Construct a linear model y = mx + b for a practical situation made of a fixed amount plus a constant rate, interpret the gradient as the rate and the y-intercept as the fixed starting amount in context, and use the model to predict an output and to find the input that gives a required output

What this dot point is asking

A linear model turns a real situation into the straight-line equation $y = mx + b$ so you can calculate with it. NESA wants this for the most common shape of real problem: something that starts at a fixed amount and then changes at a steady rate. A plumber charges a call-out fee and then so much per hour; a taxi has a flagfall and then so much per kilometre; a phone plan has a monthly fee and then so much per gigabyte. In each one the fixed amount is the $y$-intercept $b$ and the steady rate is the gradient $m$, so the situation is exactly $y = mx + b$ wearing different letters. Three skills are bundled here, and a question can test any of them. First, build the model: read the fixed amount and the rate out of the words and write the equation with letters that fit the context. Second, interpret the gradient and the $y$-intercept in context, in words and with units, the gradient as a rate and the intercept as a starting value. Third, use the model: substitute an input to predict an output, and run it the other way, substituting a known output and solving to find the input that produced it.

The deeper idea that makes all of this click is that a linear model is a recipe with a starting amount and a constant step. The intercept $b$ tells you where the quantity begins (the cost before any work, the water already in the tank), and the gradient $m$ tells you how much it changes for each one-unit increase in the input (per hour, per kilometre, per gigabyte). So interpreting $m$ as a rate and $b$ as a fixed value is not a separate fact to memorise. It is simply what the two letters mean. The marks come from a few habits: choose the right number for $m$ and for $b$, write the model with sensible variables, interpret each with the correct units, substitute carefully (multiply before adding), and, when reading backwards, solve the equation rather than guess. Marks are lost in predictable ways: swapping the rate and the fixed amount, dropping the units in an interpretation, multiplying the fixed fee by the input, and forgetting that the input often cannot be negative.

The answer

What makes a situation linear

A situation can be modelled by a straight line whenever it is built from two parts: a fixed amount that does not change, and a quantity that grows (or shrinks) by the same amount for every one-unit step. The fixed amount is there even when the input is zero; the steady rate is what the per-unit words ("per hour", "per km", "each", "for every") are telling you. If both parts are present, the model is

where $b$ is the fixed starting amount and $m$ is the constant rate. The test for "is it linear?" is simply whether the rate is constant: if every extra hour adds the same dollars, or every extra kilometre adds the same fare, the graph is a straight line and a linear model fits. If the rate itself changes (a price that doubles, interest that compounds), the situation is not linear and this model does not apply.

The reason the fixed amount lands on $b$ and the rate lands on $m$ is worth seeing once. Put the input at zero: every "per unit" charge disappears because there are zero units, and all that is left is the fixed amount, which is exactly the value of $y$ when $x = 0$, the $y$-intercept. Now increase the input by one unit: the only thing that changes is one more lot of the rate is added, so $y$ goes up by the rate for each one-unit step, which is exactly the gradient. So you never have to wonder which number is which: the standing charge is always $b$, and the per-unit charge is always $m$.

Building the model from the words

To construct the model, pull two numbers out of the description and choose letters that suit the situation.

1. Find the fixed amount. Look for the charge that applies once, regardless of the input: a call-out fee, a flagfall, a booking fee, a monthly fee, the water already in the tank. This is $b$.
2. Find the rate. Look for the "per unit" charge: dollars per hour, dollars per km, dollars per GB, litres per minute. This is $m$.
3. Name the variables. Choose a letter for the output (often $C$ for cost, $V$ for volume) and a letter for the input (often $h$ for hours, $d$ for distance, $t$ for time), and state what each one means with its unit.
4. Write $y = mx + b$ with those letters, usually as (output) $=$ (rate) $\times$ (input) $+$ (fixed amount).

For a plumber charging a $90 call-out fee plus $70 per hour, the fixed amount is $b = 90$ and the rate is $m = 70$. Letting $C$ be the cost in dollars and $h$ the hours, the model is

Writing it as "rate times input plus fixed amount" keeps it in the same shape as $y = mx + b$, so the gradient and intercept are easy to read straight back off it.

Interpreting the gradient and the y-intercept in context

"Interpret" or "explain what it represents" asks for the meaning of a number in plain words, not just the number itself. Two rules cover every case.

• The $y$-intercept $b$ is the value of the output when the input is $0$. In a cost model that is the fixed amount you pay before anything happens: the call-out fee, the flagfall, the monthly fee. State it as an amount with its unit, for example "the $90 fixed call-out fee".
• The gradient $m$ is the rate: how much the output changes for each one-unit increase in the input. State it as a rate with its units, for example "$70 per hour". A positive gradient means the output rises as the input grows; a negative gradient (a tank draining, a value depreciating) means it falls.

For the model $C = 70h + 90$, the intercept $b = 90$ is the fixed $90 call-out fee charged for $h = 0$ hours of work, and the gradient $m = 70$ is the labour rate of $70 per hour. The figure below shows both: the line meets the cost axis at the $90 fixed fee, and the rise-over-run step shows the $70 per hour rate as a rise of $140 over a run of $2$ hours. The dots are sample jobs, each sitting exactly on the line.

Using the model: predicting forwards

Once the model is built, predicting an output is just substitution. Replace the input variable with the given value and evaluate, taking care to multiply by the rate before adding the fixed amount (order of operations). For the plumber model $C = 70h + 90$, the cost of a $3$-hour job is

so a $3$-hour job costs $300. The single most common slip here is adding before multiplying, which would wrongly give $70 \times (3 + 90)$. The fixed fee is added once at the end, not folded into the hours. A linear model is reliable for predictions inside the sensible range of the input; pushing it to extreme values (a $1000$-hour job) can give an answer the situation would never actually produce.

Using the model: reading backwards

The reverse question gives you the output and asks for the input: "the bill was $440, how long was the job?" Now you cannot just substitute, because the unknown is inside the equation. Instead, substitute the known output and solve for the input using the balancing steps from solving linear equations. For $C = 70h + 90$ with a $440 bill,

Subtract the fixed fee from both sides to remove it, then divide by the rate:

So the job took $5$ hours. The structure is always the same: take off the fixed amount first (undo the $+\,b$), then divide by the rate (undo the $\times\, m$). Doing it in that order matches peeling the operations off in reverse, and it is far safer than trying values until one fits.

How exam questions ask about linear models

The wording tells you which of the three skills a part is testing:

• "Write an equation / model for this situation" or "construct a linear model" means find the fixed amount ($b$) and the per-unit rate ($m$) in the words, then write $y = mx + b$ with context letters.
• "What does the gradient represent?" wants the gradient stated as a rate with units (dollars per hour, litres per minute).
• "What does the $y$-intercept represent?" (or "how much at the start", "the cost before any...") wants the intercept as the fixed starting amount with its unit.
• "Find the cost when...", "how much after $n$...", "use the model to predict..." means substitute the input and evaluate.
• "How many hours / how far / how long..." given an output, or "the bill was $...", means substitute the output and solve for the input.
• "How much is added each..." or "the rate of..." is just asking for the gradient.
• "Which option is cheaper" means build a model for each, substitute the same input into both, and compare.

Building the model stage by stage

It helps to see a model grow on the axes the same way you would build it on paper. The stages below build the plumber model $C = 70h + 90$ on cost-versus-hours axes: set up the axes for the context, mark the fixed fee at the intercept, step out the rate, then rule the line and use it to predict.

Stage 1, set up the axes for the context

Put the input (hours worked) along the horizontal axis and the output (cost) up the vertical axis, and label each with its quantity. Nothing is plotted yet, this is the frame the model will sit in.

Stage 2, mark the fixed fee at the intercept

The fixed call-out fee is the cost when $h = 0$, so it is the $y$-intercept. Plot the single point $(0, 90)$ on the cost axis. This is where the model begins, before any hours are worked.

Stage 3, step out the rate

The gradient is the $70 per hour rate, written as a rise of $70 for a run of $1$ hour. From $(0, 90)$ move a run of $1$ hour across and a rise of $70 up to reach $(1, 160)$. That right-angled step is the rate made visible.

Stage 4, rule the line and predict

Join the points and rule the straight line; the sample jobs at $1$, $2$ and $3$ hours all lie on it. To predict a $3$-hour job, go up from $h = 3$ to the line and across to the cost axis, reading off $300, matching $C = 70 \times 3 + 90 = 300$.

What the model can and cannot tell you

A linear model is a simplification, so it is worth knowing where it holds and where it breaks. It assumes the rate is perfectly constant, which is exactly why the graph is a straight line. Real charges sometimes are not: a plumber may round part-hours up to the next half hour, a phone plan may cap the extra-data charge, a taxi may add a night surcharge. Inside the range the model is built for, those details rarely matter and the prediction is reliable. Outside it, the model can give answers the situation would never produce, so you should sanity-check an extreme result. It is also why the input usually has a sensible domain. Hours worked, kilometres driven, gigabytes used and litres pumped cannot be negative, so values of the input below $0$ have no meaning even though the equation would happily calculate them. When a question asks you to comment on the model, these are the points that earn the marks: the rate is assumed constant, and the input is restricted to values that make sense.