Unit 2 · Topic 2.5 · about 30 minutes · 7 parts · 14 scored checks, 2 of them multiple choice

Exponential Function Context and Data Modeling

Built from College Board's course framework · Every answer key independently checked

Part 1 Think first

Day 10 for a new video

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A new video had 1,000 views on its first day and 2,000 on its second. One friend says it will get about 10,000 views on day 10. Another says it will get over 500,000. Who could be right?

In short. An exponential model describes a quantity that is multiplied by the same factor over every equal-length interval of input, and you build it from an initial value and that factor, from two points, or from data with a regression.

In $f (x) = a b^{x}$ , $a$ is the value at $x = 0$ and $b$ is the growth factor for each unit of input: a 4% increase is $b = 1.04$ and a 4% decrease is $b = 0.96$ .
Two input-output pairs give a system you solve for $a$ and $b$ . If the data level off toward some value, subtract that value first and the constant ratio appears.
An equivalent form such as $(2^{7})^{d / 7}$ shows the factor over a different period, and a prediction only means something inside the domain the context allows.

Part 2 Learn it

Four short ideas, each checked as you go

2a. Equal steps in, equal ratios out

A bike-share program counts its rides each month: 1,000, then 1,200, then 1,440, then 1,728. The differences grow (200, 240, 288), so this is not linear. Divide instead: $\frac{1200}{1000} = \frac{1440}{1200} = \frac{1728}{1440} = 1.2$ .

When outputs over equal-length input intervals have a constant ratio, an exponential function models them. With whole-number inputs it is repeated multiplication: start at 1,000 and multiply by 1.2 once per month.

R (m) = 1000 (1.2)^{m}

In $f (x) = a b^{x}$ , $a$ is the value at $x = 0$ and $b$ is the growth factor for each one-unit step. The factor ties to a percent change: 1.2 means each month is 120% of the month before, an increase of 20%. A factor below 1 is decay: $b = 0.85$ keeps 85% and loses 15% each step.

So a 4% increase is $b = 1.04$ , never 0.04 and never 4.

Quick check · Fill the statement

No calculator. A town's population grows by 3% each year. One year after the count began, it had 12,360 people. Tap a blank, then tap an answer.

The model is $P (t) = a \cdot b^{t}$ , with $t$ in years, where $a$ = and $b$ = .

2b. Build it from two points, then use it

Often you know two outputs, not the start. A phone game had 6,000 players in week 1 and 13,500 in week 3. Write both into $P (w) = a b^{w}$ :

a b = 6000 a b^{3} = 13500

Divide the second equation by the first and $a$ cancels: $b^{2} = 2.25$ , so $b = 1.5$ . Then $a = \frac{6000}{1.5} = 4000$ , the players in week 0.

With several data points that are close to proportional but not exactly, a graphing calculator runs an exponential regression and gives the $a$ and $b$ that fit best.

Once you have a model, use it both ways. Put in a week to predict players; set $4000 (1.5)^{w} = 30000$ and solve, with a graph or a calculator, to find when the game reaches 30,000 (about week 5). Keep the context's domain in mind. Weeks before launch mean nothing, and no game keeps growing 50% a week forever, so a prediction far past the data deserves doubt.

Quick check · Calculate

No calculator. A jar holds 60 fruit flies on day 2 and 480 on day 5. The count is modeled by $F (d) = a b^{d}$ . What is $a$ , the number of flies on day 0?

Your answer flies

2c. Shift first, then look for the ratio

A can of soda comes out of the fridge at 4 degrees Celsius into a 22 degree room. Every 10 minutes it reads 4, 10, 14 and about 16.67 degrees. The ratios $\frac{10}{4}$ and $\frac{14}{10}$ are not equal, so the readings themselves are not exponential.

But the soda is closing a gap. Subtract each reading from 22: the gaps are 18, 12, 8 and 5.33, and each is $\frac{2}{3}$ of the one before. Adding or subtracting a constant can reveal a proportional pattern that the raw data hide.

Build the model as a transformation of $a b^{x}$ :

T (t) = 22 - 18 {(\frac{2}{3})}^{t / 10}

The 22 is a vertical shift to the room temperature the soda approaches. The minus sign flips the shrinking gap below it. Dividing $t$ by 10 makes the factor $\frac{2}{3}$ apply once every 10 minutes. If a context counts years from 2015 rather than from 0, shift horizontally: $a b^{x - 2015}$ equals $a$ when $x = 2015$ .

Quick check · Decide and justify

No calculator. Tea is poured at 90 degrees Celsius in a 20 degree room. Every 5 minutes it reads 90, 76, 64.8 and 55.84 degrees. Which model fits, with $t$ in minutes?

Which reason supports your answer?

2d. Same function, different clock

A cell culture is modeled by $N (d) = 200 (1.5)^{d}$ , with $d$ in days. The base says the culture grows by a factor of 1.5 each day. An equivalent form answers a different question:

N (d) = 200 {({1.5}^{7})}^{d / 7} \approx 200 (17.09)^{d / 7}

Now the base says the culture multiplies by about 17.09 every week, because $\frac{d}{7}$ counts weeks. For hours, $N (d) = 200 {({1.5}^{1 / 24})}^{24 d} \approx 200 (1.0170)^{24 d}$ : about 1.70% per hour. Dividing 50% by 24 gives 2.08%, which is wrong, because growth on growth adds up.

Many models use the natural base $e$ , about 2.718, written $f (t) = a e^{k t}$ . It is still an exponential with a factor per unit: $e^{k t} = {(e^{k})}^{t}$ . For $e^{0.05 t}$ the factor is $e^{0.05} \approx 1.0513$ , a growth of about 5.13% per unit, a little more than 5%.

Quick check · Match

No calculator. Each form below equals $f (d) = 400 (1.1)^{d}$ , with $d$ in days. Use ${1.1}^{7} \approx 1.95$ , ${1.1}^{30} \approx 17.4$ and ${1.1}^{1 / 24} \approx 1.004$ . Tap a form, then what its base shows.

Part 3 Watch it done

What is the car worth?

Watch each step, then do the last one yourself.

A used car is worth 20,000 dollars one year after it is bought and 16,000 dollars two years after. Its value $V (t)$ , in dollars, $t$ years after purchase, is modeled by an exponential function. Build the model and find the car's value four years after purchase.

Step 1 · Find the factor from equal steps
Years 1 and 2 are one year apart, so the factor is the ratio of the values: $b = \frac{16000}{20000} = 0.8$ . The car keeps 80% of its value each year, a 20% drop.
Step 2 · Go back to $t = 0$ for $a$
$V (1) = a b = 20000$ , so $a = \frac{20000}{0.8} = 25000$ . The model is $V (t) = 25000 (0.8)^{t}$ for $t \geq 0$ .
Why this step?
What does $a = 25000$ mean, and why is it not the 20,000 from the problem?
The value $a$ is $V (0)$ , the car's value on the day it was bought, 25,000 dollars. The 20,000 dollars is one year later, after one drop of 20%.
Your step · Calculate

No calculator. Finish it. What is the car worth four years after purchase, in dollars?

Your answer dollars

Part 4 Try it

Fit a curve to a medicine's decay

A nurse records the amount of a medicine left in a patient's bloodstream every hour after a dose. The ratios of consecutive readings sit between about 0.84 and 0.87, close to constant but not exactly, which is what real measurements look like. That makes this a job for an exponential regression. Enter the hours and the amounts as two lists and run ExpReg, which reports a model $y = a b^{x}$ .

Medicine in the bloodstream
$t$ (hours)	$A (t)$ (milligrams)
0	250
1	212
2	181
3	152
4	131
5	110
6	94

Try it · Calculate

Graphing calculator required. The amount of a medicine in the bloodstream, in milligrams, is 250, 212, 181, 152, 131, 110 and 94 at $t = 0, 1, 2, 3, 4, 5, 6$ hours. Run an exponential regression on these data. By what percent does the amount decrease each hour, according to the model? Round only at the end, and give a percent to one decimal place.

Your answer %

Try it · Calculate

Graphing calculator required. An exponential regression on medicine data (250, 212, 181, 152, 131, 110 and 94 milligrams at $t = 0$ to 6 hours) gives $A (t) = 249.88 (0.8494)^{t}$ . After how many hours does the model first give 50 milligrams? Give your answer to the nearest tenth of an hour.

Your answer hours

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 2, Modeling a Non-Periodic Context.

Free-response practice · Question 2 style

Graphing calculator required. A streaming service had 4.0 million subscribers at $t = 2$ and 6.25 million at $t = 4$ , where $t$ is years since 2020. The number of subscribers, in millions, is modeled by $S (t) = a b^{t}$ . (a) Find $a$ and $b$ . (b) Interpret $b$ as a percent change in context. (c) Rewrite $S$ in a form that shows the growth factor per month, and explain why the monthly percent is not one twelfth of the yearly percent. (d) Use the model to predict the subscribers in 2025.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $a b^{4} = 6.25$ and $a b^{2} = 4$ , so $b^{2} = 1.5625$ and $b = 1.25$ . Then $a = \frac{4}{{1.25}^{2}} = 2.56$ , so $S (t) = 2.56 (1.25)^{t}$ . (b) Each year the subscribers are 1.25 times the year before, an increase of 25% per year. (c) $S (t) = 2.56 {({1.25}^{1 / 12})}^{12 t} \approx 2.56 (1.0188)^{12 t}$ , about 1.88% per month. One twelfth of 25% is about 2.08%, but each month grows on the larger total from the month before, so a smaller monthly percent compounds to 25% over 12 months. (d) 2025 is $t = 5$ : $S (5) = 2.56 (1.25)^{5} \approx 7.81$ million subscribers.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Sets up two equations from the points and finds $b = 1.25$ and $a = 2.56$ .
States that subscribers increase by 25% each year.
Writes $S (t) = 2.56 {({1.25}^{1 / 12})}^{12 t}$ with a monthly factor of about 1.0188.
Explains that monthly growth compounds, so the monthly percent is less than 25% divided by 12.
Predicts about 7.81 million subscribers for $t = 5$ .

Part 6 Check yourself

Algae on a pond

The three case questions use the same data. Answer them from the table.

A science class measures the area of a pond covered by algae, $A (w)$ in square meters, $w$ weeks after they first notice it. The whole pond has a surface of 400 square meters.

Area covered by algae
$w$ (weeks)	$A (w)$ (square meters)
0	12
1	18
2	27
3	40.5
4	60.75

Case question 1 · Multiple choice

No calculator. Which statement describes how the area covered by algae changes from week to week?

Case question 2 · Calculate

No calculator. The class models the data with $A (w) = 12 (1.5)^{w}$ . The model can only hold while the algae fits on the pond. What is the last whole week for which the model gives an area of no more than 400 square meters?

Your answer weeks

Case question 3 · Multiple choice

No calculator. The area is $A (w) = 12 (1.5)^{w}$ with $w$ in weeks. Which expression gives the same area with the input $d$ in days?

The rest stand on their own and do not use the case.

Check yourself · Sort it

No calculator. Each statement describes a yearly change in a model $f (t) = a b^{t}$ . Tap a statement, then the base it gives.

b = 1.04

b = 0.96

b = 1.4

Check yourself · Spot the error

No calculator. A bacteria culture is modeled by $N (t) = 500 e^{0.2 t}$ , with $t$ in hours. A student finds the percent change per hour. Tap the first line with a mistake.

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

A museum counts weekly visitors, in hundreds, $w$ weeks after a new exhibit opens: 5.0, 16.2, 25.8, 33.8 and 40.2 for $w = 0$ to 4. The first differences, 11.2, 9.6, 8.0 and 6.4, fall by 1.6 each week, so a quadratic regression is the right one to run.

Weekly visitors, in hundreds
$w$ (weeks)	$V (w)$
0	5.0
1	16.2
2	25.8
3	33.8
4	40.2

From 1.14 Function Model Construction and Application · Calculate

Graphing calculator required. A museum counts weekly visitors, in hundreds, $w$ weeks after a new exhibit opens: 5.0, 16.2, 25.8, 33.8 and 40.2 for $w = 0, 1, 2, 3, 4$ . Run a quadratic regression on these data, and use it to predict the number of visitors, in hundreds, in week 6.

Your answer hundred visitors

From 1.13 Function Model Selection and Assumption Articulation · Match

No calculator. Tap a situation, then the model type it suggests.

In three lines

Constant ratios over equal input steps mean exponential: $f (x) = a b^{x}$ with $a = f (0)$ and growth factor $b$ , so 4% more is 1.04.
Build the model from a factor and a start, from two points by dividing one equation by the other, from shifted data, or with an exponential regression.
Rewrite the exponent to show the factor for another period, and use the model only where the context makes sense.

Vocabulary

Exponential model: A function $f (x) = a b^{x}$ , or a transformation of one, used for a quantity whose outputs over equal-length input intervals are proportional.
Initial value: The value $a$ in $f (x) = a b^{x}$ , which is the output at $x = 0$ .
Growth factor: The base $b$ in $f (x) = a b^{x}$ , the number the output is multiplied by for each one-unit increase in the input.
Percent change: The change a growth factor stands for in context: $b = 1.04$ is a 4% increase per unit and $b = 0.96$ is a 4% decrease.
Exponential regression: A model $y = a b^{x}$ that technology fits as closely as possible to a data set.
Natural base $e$: The number $e \approx 2.718$ , often used as the base of exponential models written $f (t) = a e^{k t}$ .

Course alignment, for teachers

AP Precalculus topic 2.5, Unit 2: Exponential and Logarithmic Functions.

Skill 1.C: Construct new functions, using transformations, compositions, inverses, or regressions, that may be useful in modeling contexts, criteria, or data, with and without technology.
Skill 3.B: Apply numerical results in a given mathematical or applied context.

This lesson maps all 9 essential knowledge statements of the topic to the part that teaches each one: 9 are checked by a scored question. The map is in the lesson file.