Linear Law Practice Questions

Q: How do I decide what to plot on each axis?

Rearrange the equation into Y=mX+cY=mX+cY=mX+c. Whatever multiplies the gradient is XXX, and the isolated variable is YYY. For y=ax2+by=ax^{2}+by=ax2+b you plot yyy against x2x^{2}x2; for y=abxy=ab^{x}y=abx you plot lg⁡y\lg ylgy against xxx.

Q: When do I need to take logarithms first?

Take lg⁡\lglg whenever an unknown constant appears as a power or the whole right side is a product of powers, such as y=axny=ax^{n}y=axn or y=abxy=ab^{x}y=abx. Logarithms bring the power down to a coefficient, which is what makes the graph a straight line.

Q: The intercept came out as \lg a. How do I find a itself?

Undo the logarithm with a power of 101010. If lg⁡a=2\lg a=2lga=2 then a=102=100a=10^{2}=100a=102=100. The same trick recovers bbb from a gradient equal to lg⁡b\lg blgb.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from reducing a simple relationship to linear form to finding two constants from a logarithmic straight-line graph. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, identify $Y$ , $X$ , the gradient and the intercept, then substitute clearly.

Resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, correctly matching your relationship to $Y=mX+c$ still earns credit even when the arithmetic slips, so always state $Y$ and $X$ before you compute.

Six practice questions

Q1[3 marks]

The variables $x$ and $y$ are related by $y=ax^{2}+b$ , where $a$ and $b$ are constants. Express this equation in the linear form $Y=mX+c$ , and state $Y$ , $X$ , the gradient and the $Y$ -intercept.

Show worked solution

The relationship is already linear if you treat $x^{2}$ as a single variable. Compare $y=ax^{2}+b$ directly with $Y=mX+c$ : the whole of $x^{2}$ plays the role of $X$ , and $y$ plays the role of $Y$ .

\underbrace{y}_{Y}=a\,\underbrace{x^{2}}_{X}+b

So you plot $y$ against $x^{2}$ . Matching term by term with $Y=mX+c$ gives the gradient and intercept:

Answer

$Y=y$ , $X=x^{2}$ , gradient $m=a$ , and $Y$ -intercept $c=b$ . The straight line of $y$ against $x^{2}$ therefore has gradient $a$ and cuts the vertical axis at $b$ .

Q2[3 marks]

The variables $x$ and $y$ are related by $y=ax^{n}$ , where $a$ and $n$ are constants. Using logarithms to base $10$ , express this equation in the linear form $Y=mX+c$ , and state what should be plotted on each axis.

Show worked solution

Because the unknown $n$ sits as a power, take $\lg$ of both sides so the power comes down. Use $\lg(pq)=\lg p+\lg q$ and $\lg p^{k}=k\lg p$ :

\lg y=\lg\big(ax^{n}\big)=\lg a+\lg x^{n}=\lg a+n\lg x

Reorder to match $Y=mX+c$ , with the variable part written first:

\underbrace{\lg y}_{Y}=n\,\underbrace{\lg x}_{X}+\lg a

Answer

Plot $\lg y$ (vertical axis, $Y$ ) against $\lg x$ (horizontal axis, $X$ ). The gradient is $m=n$ and the $Y$ -intercept is $c=\lg a$ .

Q3[4 marks]

The variables $x$ and $y$ are related by $y=ab^{x}$ , where $a$ and $b$ are constants. Express the equation in linear form, and explain how $a$ and $b$ are obtained from the straight-line graph of $\lg y$ against $x$ .

Show worked solution

Here the variable $x$ is the power, so take $\lg$ of both sides. The term $\lg b^{x}$ becomes $x\lg b$ , which is linear in $x$ :

\lg y=\lg a+\lg b^{x}=\lg a+x\lg b

Write the variable term first to match $Y=mX+c$ . Note that the plot is $\lg y$ against $x$ itself, only $y$ is logged, not $x$ :

\underbrace{\lg y}_{Y}=(\lg b)\,\underbrace{x}_{X}+\lg a

The gradient equals $\lg b$ and the $Y$ -intercept equals $\lg a$ . Reverse each with a power of $10$ to recover the constants:

b=10^{\,\text{gradient}} \qquad a=10^{\,\text{intercept}}

Answer

Gradient $=\lg b$ , so $b=10^{\text{gradient}}$ ; $Y$ -intercept $=\lg a$ , so $a=10^{\text{intercept}}$ . Read the gradient and intercept off the line, then undo the logarithm with a power of $10$ .

Q4[4 marks]

When $y=ax^{2}+b$ is reduced to linear form and $y$ is plotted against $x^{2}$ , the resulting straight line passes through the points $(1, 7)$ and $(3, 15)$ , where each point is $(x^{2}, y)$ . Find the values of $a$ and $b$ .

Show worked solution

From question 1, plotting $y$ against $x^{2}$ gives $Y=y$ , $X=x^{2}$ , gradient $a$ and intercept $b$ . Find the gradient from the two points using $m=\dfrac{Y_{2}-Y_{1}}{X_{2}-X_{1}}$ :

a=\frac{15-7}{3-1}=\frac{8}{2}=4

Now substitute one point, say $(1, 7)$ , into $y=aX+b$ with $a=4$ to find $b$ :

7=4(1)+b \;\Rightarrow\; b=3

Answer

$a=4$ and $b=3$ . Check with the other point: at $X=3$ , $y=4(3)+3=15$ , which matches the given point $(3,15)$ .

Q5[5 marks]

The variables $x$ and $y$ are related by $y=ax^{n}$ . When $\lg y$ is plotted against $\lg x$ , a straight line passes through the points $(0, 1)$ and $(2, 5)$ .

Find the values of $a$ and $n$ .

Show worked solution

From question 2, $\lg y=n\lg x+\lg a$ , so the plot of $\lg y$ against $\lg x$ has $Y=\lg y$ , $X=\lg x$ , gradient $n$ and $Y$ -intercept $\lg a$ . Find the gradient from the two points:

n=\frac{5-1}{2-0}=\frac{4}{2}=2

The point $(0, 1)$ is on the $Y$ -axis, so the $Y$ -intercept is $1$ . Since the intercept equals $\lg a$ , undo the logarithm with a power of $10$ :

\lg a=1 \;\Rightarrow\; a=10^{1}=10

Answer

$n=2$ and $a=10$ , so $y=10x^{2}$ . Check the far point: at $\lg x=2$ we have $x=100$ , and $y=10(100)^{2}=100000$ , giving $\lg y=5$ , which matches $(2,5)$ .

Q6[5 marks]

The variables $x$ and $y$ are related by $y=ab^{x}$ . When $\lg y$ is plotted against $x$ , a straight line passes through the points $(0, 2)$ and $(3, 5)$ .

Find the values of $a$ and $b$ .

Show worked solution

From question 3, $\lg y=(\lg b)x+\lg a$ , so the plot of $\lg y$ against $x$ has gradient $\lg b$ and $Y$ -intercept $\lg a$ . Find the gradient from the two points:

\lg b=\frac{5-2}{3-0}=\frac{3}{3}=1 \;\Rightarrow\; b=10^{1}=10

The point $(0, 2)$ lies on the $Y$ -axis, so the $Y$ -intercept is $2$ . Since the intercept equals $\lg a$ , undo the logarithm:

\lg a=2 \;\Rightarrow\; a=10^{2}=100

Answer

$b=10$ and $a=100$ , so $y=100(10)^{x}$ . Check the far point: at $x=3$ , $y=100(10)^{3}=100000$ , giving $\lg y=5$ , which matches $(3,5)$ .

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct linear form, the right identification of $Y$ and $X$ , and a clean value for each constant.

Method mark: did you reduce to $Y=mX+c$ correctly, taking $\lg$ only when an unknown sits as a power?
Identification mark: are $Y$ , $X$ , the gradient and the intercept each named correctly, for $y=ab^{x}$ the plot is $\lg y$ against $x$ , not against $\lg x$ ?
Substitution mark: is the gradient found as $\dfrac{Y_{2}-Y_{1}}{X_{2}-X_{1}}$ , and a point put back to find the intercept?
Answer mark: when the intercept or gradient equals $\lg a$ or $\lg b$ , did you undo it with a power of $10$ to recover the constant?
If your final number is wrong but the linear form and substitution are right, give yourself the method marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, logging $x$ when only $y$ should be logged, forgetting to undo $\lg a$ with a power of $10$ , or misreading a gradient from the graph, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How do I decide what to plot on each axis?

Rearrange the equation into $Y=mX+c$ . Whatever multiplies the gradient is $X$ , and the isolated variable is $Y$ .

For $y=ax^{2}+b$ you plot $y$ against $x^{2}$ ; for $y=ab^{x}$ you plot $\lg y$ against $x$ .

When do I need to take logarithms first?

Take $\lg$ whenever an unknown constant appears as a power or the whole right side is a product of powers, such as $y=ax^{n}$ or $y=ab^{x}$ . Logarithms bring the power down to a coefficient, which is what makes the graph a straight line.

The intercept came out as

\lg a

. How do I find

a

itself?

Undo the logarithm with a power of $10$ . If $\lg a=2$ then $a=10^{2}=100$ .

The same trick recovers $b$ from a gradient equal to $\lg b$ .

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct linear form and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always state $Y$ and $X$ first.

Source:SRC-DSKP-EN

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

Linear Law, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

Ready to get started?