Mathematics • Year 10 • Unit 2 • Lesson 17
Graphing Linear Functions — Mixed Challenge
Six mixed problems testing table of values, gradient–intercept, intercepts, point-on-line checks, and graphing horizontal and vertical lines. Spot a swapped-sign y-intercept, then design a line meeting three specific conditions.
1. Mixed problems — graph or analyse
Show enough to make the answer reproducible: equation, gradient, both intercepts, table or points. 3 marks each
1.1 For y = −3x + 9, list both intercepts and one extra point on the line.
1.2 For 2x − 3y = 12, rearrange to y = mx + c form, then state both intercepts.
1.3 Does (4, −5) lie on y = 2x − 13? Show the substitution.
1.4 Make a table for y = (2/3)x − 4 using x = 0, 3, 6, 9 (so each y is an integer). List the four points.
1.5 Sketch a graph (in words) of the horizontal line y = −2 and the vertical line x = 5. State the intercepts of each.
1.6 Two parts: (a) Find the y-intercept and x-intercept of y = (1/4)x + 2. (b) Use the intercepts to sketch (in words: state both points and direction of the line).
2. Find the mistake
A Year 10 student finds the intercepts of y = −2x + 6 and writes the working below. One line contains a sign mistake. Find it. 3 marks
Student's working:
Line 1: y-intercept: set x = 0 → y = −2(0) + 6 = 6, so (0, 6).
Line 2: x-intercept: set y = 0 → 0 = −2x + 6
Line 3: −2x = 6 → x = −3, so (−3, 0).
(a) Which line is wrong?
(b) Explain the sign mistake in one or two sentences.
(c) Show the corrected working with the correct x-intercept.
Stuck? When 0 = −2x + 6, moving 6 to the other side gives −2x = −6 (not 6). Then x = 3, not −3.3. Open-ended challenge — design a line
Many valid answers. 4 marks
3.1 Design a linear function y = mx + c that satisfies all three of: (i) negative gradient, (ii) positive y-intercept, (iii) the x-intercept is exactly x = 4.
Submit:
(a) Your equation in y = mx + c form.
(b) Both intercepts.
(c) A small table for x = 0, 2, 4 confirming the line passes through (4, 0).
(d) State the gradient and explain in one sentence why it is negative.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = −3x + 9
y-int (0, 9). x-int: 0 = −3x + 9 → x = 3, so (3, 0). Extra point at x = 1: y = 6 → (1, 6).
1.2 — 2x − 3y = 12
−3y = −2x + 12 → y = (2/3)x − 4. x-int: 0 = (2/3)x − 4 → x = 6, so (6, 0). y-int: (0, −4).
1.3 — Check (4, −5) on y = 2x − 13
y = 2(4) − 13 = 8 − 13 = −5 ✓. Yes, the point is on the line.
1.4 — y = (2/3)x − 4 table
x=0→y=−4; x=3→y=−2; x=6→y=0; x=9→y=2. Points: (0,−4), (3,−2), (6,0), (9,2).
1.5 — y = −2 and x = 5
y = −2: horizontal line passing through (0, −2) and (any x, −2). y-intercept (0, −2); no x-intercept (it never crosses the x-axis).
x = 5: vertical line through (5, 0) and (5, any y). x-intercept (5, 0); no y-intercept.
1.6 — y = (1/4)x + 2
(a) y-int (0, 2). x-int: 0 = (1/4)x + 2 → x = −8, so (−8, 0). (b) Line passes through (−8, 0) and (0, 2), sloping upward gently (m = 1/4 > 0).
2 — Find the mistake
(a) Line 3.
(b) When solving 0 = −2x + 6, moving 6 to the other side gives −2x = −6 (not 6) — the student forgot to flip the sign of the 6 when it crossed the equals sign. Then dividing by −2: x = 3, not −3.
(c) Corrected: 0 = −2x + 6 → 2x = 6 → x = 3. x-intercept is (3, 0). Quick check: y = −2(3) + 6 = 0 ✓.
3 — Open-ended (sample solution)
Sample equation: y = −x + 4.
Intercepts: y-int (0, 4) (positive ✓). x-int: 0 = −x + 4 → x = 4, so (4, 0) ✓.
Table: x=0→y=4; x=2→y=2; x=4→y=0 ✓ (passes through (4, 0)).
Gradient m = −1 (negative because as x rises by 1, y drops by 1 — line slopes downward).
Marking: 1 for an equation meeting all three criteria; 1 for both intercepts correctly stated; 1 for a table that includes (4, 0); 1 for the gradient sign explanation.