Year 7 Mathematics · AC9M7A04

Relationships Between Variables in Authentic Graphs

We are learning to describe relationships between variables represented in graphs of functions from authentic data

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Learning goalsSay it simply

We are learning to describe relationships between variables represented in graphs of functions from authentic data.

A graph of an authentic function shows how a dependent variable changes with an independent variable; axes, scale and units determine the meaning of every point.

Interpret segments through rate of change: rising, falling and horizontal sections represent positive, negative and zero rates, and steepness compares the size of those rates.

Graph evidence can support a relationship or prediction but does not by itself establish causation. Verify claims by reading coordinates and recomputing rise divided by run.

Success criteria

  • I can read coordinates, axes, scales and units on an authentic graph.
  • I can interpret segments and compare rates of change.
  • I can distinguish a graphed relationship from a causal claim.
Key conceptTeach from the board

Use these models to connect the mathematical idea to values, diagrams, coordinates, graphs or structure before moving to symbolic calculation.

Follow each example from representation and setup through calculation/reasoning, interpretation and an independent check.

Example 1

Interpret the displayed distance–time graph

  1. Read time in hours on the horizontal axis and distance in kilometres on the vertical axis.
  2. Read the plotted points (0,0), (1,4), (2,8), (3,8) and (4,14).
  3. Locate the horizontal segment from hour 2 to hour 3 at 8 km.
  4. Interpret the unchanged distance over that interval.

Final answer: The traveller was stationary for 1 hour, from hour 2 to hour 3.

Check: Both endpoints of the flat segment have distance 8 km.

Example 2

Compare rates on the displayed graph

  1. Use the same time and distance axes, units and scales for both intervals.
  2. From hour 0 to hour 2, distance increases from 0 km to 8 km.
  3. From hour 3 to hour 4, distance increases from 8 km to 14 km.
  4. Calculate 8÷2=4 km/h and 6÷1=6 km/h, then compare.

Final answer: The final rising segment is faster: 6 km/h compared with 4 km/h.

Check: The steeper segment has the greater rise per hour on the same axes and scales.

Example 3

Application problem 1

Problem: A distance–time graph rises from (0,0) to (2,6), stays flat to (4,6), then rises to (6,14). Describe what each segment means and identify which moving segment is steeper.

  1. Plan: Represent the information first, then calculate, interpret and independently check the result.
  2. Work: From 0–2 h the object moves 6 km; from 2–4 h it is stationary; from 4–6 h it moves 8 km. The last moving segment is steeper because its rate is 4 km/h versus 3 km/h.
  3. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Final answer: From 0–2 h the object moves 6 km; from 2–4 h it is stationary; from 4–6 h it moves 8 km. The last moving segment is steeper because its rate is 4 km/h versus 3 km/h.

Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Application problem 2

Problem: A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.

  1. Plan: Calculate each consecutive rise per one-hour run, then consider the observed domain and changing rate.
  2. Work: Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.
  3. Interpret: The complete model for “A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.” shows Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

Final answer: Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

Check: The complete model for “A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.” shows Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

Clean visual examplesOne-page board

Clean one-page examples

AC9M7A04 - Relationships Between Variables in Authentic Graphs
Example 1

Example 1 Interpret the displayed distance–time graph Read time in hours on the horizontal axis and distance in kilometres on the vertical axis. Read the plotted points (0,0), (1,4), (2,8), (3,8) and (4,14). Locate the horizontal segment from hour 2 to hour 3 at 8 km. Interpret the unchanged distance over that interval. Final answer: The traveller was stationary for 1 hour, from hour 2 to hour 3. Check: Both endpoints of the flat segment have distance 8 km.

Example 2

Example 2 Compare rates on the displayed graph Use the same time and distance axes, units and scales for both intervals. From hour 0 to hour 2, distance increases from 0 km to 8 km. From hour 3 to hour 4, distance increases from 8 km to 14 km. Calculate 8÷2=4 km/h and 6÷1=6 km/h, then compare. Final answer: The final rising segment is faster: 6 km/h compared with 4 km/h. Check: The steeper segment has the greater rise per hour on the same axes and scales.

Example 3

Example 3 Application problem 1 Problem: A distance–time graph rises from (0,0) to (2,6), stays flat to (4,6), then rises to (6,14). Describe what each segment means and identify which moving segment is steeper. Plan: Represent the information first, then calculate, interpret and independently check the result. Work: From 0–2 h the object moves 6 km; from 2–4 h it is stationary; from 4–6 h it moves 8 km. The last moving segment is steeper because its rate is 4 km/h versus 3 km/h. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills. Final answer: From 0–2 h the object moves 6 km; from 2–4 h it is stationary; from 4–6 h it moves 8 km. The last moving segment is steeper because its rate is 4 km/h versus 3 km/h. Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Example 4 Application problem 2 Problem: A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5. Plan: Calculate each consecutive rise per one-hour run, then consider the observed domain and changing rate. Work: Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5. Interpret: The complete model for “A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.” shows Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5. Final answer: Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5. Check: The complete model for “A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.” shows Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

Curriculum examplesCopied content

Content description: describe relationships between variables represented in graphs of functions from authentic data.

Questions and answersWith answers

Attempt each problem before opening Check answer. The set moves from core understanding to application and synthesis.

  1. 1. Which axis usually displays time in a time-dependent graph?

    Check answer

    Answer: The horizontal axis.

    Hint: Locate the independent variable.

    Why: Using the method “Locate the independent variable.” gives The horizontal axis.

  2. 2. What does a horizontal segment on a distance–time graph mean?

    Check answer

    Answer: Distance did not change, so the object was stationary.

    Hint: Compare the segment endpoints.

    Why: Using the method “Compare the segment endpoints.” gives Distance did not change, so the object was stationary.

  3. 3. A graph rises 6 metres in 3 seconds. Find its rate.

    Check answer

    Answer: 2 m/s.

    Hint: Divide vertical change by horizontal change.

    Why: Using the method “Divide vertical change by horizontal change.” gives 2 m/s.

  4. 4. Temperature falls from 18°C to 12°C in 2 hours. State the average hourly change.

    Check answer

    Answer: −3°C per hour.

    Hint: Calculate (12−18)÷2.

    Why: Using the method “Calculate (12−18)÷2.” gives −3°C per hour.

  5. 5. On two distance–time graphs with the same axis quantities, units and scales, what does a steeper rising segment show?

    Check answer

    Answer: A greater speed.

    Hint: Compare rise per unit run only after confirming the axes and scales match.

    Why: Using the method “Compare rise per unit run only after confirming the axes and scales match.” gives A greater speed.

  6. 6. A line passes through (0,5) and (4,13). Describe its start and rate.

    Check answer

    Answer: It starts at 5 and increases by 2 per x-unit.

    Hint: Find the intercept and (13−5)÷4.

    Why: Using the method “Find the intercept and (13−5)÷4.” gives It starts at 5 and increases by 2 per x-unit.

  7. 7. Can a graph of temperature and ice-cream sales prove temperature causes sales?

    Check answer

    Answer: No. The graph shows an association, but it does not control other influences or establish which variable caused the change, so a causal conclusion is unsupported.

    Hint: Consider other variables and study design.

    Why: The reasoning for “Can a graph of temperature and ice-cream sales prove temperature causes sales?” is complete because No. The graph shows an association, but it does not control other influences or establish which variable caused the change, so a causal conclusion is unsupported.

  8. 8. A tank rises 20 L in 5 min, stays flat 2 min, then falls 9 L in 3 min. Give all rates.

    Check answer

    Answer: The rates are 20÷5=+4 L/min, 0÷2=0 L/min and −9÷3=−3 L/min; the signs show filling, no change and draining.

    Hint: Calculate change divided by time for each segment.

    Why: The complete model for “A tank rises 20 L in 5 min, stays flat 2 min, then falls 9 L in 3 min. Give all rates.” shows The rates are 20÷5=+4 L/min, 0÷2=0 L/min and −9÷3=−3 L/min; the signs show filling, no change and draining.

  9. 9. A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.

    Check answer

    Answer: Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

    Hint: Calculate each consecutive rise per one-hour run, then consider the observed domain and changing rate.

    Why: The complete model for “A distance–time graph plots (0,0), (1,4), (2,8), (3,8) and (4,14), with time in hours and distance in kilometres. Describe every interval rate and explain whether the graph alone justifies a distance prediction at hour 5.” shows Across the displayed one-hour intervals the rates are 4, 4, 0 and 6 km/h; because the rate changes and the observed domain ends at hour 4, the graph alone does not justify one particular distance at hour 5.

  10. 10. A distance–time graph rises from (0,0) to (2,6), stays flat to (4,6), then rises to (6,14). Describe what each segment means and identify which moving segment is steeper.

    Check answer

    Answer: From 0–2 h the object moves 6 km; from 2–4 h it is stationary; from 4–6 h it moves 8 km. The last moving segment is steeper because its rate is 4 km/h versus 3 km/h.

    Hint: Represent the information first, then calculate, interpret and independently check the result.

    Why: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Practice and reviewReady for practice

Common mistake: Axes read without units.

Correction: State the quantities and scales.

Common mistake: Steepness compared across different axis scales.

Correction: Check scale before judging rate.

Common mistake: Graph described point by point only.

Correction: Summarise intervals and trends.

Curriculum alignmentStart here

We are learning to describe relationships between variables represented in graphs of functions from authentic data.

A graph of an authentic function shows how a dependent variable changes with an independent variable; axes, scale and units determine the meaning of every point.

Interpret segments through rate of change: rising, falling and horizontal sections represent positive, negative and zero rates, and steepness compares the size of those rates.

Graph evidence can support a relationship or prediction but does not by itself establish causation. Verify claims by reading coordinates and recomputing rise divided by run.

Success criteria

  • I can read coordinates, axes, scales and units on an authentic graph.
  • I can interpret segments and compare rates of change.
  • I can distinguish a graphed relationship from a causal claim.
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