Drawing straight-line graphs from an equation
Build a table, plot accurate points and draw the complete line with a ruler.
Core and Extended. Core equations are given as y = mx + c unless a table is supplied; rearranging other forms is taught later as Extended content.
Before you begin
- Substitute positive and negative values into an expression.
- Plot an ordered pair on a labelled grid.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is −2(−1) + 3?
What you will learn
- Generate coordinate pairs from y = mx + c.
- Draw and check a straight-line graph on a suitable scale.
- Recognise horizontal and vertical lines.
An equation describes every point on the line
For y = 2x − 1, choosing x = −1 gives y = −3, so (−1, −3) is one point on the graph. Choosing x = 0 gives (0, −1), and x = 2 gives (2, 3). Each pair satisfies the same equation. A table records the input x and its corresponding output y; do not mix a y-value with an x-value from another column.
Work through each substitution in brackets, especially for negative inputs. For y = −2x + 3 and x = −1, y = −2(−1) + 3 = 5. The minus sign is part of the multiplier. Choose x-values that give points spread across the available grid, rather than three points crowded together.
Plot, check, then join
Label both axes and choose a scale that fits the required values. Plot at least three points, although two distinct points determine a line. The third is a useful check: if it does not lie on the line through the first two, revisit the arithmetic, scales and plotting before drawing a final line.
Use a ruler to draw a straight line across the requested graph interval. Do not join points with a curved stroke or stop the graph at the first and last calculated point unless the question restricts the domain. A line's intercept at x = 0 is especially useful for checking the constant c in y = mx + c.
Horizontal and vertical equations
For y = 3, every point has vertical coordinate three, while x is free to vary. The graph is horizontal. For x = −2, every point has horizontal coordinate negative two, while y varies, so the graph is vertical. A vertical line cannot be written as y = mx + c with a finite gradient.
The equations x = 0 and y = 0 describe the y-axis and x-axis respectively. Name a line by the coordinate that stays fixed. A value table supplied with a question can also be plotted directly: use the given pairs, check the scale and decide whether the points lie on a straight line.
Pause and explain
Which point lies on y = 2x − 1?
Worked example
Draw y = −2x + 3 for −1 ≤ x ≤ 3.
Show the worked solution
- For x = −1, 0, 1, 2, 3, calculate y = 5, 3, 1, −1, −3.
- Plot the five pairs on labelled axes, using one consistent scale on each axis.
- Check that (0, 3) is the y-intercept and all points align; draw the straight segment over the stated interval.
Answer The line passes through (−1, 5), (0, 3), (1, 1), (2, −1), (3, −3).
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Make a table and draw y = 3x + 1 for x = −1, 0, 1, 2.
Give me a hint
Find each y-value before plotting the pair.
Compare my reasoning
- The y-values are −2, 1, 4 and 7.
- Plot (−1, −2), (0, 1), (1, 4), (2, 7).
- The points align and cross the y-axis at one; draw the line with a ruler.
Pairs: (−1, −2), (0, 1), (1, 4), (2, 7)
Look for these in your work
- Computed every substitution correctly.
- Used a straight line and labelled axes.
Draw y = 4 and x = −1 on the same axes. State their intersection.
Give me a hint
Identify which coordinate each equation holds constant.
Compare my reasoning
- Draw a horizontal line four units above the origin.
- Draw a vertical line one unit left of the origin.
- Their intersection has x = −1 and y = 4.
Intersection (−1, 4)
Look for these in your work
- Distinguished horizontal and vertical equations.
- Read the intersection as an ordered pair.
A tank initially holds 6 litres and gains 2 litres each minute. Draw V = 2t + 6 for 0 ≤ t ≤ 4 and explain why the negative-time part is excluded.
Give me a hint
Use time on the horizontal axis and volume on the vertical axis.
Compare my reasoning
- For t = 0, 1, 2, 3, 4, volumes are 6, 8, 10, 12, 14 litres.
- Plot the pairs and join them over the interval from zero to four minutes.
- Negative t is before the modelled filling starts, so that part is outside the stated situation.
Line segment from (0, 6) to (4, 14)
Look for these in your work
- Labelled quantities and units.
- Respected the context's restricted time interval.
Common mistakes
- Multiplying negative inputs without brackets.
- Drawing a curve through linear points.
- Confusing x = k with a horizontal line.
What can you explain now?
Give the y-intercept of y = −3x − 2.
Compare with the explanation
(0, −2)
At the y-axis x = 0, so substitution leaves y = −2.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw a new line from a table of three values tomorrow, then check a fourth point by substitution.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons; Geometry contains 15. Coordinate geometry and Geometry each have a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other five syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice and paper constructions are self-checked, not automatically graded. The checkpoints sample skills and do not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources