AC9M9A05 · YEAR 9 · ALGEBRA

Modelling Change with Linear and Quadratic Functions

ACARA v9 CONTENT DESCRIPTION “use mathematical modelling to solve applied problems involving change including financial contexts; formulate problems, choosing to use either linear or quadratic functions; interpret solutions in terms of the situation; evaluate the model and report methods and findings”
Builds on: Quadratic Functions and Equations (AC9M9A04). This unit builds on linear relationships and on quadratic functions and their graphs. The modelling cycle here, formulating, solving, interpreting, evaluating and reporting, is a method you will reuse across measurement and statistics later in Year 9.

The modelling cycle

Mathematical modelling is the craft of turning a messy real situation into a clean equation, using that equation to answer a question, and then asking honestly how well the answer fits reality. It is less about a single calculation and more about a cycle: formulate a model, solve it, interpret the solution in the words of the original problem, and finally evaluate whether the model was good enough and report what you found. This unit applies that cycle to problems involving change, including money, choosing in each case between a linear and a quadratic function.

The modelling cycle
Modelling is a loop, not a single sum: every model is formulated, solved, interpreted, evaluated and reported.
Formulate: turn the situation into an equation, choosing linear or quadratic.

Choosing linear or quadratic

The first decision is the most important one: which kind of function fits the situation. The test is how the quantity changes. If something grows or shrinks by the same amount every step, the rate of change is constant and a linear function, a straight line, is the right tool. If instead the rate of change itself changes, or the quantity is built from an area or a product of two things that both vary, a squared term appears and a quadratic function fits. Getting this choice right at the start saves all the work that follows.

A linear model for everyday money

Linear models suit most everyday money problems. Suppose a phone plan charges $15 a month plus 20 cents for each gigabyte of data. The cost in dollars is 15 plus 0.2 times the number of gigabytes, written as cost = 0.2x + 15. The 15 is the fixed starting value and the 0.2 is the constant rate per gigabyte. At 25 gigabytes the cost is $20. Because every extra gigabyte adds exactly the same 20 cents, the model is linear, and its graph is a straight line climbing steadily.

Constant rate means linear
When each extra unit costs the same, the running total climbs by equal steps and the graph is a straight line.
At 25 GB the cost is $20: cost = 0.20 × 25 + 15. Every extra gigabyte adds the same $0.20, so cost = 0.20x + 15 is a straight line.

Running a linear model forwards and backwards

Solving a linear model usually means running it forwards or backwards. A savings plan starting at $200 and adding $50 a week is modelled by amount = 50x + 200, where x counts the weeks. Forwards, after 8 weeks the balance is $600. Backwards, asking when the balance first reaches $600 means solving 50x + 200 = 600, which gives x = 8 weeks. Interpreting that solution in context, not as a bare number but as eight weeks of saving, is part of the modelling, not an afterthought.

Saving toward a goal
One linear model answers two questions: where it ends after some weeks, and which week first reaches a target.
Forwards: at x = 8, amount = 50 × 8 + 200 = $600. Eight weeks of saving reaches the goal.

Quadratic models for area

Quadratic models appear whenever an area or a falling-and-rising motion is involved. Imagine fencing a rectangular garden whose width and length add to 20 metres, so the length is 20 − x when the width is x. The area is width times length, that is x(20 − x), which expands to 20x − x², a quadratic. A table of widths shows the area rising to a peak and falling away again, reaching its largest value of 100 square metres when the width is 10 metres, which makes the garden a square. The peak sits at the vertex of the parabola; here the squared term has a negative coefficient, so the parabola opens downward and the vertex is a maximum rather than a minimum.

Maximising a garden's area
Fix the perimeter and the area becomes a product of two changing lengths, so it curves and has a single best width.
Width 10 m gives area = 20 × 10 − 10² = 100 square metres. This is the vertex (10, 100): the largest area, since the negative squared term opens the parabola downward.

Quadratic models for motion

Motion under gravity gives a second classic quadratic. A toy rocket whose height in metres is modelled by h = −5t² + 20t, with t in seconds, climbs and then falls. Evaluating the model at a few times gives heights of 15, 20 and 15 metres at 1, 2 and 3 seconds. The greatest height, 20 metres, occurs at the vertex when t = 2 seconds, and the rocket returns to the ground, height zero, at t = 4 seconds. Each of these numbers answers a real question about the flight, which is what interpreting a solution means.

A rocket's flight
Thrown straight up, height rises then falls; the squared term bends the path into a parabola with a peak.
At t = 2 s, height = −5 × 2² + 20 × 2 = 20 m. This vertex is the greatest height, 20 m, reached at t = 2 s.

Evaluate and report: a model knows its limits

The cycle is not finished until you evaluate and report. Evaluating asks whether the model is realistic: the savings line assumes you never miss a week, and the rocket model ignores air resistance, so each answer is only as trustworthy as those assumptions. Reporting means stating clearly what you did, what you found, and where the model might break down. A good model is honest about its own limits, and presenting both the answer and its caveats is the hallmark of real mathematical modelling rather than mere arithmetic.

Evaluating the assumptions
A model is a deliberate simplification, so its answers are only as good as the assumptions behind them.
On its own the model line assumes saving never stops. Reveal a realistic missed week to test that assumption.
Quick self-check
1. A taxi charges a $4 flagfall plus $2 per kilometre. Which function models the fare for x kilometres, and what type is it?
2. Which situation is best modelled by a QUADRATIC function?
3. Using amount = 50x + 200 dollars after x weeks, when does the balance first reach $600?
4. For the area model area = 20x − x² (width x), what width gives the MAXIMUM area, and why is it a maximum?
5. A rocket's height is height = −5t² + 20t metres after t seconds. What is its greatest height?
Teaching pack: a week of lessons, free to printFive ready-to-teach lessons for this unit: teacher plans, student worksheets with stretch sections, situation sort and model match cards, a mini-check and every answer, with working. Print it or save as PDF, and the week is prepared.