Mathematics • Year 8 • Unit 2 • Lesson 7
Positive and Negative Gradients
Build fluency in classifying gradient as positive, negative, zero or undefined. One worked example, one guided example with blanks, then eight independent problems from quick recognise to predicting from equations and tables.
1. I do — fully worked example
From an equation in the form y = mx + c you can name the gradient sign without drawing a single point.
Problem. Classify the gradient of y = −2x + 4 as positive, negative, zero or undefined.
Step 1 — Spot the form.
y = mx + c, with m = −2 and c = 4.
Reason: the coefficient of x is the gradient — read it straight off.
Step 2 — Read the sign of m.
m = −2 < 0
Reason: negative m means the line falls from left to right.
Step 3 — Sanity-check with two points.
x = 0 → y = 4; x = 1 → y = −2(1) + 4 = 2. y went DOWN from 4 to 2 as x went up.
Reason: y dropping as x grows confirms a negative gradient.
Answer: Negative gradient (m = −2). Line slopes downhill left-to-right.
2. We do — fill in the missing steps
Same shape as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. Classify the gradient of y = 5x − 7.
Step 1 — Spot the form: y = mx + c with m = ______ and c = ______.
Step 2 — Read the sign of m: ______ is ____________ (positive / negative / zero).
Step 3 — Sanity check:
x = 0 → y = ______; x = 1 → y = 5(1) − 7 = ______. y went ______________ (up / down / same).
Conclusion: The gradient is __________________________ (write: positive / negative / zero / undefined).
3. You do — independent practice
Show your working under each problem. First four are foundation, next two are standard, last two are extension.
Foundation — read the sign
3.1 Classify the gradient of y = 4x − 3. 1 mark
3.2 Classify the gradient of y = 7. 1 mark
3.3 Classify the gradient of x = −2. 1 mark
3.4 Classify the gradient of y = −½ x + 10. 1 mark
Standard — predict from a table
3.5 A table has x: 0, 1, 2, 3 and y: 8, 6, 4, 2. Is the gradient positive, negative or zero? Calculate m. 2 marks
3.6 Describe a line with m = −1 that passes through (0, 3). State two more points on the line and which way it slopes. 2 marks
Extension — explain and apply
3.7 A parked car's distance-from-home stays at 8 km. Sketch (or describe) the distance-time graph and state the gradient with reason. 2 marks
3.8 Sort these four lines into the correct gradient column (positive, negative, zero, undefined): (a) y = 3x + 5, (b) x = 6, (c) y = −x + 2, (d) y = −4. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (y = 5x − 7)
Step 1: m = 5, c = −7.
Step 2: 5 is positive.
Step 3: x = 0 → y = −7; x = 1 → y = −2. y went up.
Conclusion: positive gradient.
3.1 — y = 4x − 3
m = 4 > 0 → positive gradient (uphill).
3.2 — y = 7
No x term, so m = 0 → zero gradient (horizontal line).
3.3 — x = −2
x = constant → vertical line → undefined gradient (run = 0, division by zero).
3.4 — y = −½ x + 10
m = −½ < 0 → negative gradient (gently downhill).
3.5 — Table x: 0,1,2,3 / y: 8,6,4,2
y decreases by 2 each step → negative gradient, m = −2.
3.6 — Line with m = −1 through (0, 3)
Equation y = −x + 3. Two more points: (1, 2) and (−1, 4). The line slopes downhill left-to-right at 45°.
3.7 — Parked car
Distance stays at 8 km regardless of time, so the graph is a horizontal line at d = 8. Gradient = 0 (zero) because distance doesn't change with time.
3.8 — Sort the four lines
Positive: (a) y = 3x + 5 (m = 3). Negative: (c) y = −x + 2 (m = −1). Zero: (d) y = −4 (horizontal). Undefined: (b) x = 6 (vertical).