Testing parameter changes digitally | Year 9

01 / LEARN

The idea, step by step

Change one thing at a time

The parameters $m$ and $c$ in $y = m x + c$ do different jobs. To investigate $c$ , hold $m$ fixed and compare the resulting lines. To investigate $m$ , hold $c$ fixed. Changing both at once can hide the reason the graph changed.

Use a graphing program if available, or complete a table for $x = - 2, - 1, 0, 1, 2$ and plot the points by hand. Either way, record the equations and compare matching inputs. A useful investigation predicts a result, tests it, and then explains why the result occurs.

Only the intercept changes
x	y = 2x - 1	y = 2x + 3
-2	-5	-1
-1	-3	1
0	-1	3
1	1	5
2	3	7

Explain the pattern algebraically

Increasing $c$ by 4 increases every $y$ -value by 4. The gradient stays the same, so the lines remain parallel. Changing $m$ while holding $c$ fixed makes the lines meet at $(0, c)$ because the $m x$ term is zero there.

Be careful with claims about a larger gradient being 'higher'. For positive $x$ , increasing $m$ increases $m x$ ; for negative $x$ , it decreases $m x$ . One counterexample is enough to disprove a claim that says 'always'. A picture gives evidence, while substitution or algebra explains the relationship.

A common trap

Changing $m$ and $c$ together, then attributing every change to the gradient.

Instead: Keep one parameter fixed. Use matching inputs and record which parameter you changed.

02 / WORKED EXAMPLES

See the thinking

Example 1

Predict an intercept change

Compare $y = 2 x - 1$ with $y = 2 x + 3$ . What changes?

The gradient is 2 in both rules.
At every input, $(2 x + 3) - (2 x - 1) = 4$ .
Therefore the second graph is 4 units higher at matching $x$ -values and is parallel to the first.

A vertical shift up 4 units; the gradient is unchanged.

Example 2

Test a tempting claim

A student says $y = 3 x + 1$ is always above $y = x + 1$ . Test the claim.

At $x = 2$ , the values are 7 and 3, which supports the claim for that input.
At $x = - 2$ , the values are $- 5$ and $- 1$ .
The first graph is lower at $x = - 2$ , so the word 'always' makes the claim false.
Both graphs meet at $(0, 1)$ .

The claim is false;

x = - 2

gives a counterexample.

Check your understanding

If $y = - 2 x + 5$ changes to $y = - 2 x + 2$ , what happens to the graph?

Show answer

It shifts down 3 units and remains parallel.

The gradient stays $- 2$ , and every output decreases by 3.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Complete the outputs for $y = x + 2$ and $y = x - 1$ at $x = - 1, 0, 1$ .
3 marks
Predict the change from $y = 4 x - 2$ to $y = 4 x + 5$ without plotting.
2 marks
Where do $y = - x + 3$ , $y = 2 x + 3$ and $y = 5 x + 3$ all meet? Explain.
2 marks
For $y = m x + 2$ , which value $m = 1$ or $m = 3$ gives the larger output at $x = - 2$ ? Show both values.
3 marks
Design two distinct lines that meet at $(0, - 1)$ , then verify the common point.
3 marks
A graphing screen suggests $y = 0.5 x + 2$ and $y = 0.5 x + 2.1$ are the same line. Explain how to check.
3 marks

Read the worked solutions

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

Complete the outputs for $y = x + 2$ and $y = x - 1$ at $x = - 1, 0, 1$ .
3 marks
Substitute each input into both rules.
The first output is always 3 greater.
First rule: 1, 2, 3. Second rule: $- 2, - 1, 0$ .
Predict the change from $y = 4 x - 2$ to $y = 4 x + 5$ without plotting.
2 marks
The gradients are equal.
The intercept increases by $5 - (- 2) = 7$ .
A shift up 7 units; the lines are parallel.
Where do $y = - x + 3$ , $y = 2 x + 3$ and $y = 5 x + 3$ all meet? Explain.
2 marks
For $x = 0$ , the $m x$ term is zero in each rule.
All three rules then give $y = 3$ .
$(0, 3)$
For $y = m x + 2$ , which value $m = 1$ or $m = 3$ gives the larger output at $x = - 2$ ? Show both values.
3 marks
For $m = 1$ , $y = - 2 + 2 = 0$ .
For $m = 3$ , $y = - 6 + 2 = - 4$ .
$m = 1$ gives 0, which is larger than $- 4$ for $m = 3$ .
Design two distinct lines that meet at $(0, - 1)$ , then verify the common point.
3 marks
Choose different gradients with the same intercept $- 1$ .
Substituting $x = 0$ gives $y = - 1$ in both.
For example, $y = 2 x - 1$ and $y = - x - 1$ .
A graphing screen suggests $y = 0.5 x + 2$ and $y = 0.5 x + 2.1$ are the same line. Explain how to check.
3 marks
The same gradient means they are parallel.
The different intercepts show they are not the same line.
A large viewing window can hide a small separation.
Compare exact equations or substitute $x = 0$ : the outputs are 2 and 2.1, so the lines are distinct and parallel.

Checking saved progress…

Ready for the topic test?

Review the chapter, then choose a practice test.

Topic review

Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Gradient, midpoint and distance; algebraic and graphical straight-line relationships; interpreting models and testing parameters. Parallel and perpendicular lines are included as a natural school-course extension. All explanations and questions are original.

Official curriculum source ↗

Back to Linear equations and graphs