Mathematics Advanced • Year 11 • Module 4 • Lesson 1

Introduction to Exponential Functions

Build procedural fluency in identifying, classifying, and evaluating exponential functions of the form y = ax.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the definition:

A function of the form y = ax is called an exponential function provided a > ____ and a ≠ ____.

Q1.2 Fill in the table:

If a > 1, the function shows ____________ ;   if 0 < a < 1, the function shows ____________.

Q1.3 State the y-intercept of every basic exponential y = ax: (____, ____).

Stuck? Revisit lesson § Concept and § Key Terms.

2. Worked example, evaluate f(x) = 2x at x = −2, 0, 3

Follow each line. Every step has a reason on the right.

Problem. Evaluate f(x) = 2x at x = −2, x = 0, and x = 3.

Step 1, Evaluate at x = −2 using the negative-exponent rule.

f(−2) = 2−2 = 1 / 22 = 1/4

Reason: a−n = 1/an; the negative exponent gives the reciprocal, not a negative number.

Step 2, Evaluate at x = 0 using the zero-exponent rule.

f(0) = 20 = 1

Reason: any non-zero base raised to 0 equals 1; the point (0, 1) is always on y = ax.

Step 3, Evaluate at x = 3 by direct multiplication.

f(3) = 23 = 2 × 2 × 2 = 8

Reason: positive integer exponent means repeated multiplication.

Conclusion. f(−2) = 1/4,   f(0) = 1,   f(3) = 8.

3. Faded example, fill in the missing steps

Determine whether y = (1/3)x shows growth or decay, and find its y-intercept. 4 marks

Step 1, Identify the base.

Base a = ____________

Step 2, Compare the base to 1.

Since ____ < 1/3 < ____, the function shows ____________ .

Step 3, Find the y-intercept by setting x = 0.

y = (1/3)0 = ____________

Conclusion. y = (1/3)x shows ____________ with y-intercept (____, ____).

Stuck? Revisit lesson § Worked Example 2.

4. Graduated practice

Show your working in the space provided. Use exact values where possible.

Foundation, direct evaluation (4 questions)

QComputeAnswer (exact)
4.1 134
4.2 150
4.3 12−3
4.4 1(1/2)4

Standard, classify and evaluate (6 questions)

For each function, state whether it shows growth or decay, and evaluate at the listed x-values.

4.5 f(x) = 4x; evaluate f(2) and f(−1). State growth/decay.    2 marks

4.6 f(x) = (2/5)x; evaluate f(0) and f(2). State growth/decay.    2 marks

4.7 f(x) = 10x; evaluate f(−2), f(0), f(3).    2 marks

4.8 f(x) = (0.5)x; evaluate f(3). Show that (0.5)3 = 1/8.    2 marks

4.9 A model P(t) = 100 × 2t gives bacterial population after t hours. Compute P(4).    2 marks

4.10 A car's value follows V(t) = 30 000 × (0.85)t. Compute V(2) to the nearest dollar.    2 marks

Extension, reason about exponentials (2 questions)

4.11 Without using a calculator, decide which is larger: 210 or 102. Show one or two lines of working that justify your choice.    3 marks

4.12 Explain why neither y = 1x nor y = 0x nor y = (−2)x is admitted as an exponential function in the form y = ax. Give one specific reason for each.    3 marks

Stuck on 4.12? Revisit lesson § Trap 03, invalid bases.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

How did this worksheet feel?

What I'll revisit before next class:

Answers, Do not peek before attempting

Q1.1, Definition

a > 0 and a ≠ 1. (Bases must be positive and not equal to 1.)

Q1.2, Growth vs decay

If a > 1: exponential growth.   If 0 < a < 1: exponential decay.

Q1.3, y-intercept

(0, 1), because a0 = 1 for any valid base.

Q3, Faded example y = (1/3)x

Step 1: a = 1/3.
Step 2: Since 0 < 1/3 < 1, the function shows decay.
Step 3: y = (1/3)0 = 1.
Conclusion: y = (1/3)x shows decay with y-intercept (0, 1).

Q4.1–4.4, Direct evaluation

4.1: 34 = 81.   4.2: 50 = 1.   4.3: 2−3 = 1/8.   4.4: (1/2)4 = 1/16.

Q4.5, f(x) = 4x

f(2) = 42 = 16.   f(−1) = 1/41 = 1/4.   Base 4 > 1 → growth.

Q4.6, f(x) = (2/5)x

f(0) = 1.   f(2) = (2/5)2 = 4/25 = 0.16.   Base 2/5 = 0.4 is between 0 and 1 → decay.

Q4.7, f(x) = 10x

f(−2) = 1/100 = 0.01.   f(0) = 1.   f(3) = 1000.

Q4.8, f(x) = (0.5)x

(0.5)3 = 0.5 × 0.5 × 0.5 = 0.125 = 1/8. Note (0.5)3 = (1/2)3 = 13/23 = 1/8 confirms this.

Q4.9, P(t) = 100 × 2t

P(4) = 100 × 24 = 100 × 16 = 1600 bacteria.

Q4.10, V(t) = 30 000 × (0.85)t

V(2) = 30 000 × (0.85)2 = 30 000 × 0.7225 = $21 675.

Q4.11, Which is larger: 210 or 102?

210 = 1024.   102 = 100.   So 210 > 102. Reasoning: 210 = (23)3 × 2 = 83 × 2 = 512 × 2 = 1024, whereas 102 = 100. This is the typical surprise: exponential growth (variable in the exponent) outpaces polynomial growth (variable in the base) for modest sizes already.

Q4.12, Why exclude a = 1, 0, −2

a = 1: 1x = 1 for every x, so the function is the constant y = 1, not an increasing or decreasing exponential, hence we exclude it.
a = 0: 0x is 0 for x > 0 and is undefined for x ≤ 0 (e.g. 00, 0−1). The function fails to be defined on all real x.
a = −2: A negative base is not defined for many real exponents (e.g. (−2)1/2 is not real). The resulting "function" is not a function of a real variable in any useful sense.