Linear Graphs of Practical Situations
Model linear graphs of practical situations for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-01). When something changes by the same amount each step — each hour, week or minute — its graph is a straight line, like a plumber's charges, weekly savings or water filling a tank.
You will draw a linear graph from a table of values, find the starting value and the rate of change, and use the rule or the graph to predict what comes next in settings such as phone plans, taxi fares and a candle burning down.
Theory
When a quantity changes by the same amount each step, it is modelled by a straight line. This NSW Year 12 Mathematics Standard 1 guide shows how to find the starting value and the rate, write the rule, and read the graph to predict charges, savings and volumes.
Many real situations change by the same amount each step — each hour, week or kilometre. They give a straight-line graph with a rule like amount \(=\) starting value \(+\) rate \(\times\) time.
The starting value (vertical intercept) is the amount when time \(=0\): a call-out or joining fee, the water already in a tank, money already saved. The rate (gradient) is the change for one step, such as \(\$30\) per hour or 2 cm per minute.
A rising line has a positive rate; a falling line (draining, burning down) has a negative rate. Use the rule or the graph to predict. The amount is 0 (empty) where the line meets the horizontal axis.
A linear model:
The rate from a table with equal steps:
The same works from two points on the line. From \((0,3)\) to \((4,11)\): \(\dfrac{11-3}{4-0}=2\) cm per minute.
How to model a practical situation with a straight line
- Find the starting value — the amount at time 0 (work back one step if the table starts at 1).
- Find the rate — the change for one step (divide if the table goes up in bigger steps).
- Write the rule: amount \(=\) start \(+\) rate \(\times\) time (subtract for a falling line).
- Predict by substituting, or solve the rule to find when an amount is reached; give units.
| \(t\) | 1 | 2 | 3 |
|---|---|---|---|
| \(C\) (\(\$\)) | 35 | 50 | 65 |
The cost rises \(\$15\) each hour, so at \(t=0\) it is \(35-15=\$20\).
| \(C\) | \(=\) | \(20+15t\) |
| \(C\) | \(=\) | \(20+15(6)\) |
| \(\) | \(=\) | \(20+90=110\) |
\(C=20+15t\), and 6 hours costs \(\$110\).
The line starts at 10 and rises 5 cm a minute: \(D=10+5t\).
| \(\text{(a) } D\) | \(=\) | \(10+5(4)=30\) |
| \(\text{(b) } 45\) | \(=\) | \(10+5t\) |
| \(35\) | \(=\) | \(5t\) |
| \(t\) | \(=\) | \(7\) |
(a) 30 cm; (b) after 7 minutes.
Rule: \(V=60-8t\). Empty when \(V=0\).
| \(\text{(a) } V\) | \(=\) | \(60-8(3)=60-24=36\) |
| \(\text{(b) } 0\) | \(=\) | \(60-8t\) |
| \(8t\) | \(=\) | \(60\) |
| \(t\) | \(=\) | \(7.5\) |
(a) 36 L; (b) empty after 7.5 hours, the point \((7.5,0)\).
(a) 120 is the starting value (at \(w=0\)); 35 is the gradient (the weekly rate).
| \(\text{(b) } S\) | \(=\) | \(120+35(10)=470\) |
| \(\text{(c) } 400\) | \(=\) | \(120+35w\) |
| \(280\) | \(=\) | \(35w\) |
| \(w\) | \(=\) | \(8\) |
(a) She starts with \(\$120\) and saves \(\$35\) a week; (b) \(\$470\); (c) after 8 weeks.
Common pitfalls
Frequently asked questions
How do you write a linear rule from a real situation?
Find the starting value (the amount at time 0) and the rate (the change for one step). The rule is amount equals starting value plus rate times time, for example C equals 50 plus 30t.
What does the gradient mean in a real-life graph?
The gradient is the rate of change: how much the amount goes up or down for one step, such as 30 dollars per hour. A falling line has a negative rate.
What does the vertical intercept mean in a practical graph?
It is the starting value, the amount when time is 0. Examples are a call-out fee, the water already in a tank or the money already saved.
How do you know if a table of values is linear?
Check the differences between consecutive values. If they go up or down by the same amount each equal step, the relationship is linear.
How do you find when a tank will be empty?
Set the amount equal to 0 in the rule and solve. For V equals 60 minus 8t, 0 equals 60 minus 8t gives t equals 7.5, so the tank is empty after 7.5 hours.