Modelling a quantity that increases by the same percentage in each unit of time with the exponential growth model. Includes converting a percentage rate into a growth factor, distinguishing a rate from a factor, applying the model to compound interest and to population growth, and seeing why compounding outpaces adding a fixed amount.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 8 — Exponents and Exponential Functions
Exponential Growth Functions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-481 — the lesson these objectives are drawn from
Warm-up
Lesson 8.3 drew exponential functions. This lesson gives their two numbers meanings drawn from real situations.
Discussion prompt
A quantity of 100 grows by 10 per cent each year. Write down its value after one, two and three years, and say what you multiplied by each time.
Hint: Ten per cent of the current amount, not of the original.
Answer:
\[ 100 \to 110 \to 121 \to 133.1 \]
Each step multiplied by 1.1, keeping the whole and adding a tenth. The steps grow — ten, then eleven, then twelve point one — because ten per cent of a larger amount is a larger increase.
Concept
A quantity is growing exponentially if it increases by the same percentage r in each unit of time. Such growth is modelled by y equals C times the quantity one plus r, raised to t.
exponential growth — Growth in which a quantity increases by the same percentage in each unit of time, modelled by y equals C times one plus r, all raised to the power t.
The expression one plus r is called the growth factor.
Figure (svg): The exponential growth model with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Section
Section 1
Concept
In the model, C is the initial amount before any growth occurs, r is the growth rate written as a decimal, and t is the time. The bracket one plus r is the growth factor.
\[ y = C(1 + r)^t \]
Both C and r are positive for growth.
Figure (svg): The exponential growth model with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476 — the exponential growth model and its definitions
Picture it
Initial amount, rate, factor, time.
Figure (svg): The exponential growth model with each part labelled
Compare this with Lesson 8.3's y equals a times b to the x. The C plays the part of a and the growth factor the part of b, so nothing new is being introduced — only interpreted.
Worked example
Naming them before writing the model is the whole setup.
\[ \text{A catfish weighs } 0.06 \text{ g and gains } 10\% \text{ a day. Identify } C, \; r \text{ and } t. \]
Find the initial amount
Why: The weight before any growth.
\[ C = 0.06\text{ grams} \]
Find the growth rate
Why: Ten per cent, written as a decimal.
\[ r = 0.10 \]
Find the time variable
Why: The number of days.
\[ t =\text{ days} \]
Form the growth factor
Why: One plus a tenth.
\[ 1.1 \]
Figure (svg): The three quantities a growth problem supplies
\[ C = 0.06, \; r = 0.10, \; 1 + r = 1.1 \]
Verify: check the units of each
Why: C is in grams, r has no units since it is a proportion, and t is in days. The growth factor is also unitless, so multiplying grams by it repeatedly leaves grams — which is what the output should be.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Matching
Divide by a hundred, then add one.
Match the pairs
Why: Every growth factor here is slightly above one, because the quantity is kept and a little added. A factor below one would describe shrinking, which is the subject of the next lesson.
Worked example
This is Example 1 from the textbook.
\[ \text{Write a model for the catfish's weight during its first six weeks.} \]
Write the general model
Why: The formula with its four parts.
\[ y = C(1 + r) ^{t} \]
Substitute the initial amount
Why: Six hundredths of a gram.
\[ 0.06(1 + r) ^{t} \]
Substitute the rate
Why: Ten per cent as a decimal.
\[ 0.06(1 + 0.10) ^{t} \]
Simplify the bracket
Why: One plus a tenth.
\[ y = 0.06(1.1) ^{t} \]
Figure (svg): The exponential growth model with each part labelled
\[ y = 0.06(1.1)^t \]
Verify: test the model after one day
Why: One day gives 0.06 times 1.1, which is 0.066 grams — six hundredths plus ten per cent of it. The model reproduces the described growth for the first step, which is the check worth doing before using it further.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Trap
\[ y = 0.06(1 + 10)^t \]
Substitute the ten from ten per cent directly
Why: The number in the problem is ten, so ten is what gets substituted.
A growth factor of eleven means the weight multiplies by eleven each day, not that it grows by a tenth. After a week the model would give a fish weighing more than a hundred grams.
\[ y = 0.06(1 + 0.10)^t = 0.06(1.1)^t \]
Convert the percentage to a decimal by dividing by a hundred
Why: Ten per cent is ten hundredths, which is 0.10.
Testing the model after one step catches this instantly: 1.1 gives a ten per cent rise and 11 gives a tenfold one.
Faded example
Initial amount, then the factor.
Fill in the blanks
C = 0.06, \; r = 0.10 \;\Longrightarrow\; y = 0.06(1 + 0.10)^t = 0.06(1.1)^t
Why: The initial amount goes in front and the rate goes inside the bracket with the one. Substituting ten rather than 0.10 would give a factor of eleven and a wildly wrong model.
Elimination
50 000 viewers growing by 2 per cent a month.
Eliminate the wrong options
Which is the growth model?
Survives elimination: A
Why: Two per cent is 0.02, and adding one gives a factor of 1.02. This is Guided Practice 1 from the textbook, and option C is worth noticing: dropping the one turns growth into rapid decay.
Socratic
The rate alone would seem enough.
Discussion prompt
Explain what the one in the growth factor is doing. Then say what the model would describe if the one were left out.
Hint: Ask what happens to the amount you already have.
Answer:
The one keeps the amount you already have and the r adds the increase on top. Multiplying by 1.1 gives you the whole original plus a tenth of it, which is exactly what growing by ten per cent means — the original does not vanish and get replaced by the increase.
Without the one, multiplying by 0.1 each period would leave a tenth of the amount, so the quantity would shrink to a tenth every step rather than growing by a tenth. That is a decay model with a very small factor, and it is the opposite of what was described.
Section
Section 2
Concept
A phrase like increases by ten per cent gives a growth rate, which needs one added. A phrase like triples gives the growth factor directly and needs nothing added.
The Study Tip in the textbook makes exactly this distinction.
Figure (svg): Two columns distinguishing a growth rate from a growth factor
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477 — Example 3 and its Study Tip on growth factors and growth rates
Picture it
A percentage or a multiplier.
Figure (svg): Two columns distinguishing a growth rate from a growth factor
Both describe the same kind of growth and they enter the model differently. Reading which one you have been given is the single most important step in setting the model up.
Worked example
This is Example 3 from the textbook.
\[ \text{Twenty mice triple each year for five years. How many are there after five years?} \]
Read what is given
Why: Triples names the factor rather than a percentage.
\[ \text{factor } 3 \]
Substitute into the model
Why: Twenty times three to the fifth.
\[ 20(3) ^{5} \]
Evaluate the power
Why: Three to the fifth is two hundred and forty-three.
\[ 243 \]
Multiply
Why: Twenty times two hundred and forty-three.
\[ 4860 \]
Figure (svg): A population multiplied by three each year
\[ 20(3)^5 = 4860 \]
Verify: build the table year by year
Why: Twenty, sixty, one hundred and eighty, five hundred and forty, one thousand six hundred and twenty, four thousand eight hundred and sixty. The table's last entry matches the formula, and the table shows the tripling explicitly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477
Sorting
Read the wording carefully.
Sort into buckets
Sort each description by what it supplies.
The three rates are all percentages and the three factors are all multipliers. That correspondence is what the Study Tip is pointing at, and it holds in nearly every problem.
Worked example
Guided Practice 3. The same reading of the wording.
\[ \text{Thirty rabbits double each year for six years. How many are there after six years?} \]
Identify the factor
Why: Doubles means multiply by two.
\[ \text{factor } 2 \]
Substitute
Why: Thirty times two to the sixth.
\[ 30(2) ^{6} \]
Evaluate the power
Why: Two to the sixth is sixty-four.
\[ 64 \]
Multiply
Why: Thirty times sixty-four.
\[ 1920 \]
Figure (svg): The solution to Worked example a doubling population shown as a ladder of expressions, one row per algebraic move
\[ 30(2)^6 = 1920 \]
Verify: check what the corresponding rate would be
Why: A factor of two is one plus one, so the growth rate is one, or a hundred per cent — the population increases by its whole size each year. Doubling and growing by a hundred per cent are the same thing, which is a useful translation between the two languages.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477
Error analysis
The student modelled a population that triples each year.
Annotate
On: \( \begin{aligned} y &= C(1 + r)^t \\ &= 20(1 + 3)^5 \\ &= 20(4)^5 = 20\,480 \end{aligned} \)
The Study Tip's rule is worth holding on to: growth factors are usually whole numbers and growth rates are usually percentages. A whole number in a bracket next to a one should prompt a second look.
Faded example
Triples names the whole bracket.
Fill in the blanks
\text3: \quad y = 20(5)^___} = 4860
Why: The three replaces the whole bracket rather than sitting inside it with a one. Writing one plus three would give a factor of four and an answer more than four times too large.
Translation
Rate or factor, then substitute.
Match the pairs
Why: The two percentage descriptions give factors just above one, and the two multiplier descriptions give whole-number factors. The size of the factor is a quick check on whether the reading was right.
Socratic
Doubling and a hundred per cent describe the same thing.
Discussion prompt
Explain how to convert a growth factor into a growth rate and back. Then say what growth rate corresponds to tripling, and why the number surprises people.
Hint: Subtract one, or add one.
Answer:
The factor is one plus the rate, so subtracting one from a factor gives the rate and adding one to a rate gives the factor. A factor of 1.1 is a rate of 0.1, or ten per cent, and a factor of two is a rate of one, or a hundred per cent.
Tripling is a factor of three, so the rate is two — two hundred per cent. That surprises people because tripling sounds like three hundred per cent, but the original amount is kept as well as the increase. Growing by three hundred per cent would quadruple the quantity, which is exactly the error the previous section's student made.
Section
Section 3
Concept
Compound interest is interest paid on the original principal and on the interest already earned. It is exponential growth, so the same model applies with P for the principal and A for the balance.
\[ A = P(1 + r)^t \]
The letters change and the formula does not.
Figure (svg): A balance growing by compound interest over several years
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477 — Example 2, Find the Balance in an Account, and its Writing Algebra note
Picture it
Five hundred at eight per cent.
Figure (svg): A balance growing by compound interest over several years
The curve steepens because each year's interest is computed on a larger balance. That is the whole difference between compound and simple interest.
Worked example
This is Example 2 from the textbook.
\[ \text{Deposit } 500 \text{ dollars at } 8\% \text{ compounded yearly. What is the balance after } 6 \text{ years?} \]
Identify the parts
Why: Principal five hundred, rate eight per cent, six years.
Write the model
Why: The growth model with P and A.
\[ A = P(1 + r) ^{t} \]
Substitute
Why: Five hundred times 1.08 to the sixth.
\[ 500(1.08) ^{6} \]
Evaluate
Why: A calculator gives about seven hundred and ninety-three.
\[ \text{about } \$ 793 \]
Figure (svg): A balance growing by compound interest over several years
\[ 500(1.08)^6 \approx 793 \]
Verify: compare with simple interest
Why: Eight per cent of five hundred is forty a year, so six years of simple interest would give seven hundred and forty. Compounding gives fifty-three dollars more, which is the interest earned on the interest — and that gap is the whole reason compounding matters.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477
Faded example
Principal times the factor to the power of the term.
Fill in the blanks
A = 500(1 + 0.08)^6 = 500(1.08)^6 \approx 793
Why: The rate of eight per cent gives a growth factor of 1.08, applied once for each of the six years. Simple interest would have given seven hundred and forty, so the compounding is worth about fifty-three dollars.
Worked example
Guided Practice 2. Ten years at six per cent.
\[ \text{Deposit } 750 \text{ dollars at } 6\% \text{ compounded yearly. What is the balance after } 10 \text{ years?} \]
Identify the parts
Why: Principal seven hundred and fifty, rate six per cent, ten years.
Write the growth factor
Why: One plus six hundredths.
\[ 1.06 \]
Substitute
Why: Seven hundred and fifty times 1.06 to the tenth.
\[ 750(1.06) ^{10} \]
Evaluate
Why: About one thousand three hundred and forty-three.
\[ \text{about } \$ 1343 \]
Figure (svg): The solution to Worked example a longer term shown as a ladder of expressions, one row per algebraic move
\[ 750(1.06)^{10} \approx 1343 \]
Verify: check against simple interest again
Why: Six per cent of seven hundred and fifty is forty-five a year, so ten years simple would give twelve hundred. Compounding adds about a hundred and forty-three more, and the gap grows with the term — over twenty years it would be far larger still.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 477-477
Trap
\[ 500 \text{ at } 8\% \text{ for } 6 \text{ years} \]
Compute eight per cent of five hundred and multiply by six
Why: The interest is eight per cent a year, so six years of it seems like six times the annual amount.
\[ 500 + 6(40) = 740 \quad \text{(simple interest)} \]
That would be right if the interest were paid out each year and never reinvested. Compounded, each year's interest earns interest of its own, so the balance is fifty-three dollars higher.
\[ 500(1.08)^6 \approx 793 \]
Multiply by the growth factor once per period
Why: Compounding means the whole balance grows, including previously earned interest.
The word compounded is what signals the exponential model; without it the situation would be linear.
Prediction
Compound against simple interest on the same deposit.
Predict first
What happens to the gap between them as the term lengthens?
Correct: It grows, and grows faster as time goes on.
\[ 6 \text{ years}: \; 793 \text{ against } 740 \qquad 20 \text{ years}: \; 2330 \text{ against } 1300 \]
Why: Simple interest adds the same amount every year while compound interest adds a percentage of an ever-larger balance, so the annual additions themselves grow. The gap is small in the first year or two and widens steadily, which is why compounding matters most over long terms.
Elimination
1000 dollars at 5 per cent for t years.
Eliminate the wrong options
Which formula gives the balance?
Survives elimination: A
Why: Five per cent gives a factor of 1.05, applied once per year to the whole balance. Options C and D are the two conversion errors and both produce absurd results, which a single check after one year would expose.
Socratic
Both use the same rate.
Discussion prompt
Explain why compound interest produces more than simple interest at the same rate. Then say when the difference would be negligible.
Hint: Ask what the interest is computed on each year.
Answer:
Simple interest is always computed on the original principal, so the annual addition never changes. Compound interest is computed on the current balance, which includes previously earned interest — so the interest earns interest, and each year's addition is larger than the last.
The difference is negligible over a single period, since in the first year the balances are identical, and it is small when the rate is very low or the term very short. It becomes dramatic over decades: at eight per cent, thirty years of compounding gives ten times the deposit while simple interest gives less than three and a half times.
Section
Section 4
Concept
Growing by a fixed percentage means the increase itself grows, because the percentage is taken of an ever-larger amount. That is what makes the graph a curve rather than a line.
Ten per cent of a hundred is ten; ten per cent of a thousand is a hundred.
Figure (svg): A quantity growing by a fixed percentage, shown as increasing steps
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-481 — the contrast between the growth model and the linear models of Chapter 5
Picture it
Each bar ten per cent longer.
Figure (svg): A quantity growing by a fixed percentage, shown as increasing steps
The percentage is constant and the increases are not. That single observation is the difference between this chapter's models and Chapter 5's.
Worked example
A hundred, growing two ways.
\[ \text{Compare adding } 10 \text{ a year with growing by } 10\% \text{ a year, from } 100. \]
Tabulate the linear growth
Why: A hundred and ten, a hundred and twenty, a hundred and thirty.
Tabulate the percentage growth
Why: A hundred and ten, a hundred and twenty-one, a hundred and thirty-three.
Compare the first steps
Why: Both add ten in the first year.
Compare later
Why: The percentage growth pulls ahead and keeps widening.
Figure (svg): Compound growth compared with a constant addition
\[ 100 + 10t \quad \text{against} \quad 100(1.1)^t \]
Verify: compute both after twenty years
Why: Linear gives three hundred and exponential gives about six hundred and seventy-three. Starting identically and ending twice as far apart is the shape every such comparison takes, and the gap keeps widening after that.
Sorting
A fixed amount or a fixed percentage.
Sort into buckets
Sort each description by the kind of growth it describes.
The wording is the whole test: a fixed number of units means linear and a percentage or a multiplier means exponential. Reading it carefully decides which model to write.
Worked example
Example 1's model, evaluated.
\[ \text{Using } y = 0.06(1.1)^t, \text{ find the catfish's weight after } 7, 14 \text{ and } 42 \text{ days.} \]
Take seven days
Why: 0.06 times 1.1 to the seventh.
\[ \text{about } 0.12 g \]
Take fourteen days
Why: 0.06 times 1.1 to the fourteenth.
\[ \text{about } 0.23 g \]
Take forty-two days
Why: 0.06 times 1.1 to the forty-second.
\[ \text{about } 3.4 g \]
Describe the pattern
Why: The weight roughly doubles each week.
Figure (svg): A quantity growing by a fixed percentage, shown as increasing steps
\[ 0.06(1.1)^{42} \approx 3.4 \text{ g} \]
Verify: check the weekly doubling
Why: 1.1 to the seventh is about 1.95, which is nearly two — so ten per cent a day is close to doubling every week. Recognising a familiar factor hidden in an unfamiliar one makes the model much easier to think about.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Trap
A quantity of 100 grows by 10 per cent a year.
Add ten each year, since ten per cent of a hundred is ten
Why: The percentage was computed once and then reused.
After the first year the amount is a hundred and ten, so ten per cent of it is eleven rather than ten. Taking the percentage of the original every time turns exponential growth into linear growth.
\[ 100 \to 110 \to 121 \to 133.1 \]
Take the percentage of the current amount each period
Why: That is what multiplying by the growth factor does automatically.
The model's power of t applies the factor once per period, which is why it handles this correctly without any extra thought.
Faded example
Not of the original.
Fill in the blanks
Starting at 100 and growing 10% a year gives 110 after one year, then 121 after two, because 10% of 110 is 11.
Why: The second year's increase is eleven rather than ten, because the percentage applies to the larger amount. Multiplying by the growth factor does this automatically, which is why the model needs no separate instruction about it.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Linear | Exponential | |
|---|---|---|
| Each period | adds a fixed amount | multiplies by a fixed factor |
| The steps | stay the same size | grow |
| The model | y = mx + b | y = C(1 + r)^t |
Every row follows from the first. Whether the constant is added or multiplied decides the step sizes, the shape and the form of the model.
Socratic
The percentage never changes.
Discussion prompt
Explain why growing by an unchanging percentage produces increases that grow. Then say what this means about the graph's steepness far to the right.
Hint: Ask what the percentage is taken of.
Answer:
A percentage is a proportion of the current amount, so as the amount grows the same proportion is a larger number. Ten per cent of a hundred is ten and ten per cent of a thousand is a hundred, so the increases grow in step with the quantity itself.
That means the graph's steepness is proportional to its height, so far to the right — where the quantity is large — the curve rises almost vertically. It is the same observation as in Lesson 8.3, now with a reason drawn from the situation rather than from the algebra.
Section
Section 5
Concept
An exponential growth model describes a period during which the percentage stays constant. Beyond that period it predicts numbers no real quantity can reach.
The catfish model is stated for the first six weeks only.
Figure (svg): A quantity growing by a fixed percentage, shown as increasing steps
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-481 — Example 1's restriction to the first six weeks of life
Picture it
Constant percentage, for a while.
Figure (svg): A quantity growing by a fixed percentage, shown as increasing steps
A catfish does not gain ten per cent a day forever, and the textbook's problem says so by naming the first six weeks. That restriction is part of the model.
Worked example
The model's own wording limits it.
\[ \text{What does } y = 0.06(1.1)^t \text{ predict after one year, and is that plausible?} \]
Substitute t equal to 365
Why: 0.06 times 1.1 to the three hundred and sixty-fifth.
Estimate the size
Why: 1.1 to the 365th is around ten to the fifteenth.
\[ \text{about } 10 ^{13}\text{ grams} \]
Interpret
Why: That is tens of millions of tonnes.
Say what went wrong
Why: The model was stated for the first six weeks only.
Figure (svg): The solution to Worked example extend the catfish model too far shown as a ladder of expressions, one row per algebraic move
\[ 0.06(1.1)^{365} \approx 10^{13} \text{ g} \]
Verify: say what actually happens to the fish
Why: Growth slows as the fish matures, so the percentage falls away from ten per cent. Every exponential growth model in a real situation eventually stops applying, because nothing in the world can grow by a fixed percentage indefinitely.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Prediction
A model with a growth factor above one.
Predict first
What does such a model predict in the very long run?
Correct: Unbounded growth, exceeding any fixed number eventually.
Models with a ceiling exist and are studied later; the exponential model is the simplest description of early growth.
Why: Multiplying by a factor above one repeatedly passes any bound you name, so the model has no ceiling built into it. Real quantities always meet limits — food, space, population, money — so a growth model eventually stops describing the world, and knowing that is part of using one responsibly.
Worked example
The viewer model from Guided Practice 1.
\[ \text{Using } v = 50\,000(1.02)^t, \text{ predict the viewers after one year and after ten years.} \]
Take twelve months
Why: Fifty thousand times 1.02 to the twelfth.
\[ \text{about } 63 400 \]
Take a hundred and twenty months
Why: Fifty thousand times 1.02 to the hundred and twentieth.
\[ \text{about } 537 000 \]
Compare with plausibility
Why: The first is a realistic target for a local station.
Question the second
Why: Half a million viewers may exceed the local population.
Figure (svg): The solution to Worked example use a model inside its range shown as a ladder of expressions, one row per algebraic move
\[ 50\,000(1.02)^{12} \approx 63\,400 \]
Verify: ask what would limit the growth
Why: A local audience is bounded by the number of people in the area, so growth must slow as the station approaches that ceiling. A model with no ceiling in it cannot represent that, which is why the ten-year figure should be treated with suspicion.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-476
Trap
\[ v = 50\,000(1.02)^{240} \]
Report that the station will have six million viewers in twenty years
Why: The model gives a number and the arithmetic is correct.
Six million probably exceeds the population of the area. The model has no ceiling in it, so it will happily predict more viewers than there are people.
State the figure with the assumption: if two per cent monthly growth continued for twenty years.
Report the assumption alongside any long extrapolation
Why: The model describes a rate, not a guarantee that the rate persists.
This is the same caution as in Lesson 5.5, and exponential models need it more, because their errors compound too.
Faded example
The wording gives it.
Fill in the blanks
The catfish model applies during the first six weeks, so a prediction for one year lies outside its range.
Why: The problem states the period explicitly, so any prediction beyond it is an extrapolation. Naming the range when reporting a model is what distinguishes an estimate from a guess.
Hypothesis
Predict before you decide.
Predict first
The model predicts about 537 000 viewers after ten years. What is the main reason for doubt?
Correct: A local audience is limited by the population of the area.
Growth that slows as it approaches a ceiling is modelled by a different family of functions, studied in later courses.
Why: The model has no ceiling, so it will keep growing past whatever number of people actually live there. The arithmetic is fine and the rate is perfectly reasonable for a month; the problem is that no rate can persist once the audience approaches everyone available. Recognising the ceiling a situation imposes is a modelling judgement rather than a mathematical one.
Socratic
It eventually predicts absurdities.
Discussion prompt
Say why exponential growth models are used despite eventually predicting impossible numbers. Then say what to report alongside any prediction from one.
Hint: Ask what it gets right.
Answer:
It describes the early phase of growth extremely well, when a quantity is far from any limit and each period's increase really is proportional to the current amount. Populations, investments, epidemics and audiences all behave that way at first, and the model captures that phase with only two numbers.
Any prediction should be reported with the assumption behind it — that the rate continues — and with the range the model was fitted or stated over. That turns a number into a conditional statement, which is what it actually is, and it is the honest way to use a model whose long-run behaviour you know to be wrong.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Given a rate | Given a factor | |
|---|---|---|
| Typical wording | increases by 8% | triples each year |
| What you substitute | 0.08 inside the bracket with the 1 | 3 as the whole bracket |
| The growth factor | 1.08 | 3 |
The bottom row is what the model actually uses, and the two top rows are the two routes to it. Reading which route a problem supplies is the setup's only real decision.
Pattern
Whether the growth is described by a percentage or a multiplier, the same five moves cover it.
Step four costs one multiplication and catches every conversion error, since a factor of eleven and a factor of 1.1 behave completely differently in a single period.
OpenStax Intermediate Algebra 2e, §10.2 Evaluate and Graph Exponential Functions §10.2
Check
Convert the percentage.
Check your understanding
A quantity of 200 grows by 5 per cent a year. What is the model?
Answer: A
Why: Five per cent is 0.05, and adding one gives a growth factor of 1.05. Testing after one year gives two hundred and ten, which is two hundred plus five per cent of it.
Check
Doubling names the factor.
Check your understanding
A colony of 40 bacteria doubles each hour. How many are there after 5 hours?
Answer: A
Why: Doubling gives a growth factor of two, so the model is forty times two to the fifth. Two to the fifth is thirty-two, and forty times thirty-two is one thousand two hundred and eighty.
Check
Compounding applies to the whole balance.
Check your understanding
Which gives the balance on 1000 dollars at 4 per cent compounded yearly for t years?
Answer: A
Why: Four per cent is 0.04, so the growth factor is 1.04 and it is applied once per year to the whole balance including earned interest.
Real world
This is Example 1's situation. A newly hatched channel catfish weighs about 0.06 grams and gains about 10 per cent of its weight each day during its first six weeks.
Discussion prompt
Write the model, predict the weight after six weeks, and say what the model would give after a year. Then say what that tells you about the model's range.
Hint: Six weeks is forty-two days.
Answer:
\[ y = 0.06(1.1)^t \;\Longrightarrow\; y(42) = 0.06(1.1)^{42} \approx 3.4 \text{ g} \]
After six weeks the fish weighs about three and a half grams, which is a fifty-fold increase from its hatching weight and entirely plausible for a growing fry.
After a year the model gives around ten to the thirteenth grams — tens of millions of tonnes. That is not a failure of the arithmetic but of the range: the problem stated ten per cent a day for the first six weeks, and growth slows sharply as a fish matures. Every exponential growth model in nature has such a range, and stating it is part of reporting the model.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A population triples each year. What goes into the model's bracket?
Correct: 3, because tripling gives the growth factor directly.
\[ 20(3)^5 = 4860 \quad \text{against} \quad 20(4)^5 = 20\,480 \]
Why: The bracket in the model is the growth factor, and tripling says the factor is three. Writing one plus three gives four, which describes a quadrupling and produces an answer more than four times too large. The distinction is the textbook's Study Tip: growth factors are usually whole numbers and growth rates are usually percentages, so a whole number sitting inside a bracket with a one should prompt a second look. In rate terms, tripling is a rate of two hundred per cent, since the original is kept as well.
Explain it
They keep putting percentages straight into the model.
Discussion prompt
In no more than four sentences, explain how a percentage becomes a growth factor and why the one is there. Then tell them the check that catches a wrong factor.
Hint: Keep the whole, add the increase.
Answer:
A usable answer: a percentage tells you how much gets added, so ten per cent means you keep everything you had and add a tenth — which is multiplying by one point one. The one keeps the original and the point one is the increase, so you divide the percentage by a hundred and add one to get the number you multiply by.
To check, work out one period by hand and compare. If your factor turns a hundred into a hundred and ten you have it right, and if it turns a hundred into eleven hundred you have used the percentage without converting it.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The conversion is fixed by dividing by a hundred and adding one, then testing one period. Rate against factor is fixed by asking whether the problem named a percentage or a multiplier. Compound interest is fixed by remembering that the whole balance grows, including earned interest. The range is fixed by reading the period the problem states and saying so with any prediction. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the growth model with each of its four parts labelled, and beside it convert three percentages into growth factors, showing the division and the addition separately. Underneath, write two situations of your own — one describing a percentage and one describing a multiplier — and write the model for each, marking clearly which number went inside the bracket and which replaced it. In the middle, take a starting amount of a hundred and tabulate six periods of ten per cent growth beside six periods of adding ten, writing each period's increase in a third column so that the growing steps are visible. Graph both on one plane. In the lower half, evaluate one of your models at a time far beyond its stated range, write the absurd answer out, and write one sentence explaining what limits the real quantity. Finally, in the margin, write the two-line rule for telling a growth rate from a growth factor.
Your two tables should agree in the first period and separate afterwards. If they agree for longer than one period, the percentage was being taken of the original amount rather than of the current one.
Recap
Five things, and the second is the one the wording decides.
| If the question says | Your first move is |
|---|---|
| Increases by 8% each year | Growth factor 1.08 |
| Triples each year | Growth factor 3, with nothing added |
| Compounded yearly | Use A = P(1 + r)^t |
| Adds a fixed amount each year | That is linear, not exponential |
| Predict far into the future | State the assumption and the range |
Lesson 8.7 turns the model round. When a quantity loses a fixed percentage each period the factor drops below one, and the same machinery describes decay — with one sign changed and every other idea intact.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.6 Exponential Growth Functions §8.6, pp. 476-481 — everything on these slides traces back here
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