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).
In linear modelling situations:
- The vertical intercept represents an initial value
- The gradient represents a rate of change
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
The graph shows the amount of water remaining in a bucket that was initially full before a hole was made in its side.
What is the gradient of the function?
What is the $y$y-value of the $y$y-intercept?
Write an equation to represent the amount of water remaining in the bucket, $y$y, as a function of time, $x$x.
What does the slope tell you?
The amount of water remaining in the bucket after $2$2 minutes.
AThe amount of water that is flowing out of the hole every minute.
BThe time it takes for the bucket to be completely empty.
CThe time it takes the amount of water remaining in the bucket to drop by one litre.
DWhat does the $y$y-intercept tell you?
The capacity of the bucket.
AThe amount of water remaining in the bucket after $30$30 minutes.
BThe amount of water remaining in the bucket when it is empty.
CThe size of the hole.
DFind the amount of water remaining in the bucket after $54$54 minutes.
question 2
Let the height of a candle be $y$y cm. If the candle is lit, the height decreases according to the equation $y=-2t+8$y=−2t+8, where $t$t is the elapsed time in minutes.
Complete the table of values below:
Time ($t$t min) $0$0 $1$1 $2$2 $3$3 Height of candle ($y$y cm) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Draw the graph of $y=-2t+8$y=−2t+8.
Loading Graph...The height of candle and the time elapsed must be non-negative. Using part (b), for which values of $t$t does the equation $y=-2t+8$y=−2t+8 accurately model the height of the candle?
$0\le t\le2$0≤t≤2
A$0\le t\le4$0≤t≤4
B$0\le t\le8$0≤t≤8
C$0\le t\le10$0≤t≤10
D
question 3
A carpenter charges a callout fee of $\$150$$150 plus $\$45$$45 per hour.
Write an equation to represent the total amount charged, $y$y, by the carpenter as a function of the number of hours worked, $x$x.
What is the gradient of the function?
What does this gradient represent?
The total amount charged increases by $\$45$$45 for each additional hour of work.
AThe minimum amount charged by the carpenter.
BThe total amount charged increases by $\$1$$1 for each additional $45$45 hours of work.
CThe total amount charged for $0$0 hours of work.
DWhat is the value of the $y$y-intercept?
What does this $y$y-intercept represent?
Select all that apply.
The total amount charged increases by $\$150$$150 for each additional hour of work.
AThe maximum amount charged by the carpenter.
BThe callout fee.
CThe minimum amount charged by the carpenter.
DFind the total amount charged by the carpenter for $6$6 hours of work.
question 4
A diver starts at the surface of the water and begins to descend below the surface at a constant rate. The table shows the depth of the diver over $5$5 minutes.
| Number of minutes passed ($x$x) | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 |
|---|---|---|---|---|---|
| Depth of diver in meters ($y$y) | $0$0 | $1.4$1.4 | $2.8$2.8 | $4.2$4.2 | $5.6$5.6 |
What is the increase in depth each minute?
Write an equation for the relationship between the number of minutes passed ($x$x) and the depth ($y$y) of the diver.
Enter each line of work as an equation.
In the equation, $y=1.4x$y=1.4x, what does $1.4$1.4 represent?
The change in depth per minute.
AThe diver’s depth below the surface.
BThe number of minutes passed.
CAt what depth would the diver be after $6$6 minutes?
We want to know how long the diver takes to reach $12.6$12.6 meters beneath the surface.
If we substitute $y=12.6$y=12.6 into the equation in part (b) we get $12.6=1.4x$12.6=1.4x.
Solve this equation for $x$x to find the time it takes.