Is estimating y at x=5 reliable?

The data cover x=1x=1x=1 to x=4x=4x=4, so x=5x=5x=5 is a short step beyond them, an extrapolation, which is less secure than reading a value inside the range. The estimate y≈15.8y\approx15.8y≈15.8 is reasonable if the model holds near x=5x=5x=5. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, and noting the extrapolation earns the comment mark.

KBAT: Choosing Variables for Linear Law

Q: Why plot xy against x^2 and not something else?

Because multiplying y=ax+bxy=ax+\tfrac{b}{x}y=ax+xb by xxx gives xy=ax2+bxy=ax^2+bxy=ax2+b, which matches Y=mX+cY=mX+cY=mX+c exactly when Y=xyY=xyY=xy and X=x2X=x^2X=x2. That pairing makes the gradient equal to aaa and the intercept equal to bbb. Any other choice would leave a term that still curves, so the graph would not be straight.

What makes this a KBAT question

A routine linear-law question tells you what to plot: 'plot $\lg y$ against $x$ ', and you obey. A KBAT version hands you a relation such as $y=ax+\tfrac{b}{x}$ and asks you to decide which variables straighten it, and here neither $y$ against $x$ nor a simple logarithm works.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: rearranging into $Y=mX+c$ and reading a gradient are ordinary Form 4 skills, but recognising that multiplying through by $x$ turns the relation into $xy=ax^2+b$ , so you must plot $xy$ against $x^2$ , is a choice the question does not make for you. Matching the reshaped equation to $Y=mX+c$ and then identifying what the gradient and intercept represent is the real test.

The algebra of the fit is easy once the right variables are chosen.

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]

Two positive quantities $x$ and $y$ are believed to be related by $y=ax+\dfrac{b}{x}$ , where $a$ and $b$ are constants. An experiment records the values $(x,y)=(1,7),\,(2,8),\,(4,13)$ .

(a) Explain why plotting $y$ directly against $x$ does not give a straight line, and reduce the relation to linear form, stating the variables to plot and what the gradient and intercept represent. (b) Using the recorded values, obtain the straight-line equation and find $a$ and $b$ .

Show worked solution

Understand. The relation $y=ax+\tfrac{b}{x}$ mixes a term in $x$ with a term in $\tfrac{1}{x}$ , so $y$ is not a linear function of $x$ , its graph curves.

We must reshape it into the straight-line template $Y=mX+c$ by choosing new variables.

Plan. The obstacle is the $\tfrac{b}{x}$ term.

Multiplying the whole equation by $x$ clears the fraction and produces powers of $x$ we can group. Then compare the result with $Y=mX+c$ , build a table, find the gradient and intercept, and read off $a$ and $b$ .

Execute. (a) Multiply $y=ax+\tfrac{b}{x}$ by $x$ :

xy = ax^2 + b

Compare with $Y=mX+c$ : let $Y=xy$ and $X=x^2$ . Then the gradient $m=a$ and the vertical intercept $c=b$ .

So we plot $xy$ against $x^2$ ; the line's gradient gives $a$ and its intercept gives $b$ .

(b) Build the table of plotting variables from the data:

x	y	X = x²	Y = xy
1	7	1	7
2	8	4	16
4	13	16	52

Using the first and last points $(X,Y)=(1,7)$ and $(16,52)$ , the gradient is

m = \frac{52-7}{16-1} = \frac{45}{15} = 3

So $a=3$ . Find the intercept from $Y=3X+c$ using $(1,7)$ : $7=3(1)+c$ , giving $c=4$ , so $b=4$ .

The straight-line equation is

xy = 3x^2 + 4, \quad\text{so}\quad y = 3x + \frac{4}{x}

y = 3(5) + \frac{4}{5} = 15 + 0.8 = 15.8

The recorded data run from $x=1$ to $x=4$ , so $x=5$ lies just beyond the range, a short extrapolation. The estimate $y\approx15.8$ is reasonable provided the relationship $y=ax+\tfrac{b}{x}$ continues to hold near $x=5$ ; the further we go past the data, the less we can rely on it.

Check. Test the middle point $(2,8)$ , which was not used for the gradient: $X=4$ , $Y=xy=16$ , and $3(4)+4=16$ .

It lies exactly on the line, confirming $a=3$ and $b=4$ .

Finding a sensible first step

When a relation refuses to be linear, the reliable first step is to look at which term is blocking a straight line and act to remove it, rather than guessing at logarithms. In $y=ax+\tfrac{b}{x}$ the blocker is the fraction $\tfrac{b}{x}$ , so the natural move is to multiply every term by $x$ , giving $xy=ax^2+b$ .

Now stand this beside the template $Y=mX+c$ and match part for part: the constant $b$ sits where $c$ is, the coefficient $a$ multiplies $x^2$ where $m$ multiplies $X$ . That reading forces the choice $Y=xy$ and $X=x^2$ .

Choosing variables by matching to $Y=mX+c$ , not by habit, is the skill being tested. Only take logarithms when the unknown constants sit in a power or an exponent; here they do not, so a plain rearrangement is the right tool.

What markers reward

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

A clear reason that $y$ against $x$ is not linear, a term in $x$ plus a term in $\tfrac{1}{x}$ .
The correct rearrangement, multiplying by $x$ to get $xy=ax^2+b$ .
The right choice of plotting variables, $Y=xy$ and $X=x^2$ , matched to $Y=mX+c$ .
A correct table of $X$ and $Y$ values computed from the data.
Gradient and intercept found correctly, giving $a=3$ and $b=4$ .
An estimate at $x=5$ with a comment on interpolation versus extrapolation.

How a teacher helps

Choosing-variables questions reward students who match an equation to $Y=mX+c$ deliberately, and that habit grows with feedback. In a one-to-one lesson our teachers ask you to name the term that blocks a straight line, to try the rearrangement rather than reaching for logarithms by reflex, and to write the pairing $Y=xy$ , $X=x^2$ , $m=a$ , $c=b$ explicitly before plotting.

We check your table column by column and your gradient from two clear points. 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 know whether to take logarithms or just rearrange?

Look at where the unknown constants sit. If they appear as a power or an exponent, $y=ax^n$ or $y=ab^x$ , take logarithms to bring them down.

If they are ordinary coefficients, as in $y=ax+\tfrac{b}{x}$ , a plain rearrangement is enough: multiply through to clear fractions, then match to $Y=mX+c$ . Choosing the method to fit the equation is the point of the question.

Why plot

xy

against

x^2

and not something else?

Because multiplying $y=ax+\tfrac{b}{x}$ by $x$ gives $xy=ax^2+b$ , which matches $Y=mX+c$ exactly when $Y=xy$ and $X=x^2$ . That pairing makes the gradient equal to $a$ and the intercept equal to $b$ .

Any other choice would leave a term that still curves, so the graph would not be straight.

Is estimating

y

x=5

reliable?

The data cover $x=1$ to $x=4$ , so $x=5$ is a short step beyond them, an extrapolation, which is less secure than reading a value inside the range. The estimate $y\approx15.8$ is reasonable if the model holds near $x=5$ .

Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, and noting the extrapolation earns the comment mark.

Source:SRC-DSKP-EN SRC-FORMAT

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