NCEA Level 1Graphs, tables, equations and patternsAS 91947

Straight-line graphs: 7 exam questions solved step by step

What this lesson covers

  • A pattern, a table, an equation and a graph can all describe the same relationship.
  • The gradient of a straight-line graph is its rate of change: how much y changes for each one unit of x.
Key idea

7 questions, two numbers

  • 1. Gradient m = rise ÷ run.
  • 2. Intercept c = y when x = 0.
  • 3. Write y = mx + c.
  • 4. In context: meaning and units.
Your turn

Question 1 · Achieved

  • Find the rule for the number of tiles, T.
  • How many tiles in pattern 20?
Pattern, nTiles, T
15
29
313
417
Worked example

Q1: the tile rule

  • Difference = gradient.
  • Step back for c.
  1. 9 − 5 = 4, 13 − 9 = 4constant difference
  2. T = 4n + cgradient 4
  3. 5 − 4 = 1step back to n = 0
  4. T = 4n + 1
  5. n = 4: 4 × 4 + 1 = 17
  6. n = 20: 4 × 20 + 1 = 81
  7. 81 tiles in pattern 20
Your turn

Question 2 · Achieved

  • The line passes through (0, 3) and (4, 11).
  • Find its equation.
Question 2 · Achieved
Worked example

Q2: the equation of a line

  • c: where it crosses.
  • m: rise ÷ run.
  1. c = 3crosses the y-axis at 3
  2. rise = 11 − 3 = 8
  3. run = 4 − 0 = 4
  4. m = 8 ÷ 4 = 2rise ÷ run
  5. y = 2x + 3y = mx + c
  6. x = 4: 2 × 4 + 3 = 11
Your turn

Question 3 · Merit

  • Find the equation of the line.
  • Careful: look at how x changes.
xy
17
313
519
725
Worked example

Q3: x goes up in twos

  • The run is 2, not 1.
  • Use a point for c.
  1. m = 6x went up 2, not 1
  2. y up 6 when x up 2
  3. m = 6 ÷ 2 = 3rise ÷ run
  4. y = 3x + c
  5. 7 = 3 × 1 + cuse (1, 7)
  6. c = 4−3 both sides
  7. y = 3x + 4
  8. x = 7: 3 × 7 + 4 = 25
Your turn

Question 4 · Merit

  • Tiles in pattern n: T = 5n + 3
  • Which pattern uses 98 tiles?
  • Could a pattern use exactly 120?
Worked example

Q4: the rule run backwards

  • Set the rule equal.
  • n must be whole.
  1. 5n + 3 = 98T = 98
  2. 5n = 95−3 both sides
  3. n = 19÷ 5
  4. 5 × 19 + 3 = 98
  5. 5n + 3 = 120T = 120
  6. 5n = 117−3 both sides
  7. n = 23.4n must be whole
  8. pattern 19; no pattern uses 120
Your turn

Question 5 · Merit, in context

  • A candle burns down steadily.
  • After 2 h: 18 cm tall. After 5 h: 12 cm.
  • Find the rate of change, with units.
  • Find h after t hours. What is c?
Worked example

Q5: the rate of change

  • Time across, height up.
  • Units: cm per hour.
  1. (2, 18) and (5, 12)
  2. rise = 12 − 18 = −6height goes down
  3. run = 5 − 2 = 3hours
  4. m = −6 ÷ 3 = −2rise ÷ run
  5. m = −2 cm per houry units per x unit
  6. burns down 2 cm every hour
Worked example

Q5: the equation and c

  • Use a point for c.
  • c is the start value.
  1. m = −2 cm per hour
  2. h = −2t + c
  3. 18 = −2 × 2 + cuse (2, 18)
  4. 18 = −4 + c
  5. c = 22+4 both sides
  6. h = 22 − 2t
  7. t = 5: 22 − 2 × 5 = 12
  8. t = 0: h = 22the intercept
  9. 22 cm tall when it was lit
Your turn

Question 6 · Excellence, in context

  • Tables in one long row for a hui.
  • Each seats 3 per long side, 1 per end.
  • Explain why n tables seat S = 6n + 2.
  • How many tables seat 100 people?
Worked example

Q6: explain the rule

  • Explain each part.
  • Inner ends are lost.
  1. 3 + 3 = 6 side seats a table
  2. n tables: 6n side seats
  3. + 2 seats, one at each end
  4. S = 6n + 2
  5. n = 1: 6 + 2 = 8
  6. n = 3: 18 + 2 = 20
Worked example

Q6: tables for 100 people

  • Round up, not off.
  • Check both choices.
  1. S = 6n + 2
  2. 6n + 2 = 100S = 100
  3. 6n = 98−2 both sides
  4. n = 16.33…÷ 6
  5. 16 tables: 96 + 2 = 982 people short
  6. 17 tables: 102 + 2 = 104
  7. 17 tables (4 spare seats)
Your turn

Question 7 · Excellence, in context

  • A train: d km from Britomart, t min on.
  • Find the speed in km/min and km/h.
  • What does the intercept mean?
  • When is it 20 km from Britomart?
Question 7 · Excellence, in context
Worked example

Q7: the speed and the rule

  • Units: km per minute.
  • × 60 for km per hour.
  1. rise = 11 − 5 = 6 km
  2. run = 12 − 4 = 8 min
  3. m = 6 ÷ 8 = 0.75rise ÷ run
  4. 0.75 km per minute
  5. 0.75 × 60 = 45 km/h60 minutes in an hour
  6. d = 0.75t + c
  7. 5 = 0.75 × 4 + cuse (4, 5)
  8. 5 = 3 + c, c = 2
  9. d = 0.75t + 2
Worked example

Q7: the intercept and 20 km

  • c: the value at t = 0.
  • Set d = 20 and solve.
  1. d = 0.75t + 2
  2. t = 0: d = 2the intercept
  3. 2 km out when timing began
  4. 0.75t + 2 = 20d = 20
  5. 0.75t = 18−2 both sides
  6. t = 18 ÷ 0.75 = 24
  7. 0.75 × 24 + 2 = 20
  8. 24 minutes after timing began
Summary

What solved all seven

  • Rise over run, the value at zero, and what it means.
StepHow
Gradientrise ÷ run, any two points
Patterndifference is m; c is term 0
x jumpsdivide the rise by the run
Intercepty when x = 0, or use a point
Unitsy units per one x unit
Reverseset the rule equal, then solve
Contextmeaning, units and rounding

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