Interpreting linear models | Year 9

01 / LEARN

Understand the idea.

Read a rule as a story

A linear model represents a constant change per input unit. Its gradient is that rate with units attached. Its intercept is the predicted output at zero input, such as a fixed fee or starting volume.

Read the domain alongside the equation. A rule may describe only a hire period or a tank before it empties. The formula can calculate outside that interval even when those outputs have no practical meaning.

output = rate \times input + starting amount

Use a model, then question it

To predict an output, substitute the input. To find an input for a target output, set the rule equal to that target and solve. Then check the domain and whether the input must be a whole number.

Interpolation predicts between observed inputs; extrapolation predicts beyond them. Extending a straight line assumes the same rate continues, so identify what could change: a fee, capacity, supply rate or other stated condition.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A booking fee and an hourly charge

A bicycle hire costs $C = 6 + 4 h$ dollars for $h$ hours, where $0 \leq h \leq 6$ . Interpret the rule and find the cost for 3 hours.

Worked solution

1
Interpret the value at zero hours.
$C (0) = 6 + 4 (0) = 6$
The constant 6 represents the fixed charge in dollars.
2
Interpret the coefficient of time.
$hourly rate = 4 dollars per hour$
Each extra hour adds 4 dollars to the modelled cost.
3
Substitute three hours.
$C (3) = 6 + 4 (3) = 18 dollars$
Add the fixed fee to the charge for three hours.
4
Check the allowed hire time.
$0 \leq 3 \leq 6$
Three hours lies within the domain supplied with the rule.

Answer

18 dollars

EXAMPLE 02

Work backwards and check

A water tank follows $V = 180 - 12 t$ litres after $t$ minutes, until empty. When does it contain 60 L?

Worked solution

1
Set the required volume.
$60 = 180 - 12 t$
The desired output is 60 litres.
2
Solve for the time.
$\begin{aligned} - 120 & = - 12 t \\ t & = 10 minutes \end{aligned}$
Subtract 180, then divide both sides by -12.
3
Find when the tank empties.
$0 = 180 - 12 t \Rightarrow t = 15 minutes$
The stated model stops being valid when the volume reaches zero.
4
Check feasibility and substitution.
$0 \leq 10 \leq 15, 180 - 12 (10) = 60$
The time is before emptying and produces the requested volume.

Answer

10 minutes

03 / YOUR TURN

Try it yourself.

Check your understanding

In $P = 24 + 3 n$ , where $P$ is dollars and $n$ is a whole number of notebooks, what does 3 mean?

Show answer & reasoning

Each extra notebook adds 3 dollars to the cost.

For one extra notebook, $P (n + 1) - P (n) = 3$ . Therefore the coefficient $3$ means the cost increases by $3$ dollars for each additional notebook; $24$ is the fixed part.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

A delivery service uses $C = 8 + 2 d$ dollars for distance $d$ km. Find $C$ when $d = 7$ .
2 marks
For $C = 8 + 2 d$ , explain the intercept and gradient.
2 marks
A tank follows $V = 150 - 5 t$ L. Find when it empties and give a practical time domain.
3 marks
A club charges 40 dollars for membership plus 6 dollars per session. Write a cost rule and find the cost of 8 sessions.
3 marks
A plant is 12 cm high at day 0 and 18 cm high at day 3. Assuming linear growth, predict day 5 and state one limitation.
4 marks
A bus-hire model is $C = 120 + 5 p$ , where $p$ is the passenger count and the bus has 24 seats. Explain why $p = 30$ is not allowed.
2 marks

Read the worked solutions

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

A delivery service uses $C = 8 + 2 d$ dollars for distance $d$ km. Find $C$ when $d = 7$ .
2 marks
$C = 8 + 2 (7) = 22$ .
22 dollars
For $C = 8 + 2 d$ , explain the intercept and gradient.
2 marks
At $d = 0$ , $C = 8$ .
Each increase of 1 km increases the cost by $2$ .
Fixed fee of 8 dollars; 2 dollars for each kilometre.
A tank follows $V = 150 - 5 t$ L. Find when it empties and give a practical time domain.
3 marks
Set $V = 0$ : $0 = 150 - 5 t$ , so $t = 30$ .
The model starts at zero time and stops when empty.
30 minutes; $0 \leq t \leq 30$ .
A club charges 40 dollars for membership plus 6 dollars per session. Write a cost rule and find the cost of 8 sessions.
3 marks
Let $n$ be sessions.
Substitute $n = 8$ : $40 + 48 = 88$ .
$C = 40 + 6 n$ ; 88 dollars.
A plant is 12 cm high at day 0 and 18 cm high at day 3. Assuming linear growth, predict day 5 and state one limitation.
4 marks
The rate is $\frac{18 - 12}{3} = 2$ cm/day.
$H = 12 + 2 t$ , so $H (5) = 22$ .
Day 5 is beyond the observed range, so this is extrapolation.
22 cm; the growth rate may change after the observed period.
A bus-hire model is $C = 120 + 5 p$ , where $p$ is the passenger count and the bus has 24 seats. Explain why $p = 30$ is not allowed.
2 marks
A passenger count is discrete.
The vehicle's 24-seat limit restricts the input.
The rule's passenger domain is whole numbers from 0 to 24; 30 exceeds capacity.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Gradient, midpoint and distance; algebraic and graphical straight-line relationships; interpreting models and testing parameters. Parallel and perpendicular lines are included as a natural school-course extension. All explanations and questions are original.

Official curriculum source ↗

In this lesson

Understand the idea.

Read a rule as a story

Use a model, then question it

See the method in action.

A booking fee and an hourly charge

Interpret the value at zero hours.

Interpret the coefficient of time.

Substitute three hours.

Check the allowed hire time.

Work backwards and check

Set the required volume.

Solve for the time.

Find when the tank empties.

Check feasibility and substitution.

Try it yourself.

Check your understanding

Keep practising