$. For example, with $f(x) = 2x + 3$, we have $f(4) = 2(4) + 3 = 11$, which is completely different from $f \times 4$ [1].

Q9 (4 marks):

(a) $f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = \mathbf{-1}$ ✓
(b) $f(-1) = (-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = \mathbf{8}$ ✓
(c) $f(a+1) = (a+1)^2 - 4(a+1) + 3 = a^2 + 2a + 1 - 4a - 4 + 3 = \mathbf{a^2 - 2a}$ ✓
Award 1 mark each for (a), (b), and method in (c); 1 mark for correct simplified form in (c).

Q10 (4 marks):

(a) $f(a+h) = 2(a+h) + 3 = 2a + 2h + 3$
$\dfrac{f(a+h) - f(a)}{h} = \dfrac{(2a + 2h + 3) - (2a + 3)}{h} = \dfrac{2h}{h} = \mathbf{2}$ ✓

(b) The difference quotient equals 2 for all values of $a$ and $h$ [1]. This tells us that the average rate of change is constant [1], which means the graph of $f(x) = 2x + 3$ is a straight line with a slope of 2 [1].

01
Boss battle
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Five timed questions on function notation and evaluation. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

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02
Science Jump · function notation & evaluation
arcade practice

Climb platforms, hit checkpoints, and answer function evaluation questions. Quick recall, lighter than the boss.

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Tick when you've finished the practice and review.

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