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Linear functions and modelling (SACE Stage 2 General Mathematics)

Syllabus dot point

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

8 min read

Quick answer

A linear model has the form y=mx+cy = mx + c, where the gradient mm is the constant rate of change (how much yy changes per unit of xx) and the intercept cc is the starting value of yy when x=0x = 0. You build the model from the situation, then substitute to predict.

Jump to a section
  1. What this dot point is asking
  2. The linear model
  3. Building a model from words
  4. Interpreting the model in context
  5. Predicting and the limits of a model
  6. Horizontal intercepts and sign of the gradient
  7. Why it matters
  8. Exam-style questions

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 yy changes by the same amount for each unit increase in xx. The model is:

y=mx+c.y = mx + c.

Gradient and intercept

The gradient mm is the constant rate of change: m=change in ychange in x=y2−y1x2−x1m = \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{y_2 - y_1}{x_2 - x_1}. The vertical intercept cc is the value of yy when x=0x = 0, the starting amount.

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 cc.
  • A "per unit" amount (per hour, per kilometre, per item) gives the gradient mm.
A taxi fare model

Worked example

A taxi charges a flag fall of 4.50 dollars plus 2.20 dollars per kilometre. Write a model for the fare FF dollars after dd kilometres and find the fare for a 12 km trip.

The flag fall is the fixed starting cost, so c=4.50c = 4.50. The cost rises by 2.202.20 per km, so m=2.20m = 2.20.

F=2.20d+4.50.F = 2.20d + 4.50.

For d=12d = 12:

F=2.20(12)+4.50=26.40+4.50=30.90.F = 2.20(12) + 4.50 = 26.40 + 4.50 = 30.90.

The fare is 30.90 dollars.

Interpreting the model in context

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

  • The gradient 2.202.20 means the fare increases by 2.202.20 dollars for each extra kilometre.
  • The intercept 4.504.50 means the fare is 4.504.50 dollars before any distance is travelled.
Finding the gradient from two points

If you are given two data points, compute m=y2−y1x2−x1m = \dfrac{y_2 - y_1}{x_2 - x_1} first, then substitute one point into y=mx+cy = mx + c to solve for cc.

Building a model from two points

A water tank holds 8080 litres after 22 minutes of draining and 5050 litres after 55 minutes. Find a linear model for the volume VV litres in terms of time tt minutes.

Step 1: Find the gradient from the two data points

The gradient mm measures how fast the volume changes per minute. We apply the gradient formula using the two given points (t1,V1)=(2,80)(t_1, V_1) = (2, 80) and (t2,V2)=(5,50)(t_2, V_2) = (5, 50):

m=50−805−2=−303=−10.m = \frac{50 - 80}{5 - 2} = \frac{-30}{3} = -10.

The negative value confirms the tank is draining: it loses 1010 litres every minute.

Step 2: Find the vertical intercept by substituting one point

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

80=−10(2)+c  ⟹  80=−20+c  ⟹  c=100.80 = -10(2) + c \implies 80 = -20 + c \implies c = 100.

So the intercept is c=100c = 100, meaning the tank held 100100 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.V = -10t + 100.

The tank started with 100100 litres and loses 1010 litres per minute. Setting V=0V = 0 and solving gives t=10t = 10 minutes, so the tank is empty after 1010 minutes.

Final answer: V=−10t+100V = -10t + 100, valid for 0≤t≤100 \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=4t = 4 minutes is interpolation and trustworthy, but predicting at t=30t = 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=0y = 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.

Always check the sign of the gradient

A positive gradient means yy increases as xx increases; a negative gradient means yy decreases. If your model gives a decreasing quantity, such as a draining tank, the gradient must be negative.

Mixing up gradient and intercept

Do not mix up the gradient and the intercept. The "per unit" rate is always the gradient mm (it multiplies xx); the fixed one-off amount is always the intercept cc. Writing F=4.50d+2.20F = 4.50d + 2.20 for the taxi reverses them and gives a wildly wrong fare.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Calculator-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 CC dollars in terms of data dd gigabytes. (b) Interpret the gradient and the vertical intercept in context.
Show worked answer →

(a) Gradient: m=66−4512−5=217=3m = \dfrac{66 - 45}{12 - 5} = \dfrac{21}{7} = 3, so the cost rises 3 dollars per GB. Substitute (5,45)(5, 45) into C=3d+cC = 3d + c: 45=15+c45 = 15 + c, so c=30c = 30. The model is C=3d+30C = 3d + 30. (2 marks)

(b) The gradient 33 means each extra gigabyte adds 3 dollars to the monthly cost. The intercept 3030 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.

Practice3 marks
Calculator-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 VV litres after tt minutes. (b) Determine when the tank is empty.
Show worked answer →

(a) Gradient: m=50−805−2=−303=−10m = \dfrac{50 - 80}{5 - 2} = \dfrac{-30}{3} = -10, so the tank drains 1010 litres per minute. Substitute (2,80)(2, 80): 80=−10(2)+c80 = -10(2) + c, so c=100c = 100. The model is V=−10t+100V = -10t + 100. (2 marks)

(b) Empty when V=0V = 0: 0=−10t+1000 = -10t + 100, so t=10t = 10 minutes. (1 mark)

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

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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