Year 8AlgebraEquations and relationshipsYear 8 Maths

Patterns, rules and the four quadrants

What this lesson covers

A coordinate plane extends to 4 quadrants that meet at the origin (0, 0).
Linear patterns have a constant increase or decrease, can be described by the rule t = a × n + d, and can be graphed as a straight line on a coordinate plane.

Key idea

The grid goes all four ways

You have used the top right corner for years.
The axes carry on the other way as well, into negatives.
That makes four regions, meeting at the origin, (0, 0).

Key idea

Four quadrants, numbered anticlockwise

The first is the top right, where both are positive.
Then you go anticlockwise: second, third, fourth.
The signs tell you which one a point lands in.

Key idea

And the order never changes

x first, then y. Along the corridor, then up the stairs.
(−3, 2) means 3 to the left and 2 up.
(2, −3) is a completely different point.

Quadrant	Signs	Example
First	x positive, y positive	(3, 2)
Second	x negative, y positive	(−3, 2)
Third	x negative, y negative	(−3, −2)
Fourth	x positive, y negative	(3, −2)

Key idea

Now a pattern: 5, 8, 11, 14

Number the positions: 1st, 2nd, 3rd, 4th. Call that n.
The value at each position is the term. Call that t.
We want a rule that turns any n into its t.

Position n	Term t
1	5
2	8
3	11
4	14

Key idea

Find the step between terms

5 to 8 is +3. 8 to 11 is +3. 11 to 14 is +3.
The same every time, which is what makes it linear.
That constant step is the a in t = a × n + d.

From		Step
5 to 8		+3
8 to 11		+3
11 to 14		+3
Constant	so the pattern is linear	a = 3

Worked example

So is the rule just t = 3n?

It is the obvious guess, and it is worth testing.
At n = 1 it gives 3.
But the first term is 5. The rule is wrong.

Try	Working	Gives
Try t = 3n	at n = 1 that gives	3
But the first term is		5

Key idea

Wrong by the same amount every time

It is 2 short at n = 1. And at n = 2. And at n = 3.
So add 2 to it. t = 3n + 2.
That 2 is the d, and this is how you find it.

Try	Working	Gives
Try t = 3n	at n = 1 that gives	3
But the first term is		5
Short by		2
So		t = 3n + 2

Key idea

What d actually is

Run the table backwards one step, to n = 0.
3 times 0 plus 2 is 2.
So d is what the term before the first one would have been.

Position	Working	Term
n = 1	3 + 2	5
n = 0	0 + 2	2
So d is	the term before the first one	2

Key idea

Always test against the first term

Finding a is easy. Everyone gets that bit.
Forgetting to check is why t = 3n gets handed in.
One substitution, five seconds, catches it every time.

Rule	At n = 1	Verdict
t = 3n	3	wrong, first term is 5
t = 3n + 2	5	right

Worked example

Now the rule earns its keep

The 10th term: 3 times 10 plus 2, which is 32.
The 50th: 152. The 100th: 302.
You never had to write the pattern out to get there.

Position	Working	Term
n = 10	3(10) + 2	32
n = 50	3(50) + 2	152
n = 100	3(100) + 2	302

Key idea

Plotted, it is a straight line

Position along the bottom, term up the side.
(1, 5), (2, 8), (3, 11), (4, 14).
Perfectly straight, because the step never changes.

Key idea

And d is where the line starts

Follow the line back to position zero.
It crosses at 2, which is the d we found.
The step a is how steeply it climbs.

Worked example

When the pattern goes down

20, 17, 14, 11. The step is −3, so a is negative.
At n = 1, −3 plus d must be 20, so d is 23.
t = −3n + 23.

Step	Working	Gives
20, 17, 14, 11	goes down by 3	a = −3
At n = 1	−3 + d = 20	d = 23
So		t = −3n + 23

Worked example

Which eventually goes below zero

At n = 8: −24 plus 23 is −1.
The pattern has run past zero into negative terms.
And the graph drops into the fourth quadrant.

Key idea

Three patterns worth knowing

Triangular: 1, 3, 6, 10, 15. Dots stacked in a triangle.
Square: 1, 4, 9, 16, 25. And cube: 1, 8, 27, 64, 125.
All three turn up constantly. Now look at their steps.

Family	Starts	Steps
Triangular	1, 3, 6, 10, 15	differences 2, 3, 4, 5
Square	1, 4, 9, 16, 25	differences 3, 5, 7, 9
Cube	1, 8, 27, 64, 125	differences grow fast

Key idea

None of them is linear

Triangular goes up by 2, then 3, then 4, then 5.
That is not constant, so t = an + d cannot describe it.
Plotted, they curve. The rules for those come later.

Key idea

So check the differences first

Before reaching for t = an + d, take the differences.
Constant? Then it is linear and the rule applies.
Not constant? Then it does not, and forcing it gives nonsense.

Test	Meaning
Linear needs	the same difference every time
Triangular gives	2, then 3, then 4	not constant
So t = an + d	does not apply

Your turn

Your turn: four questions

Pause here and try all four.
For question 1, test your rule at n = 1 before you write it down.
Question 4 wants a reason, not just yes or no.

	Question
1	Find the rule for 7, 10, 13, 16.
2	Which quadrant holds the point (−5, 3)?
3	For t = 4n − 1, what is the 10th term?
4	Is 1, 3, 6, 10 a linear pattern? How do you know?

Show answer

1. t = 3n + 4. The difference is 3, and 3n would give 3 at n = 1 when the first term is 7, so it is 4 short. 2. The second. x is negative and y is positive. 3. 39. Substitute n = 10 to get 40 − 1. 4. No. The differences are 2, 3 and 4, which are not constant. These are the triangular numbers, and their rule is not of the form an + d.

Worked example

The answers

t = 3n + 4. The step is 3, and 3n is 4 short at n = 1.
(−5, 3) is in the second quadrant. And 4(10) − 1 is 39.
1, 3, 6, 10 is not linear. Its steps are 2, 3, 4.

	Answer
1	t = 3n + 4
2	second
3	39
4	no, steps are 2, 3, 4

Summary

What to take away

a is the constant step. d is the term before the first one.
Test every rule at n = 1. That is where t = 3n dies.
And if the differences are not constant, it is not linear.

	Remember
Four quadrants	numbered anticlockwise from the top right
a	is the constant difference
d	is what the term before the first one would be
Always test	your rule against n = 1
No constant difference	means it is not linear at all

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