Topic 2.5B · CED: Exponential Function Context and Data Modeling

Exponential Context and Data Modeling

In 2.5A every model came from exact information: a percent, a half-life, or two points. Real data is messier. Nova's actual subscriber counts don't multiply by exactly 1.5 every month. And some quantities, like a cooling cup of coffee, don't fade toward 0 at all; they level off at room temperature.

10 MIN READ5 IDEAS34 PROBLEMS18 flashcards

Practice 34 problems Next: Chapter 2.6 Competing Function Model Validation

Remember from 2.5A: an exponential model needs an initial amount $a$ and $a$ factor $b$ . This note shows how to get $a$ and $b$ from a whole data set, and how to handle data whose asymptote is not $y = 0$ .

Real Data: Nova's Actual Counts

Here are Nova's real end-of-month subscriber counts:

t (month)	0	1	2	3	4	5	6	7	8
subscribers	250	390	560	880	1270	1990	2870	4450	6490

The ratios of consecutive counts are 1.56, 1.44, 1.57, 1.44, 1.57, 1.44, 1.55, 1.46. They are not all equal, so no exponential function fits every point exactly. But they all stay between 1.44 and 1.57, close to one value. That is the sign that an exponential model is reasonable: the data changes roughly proportionally over equal intervals.

CONCEPT

Exponential Regression

No single curve $y = a \cdot b^{x}$ passes through all nine points. Exponential regression picks the $a$ and $b$ that make the curve as close as possible to ALL the points at once. The curve may not pass through any of them exactly.

It will usually miss some points a little above and some a little below. Those misses are called residuals, and 2.6 is all about them.

Worked example

Example 1. Find an exponential regression model for Nova's data on a TI-84.

01
STAT → EDIT: type the months 0 to 8 in L1 and the subscriber counts in L2.
02
STAT → CALC → ExpReg, then enter. (To also see $r$ , turn on DiagnosticOn first, from the CATALOG.)
03
Read the output: $a \approx 253.844$ , $b \approx 1.5022$ , $r \approx 0.99983$ . Write the model with at least 3 decimal places:
$y \approx 253.844\,(1.5022)^{t}, \qquad r \approx 0.99983$

CONCEPT

The Same Regression in Desmos

Step 1. Add a table: put the months in the $x_{1}$ column and the counts in the $y_{1}$ column.

Step 2. On a new line type $y_{1}$ ~ $a \cdot b^ x_{1}$ (the ~ means “fit”). Desmos finds $a$ and $b$ .

Step 3. Desmos may offer a “Log Mode” switch for this regression. With Log Mode ON you get the same numbers as the TI-84: 253.844 and 1.5022. With it OFF, Desmos fits the curve directly and gives 261.83 and 1.4948. Both are reasonable; say which one you used.

The regression curve runs through the middle of the data. Some points sit slightly above it, some slightly below.

Reading the model: Nova started with about 254 subscribers and grew by a factor of about 1.502 per month, about 50.2% monthly growth. The value $r$ is close to 1, which says the data fits an exponential shape very well. (Where exactly $r$ comes from is explained in 2.15: it measures how straight the data becomes after taking logarithms.)

For example, at $t = 3$ the model predicts about 860.4 subscribers and the actual count was 880, so the model missed by about 19.6.

COMMON MISTAKE

“The first data point is 250, so $a = 250$ .”

In a regression model, a is the value the MODEL gives at $t = 0$ (here about 253.844), not the first data value. The model is fitted to all the points at once.

“The model is great, so let's predict month 24: about 4.4 million subscribers.”

The data only covers months 0 to 8. Predicting a little past the data (month 10: about 14,849) is reasonable; predicting far past it assumes the channel keeps growing 50% a month for two years. Real growth usually slows down. Be careful when you extrapolate.

When the Asymptote Is Not y = 0

A cup of coffee at 84 °C sits in a 20 °C room. Here is its temperature every 5 minutes:

t (minutes)	0	5	10	15	20
T (°C)	84	68	56	47	40.25
change	—	−16	−12	−9	−6.75

Run the ratio test from 2.2: $68/84 \approx 0.81$ , $56/68 \approx 0.82$ , $47/56 \approx 0.84$ . Not constant. Is the coffee not exponential? It is. The problem is that the coffee cools toward 20 °C, not toward 0 °C. What decays proportionally is the DIFFERENCE between the coffee and the room:

t (minutes)	0	5	10	15	20
T − 20	64	48	36	27	20.25
ratio	—	0.75	0.75	0.75	0.75

CONCEPT

Vertically Translated Exponential Functions

$f(x) = a \cdot b^{x} + k$ is the graph of $a \cdot b^{x}$ moved up $k$ units. Its horizontal asymptote is $y = k$ instead of $y = 0$ .

Subtracting $k$ from the outputs gives back pure proportional change. For the coffee, $k = 20$ and every 5 minutes the gap to room temperature shrinks by the factor 0.75.

f(x) = a \cdot b^{x} + k

Worked example

Example 2. Build the coffee model.

01
$k$ is the value being approached: the room temperature, $k = 20$ .
02
a is the starting GAP, not the starting temperature: $a = 84 - 20 = 64$ .
03
The gap is multiplied by 0.75 every 5 minutes, so after $t$ minutes it has been multiplied t/5 times (2.5A):
$T(t) = 20 + 64\,(0.75)^{t/5}$
04
Per minute (2.4): $0.75^{1/5} \approx 0.9441$ , so the gap shrinks about 5.6% each minute:
$T(t) = 20 + 64\,(0.75)^{t/5} = 20 + 64\,\left(0.75^{1/5}\right)^{t} \approx 20 + 64\,(0.9441)^{t}$

The coffee approaches the room temperature $y = 20$ , not 0. After an hour it is still slightly warm: $T(60) \approx 22.03$ °C.

COMMON MISTAKE

“The coffee's temperature will eventually reach 0 °C.”

The asymptote is $y = k = 20$ . The model says the coffee gets closer and closer to room temperature, which is also what a real cup does. Always ask what value the quantity is approaching, and that is your $k$ .

“ $a = 84$ , the starting temperature.”

In $a \cdot b^{x}$ + k, the value at $x = 0$ is a + k, not a. So $a = 84 - 20 = 64$ : the part that is actually decaying.

Quick check

Tea cools by $T(t) = 20 + 70(0.9)^t$ . What is $T(0)$ , and what value does $T$ approach?

Finding k From a Table

In context, $k$ usually comes from the situation. Without context, look at the DIFFERENCES of consecutive outputs. For $f(x) = a \cdot b^{x} + k$ , the $k$ cancels when you subtract:

f(x+1) - f(x) = a\,b^{x+1} - a\,b^{x} = a\,(b - 1) \cdot b^{x}

So the differences themselves are an exponential pattern with the SAME factor $b$ . In particular, the first difference is

f(1) - f(0) = (a\,b + k) - (a + k) = a\,b - a = a\,(b - 1)

Worked example

Example 3. $x$ : 0, 1, 2, 3, 4 and $g(x)$ : 1, 4, 13, 40, 121. Show that $g$ is a translated exponential function and find it.

01
The x-steps are all 1. Ratios of $g$ : 4, $13/4$ , $40/13$ , $121/40$ , not constant. So $g$ is not a plain $a \cdot b^{x}$ .
02
Differences: 3, 9, 27, 81. Their ratios are $9/3 = 27/9 = 81/27 = 3$ , constant. So $g(x) = a \cdot 3^{x} + k$ with $b = 3$ .
03
First difference = $a(b - 1)$ : $3 = a(3 - 1)$ , so $a = 1.5$ .
04
$g(0) = a + k = 1$ , so $k = 1 - 1.5 = -0.5$ . Result: $g(x) = 1.5 \cdot 3^{x} - 0.5$ . Check: $g(4) = 1.5(81) - 0.5 = 121$ ✓.

COMMON MISTAKE

“The ratios of $g$ aren't constant, so $g$ is not exponential.”

That only rules out $a \cdot b^{x}$ with asymptote $y = 0$ . When the ratio test fails, check whether the DIFFERENCES have a constant ratio. If they do, the function is exponential with a vertical shift.

Worked example

Try it yourself. $x$ : 0, 1, 2, 3, 4 and $h(x)$ : 50, 26, 14, 8, 5. Write $h(x)$ in the form $a \cdot b^{x}$ + k. What is the horizontal asymptote?

Answer: Differences −24, −12, −6, −3 have ratio $1/2$ , so $b = 0.5$ . $a(0.5 - 1) = -24$ gives $a = 48$ , and $k = 50 - 48 = 2$ . $h(x) = 48(0.5)^{x} + 2$ . Check: $h(4) = 3 + 2 = 5$ ✓. The asymptote is $y = 2$ .

Quick check

A function of the form $a \cdot b^x + k$ has outputs $7, 13, 25, 49$ at $x = 0, 1, 2, 3$ . Find $b$ and $k$ .

Practice

Worked example

P1. Use Nova's regression model $y \approx 253.844(1.5022)^{t}$ to predict month 9. Is this a reasonable prediction?

Answer: $253.844(1.5022)^{9} \approx 9,888$ . Month 9 is just past the data (months 0 to 8), so the prediction is reasonable.

Worked example

P2. $x$ : 0, 1, 2, 3, 4 and $f(x)$ : 10, 16, 19, 20.5, 21.25. Find $f(x) = a \cdot b^{x} + k$ and describe what happens in the long run.

Answer: Differences 6, 3, 1.5, 0.75 have ratio $1/2$ , so $b = 0.5$ . $a(0.5 - 1) = 6$ gives $a = -12$ , and $k = 10 - (-12) = 22$ . $f(x) = -12(0.5)^{x} + 22$ : it rises toward 22 but never reaches it (like a cold drink warming to room temperature).

Worked example

P3. $x$ : 0, 1, 2, 3, 4 and $y$ : 5, 8, 14, 26, 50. Find the model.

Answer: Differences 3, 6, 12, 24 have ratio 2, so $b = 2$ . $a(2 - 1) = 3$ gives $a = 3$ , and $k = 5 - 3 = 2$ . $y = 3 \cdot 2^{x} + 2$ .

Common slips

“The first data point is 250, so $a = 250$ .”

In a regression model, a is the value the model gives at $t = 0$ (here about 253.844), not the first data value. The model is fitted to all the points at once.

“The model is great, so let's predict month 24: about 4.4 million subscribers.”

The data only covers months 0 to 8. Predicting a little past the data (month 10: about 14,849) is reasonable; predicting far past it assumes the channel keeps growing 50% a month for two years. Real growth usually slows down. Be careful when you extrapolate.
“The coffee's temperature will eventually reach 0 °C.”

The asymptote is $y = k = 20$ . The model says the coffee gets closer and closer to room temperature, which is also what a real cup does. Always ask what value the quantity is approaching, and that is your $k$ .

“ $a = 84$ , the starting temperature.”

In $a \cdot b^{x}$ + k, the value at $x = 0$ is a + k, not a. So $a = 84 - 20 = 64$ : the part that is actually decaying.
“The ratios of $g$ aren't constant, so $g$ is not exponential.”

That only rules out $a \cdot b^{x}$ with asymptote $y = 0$ . When the ratio test fails, check whether the differences have a constant ratio. If they do, the function is exponential with a vertical shift.

Lock it in

Try the flashcards

18 cards · Linear or exponential?, Exponential models

Start

Recap card

5 lines to re-read the night before.

01
Real data is rarely perfectly exponential. If the ratios over equal intervals are nearly constant, an exponential model is appropriate.
02
Exponential regression (ExpReg) finds the $a \cdot b^{x}$ closest to all the data; a is the model's value at 0, not the first data value. Far extrapolation is risky.
03
$f(x) = a \cdot b^{x} + k$ has asymptote $y = k$ . In context, $k$ is the value being approached, and a is the starting gap.
04
Without context: if the differences have a constant ratio, that ratio is $b$ , the first difference is $a(b - 1)$ , and $k = f(0) - a$ .
05
Next, in 2.6: when data could be linear, quadratic, or exponential, how do you decide? Residuals will tell us.

Practice 34 problems Next: Chapter 2.6 Competing Function Model Validation

Unlock every question

Free

Premium

Sign in to FiveWay

Exponential Context and Data Modeling

Real Data: Nova's Actual Counts

When the Asymptote Is Not y = 0

Finding k From a Table

Practice

Recap card

Unlock Premium