§-Syllabus dot point

SAGeneral MathematicsSyllabus dot point

How do we use a straight-line equation to model and predict from a real situation?

Construct and interpret linear functions of the form y = mx + c to model practical situations, identifying the meaning of the gradient and intercept.

How to build a linear model y = mx + c from a worded situation, interpret the gradient as a rate and the intercept as a starting value, and use the model to predict, with worked SACE-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking
The linear model
Building a model from words
Interpreting the model in context
Predicting and the limits of a model
Horizontal intercepts and sign of the gradient
Why it matters

What this dot point is asking

You must be able to read a practical situation, set up a linear equation that fits it, explain what the gradient and intercept mean in context, and use the model to make predictions. This is the foundation of the entire linear-relationships topic and recurs in break-even analysis, piecewise models and regression.

The linear model

A relationship is linear when $y$ changes by the same amount for each unit increase in $x$ . The model is:

y = mx + c.

Building a model from words

The key is to identify the two parameters from the wording:

A fixed starting amount, base fee, or initial value gives the intercept $c$ .
A "per unit" amount (per hour, per kilometre, per item) gives the gradient $m$ .

Interpreting the model in context

Examiners reward interpretation, not just numbers. For the taxi model:

The gradient $2.20$ means the fare increases by $2.20$ dollars for each extra kilometre.
The intercept $4.50$ means the fare is $4.50$ dollars before any distance is travelled.

Building a model from two points

A water tank holds $80$ litres after $2$ minutes of draining and $50$ litres after $5$ minutes. Find a linear model for the volume $V$ litres in terms of time $t$ minutes.

Step 1: Find the gradient from the two data points

The gradient $m$ measures how fast the volume changes per minute. We apply the gradient formula using the two given points $(t_1, V_1) = (2, 80)$ and $(t_2, V_2) = (5, 50)$ :

m = \frac{50 - 80}{5 - 2} = \frac{-30}{3} = -10.

The negative value confirms the tank is draining: it loses $10$ litres every minute.

Step 2: Find the vertical intercept by substituting one point

Now that we know $m = -10$ , we substitute one of the data points into $V = mt + c$ to find $c$ , the starting volume when $t = 0$ . Using the point $(2, 80)$ :

80 = -10(2) + c \implies 80 = -20 + c \implies c = 100.

So the intercept is $c = 100$ , meaning the tank held $100$ litres before draining began.

Step 3: Write the model and interpret it

Combining the gradient and intercept gives the complete linear model:

V = -10t + 100.

The tank started with $100$ litres and loses $10$ litres per minute. Setting $V = 0$ and solving gives $t = 10$ minutes, so the tank is empty after $10$ minutes.

Final answer: $V = -10t + 100$ , valid for $0 \le t \le 10$ .

Predicting and the limits of a model

Substitute a value to predict. Predicting inside the range of your data is interpolation (usually reliable); predicting outside it is extrapolation (less reliable, because the linear pattern may not continue). For the draining tank, predicting the volume at $t = 4$ minutes is interpolation and trustworthy, but predicting at $t = 30$ minutes is meaningless, since the volume cannot go negative - the model only holds while there is water in the tank. Always state the domain over which a linear model is valid.

Horizontal intercepts and sign of the gradient

The horizontal intercept, found by setting $y = 0$ , often answers a "when does it run out" or "break-even time" question. For an increasing quantity the gradient is positive; for a decreasing one (a draining tank, a depreciating asset) the gradient is negative. Checking that the sign of your gradient matches the direction the quantity actually moves is a quick guard against setting up the model upside down.

Why it matters

Linear modelling is the workhorse of Topic 1 and the basis for the break-even, piecewise and optimisation work that follows. The same read-the-rate, find-the-start structure appears whenever a quantity changes at a steady rate, and interpreting the gradient and intercept in context is among the most frequently examined skills in Stage 2 General Mathematics.

Exam-style practice questions

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

SACE 20234 marksCalculator-assumed. A mobile phone plan charges a fixed monthly fee plus a cost per gigabyte of data. A customer using 5 GB pays 45 dollars, and a customer using 12 GB pays 66 dollars. (a) Find a linear model for the monthly cost

C

dollars in terms of data

d

gigabytes. (b) Interpret the gradient and the vertical intercept in context.

Show worked answer →

(a) Gradient: $m = \dfrac{66 - 45}{12 - 5} = \dfrac{21}{7} = 3$ , so the cost rises 3 dollars per GB. Substitute $(5, 45)$ into $C = 3d + c$ : $45 = 15 + c$ , so $c = 30$ . The model is $C = 3d + 30$ . (2 marks)

(b) The gradient $3$ means each extra gigabyte adds 3 dollars to the monthly cost. The intercept $30$ is the fixed monthly fee paid even when no data is used. (2 marks)

Marks: two for the model via gradient and intercept, two for contextual interpretation of both parameters.

SACE 20223 marksCalculator-assumed. A water tank holds 80 litres after 2 minutes of draining and 50 litres after 5 minutes. (a) Find a linear model for the volume

V

litres after

t

minutes. (b) Determine when the tank is empty.

Show worked answer →

(a) Gradient: $m = \dfrac{50 - 80}{5 - 2} = \dfrac{-30}{3} = -10$ , so the tank drains $10$ litres per minute. Substitute $(2, 80)$ : $80 = -10(2) + c$ , so $c = 100$ . The model is $V = -10t + 100$ . (2 marks)

(b) Empty when $V = 0$ : $0 = -10t + 100$ , so $t = 10$ minutes. (1 mark)

Marks: two for the model, one for solving $V = 0$ to find $t = 10$ .