Mathematics • Year 10 • Unit 2 • Lesson 20
Linear Modelling — Skill Drill
Build fluency with the two ways to build a linear model from Lesson 20: from a word description (identify rate m and initial value c → write y = mx + c), and from two data points (compute m, then find c). Then use the model to interpolate (safe) and extrapolate (risky).
1. I do — fully worked example
Build a linear cost model from a word description.
Problem. A taxi charges a $5 flag fall plus $2.50 per kilometre. Build a linear model for total cost C in terms of distance d (km), then predict the cost for 12 km.
Step 1 — Identify the variables.
d = distance in km (independent); C = total cost in $ (dependent).
Step 2 — Identify the gradient (rate of change).
m = $2.50/km (cost per km)
Step 3 — Identify the y-intercept (initial value at d = 0).
c = 5 (the flag fall charged before the meter starts moving)
Step 4 — Write the model and interpret.
C = 2.50d + 5
Reason: cost rises by $2.50 for every km, plus a fixed $5 flag fall.
Step 5 — Predict the cost for d = 12 km.
C = 2.50(12) + 5 = 30 + 5 = $35
Answer: Model C = 2.50d + 5; cost for 12 km = $35.
2. We do — fill in the missing steps
Build a temperature model from two data points. Fill in each blank. 5 marks
Problem. The temperature of a cooling liquid was recorded: after 2 minutes T = 72 °C; after 6 minutes T = 48 °C. Build a linear model T = mt + c.
Step 1 — Compute the gradient using m = (T₂ − T₁)/(t₂ − t₁):
m = ( ____ − ____ ) / ( ____ − ____ ) = ____ / ____ = ____ °C/min
Note: negative because temperature falls over time.
Step 2 — Substitute one of the points to find c.
Using (2, 72): 72 = ____ (2) + c → c = ____
Step 3 — Write the model:
T = ____ t + ____
Step 4 — Predict T after 10 minutes (extrapolation):
T = ____ (10) + ____ = ____ °C
Step 5 — Verify (4, ____ ) using the model. Is your model consistent with the original points?
3. You do — independent practice
For each: state your variables and units, write the model, and use it to answer the asked question.
Foundation — read off rate and initial value
3.1 A plumber charges $80 call-out plus $60 per hour. Write a model C = ____ for the cost C in terms of hours h. 1 mark
3.2 A plant is 5 cm tall at planting and grows 2 cm per week. Write a model h = ____ for height after w weeks. 1 mark
3.3 A gym charges $50 joining + $25 per week. Write a model and find the cost after 10 weeks. 1 mark
3.4 A car travels at 80 km/h with a 20 km head start. Write a model for distance d in terms of time t (hours). 1 mark
Standard — from two data points
3.5 A car's value depreciates linearly from $24,000 to $18,000 over 3 years. (a) Find the gradient. (b) Write V = mt + c. (c) Predict the value after 5 years. 2 marks
3.6 Mai's bank balance was $300 after 2 weeks of saving and $540 after 5 weeks. Build a linear model B = mt + c (assuming a constant rate), then predict the balance after 10 weeks. 2 marks
Extension — interpret + interpolate vs extrapolate
3.7 Use your model from 3.5 to predict the car's value after 12 years. Is this prediction reliable? Explain in one sentence using the words "interpolation" or "extrapolation". 3 marks
3.8 For the model C = 0.10n + 20 (a $20 base + $0.10 per text message): (a) what is the cost for 150 texts? (b) If a bill is $35, how many texts were sent? (c) Interpret the gradient and y-intercept in real-world words. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (cooling liquid)
Step 1: m = ( 48 − 72 ) / ( 6 − 2 ) = −24 / 4 = −6 °C/min. Step 2: 72 = −6(2) + c → c = 84. Step 3: T = −6t + 84. Step 4: T = −6(10) + 84 = 24 °C. Step 5: at t = 4, T should be −6(4) + 84 = 60 °C. ✓ (matches the linear pattern between the two given points).
3.1 — Plumber
C = 60h + 80.
3.2 — Plant
h = 2w + 5.
3.3 — Gym
C = 25w + 50. At w = 10: C = 250 + 50 = $300.
3.4 — Car
d = 80t + 20.
3.5 — Car depreciation
(a) m = (18000 − 24000)/(3 − 0) = −$2000/year. (b) V = −2000t + 24000. (c) At t = 5: V = −10000 + 24000 = $14,000.
3.6 — Bank balance
m = (540 − 300)/(5 − 2) = 240/3 = $80/week. Using (2, 300): 300 = 80(2) + c → c = 140. B = 80t + 140. At t = 10: B = 800 + 140 = $940.
3.7 — Extrapolate to 12 years
V = −2000(12) + 24000 = $0. This is an extrapolation well beyond the data range (0–3 years) — the linear model probably breaks down (cars don't depreciate to exactly $0; they reach a "scrap value" and stop). Unreliable prediction.
3.8 — Phone bill
(a) C = 0.10(150) + 20 = 15 + 20 = $35. (b) 35 = 0.10n + 20 → 0.10n = 15 → n = 150 texts. (c) Gradient = $0.10 per text (cost per message sent); y-intercept = $20 (the fixed monthly base fee paid even with 0 texts).