4.04 Linear functions in context

Now that we have the skills and techniques to construct, draw and analyse linear functions, we will put them into context.

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).

Restrictions on the model

The linear graphs in the previous chapter extended indefinitely in both directions. However, since most physical quantities like distance, volume or time do not contain negative values, the graphs of many linear models 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. When analysing a linear model consider carefully what values make sense in the given context.

Practice questions

question 1

question 2

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 questions

question 3

question 4