Linear models in context

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A linear model describes a situation where the output increases or decreases at a constant rate as the input changes. It has the form y = mx + c. Use it when: the rate of change is constant (e.g. fixed cost per unit, steady temperature drop).

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C = 2.5d + 4. Gradient m = 2.50 (cost per km). y-intercept c = 4 (fixed booking fee — the cost when d = 0).

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Model: C = 0.15t + 10. When t = 40: C = 0.15(40) + 10 = 6 + 10 = $16.

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They have swapped m and c. m (gradient) = the rate — the amount added per unit (per km, per hour, etc.). c (y-intercept) = the fixed starting value — the value when the variable equals 0.

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The gradient is the rate of change — how much y changes for each 1-unit increase in x. Examples: • m = 3 km/h → speed of 3 km per hour. • m = −50 → value decreases by 50 per unit. Always state the units when interpreting.

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The y-intercept is the initial value — the value of y when x = 0. Examples: • c = 200 → 200 items in stock at the start. • c = 15 → the temperature was 15°C at time 0. It is the starting point before any change occurs.

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Gradient m = 8: the cost increases by $8 per hour. y-intercept c = 25: the initial cost (before any time passes) is $25 — a fixed/setup fee.

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IB requires contextual interpretation — the gradient must be described in terms of the variables in the problem. For example: "The cost increases by $50 per kilogram." Just stating the number "50" earns no credit for an interpretation question.

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1. The rate of change (→ this becomes m). 2. An initial value or a specific data point (→ this lets you find c). If two data points are given, find m first using the gradient formula, then find c.

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V = −60t + 800. m = −60 (rate of decrease — negative because draining). c = 800 (starting volume at t = 0).

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m = (300 − 180)/(7 − 3) = 120/4 = 30. C = 30d + c. Use (3, 180): 180 = 30(3) + c → c = 90. Model: C = 30d + 90 (daily rate $30, fixed fee $90).

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A decrease means a negative gradient: m = −3. When a quantity is falling or decreasing, the gradient must be negative. Always check the direction of change before assigning the sign to m.

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Substitute the given input value for x into the model equation and calculate y. Example: If C = 12t + 30 and t = 4, then C = 12(4) + 30 = 78.

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Interpolation: predicting within the range of the original data — generally reliable. Extrapolation: predicting outside the range of the original data — less reliable; the model may not hold. IB questions often award 1 mark for commenting on reliability.

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Set P = 0: 0 = −3t + 120 → t = 40 years. This is extrapolation (t = 40 is beyond the data range of 0–20) — the prediction is less reliable.

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No — IB always requires a reason for reliability judgements. A correct answer gives: (a) whether it is interpolation or extrapolation, and (b) a reason (e.g. "within the data range" or "outside the data range — the trend may not continue").