§-Syllabus dot point

NSWMaths Standard 2Syllabus dot point

How do we decide whether a real situation is best modelled by a linear, quadratic, exponential or reciprocal function?

Compare linear and non-linear models of real-world data and select the most appropriate model

A focused answer to the HSC Maths Standard 2 dot point on selecting an appropriate model. The four model shapes shown side by side, the difference-ratio-product diagnostic tests, when a non-linear model fits better than a straight line, and worked Australian examples for choosing the right model.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to look at a table or graph of data and decide what kind of relationship it shows. Is it linear (a straight line), or is it one of the three standard curved shapes: quadratic, exponential, or reciprocal? Once you decide, you write the matching model. The diagnostic rules (the quick checks that tell the shapes apart) are short and worth memorising. A "which model fits, justify your answer" question almost always just wants you to run one of three quick tests on a table.

The deeper skill is knowing when a straight line is not enough. Many real quantities look roughly straight over a short stretch but are really curved. A population growing by a constant percentage is not growing by a constant amount. A fixed cost shared among more people falls fast at first and then levels off. You earn full marks by spotting that a non-linear (curved) model fits better and saying why. A "looked like a line" guess does not.

The answer

The four model shapes at a glance

Each model has a signature shape and a signature diagnostic. The panels below put all four on identical axes so the differences are visible side by side. The data points sit exactly on each curve; the caption names the test that confirms it.

Linear. A straight line, $y = mx + c$ . Equal steps in $x$ produce equal steps in $y$ , so the first differences are constant. This is the only model with no curvature at all.

Quadratic. A single smooth bend (a parabola), $y = ax^2 + bx + c$ . The first differences are not constant, but the differences of those differences, the second differences, are constant. One turning point is the visual tell.

Exponential. A curve that climbs slowly then ever more steeply (or, for decay, falls fast then flattens), $y = ab^x$ . Equal steps in $x$ multiply $y$ by the same factor, so the ratios of consecutive $y$ values are constant (and that constant ratio is the base $b$ ).

Reciprocal. A curve that falls steeply then flattens towards the axes, $y = \dfrac{k}{x}$ . As $x$ rises, $y$ shrinks towards zero, and the product $xy$ is constant (it always equals $k$ ). This is the shape of a fixed total shared among more and more parts.

The diagnostic checklist

Given a table of $(x, y)$ pairs with $x$ equally spaced, run these in order until one works. Each is a one-line calculation.

First differences $y_{i+1} - y_i$ . If they are constant, the model is linear, and that constant is the gradient $m$ .
Ratios $\dfrac{y_{i+1}}{y_i}$ . If they are constant, the model is exponential, and that constant is the base $b$ .
Products $x_i \, y_i$ . If they are constant, the model is reciprocal, and that constant is $k$ .
If none of the above, try second differences (differences of the first differences). If those are constant, the model is quadratic.

The order matters. Linear is the simplest model, and exponential and reciprocal are the most common curved cases in Standard 2. So run the quick test first and stop as soon as something is constant.

When a non-linear model fits better than a line

A straight line is the right answer only when the change is by a constant amount each step. The moment the change is by a constant percentage, a constant product, or accelerates, a non-linear model fits better, and a question often hinges on you seeing this:

Constant percentage change means exponential, not linear. A population growing $10\%$ a year adds more people every year (because $10\%$ of a bigger number is bigger), so the increments grow and the first differences are not constant. A line would undershoot in later years; the exponential captures the accelerating climb. The same goes for compound interest and depreciation.
A fixed total shared out means reciprocal, not linear. Splitting a $50 bill among more friends, or a fixed-length trip taking less time at higher speed, gives a cost (or time) that falls fast at first and then barely changes. A line cannot bend like that, and would wrongly go negative; the reciprocal flattens towards the axis exactly as the data does.
An accelerating, single-bend rise means quadratic. Area against side length, or a revenue-versus-price curve with one peak, bends once. First differences grow steadily (constant second differences), which a straight line cannot reproduce.
Watch the danger of a short window. Over three or four points many curves look almost straight, so a "looks linear" eyeball is unreliable. The numerical test is what decides it: compute the differences and the ratios and let the constant one win.

Shape clues from a graph

If you are given a graph rather than a table, the shape narrows it down quickly:

Straight line: linear.
Single smooth bend with one turning point: quadratic.
Rises slowly then very steeply (or falls fast then flattens) with no turning point: exponential.
Two branches hugging both axes, never crossing them: reciprocal.

For a rising curve with no turning point, it is hard to tell quadratic from exponential by eye. When in doubt, go back to the table test (second differences versus ratios).

Practical clues from context

The wording of a problem usually hints at the model before you compute anything:

A flat fee plus a charge per unit: linear.
Compound growth or decay (interest, depreciation, a population growing by a fixed percentage): exponential.
A fixed total split among a variable number of parts (a bill shared, fuel for a fixed trip, time for a fixed job): reciprocal.
A projectile, an area, or a quantity to be maximised at one peak: quadratic.

When two models seem to fit

Sometimes a small set of data fits more than one model fairly well. The question usually steers you. For example, "the growth percentage is constant" points straight at exponential. If you must compare the fit yourself, work out the residuals (a residual is the gap between a data point and the value the model predicts). The better model has smaller residuals across the whole set of data, not just at one or two points. When the fits are very close, pick the simpler model, because the right model is the simplest one that explains the data.

How exam questions ask about choosing a model

The phrasings are a small, recognisable set:

"Which model best fits, linear or exponential? Justify." Run the first-difference test and the ratio test, state which one is constant, and conclude. Show both tests so the justification is complete.
"What type of relationship / model does this table show?" Work down the diagnostic checklist (differences, ratios, products) and name the first constant one, then write the equation with its constant.
"Write the model / equation." After identifying the type, read the constant off your test: $m$ from the difference, $b$ from the ratio, $k$ from the product, then find the remaining parameter from a known point.
"Explain why a linear model is not appropriate here." Point to the failed test: the first differences are not constant, so the data is not linear, and name what is constant instead.
"Use your model to predict ..." Substitute the required value, but flag if it is far outside the data (extrapolation is risky, especially for exponentials, which grow without bound).
"Sketch the data and state the model." Plot the points, judge the shape against the four signatures above, then confirm with the matching numerical test.

Identifying the right model for three Australian datasets

Three situations: ATO income tax, a regional town's population, and splitting a pizza bill. In each we apply a diagnostic test to choose between linear, exponential and reciprocal.

Classify Australian income tax

When a rate is constant within a range, the relationship between two quantities is linear on that interval. The ATO 2024-25 rates are $0\%$ on the first $ $18\,200$ , $16\%$ from $ $18\,201$ to $ $45\,000$ , and $30\%$ from $ $45\,001$ to $ $135\,000$ .

Within each bracket the marginal rate is constant, so tax owed is a straight-line function of income on that piece. However, the slope changes at each cutoff, so the overall relationship is not a single straight line. The correct classification is a piecewise linear function with a kink at each bracket boundary.

Identify the model for a town's population

Test the population data for linearity using first differences: For a linear model the absolute increase each period must be constant. Computing the year-on-year changes for the town's population reveals whether this holds.

Year	Population	First difference
2020	$8000$
2021	$8240$	$240$
2022	$8487$	$247$
2023	$8742$	$255$
2024	$9004$	$262$

The differences are not constant (they grow), so the data is not linear.

Test the same population data for exponential growth using ratios: For an exponential model the ratio of consecutive terms must be constant. Dividing each year's population by the previous year's gives the growth factor.

\frac{8240}{8000} = 1.030, \quad \frac{8487}{8240} \approx 1.030, \quad \frac{8742}{8487} \approx 1.030, \quad \frac{9004}{8742} \approx 1.030

The ratio is constant at $1.030$ , confirming an exponential model. The equation is $P = 8000(1.03)^t$ , representing $3\%$ growth per year. This is the textbook case of constant-percentage change beating a line.

Identify the pizza-splitting model

When a fixed total is divided equally among $n$ people, each person's share decreases as $n$ increases. Testing the product of people and cost per person:

n \times c = 50 \text{ for every } n, \qquad \text{so } c = \frac{50}{n}

Because the product is constant at $50$ , this is a reciprocal (inverse) relationship, not a linear one. A common HSC error is to write $c = 50n$ instead. The sanity check is simple: more people sharing means a lower cost each, so $c$ must decrease as $n$ increases, which only the reciprocal does.

Final answer: income tax is piecewise linear; the town's population follows the exponential model $P = 8000(1.03)^t$ at $3\%$ per year; the cost per person when splitting a $50 pizza is the reciprocal model $c = \dfrac{50}{n}$ .

Common traps

Calling exponential growth "fast linear": Exponential is multiplicative; linear is additive. They look alike over a few points but diverge dramatically over a long timeframe, so always run the ratio test before settling on a line.
Confusing constant difference with constant ratio: Linear has constant first differences (you add the same amount); exponential has constant ratios (you multiply by the same amount). State which one you checked, because that is the justification.
Missing the reciprocal hint of "fixed total": Whenever the problem splits a fixed total into a variable number of parts, the model is reciprocal, and the product $xy$ is the constant to test.
Writing the reciprocal upside down: A fixed total shared out gives $c = \dfrac{k}{n}$ , not $c = kn$ . Check the direction: more parts must mean less each.
Ignoring units: A linear model in dollars per year is a different model from one in dollars per month. State the units of $x$ and $y$ .
Choosing complexity over fit: If a line fits, do not fit a quadratic just to use more algebra. The right model is the simplest one that matches the data.

Exam technique

For a "which model" question, set out the actual test: tabulate first differences, then ratios (then products or second differences if needed), and write a sentence naming which one is constant, because the named test is what earns the justification mark. Read the model's constant straight off the test ( $m$ from the difference, $b$ from the ratio, $k$ from the product), then pin down any remaining parameter from a given point and quote the full equation. If a graph is given, match it to one of the four shape signatures, then confirm numerically rather than trusting the eye, especially for a rising curve where exponential and quadratic look similar. When asked to predict, substitute into your model but warn that extrapolating far past the data is unreliable, and never fit a more complex model than the data needs.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style4 marksA table shows population data over five years. Year

0

has

1000

, year

1

has

1100

, year

2

has

1210

, year

3

has

1331

, year

4

has

1464

. Which model best fits, linear or exponential? Justify your answer and write the model.

Show worked answer →

For linear, check first differences: $1100 - 1000 = 100$ , $1210 - 1100 = 110$ , $1331 - 1210 = 121$ , $1464 - 1331 = 133$ . Differences are increasing, not constant, so not linear.

For exponential, check ratios: $\frac{1100}{1000} = 1.10$ , $\frac{1210}{1100} = 1.10$ , $\frac{1331}{1210} = 1.10$ , $\frac{1464}{1331} \approx 1.10$ . Ratios are constant at $1.10$ , so the model is exponential.

Model: $P = 1000 (1.10)^t$ . This is $10\%$ growth per year.

Markers reward both first-difference and ratio checks, the conclusion that ratios are constant, and the model with correct base $1.10$ .

2021 HSC-style3 marksA table shows fuel cost for a trip of fixed length at different speeds.

40

km/h costs $18,

60

km/h costs $12,

80

km/h costs $9,

120

km/h costs $6. What type of model fits this data?

Show worked answer →

Check the product of speed and cost: $40 \times 18 = 720$ , $60 \times 12 = 720$ , $80 \times 9 = 720$ , $120 \times 6 = 720$ .

The product is constant at $720$ , so the relationship is inverse: cost $\times$ speed $= 720$ , that is $C = \frac{720}{s}$ .

Markers reward checking the product, identifying it as constant, and writing the reciprocal model with the constant explicitly.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA gym records the total cost of membership, in dollars, after

t

months: month

0

is $60, month

1

is $95, month

2

is $130, month

3

is $165, month

4

is $200. Show that a linear model fits, write the equation, and use it to find the total cost after

10

months.

Show worked solution →

Test the first differences: The $t$ values are equally spaced, so subtract each cost from the next: $95 - 60 = 35$ , $130 - 95 = 35$ , $165 - 130 = 35$ , $200 - 165 = 35$ .
Conclude the model: The first differences are constant at $35$ , so the data is linear and the gradient is $m = 35$ dollars per month.
Find the intercept and write the equation: At $t = 0$ the cost is $60, so $c = 60$ (the joining fee). The model is $C = 35t + 60$ .
Predict at $t = 10$: Substitute: $C = 35 \times 10 + 60 = 410$ .
State the answer: A constant first difference of $35 confirms a linear model $C = 35t + 60$ , and after $10$ months the total cost is $410.00.

foundation3 marksA car's value, in dollars, is recorded each year: year

0

is $30\,000, year

1

is $25\,500, year

2

is $21\,675, year

3

is $18\,423.75. Identify the model, write its equation, and find the value after

5

years.

Show worked solution →

Test the ratios: The first differences ( $-4500$ , then $-3825$ , then $-3251.25$ ) are not constant, so try ratios of consecutive values: $\frac{25\,500}{30\,000} = 0.85$ , $\frac{21\,675}{25\,500} = 0.85$ , $\frac{18\,423.75}{21\,675} = 0.85$ .
Conclude the model: The ratio is constant at $0.85$ , so the data is exponential with base $b = 0.85$ . Because $1 - 0.85 = 0.15$ , the car loses $15\%$ of its value each year.
Write the equation: The starting value is $30,000, so $V = 30\,000(0.85)^t$ .
Predict at $t = 5$: Substitute: $V = 30\,000 \times 0.85^5 = 13\,311.16$ (to the nearest cent).
State the answer: A constant ratio of $0.85$ confirms exponential decay $V = 30\,000(0.85)^t$ , and after $5$ years the car is worth $13,311.16.

foundation2 marksA water tank is being filled. The volume, in litres, is recorded each minute: minute

0

50

, minute

1

58

, minute

2

66

, minute

3

74

. Show that a linear model fits, then use it to find the volume after

6

minutes.

Show worked solution →

Test the first differences: The minute values are equally spaced, so subtract each volume from the next: $58 - 50 = 8$ , $66 - 58 = 8$ , $74 - 66 = 8$ .
Conclude the model: The first differences are constant at $8$ , so the data is linear with gradient $m = 8$ litres per minute, and at minute $0$ the volume is $50$ , so the model is $V = 8t + 50$ .
Predict at $t = 6$: Substitute: $V = 8 \times 6 + 50 = 98$ .
State the answer: A constant first difference of $8$ confirms the linear model $V = 8t + 50$ , so after $6$ minutes the tank holds $98$ litres.

foundation2 marksA post is being shared online and the number of shares is counted each hour: hour

0

100

, hour

1

300

, hour

2

900

, hour

3

2700

. Identify the type of model and write its equation.

Show worked solution →

Try first differences first: $300 - 100 = 200$ , $900 - 300 = 600$ , $2700 - 900 = 1800$ . These are not constant, so the data is not linear.
Test the ratios: Divide each count by the one before: $\frac{300}{100} = 3$ , $\frac{900}{300} = 3$ , $\frac{2700}{900} = 3$ .
Conclude the model: The ratio is constant at $3$ , so the data is exponential with base $b = 3$ (the shares triple each hour).
State the answer: A constant ratio of $3$ confirms an exponential model, and with $100$ shares at the start the equation is $S = 100(3)^t$ .

core4 marksA fixed building job is shared among workers. The table shows the number of workers

n

and the days

d

to finish:

4

workers take

15

days,

6

take

10

days,

8

take

7.5

days,

12

take

5

days. Determine the type of model, justify your choice, write its equation, and find how long

5

workers would take.

Show worked solution →

Spot the context clue: A fixed total job split among a variable number of workers points to a reciprocal (inverse) model, so test the product $n \times d$ .
Test the product: $4 \times 15 = 60$ , $6 \times 10 = 60$ , $8 \times 7.5 = 60$ , $12 \times 5 = 60$ .
Conclude the model: The product is constant at $60$ , so the relationship is reciprocal with $k = 60$ , and the equation is $d = \dfrac{60}{n}$ . It cannot be linear: more workers give fewer days, so $d$ must decrease as $n$ rises.
Predict for $5$ workers: Substitute $n = 5$ : $d = \dfrac{60}{5} = 12$ .
State the answer: A constant product of $60$ confirms the reciprocal model $d = \dfrac{60}{n}$ , so $5$ workers would take $12$ days. Check: $5 \times 12 = 60$ , matching the constant.

core4 marksA ball is thrown and its height, in metres, is recorded each second:

t = 0

gives

0

t = 1

gives

25

t = 2

gives

40

t = 3

gives

45

t = 4

gives

40

. Explain why the data is neither linear nor exponential, then identify the correct model.

Show worked solution →

Test for linear (first differences): $25 - 0 = 25$ , $40 - 25 = 15$ , $45 - 40 = 5$ , $40 - 45 = -5$ . The first differences are not constant, so the data is not linear.
Test for exponential (ratios): $\frac{25}{0}$ is undefined (division by zero at the start), and $\frac{40}{25} = 1.6$ while $\frac{45}{40} = 1.125$ are not equal, so the data is not exponential.
Test for quadratic (second differences): Take differences of the first differences: $15 - 25 = -10$ , $5 - 15 = -10$ , $-5 - 5 = -10$ .
Conclude the model: The second differences are constant at $-10$ , so the data is quadratic (a single bend with one turning point, here a peak between $t = 3$ and $t = 4$ ).
State the answer: Differences that change rule out linear, an undefined and non-constant ratio rules out exponential, and a constant second difference of $-10$ confirms a quadratic model, the expected shape for the height of a thrown ball.

core4 marksA streaming channel's subscribers are recorded each month: month

0

2000

, month

1

2400

, month

2

2880

, month

3

3456

. The channel claims linear growth. By testing both first differences and ratios, decide whether linear or exponential is the better model, and write it.

Show worked solution →

Test the first differences: $2400 - 2000 = 400$ , $2880 - 2400 = 480$ , $3456 - 2880 = 576$ . The differences grow, so the growth is not by a constant amount and the data is not linear; the channel's claim is wrong.
Test the ratios: $\frac{2400}{2000} = 1.2$ , $\frac{2880}{2400} = 1.2$ , $\frac{3456}{2880} = 1.2$ .
Conclude the model: The ratio is constant at $1.2$ , so the data is exponential with base $b = 1.2$ , i.e. $20\%$ growth per month. Starting from $2000$ , the model is $S = 2000(1.2)^t$ .
State the answer: Growing first differences rule out linear and a constant ratio of $1.2$ confirms exponential, so the better model is $S = 2000(1.2)^t$ at $20\%$ growth per month. The constant percentage, not a constant headcount, is the giveaway.

core4 marksA council buys synthetic turf for a square play area. The total turf cost, in dollars, is recorded against the side length

s

in metres:

2

m costs $160,

4

m costs $640,

6

m costs $1440,

8

m costs $2560. A planner claims the cost is linear in

s

. By testing first and second differences, decide whether linear or quadratic fits, then predict the cost for a

10

m side.

Show worked solution →

Test the first differences: The side lengths are equally spaced (steps of $2$ m), so subtract each cost from the next: $640 - 160 = 480$ , $1440 - 640 = 800$ , $2560 - 1440 = 1120$ . They are not constant, so the data is not linear and the planner's claim is wrong.
Test the second differences: Take the differences of those first differences: $800 - 480 = 320$ , $1120 - 800 = 320$ .
Conclude the model: The second differences are constant at $320$ , so the data is quadratic. This makes sense: the area of a square is $s^2$ , so doubling the side roughly quadruples the area and the cost.
Predict at $s = 10$: The pattern is cost $= 40 s^2$ (check: $40 \times 2^2 = 160$ , $40 \times 8^2 = 2560$ ). At $s = 10$ : $40 \times 10^2 = 4000$ .
State the answer: Growing first differences rule out linear and a constant second difference of $320$ confirms a quadratic model, so a $10$ m side would cost $4000.

exam5 marksFor a fixed

240

km drive, the time taken (in hours) is recorded at several average speeds:

40

km/h takes

6

60

km/h takes

4

80

km/h takes

3

120

km/h takes

2

h. Identify and justify the model, write its equation, predict the time at

100

km/h, and explain why a linear model would give a poor prediction here.

Show worked solution →

Identify the model: A fixed distance covered at a variable speed is a fixed total shared out, so test the product $s \times t$ : $40 \times 6 = 240$ , $60 \times 4 = 240$ , $80 \times 3 = 240$ , $120 \times 2 = 240$ .
Conclude: The product is constant at $240$ (the distance), so the model is reciprocal, $t = \dfrac{240}{s}$ .
Predict at $s = 100$: Substitute: $t = \dfrac{240}{100} = 2.4$ hours, which is $2$ hours and $0.4 \times 60 = 24$ minutes.
Show why linear fails: A straight line through the first and last points, $(40, 6)$ and $(120, 2)$ , has gradient $\frac{2 - 6}{120 - 40} = -0.05$ and equation $t = -0.05s + 8$ . At $s = 100$ that predicts $t = -0.05 \times 100 + 8 = 3$ hours, well above the true $2.4$ hours, because the real data curves (falls fast then flattens) while a line cannot bend.
State the answer: A constant product of $240$ gives the reciprocal model $t = \dfrac{240}{s}$ ; the time at $100$ km/h is $2.4$ hours, and a linear fit overstates it as $3$ hours by ignoring the curve.

exam5 marksAfter a single dose, the mass of a drug in the bloodstream (in milligrams) is measured each hour: hour

0

80

, hour

1

60

, hour

2

45

, hour

3

33.75

. Show the data is exponential rather than linear, state the percentage eliminated each hour, write the model, and predict the mass after

5

hours.

Show worked solution →

Rule out linear: First differences are $60 - 80 = -20$ , $45 - 60 = -15$ , $33.75 - 45 = -11.25$ . They are not constant, so the data is not linear.
Confirm exponential: Test ratios: $\frac{60}{80} = 0.75$ , $\frac{45}{60} = 0.75$ , $\frac{33.75}{45} = 0.75$ . The ratio is constant at $0.75$ , so the data is exponential with base $b = 0.75$ .
Interpret the base: Since $1 - 0.75 = 0.25$ , the body eliminates $25\%$ of the remaining drug each hour (so $75\%$ stays).
Write the model and predict: The starting mass is $80$ mg, so $A = 80(0.75)^t$ . At $t = 5$ : $A = 80 \times 0.75^5 = 18.98$ mg (to $2$ decimal places).
State the answer: A constant ratio of $0.75$ confirms exponential decay $A = 80(0.75)^t$ , with $25\%$ eliminated hourly, leaving about $18.98$ mg after $5$ hours. Falling by a constant percentage, not a constant amount, is why a line does not fit.

exam5 marksA student must decide between two models for the data

(1, 12)

(2, 18)

(3, 24)

(4, 30)

. Model A is linear,

y = 6x + 6

. Model B is exponential,

y = 8(1.5)^x

. Using residuals, decide which model fits better, then use the better model to predict

y

when

x = 8

Show worked solution →

Find Model A's predictions and residuals: $y = 6x + 6$ gives, at $x = 1, 2, 3, 4$ : $6 + 6 = 12$ , then $18$ , then $24$ , then $30$ . The residuals (data minus prediction) are $12 - 12 = 0$ , $18 - 18 = 0$ , $24 - 24 = 0$ , $30 - 30 = 0$ .
Find Model B's predictions and residuals: $y = 8(1.5)^x$ gives $8 \times 1.5 = 12$ , $8 \times 1.5^2 = 18$ , $8 \times 1.5^3 = 27$ , $8 \times 1.5^4 = 40.5$ . The residuals are $12 - 12 = 0$ , $18 - 18 = 0$ , $24 - 27 = -3$ , $30 - 40.5 = -10.5$ .
Compare: Model A has residuals of $0$ at every point, while Model B drifts to $-3$ and $-10.5$ . The smaller residuals across the whole dataset belong to Model A, the linear model. (The first differences $6, 6, 6$ confirm linear directly.)
Predict with the better model at $x = 8$: $y = 6 \times 8 + 6 = 54$ .
State the answer: Model A fits better (residuals all $0$ versus Model B's growing gaps), so the prediction at $x = 8$ is $y = 54$ .

exam6 marksTwo regional towns are studied from the same start year (

t = 0

, in years). Town A's population is recorded as

12\,000

12\,500

13\,000

13\,500

; Town B's is recorded as

8000

8480

8989

9528

(to the nearest person). (a) Identify the model for each town, justify each choice, and write both equations. (b) Although Town A starts larger, Town B grows faster. Find the first whole year in which Town B's population exceeds Town A's.

Show worked solution →

(a) Test Town A: The years are equally spaced, so check first differences: $12\,500 - 12\,000 = 500$ , $13\,000 - 12\,500 = 500$ , $13\,500 - 13\,000 = 500$ . They are constant, so Town A is linear, $A = 12\,000 + 500t$ (a fixed $500$ people added per year).
Test Town B: The first differences ( $480$ , $509$ , $539$ ) grow, so it is not linear; test ratios instead: $\frac{8480}{8000} = 1.06$ , $\frac{8989}{8480} \approx 1.06$ , $\frac{9528}{8989} \approx 1.06$ . The ratio is constant at $1.06$ , so Town B is exponential, $B = 8000(1.06)^t$ (growth of $6\%$ per year).
(b) Compare the models year by year: Town B catches up because a constant percentage beats a constant amount. Evaluate near the crossover: at $t = 15$ , $A = 12\,000 + 500 \times 15 = 19\,500$ while $B = 8000 \times 1.06^{15} \approx 19\,172$ , so B is still behind. At $t = 16$ , $A = 12\,000 + 500 \times 16 = 20\,000$ while $B = 8000 \times 1.06^{16} \approx 20\,323$ , so B has overtaken A.
State the answer: Town A is linear, $A = 12\,000 + 500t$ ; Town B is exponential, $B = 8000(1.06)^t$ . Town B first exceeds Town A in year $16$ (about $20\,323$ versus $20\,000$ ), the year the accelerating curve passes the straight line.