Mathematics • Year 8 • Unit 2 • Lesson 6
Linear vs Non-Linear — Mixed Challenge
Pull together everything from Lesson 6: first differences, constant-gradient property, equation form y = mx + c, and the non-linear counterexamples. Six mixed problems, one "find the mistake", and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 6. Decide which test applies before you start writing. Show your working. 3 marks each
1.1 Test for linearity using first differences. x: 0, 1, 2, 3, 4 y: 7, 10, 13, 16, 19. State the gradient if linear.
1.2 A line passes through (1, 4), (3, 10) and (5, 16). Use the constant-gradient property — show that the gradient from (1,4)→(3,10) equals the gradient from (3,10)→(5,16). What is m?
1.3 Decide whether each equation is linear or non-linear. Tick one column for each.
(a) y = 5x − 1 (b) y = x² + 2 (c) y = ½x (d) y = 1/x (e) y = 7
1.4 A pattern of dots: 1 dot, 4 dots, 9 dots, 16 dots, 25 dots, … Is this a linear pattern? Justify using first differences.
1.5 An online video subscription is $12 fixed setup plus $9 per month. Write the equation linking total cost C and months n, build a 3-row table (n = 1, 2, 3), and confirm linearity with first differences.
1.6 Without drawing the graph, decide whether y = −4x + 10 is linear and predict whether its gradient is positive or negative. Justify both answers from the equation alone.
2. Find the mistake
Another student has tried to decide whether the following table is linear. Exactly one line of their reasoning is wrong. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Table. x: 0, 1, 2, 3, 4 y: 5, 8, 11, 13, 16.
Student's working:
Line 1: First differences: 8 − 5 = 3, 11 − 8 = 3, 13 − 11 = 2, 16 − 13 = 3.
Line 2: The first two differences are both 3.
Line 3: Two matching differences in a row is enough — the relationship is linear.
Line 4: So the gradient is m = 3.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the correct conclusion — is the table linear or not, and why?
Stuck? Revisit lesson § "Common Pitfalls" — "checking only 2 first differences" is exactly this trap.3. Open-ended challenge — design a non-linear pattern
This question has more than one valid answer. 4 marks
3.1 Design a non-linear table of five y-values for x = 0, 1, 2, 3, 4 where:
(i) Every y-value is positive.
(ii) Every y-value is bigger than the previous one (y is increasing).
(iii) The first differences are NOT all equal — so the table fails the linearity test.
For your table:
(a) Write your five y-values.
(b) List the four first differences and confirm they are not all equal.
(c) Describe in one sentence the kind of pattern your differences follow (e.g. they grow, they shrink, they alternate, …).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y: 7, 10, 13, 16, 19
Differences: 3, 3, 3, 3 — all equal. Linear, m = 3.
1.2 — Constant-gradient check
(1,4)→(3,10): m = (10 − 4)/(3 − 1) = 6/2 = 3. (3,10)→(5,16): m = (16 − 10)/(5 − 3) = 6/2 = 3. Same gradient on both pairs, so the line is straight. m = 3.
1.3 — Linear or not?
(a) y = 5x − 1 — linear (form y = mx + c).
(b) y = x² + 2 — non-linear (x squared).
(c) y = ½x — linear (m = ½, c = 0).
(d) y = 1/x — non-linear (x in the denominator).
(e) y = 7 — linear (m = 0, horizontal line).
1.4 — Dot pattern 1, 4, 9, 16, 25
Differences: 3, 5, 7, 9 — not all equal. Not linear (this is y = x², a square-number pattern).
1.5 — Subscription
C = 9n + 12. Table: n = 1, C = 21; n = 2, C = 30; n = 3, C = 39. First differences: 9, 9 — both equal. Linear, m = 9 ($/month).
1.6 — y = −4x + 10
Linear — it is in the form y = mx + c with m = −4 and c = 10, and x is only to the power of 1. Since m = −4 < 0, the gradient is negative (the line falls left-to-right).
2 — Find the mistake
(a) The mistake is on Line 3 (also makes Line 4 wrong).
(b) Two matching differences is NOT enough — you must check that ALL differences are equal. Here the four differences are 3, 3, 2, 3, so the relationship is not linear.
(c) Correct conclusion: the table is not linear because the 2 breaks the constant pattern. No single gradient value works for all the data.
3 — Design a non-linear pattern (sample solution)
Many valid answers. One example: y = x² + 1, giving the table
x: 0, 1, 2, 3, 4 y: 1, 2, 5, 10, 17.
(a) y-values: 1, 2, 5, 10, 17 — all positive, all increasing.
(b) First differences: 2 − 1 = 1, 5 − 2 = 3, 10 − 5 = 5, 17 − 10 = 7. Not all equal — so fails the linearity test.
(c) The differences themselves grow steadily (1, 3, 5, 7 — odd numbers), so this is a parabola-style growth.
Marking: 1 mark for a valid increasing positive table; 1 mark for correctly listing the first differences; 1 mark for showing they are not all equal; 1 mark for a sensible one-sentence description of the difference pattern.