Constructing and interpreting straight-line models | Year 11

01 / LEARN

Understand the idea.

Turn observations into a rule

A linear model represents a constant rate of change. Name the variables and units, find the rate from two observations, then substitute either observation into $y = m x + b$ to determine the starting constant.

Check the fitted rule against every supplied observation. Two points always determine a line when their inputs differ; they do not establish that an actual process will keep changing at that rate.

y = m x + b

A rule needs conditions

A model includes the inputs for which it makes sense. For a filling container, time begins at the stated start and the linear rule ends when capacity is reached. Continuing the formula beyond that limit predicts impossible volumes.

The intercept represents the output at input zero, which need not be zero. Whole item counts require non-negative integer inputs; measurements such as elapsed time can often vary continuously within their physical limits.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Filling a container

A container holds 12 L at $t = 0$ and 36 L after 3 min. It fills steadily to its 60 L capacity. Model its volume.

Worked solution

1
Calculate the constant filling rate.
$m = \frac{36 - 12}{3 - 0} = 8 L/min$
Steady filling adds the same volume in each minute.
2
Include the initial volume.
$V = 12 + 8 t$
The container already holds twelve litres when the clock starts.
3
Find when capacity is reached.
$\begin{aligned} 60 & = 12 + 8 t \\ t & = \frac{48}{8} = 6 \min \end{aligned}$
The filling rule stops at the stated sixty-litre capacity.
4
State the model and its domain.
$V = 12 + 8 t, 0 \leq t \leq 6$
Here volume is in litres and time in minutes; the rule gives thirty-six litres at three minutes.

Answer

$V = 12 + 8 t$ , for $0 \leq t \leq 6$ min.

EXAMPLE 02

A descending line

Find the line through $(2, 9)$ and $(6, 1)$ .

Worked solution

1
Calculate the gradient.
$m = \frac{1 - 9}{6 - 2} = \frac{- 8}{4} = - 2$
The output decreases eight over an input increase of four.
2
Substitute the first point.
$\begin{aligned} 9 & = - 2 (2) + b \\ b & = 13 \end{aligned}$
The constant must make the line pass through the given point.
3
State the rule and check the second point.
$\begin{aligned} y & = - 2 x + 13 \\ - 2 (6) + 13 & = 1 \end{aligned}$
Both observations satisfy the resulting equation.

Answer

$y = - 2 x + 13$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A plant starts at 15 cm and grows 2 cm per week in a model. Write its rule.

Show answer & reasoning

$H = 15 + 2 t$ cm, where $t$ is weeks since the first measurement.

The height is initially $15 cm$ and increases by $2 cm$ per week, so $H = 15 + 2 t$ , where $t$ is weeks since the first measurement.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the line through $(0, 6)$ and $(4, 18)$ .
2 marks
Find the line through $(3, 8)$ and $(7, 20)$ .
2 marks
A machine has a setup cost of 250 dollars and costs 7 dollars per item. Model total cost.
2 marks
A tank contains 80 L and loses 5 L/min. Write the rule and physical time domain.
2 marks
Explain why $d = 12 t$ passes through the origin when $d$ is distance travelled and $t$ is elapsed time.
2 marks
Are $(0, 2), (1, 5), (2, 10)$ exactly linear? Explain.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the line through $(0, 6)$ and $(4, 18)$ .
2 marks
Gradient $= \frac{(18 - 6)}{4} = 3$ .
The first point gives intercept 6.
$y = 3 x + 6$ .
Find the line through $(3, 8)$ and $(7, 20)$ .
2 marks
Gradient $= \frac{12}{4} = 3$ .
$8 = 3 (3) + b$ gives $b = - 1$ .
$y = 3 x - 1$ .
A machine has a setup cost of 250 dollars and costs 7 dollars per item. Model total cost.
2 marks
Separate fixed cost 250 from variable cost $7 n$ .
$C = 250 + 7 n$ , for non-negative integer $n$ .
A tank contains 80 L and loses 5 L/min. Write the rule and physical time domain.
2 marks
Set $V = 0$ to find emptying time 16 min.
Exclude negative volume.
$V = 80 - 5 t$ , $0 \leq t \leq 16$ .
Explain why $d = 12 t$ passes through the origin when $d$ is distance travelled and $t$ is elapsed time.
2 marks
Substitute $t = 0$ : $d = 0$ .
There is no starting-distance term.
At elapsed time zero, distance travelled is zero.
Are $(0, 2), (1, 5), (2, 10)$ exactly linear? Explain.
2 marks
Equal $x$ -steps give increases 3 and 5.
A straight line needs a constant increase.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

In this lesson

Understand the idea.

Turn observations into a rule

A rule needs conditions

See the method in action.

Filling a container

Calculate the constant filling rate.

Include the initial volume.

Find when capacity is reached.

State the model and its domain.

A descending line

Calculate the gradient.

Substitute the first point.

State the rule and check the second point.

Try it yourself.

Check your understanding

Keep practising