# Modelling with Functions

> Direct and inverse variation, solving equations with graphs, and choosing a function to model real data.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/modelling-with-functions/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The lemonade stall
Chloe sells lemonade at the Saturday market. She notices that every cup costs her 50 cents in lemons and sugar, and she pays 20 dollars for the stall. She also notices that when she raises the price, fewer people buy.

Chloe writes it all as functions: cost, number of sales, profit. Now she can test prices on paper before trying them on real customers. That is **modelling**: turning a real situation into a function, using it to predict, then checking if the prediction makes sense.

**In short:** Modelling with functions means choosing a function that matches a real situation, like a straight line for a constant rate or $y = \dfrac{k}{x}$ for inverse variation. Find the constants from known values, then check that the model's predictions make sense.

## Starting point: Years 9 and 10

### Direct variation

$y$ is **directly proportional** to $x$ when $y = kx$. The graph is a straight line through the origin. $k$ is the **constant of variation**.

### Worked example: Finding k
The cost of petrol varies directly with the litres bought. 40 L costs 80 dollars.

$80 = k \times 40$, so $k = 2$ dollars per litre. 55 L costs $2 \times 55 = 110$ dollars.

Variation can also be with a power: the stopping distance of a car varies with the **square** of its speed, $d = kv^2$. Double the speed and the distance becomes four times as long.

### Inverse variation

$y$ is **inversely proportional** to $x$ when $y = \dfrac{k}{x}$. Double one, the other halves.

Inverse variation can use a power too. Light intensity varies inversely with the **square** of the distance: $I = \dfrac{k}{d^2}$. If $I = 36$ at 2 m, then $k = 36 \times 4 = 144$, and at 6 m, $I = \dfrac{144}{36} = 4$. Tripling the distance makes the light 9 times weaker.

The same method works for any power, such as $W = kr^3$: find $k$ from the known pair, then use the formula. Put every measurement in the same units first, for example $5000 \text{ cm}^2 = 0.5 \text{ m}^2$ (because $1 \text{ m}^2 = 10\,000 \text{ cm}^2$).

### Reading a model

A model has **parameters** (the fixed numbers) and **variables**. In $C = 0.5n + 20$ for Chloe's cost, 20 is the fixed cost and 0.5 is the cost per cup.

### Key idea: The modelling cycle
1. Pick the variables and what they mean.
2. Choose a type of function that fits the pattern.
3. Use data to find the parameters.
4. Use the model to predict.
5. Check: does the answer make sense? What are the limits?

### Watch out: Models have limits
A model may say Chloe sells $-10$ cups at a very high price. Negative cups make no sense. Always state a sensible domain.

**Mind map**

```text
Modelling
  Variation
    Direct: y = kx
    Power: y = kx²
    Inverse: y = k/x
  Pick a shape
    Constant rate: linear
    Max or min: quadratic
    Grows by a percent: exponential
    Repeats: trigonometric
  Graphs solve equations
    Intersection points
  Check
    Sensible domain
    Compare with data
```

### Year 11: Solving equations with graphs
The solutions of $f(x) = g(x)$ are the x-values where the graphs of $y = f(x)$ and $y = g(x)$ cross.

**Example.** How many solutions does $x^2 = 2^x$ have? Sketch both. They cross at $x = 2$ and $x = 4$, and once more for a negative $x$ near $-0.77$. So there are three solutions.

To solve $x^3 - 3x = 1$ approximately, sketch $y = x^3 - 3x$ and the line $y = 1$ and read off the crossings. Graphing software gives the values to any accuracy.

### Year 11: Profit and break-even
**Revenue** is price times quantity. **Profit** is revenue minus cost.

If Chloe sells cups at 2 dollars, revenue is $R = 2n$ and cost is $C = 0.5n + 20$. Profit $P = R - C = 1.5n - 20$. **Break-even** is when $P = 0$: $n = 13.3$, so she needs 14 cups to make a profit.

### Year 12: Choosing and fitting a model
Look at the pattern in the data to pick the type of function:

| Pattern | Model |
|---|---|
| Same amount added each step | linear $y = mx + c$ |
| Rises then falls, or a best value | quadratic |
| Same percentage change each step | exponential $y = Ae^{kt}$ |
| Repeats in cycles | trigonometric $y = a\sin(b(t - c)) + d$ |

**Example.** Chloe finds that at price $p$ dollars she sells $n = 100 - 20p$ cups. Profit is

$$P = (p - 0.5)(100 - 20p) - 20$$

This is a quadratic that opens downward. The product $(p - 0.5)(100 - 20p)$ is zero at $p = 0.5$ and $p = 5$, and taking away 20 does not move the vertex sideways. So the best price is halfway: $p = 2.75$ dollars. Calculus gives the same result by setting $P'(p) = 0$.

**Using transformations to fit.** Daily temperature can be modelled by a sine curve: the centre line is the average, the amplitude is half the difference between high and low, and the period is 24 hours. Each parameter has a real meaning.

**Fitting a quadratic through three points.** Put each point into $y = ax^2 + bx + c$ to get three equations. Through $(0, 1)$, $(1, 4)$ and $(2, 5)$: $c = 1$, $a + b + 1 = 4$ and $4a + 2b + 1 = 5$. So $a + b = 3$ and $2a + b = 2$, giving $a = -1$, $b = 4$, and $y = -x^2 + 4x + 1$. Its greatest value is at $x = 2$.

**Piecewise models.** Some prices change rule part way: a hire costs 30 dollars for up to 3 hours, then 10 dollars for each extra hour. Write a rule for each part ($C = 30$ for $0 < t \leq 3$, and $C = 30 + 10(t - 3)$ for $t > 3$), then compare with other plans part by part.

## Quick summary

- Direct variation: $y = kx$. Inverse variation: $y = \dfrac{k}{x}$. Find $k$ from one known pair.
- A model has parameters with real meanings, such as a fixed cost and a rate.
- Pick the function type from the pattern: constant rate, maximum, percentage growth or cycles.
- Solutions of $f(x) = g(x)$ are where the two graphs meet.
- Always check that predictions make sense and state a sensible domain.

## Quiz

Try these questions from the chapter quiz.

1. $y$ varies directly with $x$, and $y = 12$ when $x = 3$. Find $y$ when $x = 5$.
   - A. 15
   - B. 20
   - C. 36
   - D. 7.2
   - Answer: B. 20. $k = 4$, so $y = 4 \times 5 = 20$.

2. The graph of a direct variation $y = kx$ is:
   - A. a straight line through the origin
   - B. a parabola
   - C. a hyperbola
   - D. a horizontal line
   - Answer: A. a straight line through the origin. $y = kx$ has gradient $k$ and intercept 0.

3. $d = kv^2$. If $v$ doubles, $d$ becomes:
   - A. double
   - B. half
   - C. four times as big
   - D. the same
   - Answer: C. four times as big. $(2v)^2 = 4v^2$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/modelling-with-functions/#quiz