4.04 Modelling linear functions

We can now use our knowledge of linear relationships to model and solve problems in real-world situations.

Before we begin, there are a few differences to be aware of between the linear relationships we have seen so far on the coordinate plane, and the ones we use to model real-world situations.

Independent and dependent variables

The first thing to notice is that the independent and dependent variables, $x$x and $y$y, will often represent physical quantities such as time, distance, cost, mass or temperature. Instead of being labelled $x$x and $y$y, they may be labelled with letters or names that better represent those quantities.

For example, if we were modelling the rate of fuel being used in a car, the independent variable may represent the distance travelled, in kilometres, and the dependent variable may represent the volume of fuel, in litres, in the car's fuel tank. Instead of $x$x and $y$y, we might use $d$d and $V$V as our variables. Each of these variables has units of measurement associated with them.

Gradient and vertical intercept

As we know already, the two key features of a linear function, or straight-line graph, are the gradient and the $y$y-intercept. In linear modelling situations, the $y$y-intercept is often referred to as the vertical intercept, because the vertical axis may be labelled with a variable other than $y$y.

The vertical intercept represents an initial value. In our example above, the vertical intercept is $63$63 litres. It represents the volume of fuel in a full tank, before the car began its journey.

In a real-world context the gradient represents a rate of change. Using our example above, the gradient would represent the volume of fuel used per distance travelled. In other words, the gradient is a measure of the car's fuel consumption.

The graph above is decreasing as the fuel is being used, so it has a negative gradient. If we divide the 'rise' of $-63$−63 by the 'run' of $900$900 we get a gradient of $-0.07$−0.07. This means fuel is being consumed at a rate of $0.07$0.07 litres per kilometre (or $7$7 litres per $100$100 kilometres).

Changes to the coordinate plane

Because most physical quantities like distance, volume or time do not contain negative values, the graphs of most linear models tend to exist only in the first quadrant of the coordinate plane (like the example above). This is not always the case though. Temperature is a physical quantity that can have negative values.

Practice questions

Question 1

Question 2

question 3

Contextual linear relationships in a table of values

Real world situations often involve the collection of data, which is commonly displayed in a table of values. The independent variable (often but not always called $x$x) is conventionally displayed in the first row of the table, followed by the dependent variable (often but not always called $y$y), in the second row.

We may first want to decide whether the values in the table represent a linear relationship. To do this, we could plot the values on a coordinate plane. If we can draw a straight line through all of the points, then we have a linear relationship. We could also check directly from the table by checking if as the $x$x-values increase by a constant amount, the $y$y-values also change by a constant amount.

Once we know that the relationship between the variables is linear, we can work out the gradient and vertical intercept, and express the relationship as a linear equation (or function).

Worked example

Consider the points in the table. The time ($x$x) is measured in minutes and temperature in degrees Celsius($^\circ$°C).

(a) Determine the gradient for the linear function represented by the table of values.

Think: The $x$x-values increase by the same amount. We can determine how the $y$y-values change for a given change in $x$x in order to calculate the gradient.

Do:

We see a constant change in $y$y (the rise) of $6$6 for an equivalent change in $x$x (the run) of $2$2. Therefore, we can calculate the gradient as follows:

(b) Interpret the value of the gradient found in part (a).

Think: The gradient gives the rate of change of the dependent variable($y$y) with respect to the independent variable($x$x).

Do: The rate of change of the linear function is $3$3 $^\circ$°C/min. That is, as the time increases by $1$1 minute the temperature increases by $3$3 $^\circ$°C. 

(c) Determine the equation of the line, passing through the points given in the table, in gradient-intercept form.

Think: From part (a) we have that the relationship between $x$x and $y$y is of the form: $y=3x+c$y=3x+c. For the first value in the table, when $x=2$x=2, $y=10$y=10, what would we have to add to $3\times2$3×2 to obtain $10$10?

Do: $6+4=10$6+4=10, so $c=4$c=4. Hence, the linear rule that fits the table of values is $y=3x+4$y=3x+4.

Reflect: Check the rule matches the other sets of points in the table. Can you also see the $y$y-intercept would be $4$4 by continuing the pattern in the table to the left?

Practice question

Question 4

Travel graphs

Travel graphs are used to show an object's change in distance over time. The horizontal axis shows time and the vertical axis shows distance.

We can determine how fast something is going based on how steep the line is. If a certain part of the line is very steep then the object is moving faster. If the line is horizontal (flat), then time is passing but the distance isn't changing, which means that the object isn't moving. If the line is sloping down this means the object is travelling back to the origin.

Break-even point

It's really important for businesses to make a profit! Otherwise, they won't be around for very long. So businesses should know their break-even point, the amount of money they need to take in to cover all their expenses. At this point there is no profit or loss, and their income is equal to their expenses.

To find this point, we use break-even analysis. Break-even analysis looks at the graphs of cost and revenue together to determine where they cross. This point of intersection is the break-even point where income equals expenses.

Here is an example of a break-even analysis for a single day of operation for the company Lovely Lemonade.

The horizontal axis is the number of drinks, or units, they sell. The expenses line (red) starts off higher than the income line (green). So for a low number of units sold, Lovely Lemonade will lose money. For example, if they only sell $1000$1000 units, the green line tells us they earn $\$200$$200, the red line tells us they spend $\$600$$600, and so overall they make $\$200-\$600=-\$400$$200−$600=−$400. The negative sign means overall they lose $\$400$$400.

But eventually, the income line becomes higher than the expenses line, so for a high number of units sold, Lovely Lemonade will make money. For example, if they sell $5000$5000 units, the graph tells us that they will earn $\$1100$$1100 but only spend $\$1000$$1000. Overall they make $\$1100-\$1000=\$100$$1100−$1000=$100, and the positive sign means a profit of $\$100$$100.

The important point is when the two lines meet - this is the break-even point. In this example we can see that selling $4000$4000 units means they don't make any profit, but they don't lose any money either - both their income and their expenses are $\$900$$900. Reaching this amount should be an important first goal for Lovely Lemonade every single day!

We can also use break-even analysis to compare two different pricing plans for the same product or service to see which one offers the best deal for a particular situation.

Here's an example comparison between two energy plans provided by rival companies Thorgate and Callisto:

In this scenario the break-even point at $\left(40,10\right)$(40,10) tells us that both plans charge $\$10$$10 for $40$40 kWh. For energy amounts less than the break-even point we can see that Thorgate is cheaper, and for energy amounts more than the break-even point Callisto is cheaper. Knowing how much energy someone plans on using can then determine the best plan for them.

Practice questions

Question 7

Question 8