Mathematics • Year 10 • Unit 2 • Lesson 17
Graphing Linear Functions — Skill Drill
Build fluency with the three graphing tools from Lesson 17: table of values, gradient–intercept method, and intercepts. Then check that random points satisfy the equation.
1. I do — fully worked example
Graphing y = 2x + 1 using a table of values.
Problem. Make a table of values for y = 2x + 1 using x = −1, 0, 1, 2 and list the four (x, y) points.
Step 1 — Choose convenient x-values.
x: −1, 0, 1, 2
Step 2 — Substitute each x into y = 2x + 1.
x = −1: y = 2(−1) + 1 = −1 → (−1, −1)
x = 0: y = 2(0) + 1 = 1 → (0, 1)
x = 1: y = 2(1) + 1 = 3 → (1, 3)
x = 2: y = 2(2) + 1 = 5 → (2, 5)
Step 3 — Sanity check using gradient.
y goes up by 2 as x goes up by 1 → matches m = 2 ✓
Answer: Points (−1, −1), (0, 1), (1, 3), (2, 5); gradient 2, y-intercept (0, 1).
2. We do — fill in the missing steps
Use the gradient–intercept method to plan a graph of y = −(1/2)x + 3. Fill in each blank. 5 marks
Problem. Plan the graph of y = −(1/2)x + 3 by stating the y-intercept and a second point reached using the gradient.
Step 1 — Identify the y-intercept (when x = 0): y = ____. So the first point is ( 0, ____ ).
Step 2 — Identify the gradient: m = ____ = rise/run = ____ / ____.
Step 3 — From the y-intercept, move ____ right and ____ down to reach a second point ( ____ , ____ ).
Step 4 — Find the x-intercept (when y = 0):
0 = −(1/2)x + 3 → x = ____. So the x-intercept is ( ____ , 0 ).
Step 5 — Verify a third point. Check (4, ____ ) satisfies y = −(1/2)x + 3.
3. You do — independent practice
For each: list the three key features (gradient, y-intercept, x-intercept) and at least two (x, y) points on the line.
Foundation — read off the equation
3.1 y = 3x + 2. State m, the y-intercept, and the point at x = 1. 1 mark
3.2 y = −x + 4. State m, the y-intercept, and the x-intercept. 1 mark
3.3 Find the y-intercept of y = 3x − 5. 1 mark
3.4 Find the x-intercept of y = 4x − 8. 1 mark
Standard — table of values and intercepts
3.5 Make a table for y = −2x + 6 using x = 0, 1, 2, 3 and list the four points. 2 marks
3.6 Find both intercepts of y = (1/2)x − 3 and state the gradient. 2 marks
Extension — check points and rearrange
3.7 Does (3, 4) lie on y = 2x − 2? Show your substitution. Then find one point that does lie on the line. 3 marks
3.8 Rearrange 3x + 2y = 12 to y = mx + c, then find the gradient and both intercepts. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (y = −(1/2)x + 3)
Step 1: y = 3; (0, 3). Step 2: m = −1/2 = −1/2 (or equivalently 1 down per 2 right). Step 3: move 2 right and 1 down → ( 2, 2 ). Step 4: x = 6; ( 6, 0 ). Step 5: y at x = 4 → y = −(1/2)(4) + 3 = 1, so (4, 1) ✓.
3.1 — y = 3x + 2
m = 3, y-intercept (0, 2), at x = 1 → y = 5, point (1, 5).
3.2 — y = −x + 4
m = −1, y-int (0, 4), x-int: 0 = −x + 4 → x = 4, so (4, 0).
3.3 — y = 3x − 5
y-int: (0, −5).
3.4 — y = 4x − 8
x-int: 0 = 4x − 8 → x = 2; point (2, 0).
3.5 — y = −2x + 6 table
(0, 6), (1, 4), (2, 2), (3, 0). (Pattern: y drops by 2 each time x rises by 1 — matches m = −2.)
3.6 — y = (1/2)x − 3
m = 1/2; y-int: (0, −3); x-int: 0 = (1/2)x − 3 → x = 6, so (6, 0).
3.7 — Does (3, 4) lie on y = 2x − 2?
y = 2(3) − 2 = 4. Yes, (3, 4) IS on the line. Another point: at x = 0, y = −2 → (0, −2). Or at x = 5, y = 8 → (5, 8).
3.8 — 3x + 2y = 12
2y = −3x + 12 → y = (−3/2)x + 6. Gradient m = −3/2; y-int (0, 6); x-int 0 = (−3/2)x + 6 → x = 4, so (4, 0).