Mathematics • Year 8 • Unit 2 • Lesson 7
Gradient Sign — Mixed Challenge
Pull together everything from Lesson 7: reading m from equations, predicting from tables, distinguishing zero from undefined, and connecting sign to direction. 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 7. Decide which test applies before you start writing. Show your working. 3 marks each
1.1 For each equation, state the gradient sign: (a) y = 6x − 2, (b) y = −x + 4, (c) y = 9, (d) x = −1.
1.2 A table shows x: 0, 1, 2, 3 and y: 12, 9, 6, 3. State the gradient sign and calculate m.
1.3 Sketch four lines on one set of axes, all through (0, 0), with gradients m = 2, m = −1, m = 0, and undefined. Label each.
1.4 Match each scenario to its gradient type. (i) Constant-speed car going forward; (ii) Parked car; (iii) Returning home; (iv) Elevator drawn on a height-vs-horizontal-position graph (going straight up).
1.5 Without graphing, decide whether y = −¼ x + 8 has a positive or negative gradient. Then find y at x = 0 and at x = 4 and use those to confirm.
1.6 A line through (1, 6) and (4, 0) — without using the gradient formula, state whether the gradient is positive or negative and why. Then estimate roughly how steep the line is in one sentence.
2. Find the mistake
Another student has tried to classify the gradient of the line x = 5. Exactly one line of their reasoning is wrong. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — classify the gradient of x = 5:
Line 1: x = 5 means x stays at 5 always.
Line 2: If x stays the same, then nothing changes.
Line 3: So the line is flat (horizontal).
Line 4: So the gradient is m = 0 (zero).
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the corrected reasoning and the correct gradient classification.
Stuck? Revisit lesson § "Zero and Undefined Gradient" — x = constant means VERTICAL, not horizontal.3. Open-ended challenge — design one of each
This question has many valid answers. 4 marks
3.1 Invent four lines that all pass through the point (2, 3) — one of each gradient type: positive, negative, zero and undefined.
For each line, give:
(i) Its equation.
(ii) Its gradient type (and value of m if defined).
(iii) One sentence describing how the line behaves on a graph.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sign of m
(a) y = 6x − 2 → m = 6, positive. (b) y = −x + 4 → m = −1, negative. (c) y = 9 → m = 0, zero. (d) x = −1 → vertical, undefined.
1.2 — Table x:0,1,2,3 / y:12,9,6,3
y drops by 3 each step → negative gradient, m = −3.
1.3 — Four lines through (0,0)
m = 2 → y = 2x, steep uphill, passes (1,2). m = −1 → y = −x, downhill at 45°, passes (1, −1). m = 0 → y = 0 (the x-axis itself). Undefined → x = 0 (the y-axis itself).
1.4 — Match scenarios
(i) Constant-speed forward: positive (distance increases with time). (ii) Parked car: zero (distance constant). (iii) Returning home: negative (distance from home decreases). (iv) Elevator going straight up on height-vs-horizontal-position: undefined (vertical line).
1.5 — y = −¼ x + 8
m = −¼ < 0 → negative gradient. Check: at x = 0, y = 8; at x = 4, y = −¼(4) + 8 = 7. y went DOWN from 8 to 7, confirming negative.
1.6 — Line through (1, 6) and (4, 0)
As x increases from 1 to 4, y decreases from 6 to 0. y falling as x rises → negative gradient. Roughly: y drops 6 for a run of 3, so it falls about 2 units for every 1 unit across — moderately steep downhill.
2 — Find the mistake
(a) The mistake first appears on Line 3 (and is carried into Line 4).
(b) x = 5 is a vertical line, not horizontal. The student confused "x doesn't change" with "y doesn't change". A horizontal line is y = constant (e.g. y = 5), not x = constant.
(c) Corrected reasoning: x = 5 is a vertical line; every point on it has x = 5 and y can be anything. The run between any two points is 0, so m = rise/0 is undefined (division by zero). The correct classification is undefined gradient.
3 — Four lines through (2, 3) (sample solution)
Many valid combos. One set:
Positive (m = 2): y = 2x − 1. Check: at x = 2, y = 3 ✓. Slopes uphill, passing (2,3) and (3,5).
Negative (m = −1): y = −x + 5. Check: at x = 2, y = 3 ✓. Slopes downhill at 45°, passing (2,3) and (3,2).
Zero (m = 0): y = 3. Horizontal line through (2,3); y stays at 3 for every x.
Undefined: x = 2. Vertical line through (2,3); x stays at 2 for every y.
Marking: 1 mark each for valid lines of each of the four gradient types passing through (2,3), with equation + classification + a short description.