Interpreting linear versus quadratic financial models | Year 9

01 / LEARN

The idea, step by step

Look at how the quantity changes

A linear cost $C = 20 + 5 n$ increases by 5 dollars for every extra item. A quadratic cost $C = n^{2} + 5 n + 20$ does not have a constant increase: the extra squared term changes by larger amounts as n grows.

A table at equal input steps helps reveal the difference. For the quadratic rule, outputs at $n = 0, 1, 2, 3$ are $20, 26, 34, 44$ ; first differences are $6, 8, 10$ . Constant second differences are a useful clue, not a reason to ignore the context.

Explain the product in a model

Revenue equals number sold times price per item. If both depend on a variable, their product can be quadratic. For example, $R = n (18 - n) = 18 n - n^{2}$ . The model assumes a stated relationship between count and price; it does not guarantee that customers will behave that way.

Compare models at relevant inputs and use a reasonable domain. A quadratic revenue rule might eventually predict negative prices, so it cannot be used for arbitrary counts. Choose a model because its assumptions and shape fit the supplied situation.

A common trap

Every cost rule is linear because money is measured in dollars.

Instead: The type of relationship depends on the rule and its changes, not the unit.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Compare $C_{A} = 20 + 5 n$ and $C_{B} = n^{2} + 5 n + 20$ at $n = 0, 1, 2, 3$ .

Substitute each count into both rules.
A has constant first difference $5$ .
B has first differences $6, 8, 10$ , so its rate of increase is not constant.

20, 25, 30, 35

; B:

20, 26, 34, 44

Example 2

Worked example

Expand $R = n (18 - n)$ and find the revenue at $n = 4$ .

Distribute $n$ to each term.
$R (4) = 4 (14) = 56$ .
At $n = 4$ , the modelled price is $14$ dollars per item.

R = 18 n - n^{2}

; AUD 56.00

Check your understanding

Is $C = 8 n^{2} + 15$ linear or quadratic?

Show answer

Quadratic

The rule contains a non-zero squared term.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Classify $C = 12 + 7 n$ and state the extra cost per item.
2 marks
Classify $C = n^{2} + 2 n + 8$ .
2 marks
Find $C = n^{2} + 2 n + 8$ for $n = 0, 1, 2, 3$ .
3 marks
Expand the revenue rule $R = n (20 - n)$ .
2 marks
For $R = n (20 - n)$ , find revenue at $n = 6$ and the price per item.
3 marks
Why is $n = 25$ unsuitable for a selling-price model $R = n (20 - n)$ ?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Classify $C = 12 + 7 n$ and state the extra cost per item.
2 marks
The input appears to the first power and the coefficient is $7$ .
Linear; AUD 7.00 per extra item.
Classify $C = n^{2} + 2 n + 8$ .
2 marks
The coefficient of $n^{2}$ is $1$ , not zero.
Quadratic
Find $C = n^{2} + 2 n + 8$ for $n = 0, 1, 2, 3$ .
3 marks
Substitute each input.
The first differences are $3, 5, 7$ .
$8, 11, 16, 23$
Expand the revenue rule $R = n (20 - n)$ .
2 marks
Distribute $n$ over the bracket.
$R = 20 n - n^{2}$
For $R = n (20 - n)$ , find revenue at $n = 6$ and the price per item.
3 marks
Price $= 20 - 6 = 14$ .
Revenue $= 6 \times 14 = 84$ .
Revenue AUD 84.00; price AUD 14.00
Why is $n = 25$ unsuitable for a selling-price model $R = n (20 - n)$ ?
3 marks
A sale price of negative five dollars does not match the stated selling context.
Restrict the model to a sensible range such as whole numbers $0 \leq n \leq 20$ .
It gives a negative price, $20 - 25 = - 5$ dollars.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Year 9 financial modelling within AC9M9A05 and AC9M9M05; supplied hypothetical rates and costs, not financial advice.

Official curriculum source ↗

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