Linear modelling
Identify gradient and y-intercept from y = mx + c, construct linear models from data points, and write equations to model real-world situations including interpreting the gradient as a rate of change.
Worked examples
Identifying gradient and y-intercept in context
Straightforward
Problem
A parking station charges according to , where is the total charge in dollars and is the number of hours parked. Identify the gradient and y-intercept and interpret each in the context of the parking station.
1
Write the equation in form and identify and .
is already in the form with and .
2
Interpret the gradient in context.
The gradient is . Since is in dollars and is in hours, the gradient means the parking cost increases by for each additional hour — that is, the hourly rate is per hour.
3
Interpret the y-intercept in context.
The y-intercept is . When , . This means there is an entry fee charged regardless of how long the car is parked.
Answer
Gradient : the hourly parking rate is per hour. Y-intercept : there is a fixed entry fee of .
Constructing a linear model from two data points
Moderate
Problem
A catering company charges for 20 guests and for 40 guests. Assuming the cost is linear, construct a model in the form , where is the total cost in dollars and is the number of guests.
1
Identify the two data points.
and
2
Calculate the gradient using the gradient formula.
3
Substitute one data point and into to find .
Using :
4
Write the complete linear model.
Answer
. The cost per guest is and the fixed charge is .
Writing and interpreting a linear model with a stated limitation
Challenging
Problem
A gym charges a joining fee of and a monthly membership fee of . Write a linear equation for the total amount paid after months of membership. Interpret the gradient and state one limitation of the model.
1
Identify the fixed cost and the rate of change.
Fixed joining fee (y-intercept) . Monthly fee (gradient) per month.
2
Write the linear equation in the form .
3
Interpret the gradient in context.
The gradient is . This means the total amount paid increases by for each additional month of membership — that is, the rate of change is per month.
4
State one limitation of the model.
The model assumes the monthly fee remains constant at indefinitely. In practice, the gym may change its pricing, offer promotional rates, or the member may cancel — so the model may not accurately represent costs beyond a certain period.
Answer
. The gradient of means the total cost rises by each month. One limitation: the model assumes the monthly fee never changes, which may not hold over many years.
Practise
Q1·Straightforward
State the gradient of the linear equation .
Explanation
The equation is in the form where and . The gradient is .
Q2·Straightforward
A mobile phone plan has a monthly cost modelled by , where is the monthly cost in dollars and is the number of messages sent. State the monthly access fee (the y-intercept).
Explanation
The equation is . When , . The monthly access fee is , which is the y-intercept.
Q3·Straightforward
A water tank is being filled. Its volume is modelled by , where is the volume in litres and is the time in minutes. What does the gradient of represent?
Explanation
In , the gradient is . Since is in litres and is in minutes, the gradient means the volume increases by litres each minute — that is, the tank fills at litres per minute.
Q4·Straightforward
A car rental company charges according to , where is the total cost in dollars and is the distance driven in km. What is the fixed daily charge (in dollars) if no distance is driven?
Explanation
When no distance is driven, the total cost is . This is the fixed daily charge (the y-intercept).
Q5·Moderate
A linear relationship passes through the points and . Calculate the gradient.
Explanation
Q6·Moderate
A delivery service charges for a 2 km trip and for a 5 km trip. Assuming a linear relationship between cost and distance, calculate the cost per kilometre (the gradient).
Explanation
Q7·Moderate
A linear model passes through the points and . Find the y-intercept .
Explanation
First, find the gradient:
Substitute the point into :
Q8·Moderate
A plumber charges a fixed call-out fee plus an hourly rate. For 2 hours the charge is ; for 5 hours the charge is . Find the hourly rate (gradient) in dollars.
Explanation
Q9·Challenging
A mobile phone plan charges a fixed monthly fee of plus per message sent. In a particular month, a user sends 150 messages. Calculate the total monthly cost in dollars.
Explanation
The linear model is where is the number of messages.
Q10·Challenging
An electricity provider charges a daily supply fee of plus per kilowatt-hour (kWh) used. A household uses 18 kWh in one day. Calculate the electricity cost for that day in dollars.
Explanation
The linear model is where is the energy used in kWh.
Q11·Challenging
The total cost of printing business cards is modelled by , where is the total cost in dollars and is the number of cards printed. Which statement correctly identifies a limitation of this model?
Explanation
The gradient applies for every card printed — the model assumes a constant cost per card. In reality, printers often offer lower rates for larger orders (bulk discounts), so the model overestimates the cost for large quantities. This is the key limitation.
Q12·Challenging
The value of a car is modelled by , where is the value in dollars and is the age of the car in years. After how many years will the car's value reach ?
Explanation
Substitute :