Modelling with linear functions
Learning Objectives
- Translate a situation with a constant rate of change into a linear model \(f(x) = ax + b\).
- Identify the rate as the slope \(a\) and the initial value as the intercept \(b\).
- Answer 'when does it reach …?' with an equation and 'when is it above/below …?' with an inequality.
- Find the break-even point by comparing two linear models.
- Recognise the domain on which a linear model is meaningful.
Putting the machinery to work
Over this unit and the last we built a complete toolkit for the linear function: its graph and slope, its sign, the equation \(ax + b = 0\), and the inequality \(ax + b > 0\). We now point all of it at real situations.
The signal that a situation is linear is the one we first met with arithmetic progressions: a quantity that starts at some value and then changes by the same amount for each unit of something else. The starting value and the constant rate are all you need.
The modelling recipe
To build a linear model:
Name the variable \(x\) and say what it counts (items, minutes, kilometres…).
Find the rate \(a\): how much the quantity changes per unit of \(x\). This is the slope.
Find the initial value \(b\): the quantity when \(x = 0\). This is the intercept.
Write \(f(x) = ax + b\).
The question then chooses the tool. "When does the quantity equal a given value?" is an equation. "When is it above or below a value?" is an inequality. "When do two options cost the same?" compares two models — still a single first-degree equation once you set them equal.
A cost model
A workshop charges a flat €40 to set up a print run, then €6 per poster. Let \(x\) be the number of posters. The cost is
\[C(x) = 6x + 40,\]
a linear model with rate \(a = 6\) (euros per poster) and initial value \(b = 40\) (the setup, the cost when \(x = 0\)).
A value. The cost of \(10\) posters is \(C(10) = 6(10) + 40 = €100\).
A target (equation). How many posters for a budget of €100? Solve \(6x + 40 = 100\): \(6x = 60\), so \(x = 10\).
A threshold (inequality). To keep the cost under €130, solve \(6x + 40 < 130\): \(6x < 90\), so \(x < 15\) — at most \(14\) posters.
Break-even: comparing two options
A climbing gym offers two ways to pay. A membership costs €60 to join plus €4 per visit; a drop-in costs €10 per visit with no joining fee. With \(x\) the number of visits,
\[A(x) = 4x + 60 \quad\text{(membership)}, \qquad B(x) = 10x \quad\text{(drop-in)}.\]
When do they cost the same? Set \(A(x) = B(x)\): \(4x + 60 = 10x\), so \(60 = 6x\) and \(x = 10\). At \(10\) visits both cost €100 — the break-even point.
Which is cheaper? The membership beats the drop-in when \(A(x) < B(x)\): \(4x + 60 < 10x\), i.e. \(60 < 6x\), so \(x > 10\). Visit more than ten times and the membership wins; visit fewer and the drop-in does. Setting two linear models equal turns a comparison into a single first-degree equation, and asking which is smaller turns it into a single inequality.
A draining tank (a falling quantity)
A tank holds \(480\) litres and drains at a steady \(8\) litres per minute. After \(t\) minutes the volume is
\[V(t) = 480 - 8t,\]
with initial value \(b = 480\) and a negative rate \(a = -8\) (the volume falls).
A value. After \(15\) minutes, \(V(15) = 480 - 120 = 360\) litres.
Empty (equation). \(V(t) = 0\) gives \(480 - 8t = 0\), so \(t = 60\) minutes.
Below a level (inequality). The volume drops below \(160\) litres when \(480 - 8t < 160\), i.e. \(-8t < -320\); dividing by \(-8\) reverses it, \(t > 40\). After the \(40\)-minute mark the tank is below \(160\) litres.
A model is a tool, not a fact
The tank model says \(V(t) = 480 - 8t\). Put \(t = 100\) and it returns \(V = -320\) litres — a negative volume, which is nonsense. The arithmetic is fine; the modelling has run past its limit. The tank empties at \(t = 60\), and from then on there is simply no water and no draining. The model is faithful only on the interval \([0, 60]\).
Every linear model carries such a domain. The poster cost \(C(x) = 6x + 40\) assumes the price per poster never changes — no bulk discount, no limit on paper — and that \(x\) is a whole number that is not negative. A model earns its keep by being simple, and the price of that simplicity is that it describes reality only within a range. Part of using one well is knowing where that range ends.
Exercises
Exercise 1
A printing service charges €15 to set up plus €2 per poster, so the cost of \(x\) posters is \(C(x) = 2x + 15\) (euros).
Exercise 2
A gym membership costs \(A(x) = 4x + 60\) euros for \(x\) visits (a €60 fee plus €4 each), while drop-in costs \(B(x) = 10x\) euros (€10 each).
Exercise 3
A tank of \(480\) litres drains at \(8\) litres per minute, so after \(t\) minutes the volume is \(V(t) = 480 - 8t\).
Exercise 4
Decide whether each statement about linear models is true or false.
Exercise 5
A linear model fits when a quantity changes by the same amount per step. Decide whether each situation is linear.
Summary
A situation with a constant rate of change and an initial value is modelled by \(f(x) = ax + b\): the rate is the slope \(a\), the initial value is the intercept \(b\).
The question picks the tool: a target value → an equation; an above/below threshold → an inequality; a comparison of two options → set the two models equal (one equation) or compare them (one inequality).
The break-even point is where two models meet; on either side of it, one option is cheaper.
A model holds only on a meaningful domain. When it predicts a negative volume or assumes a rate that would really change, it has reached the edge of where it describes reality.