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Module 4 · L1 of 15 ~35 min ⚡ +95 XP available

Introduction to Exponential Functions

Exponential functions grow or decay by a constant multiplicative factor. Like bacteria doubling every hour or radioactive material halving over time, these functions model processes where the rate of change is proportional to the current amount.

Today's hook — A sheet of paper is 0.1 mm thick. You fold it in half 10 times. How tall is the stack? Most people guess under 10 cm. The real answer is $2^{10} = 1024$ layers — about 10 cm. Fold 42 times and it reaches the Moon. That explosive growth is exactly what this lesson is about.

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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.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

A piece of paper is folded in half 10 times. Roughly how many layers thick is it? Pick a number first, then explain why you chose it — no formula needed yet.

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The two moves

+5 XP to read

Every exponential question in this module comes down to one decision: identify the base, then decide if it causes growth or decay. Everything else follows automatically.

For $y = a^x$: if $a > 1$ the function grows as $x$ increases; if $0 < a < 1$ the function decays. Both always pass through $(0, 1)$ since $a^0 = 1$ for any valid base.

$$y = a^x \quad a > 0,\ a \neq 1$$

Base > 1 (growth)

$y$ increases without bound as $x \to +\infty$ and approaches 0 as $x \to -\infty$.

0 < Base < 1 (decay)

$y$ approaches 0 as $x \to +\infty$ and grows without bound as $x \to -\infty$.

Negative exponents

$a^{-n} = \dfrac{1}{a^n}$ — a negative exponent creates a reciprocal, not a negative value.

What you'll master

Know

Key facts

The definition of an exponential function $y = a^x$
The conditions on the base: $a > 0$, $a \neq 1$
That $a^0 = 1$ for all valid bases

Understand

Concepts

Why $a > 1$ produces growth and $0 < a < 1$ produces decay
The domain (all real $x$) and range ($y > 0$) of $y = a^x$
How exponential growth differs from polynomial growth

Can do

Skills

Evaluate $a^x$ for positive, negative and fractional values of $x$
Classify a given base as growth or decay
Apply exponential models to real-world problems

Key terms

Exponential functionA function of the form $y = a^x$ where $a > 0$ and $a \neq 1$.

BaseThe constant $a$ in $y = a^x$; determines whether the function shows growth or decay.

Exponential growthOccurs when $a > 1$: as $x$ increases, $y$ increases without bound.

Exponential decayOccurs when $0 < a < 1$: as $x$ increases, $y$ approaches zero.

Horizontal asymptoteThe line $y = 0$ that the basic exponential curve approaches but never crosses.

Domain / RangeFor $y = a^x$: domain is all real $x$; range is $y > 0$.

What is an exponential function?

core concept

An exponential function has the variable in the exponent rather than the base. This creates dramatically different behaviour from polynomial functions. For $y = a^x$ with $a > 1$, each unit increase in $x$ multiplies $y$ by $a$ — that is growth by a constant factor, not a constant amount.

$$y = a^x \quad \text{where } a > 0,\ a \neq 1$$

The function always passes through $(0, 1)$ because $a^0 = 1$ for any valid base. The $x$-axis ($y = 0$) is a horizontal asymptote — the graph gets infinitely close but never touches it. Exponential functions model population growth, compound interest, radioactive decay, and Newton's Law of Cooling.

Why the restriction $a \neq 1$? If $a = 1$, then $1^x = 1$ for every value of $x$ — that is just the constant function $y = 1$, not interesting or exponential. If $a \le 0$, then $a^{1/2}$ may not be real. Both restrictions keep the function well-defined for all real $x$.

An exponential function has the variable in the exponent: $y = a^x$ where $a > 0$, $a \neq 1$; Always passes through $(0, 1)$ since $a^0 = 1$

Pause — copy the exponential function definition ($y = a^x$, $a > 0$, $a \neq 1$) and the universal fixed point ($a^0 = 1$, so all pass through $(0,1)$) into your book.

Did you get this? True or false: the function $y = (-2)^x$ is a valid exponential function because $|-2| > 1$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · EVALUATING

Evaluate $f(x) = 2^x$ at $x = -2,\ 0,\ 3$.

$f(-2) = 2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$

Negative exponent means reciprocal: $a^{-n} = \frac{1}{a^n}$.

PROBLEM 2 · GROWTH OR DECAY?

Determine whether $y = \left(\dfrac{1}{3}\right)^x$ shows growth or decay, and find the $y$-intercept.

Base $a = \dfrac{1}{3}$

Identify the base of the exponential function.

PROBLEM 3 · REAL-WORLD MODEL

A population of bacteria doubles every hour, modelled by $P = 100 \times 2^t$. Find the population after 5 hours.

$P = 100 \times 2^t$ where $t$ is in hours

The initial population is 100 and the growth factor is 2 per hour.

Quick check: What is $f(-3)$ for $f(x) = 2^x$?

Common errors · the 3 traps that cost marks

Trap 01

Confusing exponential growth with polynomial growth

$2^x$ grows much faster than $x^2$. For large $x$, exponential functions eventually outpace any polynomial. Students sometimes think $x^2$ grows faster because the notation looks similar.

Trap 02

Forgetting that $a^{-n} = \frac{1}{a^n}$, not $-a^n$

Negative exponents create reciprocals, not negative answers. $2^{-3} = \frac{1}{8}$, not $-8$. This is a common arithmetic error when evaluating exponential functions.

Trap 03

Including $a = 1$ or $a \le 0$ as valid bases

$y = 1^x$ is just the constant $y = 1$, not exponential. For $a \le 0$, the function is not defined for all real $x$ (e.g. $(-2)^{1/2}$ is not real).

Fill in the blank: For the exponential function $y = a^x$ to be valid, we require $a$ ___ 0 and $a$ ___ 1.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Evaluate $3^4$ and $3^{-2}$.

Does $y = 5^x$ show growth or decay? Explain.

Find the $y$-intercept of $y = \left(\dfrac{2}{3}\right)^x$.

Evaluate $f(x) = 10^x$ at $x = -1,\ 0,\ 2$.

A substance decays by half each year. Starting at 800 g, how much remains after 3 years? Write the model and evaluate it.

Odd one out — Three of these represent valid exponential functions. Which one does NOT?

Two truths, one lie — Three statements about $y = a^x$ where $a > 1$. Which one is the lie?

Revisit your thinking

Earlier you guessed how thick a paper would be after 10 folds. The answer is $2^{10} = 1024$ layers — about 10 cm. That doubling pattern is exactly what an exponential function with base 2 captures. Whether the base is greater than 1 (growth) or between 0 and 1 (decay), the same structure $y = a^x$ governs the behaviour.

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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. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. Evaluate $f(x) = \left(\dfrac{1}{2}\right)^x$ at $x = -2,\ 0,\ 4$. Show your working. (3 marks)

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ApplyBand 43 marks

Q2. The value of a car depreciates by 15% per year, modelled by $V = 30\,000 \times (0.85)^t$. Find the value after 2 years and after 4 years, to the nearest dollar. (3 marks)

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AnalyseBand 53 marks

Q3. For the function $f(x) = a^x$ where $a > 1$, explain why $f(x) > 0$ for all real $x$, and why $f(x) \to 0$ as $x \to -\infty$. (3 marks)

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Comprehensive answers (click to reveal)

Drill answers: 1) $3^4 = 81$; $3^{-2} = \frac{1}{9}$ · 2) Growth; base $5 > 1$ · 3) $(0, 1)$ · 4) $f(-1) = \frac{1}{10}$; $f(0) = 1$; $f(2) = 100$ · 5) $M = 800 \times \left(\frac{1}{2}\right)^3 = 100$ g

Q1 (3 marks): $f(-2) = \left(\frac{1}{2}\right)^{-2} = 2^2 = 4$ [1]. $f(0) = 1$ [1]. $f(4) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$ [1].

Q2 (3 marks): $V(2) = 30\,000 \times (0.85)^2 = 30\,000 \times 0.7225 = \$21\,675$ [1.5]. $V(4) = 30\,000 \times (0.85)^4 \approx 30\,000 \times 0.5220 \approx \$15\,660$ [1.5].

Q3 (3 marks): Any positive number raised to any real power is positive, so $a^x > 0$ for all $x$ [1]. As $x \to -\infty$, write $a^x = \frac{1}{a^{-x}}$ where $-x \to +\infty$ [1]. Since $a > 1$, $a^{-x} \to \infty$, so $\frac{1}{a^{-x}} \to 0$ [1].

Boss battle · Exponential Basics

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90%+ speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering exponential function questions. A lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← M3 Lesson 15 Lesson 2 · Graphs of Exponential Functions →

Module overview · Maths Advanced · Checkpoint 1