IGCSE Additional Mathematics 0606 · Topic 8

IGCSE Additional Mathematics: Straight Line Graphs Practice Questions

Linear law reduces a non-linear relationship to the form Y = mX + c so that constants can be found from a straight line graph. Take logarithms for power and exponential laws, or divide through for simpler cases.

Cambridge IGCSE Additional Mathematics (0606) · Topic 8: Straight Line Graphs

Topic 8 of Cambridge IGCSE Additional Mathematics 0606 tests one skill repeatedly: rearranging a relationship until it matches Y = mX + c, then saying exactly what to plot. Naming the axes is where the marks sit. The questions below cover both standard forms.

What you need to know for Straight Line Graphs

IGCSE Additional Mathematics Straight Line Graphs questions and answers

4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[4 marks]

The variables x and y are related by y = axn, where a and n are constants. Explain how a straight line graph can be drawn, stating what should be plotted on each axis.

Show the worked answer
Answer: Plot lg y against lg x. Gradient n and vertical intercept lg a.
  1. Take logarithms of both sides: lg y = lg(axn).
  2. Apply the product law: lg y = lg a + lg(xn).
  3. Apply the power law: lg y = n lg x + lg a.
  4. This matches Y = mX + c with Y as lg y and X as lg x. Plot lg y on the vertical axis against lg x on the horizontal. The gradient is n and the vertical intercept is lg a.
How the marks are awarded. 1 mark for taking logarithms. 1 mark for reaching lg y = n lg x + lg a. 1 mark for stating what to plot on each axis. 1 mark for identifying gradient as n and intercept as lg a.
Where students lose the mark. Saying plot y against x. The point of linear law is that the transformed quantities are plotted, not the originals.
Question 2[4 marks]

A graph of lg y against lg x is a straight line with gradient 0.5 and vertical intercept 0.6. Given that y = axn, find a and n.

Show the worked answer
Answer: n = 0.5 and a = 3.98
  1. From lg y = n lg x + lg a, the gradient equals n, so n = 0.5.
  2. The vertical intercept equals lg a, so lg a = 0.6.
  3. Convert from logarithmic form: a = 100.6.
  4. a = 3.981, which is 3.98 to 3 significant figures.
How the marks are awarded. 1 mark for n = 0.5. 1 mark for lg a = 0.6. 1 mark for a = 100.6. 1 mark for 3.98.
Where students lose the mark. Giving a = 0.6. The intercept is the logarithm of a, so it must be converted back using a power of 10.
Question 3[4 marks]

The variables x and y are related by y = abx. Show how a straight line graph can be obtained, and state the gradient and intercept.

Show the worked answer
Answer: Plot lg y against x. Gradient lg b and intercept lg a.
  1. Take logarithms of both sides: lg y = lg a + lg(bx).
  2. Apply the power law to the second term: lg(bx) = x lg b.
  3. So lg y = (lg b)x + lg a.
  4. Comparing with Y = mX + c, plot lg y against x. The gradient is lg b and the vertical intercept is lg a. Note that x is plotted directly, not lg x.
How the marks are awarded. 1 mark for taking logarithms. 1 mark for x lg b. 1 mark for stating lg y is plotted against x. 1 mark for gradient lg b and intercept lg a.
Where students lose the mark. Plotting lg y against lg x. In an exponential law the variable is in the index, so x itself goes on the horizontal axis.
Question 4[3 marks]

The variables are related by y = Ax2 + Bx. Explain how to obtain a straight line graph without using logarithms.

Show the worked answer
Answer: Plot y divided by x against x. Gradient A and intercept B.
  1. Divide every term by x, which is valid provided x is not zero: y divided by x = Ax + B.
  2. This is already in the form Y = mX + c, with Y as y divided by x and X as x.
  3. Plot y divided by x on the vertical axis against x on the horizontal axis. The gradient is A and the vertical intercept is B.
How the marks are awarded. 1 mark for dividing through by x. 1 mark for stating what to plot on each axis. 1 mark for identifying gradient A and intercept B.
Where students lose the mark. Reaching for logarithms automatically. Logarithms are needed only when the unknown constant is an index or a base.

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Straight Line Graphs FAQs

What is linear law?

Linear law is the technique of rearranging a non-linear relationship into the form Y equals mX plus c, so that plotting the transformed variables produces a straight line. The gradient and intercept of that line then give the unknown constants.

When do I plot lg y against lg x, and when against x?

For a power law such as y equals a times x to the power n, take logs of both sides and plot lg y against lg x. For an exponential law such as y equals a times b to the power x, plot lg y against x, because the variable is in the index.

How do I recover the constants from the graph?

Compare your transformed equation with Y equals mX plus c. If the intercept equals lg a, then a is 10 raised to the intercept value. If the gradient equals lg b, then b is 10 raised to the gradient. A gradient that equals n directly needs no conversion.

Do I always need logarithms for linear law?

No. Logarithms are needed only when an unknown constant appears as an index or a base. A relationship such as y equals Ax squared plus Bx becomes linear simply by dividing through by x, giving y over x plotted against x.

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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.