Overview
What you’ll learn
- Read values and rates from conversion and distance–time graphs, quoting units.
- Find acceleration, deceleration and total distance from a speed–time graph (Extended).
- Complete a table of values and plot a curve accurately with a smooth line.
- Read roots and line–curve intersections off a graph, and draw straight lines to solve related equations.
- Sketch parabolas with roots, symmetry and turning points, and describe cubic, reciprocal and exponential shapes with asymptotes (Extended).
Syllabus checklist (0580)
Statements are paraphrased and grouped by tier. Core is studied by everyone; Extended content is examined on Papers 2 and 4 only.
Core — every candidate
- 2.9Interpret and use conversion graphs and travel (distance–time) graphs, with gradient as a rate of change.
- 2.10Draw and read graphs of linear functions, quadratic functions and the reciprocal function a⁄x from tables of values; Extended candidates also draw sums of powers and exponential graphs.
- 2.10Use graphs to solve equations, including finding roots and intersections with drawn lines.
Extended — Extended candidates only
- 2.9Work with speed–time graphs, including acceleration and deceleration as gradients, and distance as the area under the graph (linear sections only). (Extended)
- 2.11Sketch and interpret linear and quadratic graphs (roots and symmetry), and — at Extended — cubic, reciprocal and exponential graphs, with turning points and asymptotes.
In 30 seconds
- 1On a conversion graph through the origin, the gradient is the exchange rate — read it from two clear grid points, not one estimated point.
- 2On a distance–time graph, gradient = speed; a horizontal section means stationary; steepness up or down shows direction of travel.
- 3Speed–time graphs (Extended): gradient = acceleration; area under the graph = distance; split the area into rectangles and triangles.
- 4Tables of values drive plotting: substitute each x, plot the points, join with a smooth curve — never a zig-zag.
- 5Roots are where the curve crosses the x-axis (y = 0); to solve f(x) = k, draw the horizontal line y = k and read the x-values of the intersections.
- 6A parabola y = ax² + bx + c is symmetric about the vertical line through its turning point, midway between the roots.
- 7Sketches need shape, labelled intercepts and turning points — not plotted tables. y = 1⁄x hugs the axes as asymptotes; y = abˣ passes through (0, a·b⁰).
Notes · 1 of 4
Conversion and distance–time graphs
Gradient is a rate — read it from the grid, quote the units.
A conversion graph maps one unit to another with a straight line, usually through the origin. If the line from (0, 0) passes through (50, 45) — dollars to euros, say — then $20 reads across-and-up to €18. The gradient 45⁄50 = 0.9 is the exchange rate itself: one dollar is 0.9 euros. Going the other way, divide: €45 → $50.
A distance–time graph tells a journey. Distance (vertical) against time (horizontal): a rising straight section means steady motion, with gradient = speed; a flat section means the object is stationary; a falling section means it is coming back. A 60 km trip in 1.5 hours has an average speed of 60 ÷ 1.5 = 40 km/h — for a whole journey, use total distance over total time, including the stops.
Notes · 2 of 4
Speed–time graphs
Gradient for acceleration, area for distance. (Extended)
Extended. Swap the roles: on a speed–time graph the gradient is now acceleration (how quickly speed changes per second). A speed rising from 0 to 12 m/s in 6 seconds gives an acceleration of 12 ÷ 6 = 2 m/s²; a fall from 12 to 0 in 3 seconds is a deceleration of 12 ÷ 3 = 4 m/s².
The area under the graph is the distance travelled. Split it into rectangles and triangles. For a journey that accelerates to 12 m/s in 6 s, cruises at 12 m/s for 10 s, then brakes to rest in 3 s: the triangle is ½ × 6 × 12 = 36 m, the rectangle is 12 × 10 = 120 m, and the last triangle is ½ × 3 × 12 = 18 m. Total distance = 174 m over 19 seconds — average speed 174 ÷ 19 ≈ 9.16 m/s (3 s.f.).
Notes · 3 of 4
Plotting from tables of values
Substitute, plot, join smoothly — then read the answers off.
To draw y = x² − 4x + 3 for −1 ⩽ x ⩽ 5, build a table: substitute each x to get 8, 3, 0, −1, 0, 3, 8. Plot the seven points with small crosses and join them with one smooth curve through the bottom — parabolas do not have corners. The same recipe works for lines (two points suffice), and for the reciprocal y = a⁄x, where you must plot plenty of points near x = 0 to capture the arms.
The finished graph is a calculator. Roots are the x-values where it crosses the x-axis: y = x² − 4x + 3 crosses at x = 1 and x = 3. To solve a related equation, rearrange so one side is your graph: to solve x² − 4x + 3 = 4, draw the horizontal line y = 4 and read where it cuts the curve — x ≈ −0.24 and 4.24. To solve x² − 4x + 3 = x + 1, draw the line y = x + 1; the x-coordinates of the crossings are the solutions.
Notes · 4 of 4
Sketching the classic curves
Shape, intercepts, turning points, asymptotes — no tables needed. (Extended for non-quadratics)
A sketch shows the important features without plotting dozens of points. For a parabola y = ax² + bx + c: mark where it crosses the x-axis (the roots), the y-intercept, the vertical line of symmetry halfway between the roots, and the turning point. From completed-square form you get the turning point for free: y = (x − 2)² + 1 has its minimum at (2, 1); the negative coefficient in y = −x² + 4 flips the parabola upside-down with maximum (0, 4).
Extended. A cubic y = (x + 2)(x − 1)(x − 4) crosses the x-axis at its three roots −2, 1, 4 and wiggles between them; a negative leading coefficient flips it end-to-end. The reciprocal y = 1⁄x lives in two arms — first and third quadrants when a is positive — never touching either axis: the x-axis and y-axis are its asymptotes. An exponential y = 2ˣ starts flat, always positive, and passes through (0, 1) and (1, 2); y = abˣ passes through (0, a) because b⁰ = 1.
Revise
Key terms
The words examiners expect you to use precisely.
- Gradient as a rate
- The slope of a practical graph measures change per unit.45⁄50 = 0.9 euros per dollar.
- Stationary
- A horizontal section of a distance–time graph.Flat for 1 hour means no movement.
- Acceleration (Extended)
- The gradient of a speed–time graph.0 to 12 m/s in 6 s → 2 m/s².
- Area under the graph (Extended)
- The distance travelled, read as rectangles and triangles.36 + 120 + 18 = 174 m.
- Root
- An x-value where a graph crosses the x-axis.y = x² − 4x + 3 has roots 1 and 3.
- Line of symmetry
- The vertical line through a parabola’s turning point.Roots −1 and 3 give symmetry x = 1.
Show 2 more key termsShow fewer key terms
- Asymptote (Extended)
- A line a curve approaches but never meets.y = 1⁄x hugs both axes.
- Exponential graph (Extended)
- A curve of the form y = abˣ with constant ratio growth.y = 2ˣ passes through (0, 1) and (1, 2).
Revise
Common mistakes
Where students most often lose marks in this topic — and what to do instead.
Mistake: Reading a conversion with one estimated point.
Grid estimates compound; two clear points fix the rate.
Instead: Use the labelled point (50, 45) to get 0.9 per unit, then scale.
Mistake: Ignoring stops when averaging a journey.
Average speed uses the whole journey time, rests included.
Instead: Total distance ÷ total time: 80 km ÷ 5 h = 16 km/h.
Show 10 more mistakesShow fewer mistakes
Mistake: Mixing up the roles on a speed–time graph.
Gradient is acceleration, not speed; area is distance, not speed.
Instead: Steepness answers “how fast is speed changing?”; area answers “how far?”.
Mistake: Forgetting the ½ in triangle areas.
A braking section of 3 s at up to 12 m/s covers ½ × 3 × 12 = 18 m, not 36 m.
Instead: Split each non-rectangular region into triangles explicitly.
Mistake: Joining plotted points with straight segments.
Curves bend smoothly; a zig-zag is not a function graph.
Instead: Plot the table points, then draw one smooth curve through them.
Mistake: Drawing y = k as a vertical line when solving graphically.
The helper for x² + x − 6 = −2 is horizontal: y = −2.
Instead: Match the form: constant on its own means a horizontal line.
Mistake: Placing a parabola’s turning point at a root.
The turning point sits on the line of symmetry, between the roots.
Instead: Roots −1 and 3 give symmetry x = 1 and minimum (1, −4) for (x + 1)(x − 3).
Mistake: Letting a reciprocal sketch touch the axes.
1⁄x is never 0 and never defined at x = 0.
Instead: Draw the arms approaching both axes as asymptotes.
Mistake: Forgetting b⁰ = 1 on exponential curves.
Every y = abˣ passes through (0, a), not (0, 0).
Instead: 2ˣ at x = 0 is 1; at x = 1 it is 2.
Mistake: Treating a trapezium area under a speed–time graph as base × height.
The slanted ends mean the area is not a rectangle.
Instead: Use ½(sum of the parallel sides) × height: ½(6 + 14) × 8 = 80 m.
Mistake: Reading the roots of a cubic from where it crosses the y-axis.
Roots are the solutions of y = 0, which live on the horizontal axis.
Instead: y = x³ − 4x crosses the x-axis at −2, 0 and 2 — that is where the roots are.
Mistake: Treating y = 2x as if it doubled at each step.
2x adds x each step; 2ˣ doubles each step.
Instead: 2ˣ gives 2, 4, 8, 16 at x = 1, 2, 3, 4 — the point (4, 16) is on the exponential, not the line.
Check yourself
Quick quiz
24 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Read a conversion graph using the unit rate
A conversion line joins (0, 0) to (50, 45). How many euros correspond to $20?
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Practise
Topical questions
14 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this topic, in three tiers, each with a model answer, mark scheme and examiner tip.
Practise topical questionsExam practice
Practice paper
A 30-mark paper set out like a Cambridge paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Topic 9 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Topic 9: conversion and distance–time graphs, speed–time graphs with distance as area, plotting curves from tables of values, solving equations graphically, and sketching quadratics, cubics, reciprocals and exponentials.
Mark scheme
PROTopic 9 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Which papers examine this topic?
Paper 1 / 3
Core: 80 marks each, 1 h 30 each
Paper 1 non-calculator, Paper 3 calculator
Paper 2 / 4
Extended: 100 marks each, 2 h each
Paper 2 non-calculator, Paper 4 calculator
Accuracy
3 s.f., angles to 1 d.p.
Unless a question states otherwise
Want a full paper? Try the Paper 1-style Core, Paper 2-style non-calculator or Paper 4-style calculator sets.
Questions
Frequently asked questions
How many points do I need to plot a quadratic?
Use the table the question gives you — usually seven values. Plot each one carefully, then join with a single smooth curve. Two extra checks help: the curve must be symmetric about its middle, and it should pass through every plotted point, not dodge them.
When the question says “solve graphically”, what am I allowed to do?
Draw the helper line (horizontal for an equation like curve = number, or the given straight line for curve = line), mark the intersection x-values, and read them off to the requested accuracy. You are not expected to rearrange and solve algebraically — the marks are for the line and the readings.
Why is average speed lower than the riding speed in journey graphs?
Because the clock keeps running during stops. Average speed is total distance divided by total time; a flat section adds time but no distance, pulling the average down.
What must a sketch include to earn full marks?
Correct shape, labelled intercepts (roots and y-intercept), the line of symmetry or turning point for a parabola, and — at Extended — asymptotes for reciprocals and the (0, a) point for exponentials. Sketches are judged on features, not on plotted accuracy.
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Topic 10: Functions and differentiation