Week 13 · Lecture outline

Week 13 — Lecture Outline · Exponential Functions

College Algebra · MATH 120 Fall 2026 · Prof. Calloway Fictional sample

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 8 — Analyze and apply exponential and logarithmic functions and equations, including growth/decay models and compound interest.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/numerical/graphical representations)
Meeting pattern: This week has ONE 75-minute session (Tue Nov 24 only — Thanksgiving Nov 26–27). The outline below covers ~150 min (a normal week's two sessions); trim Segments 5–8 for delivery and direct students to the tutorial and readings for the full coverage. Scale to your own pattern.

Week at a Glance


The week's big question	"What happens when the variable moves to the exponent — and how does that one change produce growth, decay, compound interest, and the natural base e?"
By the end of the week, students can…	(1) evaluate f(x) = a·bˣ at any x-value and classify it as growth or decay; (2) state the graph features — y-intercept (0, a), horizontal asymptote y = 0, always-increasing or always-decreasing; (3) recognize e ≈ 2.718 and apply A = P(1 + r/n)^(nt) and A = Peʳᵗ; (4) set up and interpret a growth or decay application (compound interest, population, depreciation, half-life).
Key vocabulary	exponential function, base, growth factor, decay factor, y-intercept, horizontal asymptote, the natural base e, compound interest, compounding periods, continuous compounding, growth rate, decay rate, initial value
Materials	slides (Deck 13), readings + video links, Desmos (or GeoGebra), a calculator, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and tutorial
Timing note	8 segments, ~150 min total. Session 1 (Tue) = Segments 1–5 (focus: definition, evaluate, classify, graph). Session 2 would be Segments 6–8 (e, applications, technology) — assign these via tutorial this week.
Holiday flag	Thanksgiving falls Thu–Fri Nov 26–27. Remind students at the start of class to plan ahead: one session, all work due Sunday Nov 29.

Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "How much would $1,000 double to if you left it in an account earning 6% interest for 12 years? Take a guess." Wait for a few answers — typically $1,200 or $1,500.

"Actually, with compound interest it grows to about $2,012 — more than double. The reason is exponential growth: the interest earns interest, and it accelerates. This week we're going to understand why."

"Every function we've studied has had the variable in the base: x², x³, 1/x. This week we flip it — the variable goes into the exponent. That one change produces the most important curves in science and economics: population growth, compound interest, radioactive decay, the spread of disease."
"The math is actually simpler than you think. One function, one shape, one set of rules."

The promise (write on the board): "By the end of this week you'll be able to look at any exponential function, read its graph instantly — y-intercept, asymptote, growing or shrinking — and apply it to a real compound-interest problem."

Why it matters: "Exponential functions aren't exotic math. They're the language the world uses to describe anything that grows or shrinks by a fixed percentage — and that includes most of finance, biology, and physics."

Segment 2 — Defining the Exponential Function (20 min)

Plain language first. An exponential function has the form:

f(x) = a · bˣ
where a ≠ 0, b > 0, b ≠ 1.

The a is the initial value (what you start with when x = 0). The b is the base — the growth or decay factor applied at every step. The x is the exponent; it's the variable.

Why b > 0 and b ≠ 1?
- If b were negative, bˣ would alternate between positive and negative for integer x and be undefined for many real x — not a usable function.
- If b = 1, then 1ˣ = 1 for every x, giving f(x) = a — a constant, not exponential.
- So b must be a positive number other than 1.

Growth vs. decay — the base decides:
- b > 1: as x increases, bˣ increases → exponential growth. The bigger x gets, the bigger f(x) gets, faster and faster.
- 0 < b < 1: as x increases, bˣ decreases → exponential decay. The bigger x gets, the smaller f(x) gets, gliding toward 0.

Memory hook: "If the base is bigger than 1, it grows. If the base is a fraction (between 0 and 1), it decays. The base is the whole story."

One fully worked example (evaluating — every step):

Let f(x) = 3 · 2ˣ. Evaluate at x = 0, x = 2, x = −1, x = −3.
- f(0) = 3 · 2⁰ = 3 · 1 = 3
- f(2) = 3 · 2² = 3 · 4 = 12
- f(−1) = 3 · 2⁻¹ = 3 · (1/2) = 3/2 = 1.5
- f(−3) = 3 · 2⁻³ = 3 · (1/8) = 3/8 = 0.375

Note: negative inputs give small positive outputs (approaching 0); the function never reaches 0 and never goes negative.

Another example (fractional base — decay):

g(x) = 4 · (1/2)ˣ. Evaluate at x = 0, x = 3.
- g(0) = 4 · 1 = 4
- g(3) = 4 · (1/2)³ = 4 · (1/8) = 4/8 = 0.5

Decay: starts at 4, shrinking with each step.

Segment 3 — The −4² Cousin: Common Evaluation Errors (12 min)

Name the misconceptions right away.

❌ "The base is the exponent." — Students write f(3) = 3² = 9 instead of f(3) = 2³ = 8.
✅ Cure: The variable is ALWAYS in the exponent. The base b is fixed. Write f(x) = 2ˣ and circle the 2 as constant, the x as variable.

❌ "f(x) = 2ˣ is the same as g(x) = x²." — They look related. They're completely different functions.
✅ Cure: Evaluate at x = 3: 2³ = 8 but 3² = 9. At x = 10: 2¹⁰ = 1024 but 10² = 100. The exponential function blows past the polynomial for large x.

❌ "A decay function has a negative base." — b = −2 is NOT a decay function; it's not even a valid exponential function.
✅ Cure: Decay means 0 < b < 1. The base is still positive. Example: b = 0.5, b = 0.8, b = 1/3.

❌ "f(x) = 2·3ˣ has y-intercept at (0, 2·3) = (0, 6)." — Multiplying a by b instead of evaluating at x = 0.
✅ Cure: Always substitute x = 0 first: f(0) = 2 · 3⁰ = 2 · 1 = 2. The y-intercept is (0, a) always.

Write on board: "Step 1 always: identify a and b. Step 2: substitute x = 0 for the y-intercept. Step 3: compare b to 1 for growth or decay."

Segment 4 — Graph Features (25 min)

Plain language first. The graph of f(x) = a · bˣ has a shape you can read like a recipe card. Four features define it:

y-intercept = (0, a). Plug in x = 0 and you get a · b⁰ = a · 1 = a. Always.
Horizontal asymptote: y = 0. As x → −∞ (for growth) or x → +∞ (for decay), the graph slides toward the x-axis without touching it. The function is always positive.
Always positive. Because b > 0 and a is a real number, bˣ is always positive, so the graph is always above or below the x-axis (same side as a).
Monotone: growth functions are strictly increasing (left to right, always rising); decay functions are strictly decreasing (left to right, always falling).

One fully worked graphing example:

f(x) = 2 · 3ˣ
- a = 2, b = 3. Since b = 3 > 1 → growth → increasing.
- y-intercept: f(0) = 2 · 1 = 2 → point (0, 2).
- Asymptote: y = 0 (as x → −∞, 3ˣ → 0, so f(x) → 0).
- Check a few points: f(1) = 2·3 = 6; f(−1) = 2/3 ≈ 0.67; f(2) = 2·9 = 18.
- Sketch: starts near 0 on the left, crosses the y-axis at (0, 2), curves steeply upward to the right.

Decay example:

h(x) = 8 · (1/2)ˣ
- a = 8, b = 1/2. Since 0 < b < 1 → decay → decreasing.
- y-intercept: h(0) = 8 · 1 = 8 → point (0, 8).
- Asymptote: y = 0 (as x → +∞, (1/2)ˣ → 0).
- Check: h(1) = 4; h(2) = 2; h(3) = 1; h(4) = 0.5 — shrinking fast.
- Sketch: starts high on the left, y-intercept at (0, 8), decays toward the x-axis.

Interaction — classify and read (quick, ~8 min): Put four graphs on the slide (two growth, two decay, different a-values). Students:
1. Growth or decay? (look at the base or the shape)
2. State the y-intercept.
3. State the equation of the asymptote.

Debrief: the asymptote is always y = 0 here (transformations come in pre-calc), and the y-intercept is always (0, a).

Segment 5 — The Natural Base e (15 min) · Session 1 closes (~80 min)

Plain language first. There is one special base that shows up throughout science, finance, and engineering: the number e.

e ≈ 2.71828…
It is irrational (like π) — its decimal never ends and never repeats.

Where does e come from? You meet it in compound interest. As you compound more and more frequently — annually, monthly, daily, every second — the formula
A = P(1 + 1/n)ⁿ
approaches a limit as n → ∞. That limit is exactly e. Formally:

lim_{n→∞} (1 + 1/n)ⁿ = e ≈ 2.71828

You don't need to compute this limit; just remember: e is the base that emerges when compounding is continuous.

Key facts to memorize:
- e⁰ = 1 (any base to the zero power = 1).
- e¹ = e ≈ 2.718.
- eˣ is an exponential growth function (since e > 1).

The function f(x) = eˣ has the same features as any a·bˣ with a = 1, b = e: y-intercept (0, 1), horizontal asymptote y = 0, always increasing.

Memory hook: "e is the 'natural' base because nature and finance — anything that grows continuously — naturally end up using it."

Segment 6 — Compound Interest Formulas (20 min) · Session 2 opens

Hook back in: "Now that we have the exponential function and the special base e, we can tackle the most important application in this half of the course: compound interest."

Two formulas — know which is which:

Periodic compounding: A = P(1 + r/n)^(nt)
Continuous compounding: A = Peʳᵗ

Where:
- A = the final amount (what you end up with)
- P = the principal (starting amount)
- r = the annual interest rate as a decimal (6% → 0.06)
- n = the number of compounding periods per year (annual = 1, semi-annual = 2, quarterly = 4, monthly = 12, daily = 365)
- t = time in years

One fully worked example (periodic compounding):

$1,000 invested at 6% annual interest, compounded annually for 2 years.
- P = 1000, r = 0.06, n = 1, t = 2
- A = 1000(1 + 0.06/1)^(1·2) = 1000(1.06)² = 1000(1.1236) = $1,123.60

Compare to simple interest: 1000 · (1 + 0.06 · 2) = 1000 · 1.12 = $1,120 — compound interest earns $3.60 more, because the first year's interest earns interest in year 2.

One fully worked example (continuous compounding):

$1,500 invested at 5% annual interest, compounded continuously for 4 years.
- P = 1500, r = 0.05, t = 4
- A = 1500 · e^(0.05·4) = 1500 · e^(0.20) ≈ 1500 · 1.2214 ≈ $1,832.10

When to use which: n is given → periodic formula. "Compounded continuously" → eʳᵗ formula.

Segment 7 — Growth and Decay Applications (20 min)

Plain language first. The formula f(x) = a · bˣ models any quantity that changes by a constant percentage at each step:
- Population growth: a = initial population, b = 1 + growth rate
- Depreciation/decay: a = initial value, b = 1 − decay rate
- Half-life: b = 1/2, x = number of half-life periods

One fully worked example (population growth):

A city's population is 800 people and grows at 3% per year.
- Model: P(t) = 800 · (1.03)ᵗ
- After 10 years: P(10) = 800 · (1.03)¹⁰ ≈ 800 · 1.3439 ≈ 1,075 people

The base b = 1.03 > 1, confirming growth.

Decay example:

A radioactive substance has 100 grams, and each year half remains.
- Model: A(t) = 100 · (1/2)ᵗ
- After 3 years: A(3) = 100 · (1/2)³ = 100/8 = 12.5 grams

Common error to name: Students plug in t = 1 and compute the rate rather than evaluate the full model. Remind: always let t = the given time and compute A directly.

Segment 8 — Technology Workflow + AI-Critique, Callback & Hand-off (10 min) · Session 2 closes (~75)

Technology workflow — check an exponential function in Desmos (exact steps):
1. Open desmos.com/calculator (free, no login).
2. Type f(x) = 2*3^x on line 1. Watch the curve: starts low on the left, crosses (0, 2), rises steeply.
3. Type y = 0 on line 2 — Desmos draws the asymptote.
4. Click any x-value on the curve to read f(x) — confirm your by-hand calculation.
5. Try g(x) = 8*(0.5)^x — the decay curve, starting at (0, 8) and falling right.

The visual cements two ideas: (1) exponential functions are never negative, and (2) the asymptote is real — the curve never touches y = 0.

AI-critique moment (students verify, not consume):

Paste this to an approved chatbot: "Evaluate f(x) = 3·2ˣ at x = −2 and tell me whether the function represents growth or decay. Explain your steps."
Then check its work by hand: f(−2) = 3 · 2⁻² = 3 · (1/4) = 3/4 = 0.75. Common errors: the model misreads 3·2 as 6 (treating it as a product before applying the exponent), or calls b = 3 instead of b = 2 (confusing a and b), or calls it decay because the answer is small. Your job: the tool drafts, you judge.

Callback + tease:
- Callback: "We spent Weeks 1–2 on exponent rules. Now those same rules let us evaluate and simplify exponential expressions. They weren't just practice — they were preparation for this."
- Tease next week: "Week 14 flips the question: if bˣ = some value, what is x? That's the logarithm — the inverse of the exponential — and it unlocks a whole new class of equations."

Hand-off (the week's graded work):
- Lecture Tutorial 13 (AI tutor, share-link submission) — evaluating, classifying, graph features, e, compound interest.
- Quiz 13 (end of week, no AI) and Discussion 13 ("Exponential in the Wild").
- Assignment 13 — AI-coached, self-scored.

Instructor FAQ — Common Stumbles

Student says / does	Quick cure
"Is f(x) = 2ˣ growth or decay? 2 is small."	Growth — any b > 1 produces growth. 2 > 1, so as x increases, f(x) increases.
Writes f(2) = 2·3² = 36 for f(x) = 3·2ˣ	The variable is in the exponent; b = 2 is the base. f(2) = 3·2² = 3·4 = 12, not 3²·something.
"The y-intercept of f(x) = 5·4ˣ is (0, 20)."	Substitute x = 0: f(0) = 5·4⁰ = 5·1 = 5. Never multiply a·b for the y-intercept.
"The horizontal asymptote is y = a."	The asymptote is y = 0 — the x-axis. As x → ±∞, bˣ → 0 (not to a).
"Is b = −0.5 a decay function?"	No — b must be positive. 0 < b < 1 for decay. b = 0.5 is decay; b = −0.5 is not a valid exponential base.
"What's the difference between n and t in A = P(1+r/n)^(nt)?"	n = compounding periods per year (how often); t = years (how long). If quarterly for 3 years: n = 4, t = 3, exponent = nt = 12.
"Continuous and annual compounding give the same answer."	Very close for small r, t — but continuous compounding always gives a slightly higher A. Show with $1000 at 6% for 10 years: periodic ≈ $1790.85 (n=1), continuous ≈ $1822.12.

Scope flag

This outline stays within Objective 8's Week 13 scope: evaluating f(x) = a·bˣ, classifying growth/decay, reading graph features, recognizing e, and applying A = P(1+r/n)^(nt) and A = Peʳᵗ. Solving exponential equations (Week 15) and logarithms (Week 14) are out of scope this week — do not introduce them; they return next week.

~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com