Interpreting linear models | Year 9

01 / LEARN

The idea, step by step

Read a rule as a story

A linear model describes a quantity changing by a constant amount per unit of another quantity. The gradient is that rate of change, including its units. The intercept describes the predicted value when the input is zero; it may represent a starting amount, a fixed fee or an initial position.

Decide whether zero input makes sense. A line fitted to temperatures over a short afternoon does not automatically describe every hour of every day. The equation alone does not state the practical range in which it is useful.

output = rate \times input + starting amount

Use a model, then question it

To predict an output, substitute the input with correct units. To find an input from a target output, form an equation and solve it. Check the result against restrictions: a number of tickets is a whole number, while a measured time may be fractional.

Interpolation predicts between observed values; extrapolation predicts beyond them and needs more caution. Always identify assumptions, such as a constant rate, no extra fees, or no change in the filling/leaking rate. A mathematically valid negative time may have no practical meaning.

A common trap

Extending a tank model beyond empty and accepting negative water volume.

Instead: Find the practical domain. Once the tank is empty, the same linear rule is no longer a valid physical model.

02 / WORKED EXAMPLES

See the thinking

Example 1

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.

The intercept 6 is a fixed charge of 6 dollars.
The gradient 4 means an additional 4 dollars per hour.
For $h = 3$ , $C = 6 + 4 (3) = 18$ .
Three hours lies in the stated domain.

18 dollars

Example 2

Work backwards and check

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

Set $V = 60$ : $60 = 180 - 12 t$ .
Subtract 180: $- 120 = - 12 t$ , so $t = 10$ .
The tank empties at $t = 15$ , so 10 minutes is valid.

10 minutes

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

Each extra notebook adds 3 dollars to the cost.

The coefficient of the input is the increase in output per input unit.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF 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 $(18 - 12) / 3 = 2$ 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.

Checking saved progress…

Continue learning

Testing parameter changes digitally

Next lesson

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 ↗

Back to Linear equations and graphs