KBAT: Quadratic Modelling Problems

Q: What makes a quadratic question KBAT rather than routine?

A routine question gives you the function; a KBAT question gives you a situation and expects you to build the function, choose a method, and interpret the result. The algebra is standard Form 4 quadratics, the higher-order part is the modelling and reasoning around it.

Q: Do I have to complete the square, or can I use the formula?

Either is accepted if the working is shown. Completing the square is often cleanest for a maximum or minimum because it gives the vertex directly. The axis of symmetry x=−b2ax=\frac{-b}{2a}x=2a−b is also valid, markers reward a correct, visible method.

Q: How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. A well-structured attempt earns marks even when the final number is wrong, so always show the model, the method and the interpretation.

Q: What is the most common mistake in quadratic modelling?

Skipping the definition of the variable and jumping into algebra. Without 'let xxx be…', later lines lose meaning and marks. The second common slip is forgetting to interpret, writing x=6x=6x=6 when the question asked for the maximum area, 72 m272\ \text{m}^{2}72 m2.

What makes this a KBAT question

A routine quadratic question hands you the function and asks for its roots or its vertex. A KBAT modelling question does the opposite: it describes a situation, a plot of land, a thrown ball, a profit, and expects you to construct the quadratic yourself, decide what its vertex or intercepts mean, and answer a question posed in ordinary language.

That is higher-order thinking, Kemahiran Berfikir Aras Tinggi: the individual skills of expanding a product, completing the square and reading a maximum are all familiar, but you apply them in an unfamiliar order and setting. Nothing tells you which technique to use, you choose.

In Add Math this rewards students who understand what a quadratic represents, not only how to manipulate one.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Work through it yourself before reading the solution.

Q1[7 marks]

A rectangular vegetable plot is to be fenced. One long side runs along an existing straight wall and needs no fence.

The total length of fencing available for the other three sides is $24$ m. Let $x$ metres be the length of each side perpendicular to the wall.

(a) Show that the enclosed area $A$ m $^{2}$ can be written as $A=24x-2x^{2}$ . (b) Find the value of $x$ that gives the maximum area, and state that maximum area.

(c) The gardener claims that a square plot fenced on three sides would enclose more area with the same fencing. Determine whether the claim is correct.

Show worked solution

Understand. There are $24$ m of fence for three sides; the wall is the fourth side.

The two sides perpendicular to the wall each have length $x$ , so the single side parallel to the wall uses the rest of the fence: $24-2x$ . We need the area as a function of $x$ , its maximum, and a comparison with the square case.

Plan. Write the area as length $\times$ width, then complete the square to find the vertex (the maximum, because the $x^{2}$ coefficient is negative).

For part (c), form the square condition, find its area, and compare.

Execute and check. (a) The side parallel to the wall is $24-2x$ , so

A = x(24-2x) = 24x - 2x^{2}

which is the required model. (b) Complete the square:

A = -2x^{2}+24x = -2(x^{2}-12x) = -2[(x-6)^{2}-36] = -2(x-6)^{2}+72

The coefficient of $(x-6)^{2}$ is negative, so the vertex is a maximum. Hence $x=6$ m gives the maximum area $A=72\ \text{m}^{2}$ .

(The side parallel to the wall is then $24-2(6)=12$ m.)

x = 24-2x \;\Rightarrow\; 3x = 24 \;\Rightarrow\; x = 8

The square would measure $8\ \text{m}\times 8\ \text{m}$ , giving an area of $64\ \text{m}^{2}$ . Since $64<72$ , the square encloses less area, so the claim is incorrect.

Check. The width must satisfy $24-2x>0$ , so $0<x<12$ ; the answer $x=6$ lies inside this range.

At the ends $x\to 0$ and $x=12$ the area falls to $0$ , which is consistent with a single maximum at $x=6$ .

Finding a sensible first step

When a modelling question looks unfamiliar, resist the urge to hunt for a formula. The reliable first step is to name a variable and write down every quantity in terms of it.

Read the problem once for the story, then again with a pen: label what is fixed, what can change, and what you are asked to maximise, minimise or find. Assign a letter, usually $x$ , to the quantity you control, then express the others using it.

The moment you can write the target (area, height, cost) as a single expression in $x$ , the modelling is done and the question becomes ordinary Add Math. If you are stuck, sketch the situation and put your letter on the diagram; a picture almost always reveals the relationship you need to write down.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a quadratic-modelling question a marker looks for:

A clearly defined variable, for example 'let $x$ be the width in metres', so every later line has meaning.
The correct model shown explicitly, such as $A=x(24-2x)$ , before any simplifying.
Completing the square (or another valid method) with the working visible, not just a stated vertex.
The maximum or minimum read correctly, with a reason it is a maximum, the negative coefficient of $x^{2}$ .
The answer interpreted back into the context, with units: $72\ \text{m}^{2}$ , not a bare number.
A check that the answer is sensible and lies within the allowed range of $x$ .

How a teacher helps

Modelling improves fastest with feedback on the step students skip, the translation from words into a quadratic. In a one-to-one lesson our teachers slow that step down: we read the question together, label the diagram, and write the model line before touching any algebra, so the habit transfers to new contexts.

We also rehearse the interpretation at the end, because markers award method marks for reasoning shown, not answers guessed. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

What makes a quadratic question KBAT rather than routine?

A routine question gives you the function; a KBAT question gives you a situation and expects you to build the function, choose a method, and interpret the result. The algebra is standard Form 4 quadratics, the higher-order part is the modelling and reasoning around it.

Do I have to complete the square, or can I use the formula?

Either is accepted if the working is shown. Completing the square is often cleanest for a maximum or minimum because it gives the vertex directly.

The axis of symmetry $x=\frac{-b}{2a}$ is also valid, markers reward a correct, visible method.

How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. A well-structured attempt earns marks even when the final number is wrong, so always show the model, the method and the interpretation.

What is the most common mistake in quadratic modelling?

Skipping the definition of the variable and jumping into algebra. Without 'let $x$ be…', later lines lose meaning and marks.

The second common slip is forgetting to interpret, writing $x=6$ when the question asked for the maximum area, $72\ \text{m}^{2}$ .

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Modelling With Quadratic Functions