KBAT: Linear Law & Experimental Data

Q: What do the gradient and intercept represent here?

After writing xy=ax2+bxy=ax^2+bxy=ax2+b, the graph of xyxyxy against x2x^2x2 has gradient aaa and YYY-intercept bbb. So the gradient gives one constant and the intercept gives the other, they are not interchangeable, and mixing them up is a frequent error.

Q: Why isn't the gradient the same as a value of y?

The gradient equals the constant aaa alone. A value like y(1)=a+by(1)=a+by(1)=a+b combines both constants. In this problem the gradient is 222 but y(1)=5y(1)=5y(1)=5, so they cannot be equal, a good reminder to interpret a graph by what its axes actually represent.

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 correct straight-line form, a visible gradient calculation and a substituted intercept can score well even if a final value is slightly off.

What makes this a KBAT question

A routine linear-law question tells you exactly what to plot,'plot $xy$ against $x^2$ ', and you follow the instruction. A KBAT question hands you a non-linear relationship and some experimental data, and expects you to decide the straight-line form yourself, work out what the gradient and intercept represent, and then read real constants back out of the graph.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: rearranging into $Y=mX+c$ , finding a gradient, and reading an intercept are all familiar, but you assemble them in an unfamiliar order and choose the plan. In Add Math this rewards students who understand why we linearise data, to test a relationship and extract its constants, not only how to draw a line of best fit.

One worked problem, in the style of Paper 2

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

Q1[8 marks]

In an experiment, corresponding values of two variables $x$ and $y$ were recorded. It is known that $x$ and $y$ are related by $y=ax+\dfrac{b}{x}$ , where $a$ and $b$ are constants.

When $xy$ is plotted against $x^2$ , a straight line is obtained that passes through the points $(2,\,7)$ and $(5,\,13)$ . (a) Find the values of $a$ and $b$ .

(b) Hence find the value of $y$ when $x=2$ . (c) A classmate claims that the gradient of the line equals the value of $y$ when $x=1$ .

Determine whether the claim is correct, giving a reason.

Show worked solution

Understand. The variables are related in a non-linear way.

Reducing the relationship to straight-line form lets us read the constants $a$ and $b$ from a graph. We are told the graph of $xy$ against $x^2$ is a straight line and given two points on it; we need $a$ and $b$ , then $y$ at $x=2$ , then a judgement on the claim.

Plan. Turn $y=ax+\frac{b}{x}$ into the form $Y=mX+c$ by multiplying through by $x$ .

Match the plotted axes to $X$ and $Y$ , so the gradient gives $a$ and the vertical intercept gives $b$ . Use the two given points to find the gradient, then substitute one point to find the intercept.

Execute and check. (a) Multiply both sides by $x$ :

xy = ax^2 + b

Comparing with $Y=mX+c$ , we plot $Y=xy$ against $X=x^2$ ; the gradient is $a$ and the $Y$ -intercept is $b$ . Using the points $(2,7)$ and $(5,13)$ :

a=\frac{13-7}{5-2}=\frac{6}{3}=2

Substitute $(2,7)$ into $Y=aX+b$ to find $b$ : $7=2(2)+b$ , so $b=3$ . Hence $a=2$ and $b=3$ , and the relationship is $y=2x+\frac{3}{x}$ .

(b) When $x=2$ :

y=2(2)+\frac{3}{2}=4+1.5=5.5

Since $2\neq 5$ , the claim is incorrect. The gradient equals only the constant $a$ , the coefficient of $x$ ; the value $y(1)=a+b$ also includes the intercept $b$ .

Confusing the gradient with a value of $y$ is a common misreading of a linear-law graph.

Check. Test the second point on the relationship: at $x^2=5$ , $xy=ax^2+b=2(5)+3=13$ , which matches the given point $(5,13)$ .

The intercept is consistent too: when $X=x^2=0$ , $Y=b=3$ .

Finding a sensible first step

When a linear-law question does not tell you what to plot, the first step is always to force the relationship into the shape $Y=mX+c$ . Look at the equation and ask what algebra will make one side linear in some group of terms: dividing by $x$ , multiplying by $x$ , or taking logarithms are the usual moves.

Here multiplying by $x$ turns $y=ax+\frac{b}{x}$ into $xy=ax^2+b$ , which is linear if we treat $xy$ and $x^2$ as single quantities. Once the form is settled, label clearly: $Y=xy$ , $X=x^2$ , gradient $=a$ , intercept $=b$ .

Writing this mapping down before you touch the numbers keeps the gradient and intercept attached to the right constant and protects every later mark.

What markers reward

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

A correct reduction to straight-line form, shown explicitly, for example $xy=ax^2+b$ .
A clear statement of what is plotted on each axis: $Y=xy$ against $X=x^2$ .
The gradient computed from two points, with the calculation $\frac{13-7}{5-2}$ visible, not just the answer $2$ .
The gradient and intercept matched to the correct constants, gradient $=a$ , intercept $=b$ .
The intercept found by substituting a point, not guessed off the axis.
The final constants used in context, and any interpretation stated in words rather than left as bare numbers.

How a teacher helps

Linear law improves fastest when a student practises the mapping step out loud,'so $Y$ is $xy$ , $X$ is $x^2$ , the gradient is $a$ ', until it becomes automatic. In a one-to-one lesson our teachers pause exactly there, before the arithmetic, because that is where marks are won or lost.

We also rehearse reading a graph honestly: what the gradient means, what the intercept means, and why they are different constants. 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

How do I decide what to plot in a linear-law question?

Rearrange the relationship until one side is linear in a group of terms, matching $Y=mX+c$ . For $y=ax+\frac{b}{x}$ , multiplying by $x$ gives $xy=ax^2+b$ , so you plot $xy$ against $x^2$ .

The move you choose (divide, multiply or take logs) depends on where the awkward term sits.

What do the gradient and intercept represent here?

After writing $xy=ax^2+b$ , the graph of $xy$ against $x^2$ has gradient $a$ and $Y$ -intercept $b$ . So the gradient gives one constant and the intercept gives the other, they are not interchangeable, and mixing them up is a frequent error.

Why isn't the gradient the same as a value of

y

The gradient equals the constant $a$ alone. A value like $y(1)=a+b$ combines both constants.

In this problem the gradient is $2$ but $y(1)=5$ , so they cannot be equal, a good reminder to interpret a graph by what its axes actually represent.

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 correct straight-line form, a visible gradient calculation and a substituted intercept can score well even if a final value is slightly off.

Source:SRC-DSKP-EN SRC-FORMAT

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

KBAT: Linear Law With Experimental Data