Interpreting linear versus quadratic financial models | Year 9

01 / LEARN

Understand the idea.

Look at how the quantity changes

A linear model has a constant increase for equal input steps. In $C = 20 + 5 n$ , each extra item adds $5$ dollars. Adding a non-zero $n^{2}$ term changes this pattern because successive squares increase by different amounts.

For $C = n^{2} + 5 n + 20$ , outputs at counts $0, 1, 2, 3$ are $20, 26, 34, 44$ . Their first differences are $6, 8, 10$ and second differences are $2, 2$ . Equal input steps are essential when using differences to compare models.

Explain the product in a model

Revenue equals the number sold multiplied by the price of each item. If count and price both change with $n$ , their product may be quadratic. For $R = n (18 - n)$ , expansion exposes the squared term $- n^{2}$ .

Interpret the rule with its assumptions and domain. Here a whole-number count must also keep the modelled price $18 - n$ non-negative. A finite table can support a pattern, while the supplied formula establishes the model type.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Evaluate the linear cost at each count.
$\begin{aligned} C_{A} (0) & = 20, & C_{A} (1) & = 25, \\ C_{A} (2) & = 30, & C_{A} (3) & = 35 \end{aligned}$
Substitute each input into 20 plus 5n.
2
Evaluate the quadratic cost at the same counts.
$\begin{aligned} C_{B} (0) & = 20, & C_{B} (1) & = 26, \\ C_{B} (2) & = 34, & C_{B} (3) & = 44 \end{aligned}$
The matching inputs allow a fair comparison; the added cost is n squared.
3
Compare first differences.
$\begin{aligned} Δ C_{A} : & 5, 5, 5 \\ Δ C_{B} : & 6, 8, 10 \end{aligned}$
Only A increases by a constant amount per extra item.
4
Connect the pattern to the formulas.
$C_{B} - C_{A} = n^{2}, Δ^{2} C_{B} = 2, 2$
The non-zero squared term explains B's changing increases and constant second differences.

Answer

A: $20, 25, 30, 35$ ; B: $20, 26, 34, 44$ .

EXAMPLE 02

Worked example

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

Worked solution

1
Distribute the count across the bracket.
$R = n (18 - n) = 18 n - n^{2}$
Both terms in the price are multiplied by n.
2
Find the price at the required count.
$18 - 4 = 14 dollars per item$
The model links a count of four to a 14-dollar price.
3
Calculate the revenue.
$R (4) = 4 (14) = 56 dollars$
Revenue is four sales multiplied by the modelled price per sale.
4
Check the expanded rule.
$18 (4) - 4^{2} = 72 - 16 = 56$
The factored and expanded forms agree, giving AUD 56.00.

Answer

$R = 18 n - n^{2}$ ; AUD 56.00

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Quadratic

$C = 8 n^{2} + 15$ is quadratic because the highest power is $2$ with non-zero coefficient $8$ . It cannot be rewritten as a constant rate times $n$ plus a fixed amount.

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Look at how the quantity changes

Explain the product in a model

See the method in action.

Worked example

Evaluate the linear cost at each count.

Evaluate the quadratic cost at the same counts.

Compare first differences.

Connect the pattern to the formulas.

Worked example

Distribute the count across the bracket.

Find the price at the required count.

Calculate the revenue.

Check the expanded rule.

Try it yourself.

Check your understanding

Keep practising