Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- (Lesson 5) A line through the origin passes through (4, 36). What is its rate?
Answer
9. The rise is 36 and the run is 4: 36 ÷ 4 = 9. - (Year 8, Unit 3, lesson 3) Work out −3 − 5.
Answer
−8. Start at −3 and go 5 more to the left. - (Year 7, Unit 3, lesson 6) Solve 4n + 7 = 31.
Answer
n = 6. Take away 7 to get 4n = 24, then divide by 4.
Words used in this lesson
- start value
- the value when the time is zero. For example: 30 litres at the start.
- rate of change
- how much the value changes in one step. For example: −4 litres per hour.
- negative rate
- a rate for a value that is falling. For example: −2 litres per hour.
- slope
- how steeply a line goes up or down. For example: the line slopes down.
- reaches zero
- becomes 0. For example: the tank is empty.
- relative to
- measured from. For example: metres relative to the surface.
A line that does not start at zero
A water tank on a farm near Nakuru holds 60 litres at 8 am. A leak drains it at 5 litres each hour. Predict first: sketch the graph of litres against hours. Does it start at the origin? Which way does it slope? When does it reach the horizontal axis?
The figure lets the rate go below zero. Set a start value and a rate, and read the value after 5 hours.
Start 20, rate −3 per hour: after 5 hours the value is 20 + 5 × (−3) = 5. The line falls 3 each hour and reaches zero after about 6.67 hours.
Show me the steps
- Start 20, rate −3 per hour: 20 + 5 × (−3) = 5 after 5 hours. The line falls, and reaches zero after 20 ÷ 3 = 6.67 hours to two places.
- Start 20, rate 3 per hour: 20 + 5 × 3 = 35 after 5 hours. The line rises 3 each hour and never reaches zero.
- Start 40, rate 0: the value stays at 40. A rate of 0 is a flat line.
- Start 10, rate −2 per hour: 10 + 5 × (−2) = 0 after 5 hours. The line reaches the horizontal axis exactly then.
A steady rate and a start value make a straight line: value = start + rate × time. The start value is where the line crosses the vertical axis, at time 0.
The rate is rise ÷ run between any two points, and it is negative when the value falls: the line slopes down, and each step across makes the value go down by the size of the rate.
A line that does not start at zero is not proportional. y ÷ x changes along it, but rise ÷ run is the same for every pair of points.
Back to the tank: it starts at 60 and the rate is −5 per hour, so after 8 hours it holds 60 + 8 × (−5) = 20 litres, and it is empty when 60 ÷ 5 = 12 hours have passed. The line slopes down from (0, 60) to (12, 0).
Rise over run, with a start value
Here the line is y = 10 + 5x. Choose two points anywhere on it and watch rise ÷ run beside y ÷ x.
From (1, 15) to (4, 30) the line rises 15 over a run of 3: 15 ÷ 3 = 5. The rate is 5 at any pair of points, but y ÷ x at the second point is 30 ÷ 4 = 7.5.
Show me the steps
- (1, 15) to (4, 30): rise 15, run 3, 15 ÷ 3 = 5. At (4, 30), y ÷ x is 7.5.
- (0, 10) to (4, 30): rise 20, run 4, 20 ÷ 4 = 5. The same rate, starting from the start point. At (4, 30), y ÷ x is 7.5 again.
- (1, 15) to (2, 20): rise 5, run 1, 5 ÷ 1 = 5. At (2, 20), y ÷ x is 10, twice the rate.
- (4, 30) to (8, 50): rise 20, run 4, 20 ÷ 4 = 5. At (8, 50), y ÷ x is 6.25. The rate never changed; y ÷ x did.
A common mistake: “The rate is y ÷ x.” That works only for a line through the origin. On y = 10 + 5x, at x = 2 the value is 20 and 20 ÷ 2 = 10, but the line rises 5 for each step across. Use rise ÷ run between two points, or the change in one step.
Worked example: rate and start from two points
A straight line passes through (2, 11) and (6, 27). Find its rate and its start value, and write the rule.
- Rate = rise ÷ run = (27 − 11) ÷ (6 − 2) = 16 ÷ 4 = 4.
- Go back from x = 2 to x = 0, which is 2 steps: 11 − 2 × 4 = 3. The start value is 3.
- Rate 4, start 3, so value = 3 + 4x. Check at x = 6: 3 + 4 × 6 = 27.
Using the tank. How many litres are in the tank after 8 hours, and when is it empty?
- After 8 hours: 60 + 8 × (−5) = 60 − 40 = 20 litres.
- Empty when 60 − 5h = 0, so h = 60 ÷ 5 = 12.
- 20 litres after 8 hours, and empty after 12 hours.
Your go. The first step is given. A candle is 30 cm tall and burns down 2 cm each hour. How tall is it after 7 hours, and after how many hours does it go out?
- The rate is −2 cm per hour and the start value is 30 cm.
- Now find the height after 7 hours, then when it reaches 0, and check.
Check the answer
After 7 hours: 30 + 7 × (−2) = 30 − 14 = 16 cm. It goes out when 30 − 2h = 0, so h = 30 ÷ 2 = 15 hours. Check: 30 − 2 × 15 = 0.
Check before you go on
1. A cyclist is 50 km from a town and rides towards it at 20 km per hour. How far from the town is she after 2 hours?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. Zeynep has 150 rupees and saves 40 rupees each week. How many rupees does she have after 6 weeks?
3. A straight line passes through (0, 8) and (4, 32). What is its rate?
Level 2 · Building
4. A plant is 15 cm tall and grows 2 cm each week. After how many weeks is it 41 cm tall?
5. A temperature is measured at 2, 5 and 8 hours as 19, 10 and 1 degrees Celsius. What is the rate of change in degrees Celsius per hour?
6. A tank holds 72 litres and drains at 6 litres per minute. After how many minutes is it empty?
Level 3 · Stretch
7. A diver is at −4 m, which is 4 m below the surface, and rises at 2 m per minute. What is her height after 3 minutes, in metres relative to the surface?
8. A quantity falls by 4 each hour, and its value at 2 hours is 17. What was its value at 0 hours?
Something to think about
A candle 20 cm tall burns down 2 cm each hour, and a second candle 14 cm tall burns down 1 cm each hour. When are they the same height? Decide how you would find out, then say how tall they are then.
One way to think about it
Write each height as start + rate × hours: 20 − 2h and 14 − h. They are equal when 20 − 2h = 14 − h, which gives h = 6. At 6 hours the first is 20 − 12 = 8 cm and the second is 14 − 6 = 8 cm. On a graph, the two lines cross at (6, 8).
Exit ticket
9. A straight line passes through (2, 30) and (5, 18). What is its rate?