Mathematics • Year 8 • Unit 2 • Lesson 14

Sketching Linear Graphs

Build fluency with the three sketching methods: two-point, gradient-intercept and intercept. One worked example, one guided fill-in, then eight independent sketches.

Build · I Do / We Do / You Do

1. I do — fully worked example

Sketch y = 2x + 1 using the gradient-intercept method (fastest when the equation is in y = mx + c form).

Problem. Sketch y = 2x + 1 on a set of axes.

Step 1 — Read m and c.

m = 2, c = 1.

Reason: in gradient-intercept form, the coefficient of x is m and the constant is c.

Step 2 — Plot the y-intercept first.

Plot the point (0, 1) on the y-axis.

Step 3 — Use the gradient to find a second point.

m = 2 = 2/1. From (0, 1): run 1 right, rise 2 up → (1, 3).

Reason: gradient = rise/run. Treat m = 2 as 2/1.

Step 4 — Join the two points with a straight line and label.

Draw a line through (0, 1) and (1, 3). Label it y = 2x + 1.

Check: at x = 2, y = 5. The point (2, 5) should lie on your line. ✓

Stuck? Revisit lesson § Card 3 (Gradient-Intercept Method) — plot (0, c) first, then use m = rise/run to step to the next point.

2. We do — fill in the missing steps

Sketch 2x + 3y = 6 using the intercept method. Fill in each blank. 4 marks

Step 1 — y-intercept: set x = 0.

2( ____ ) + 3y = 6 → 3y = ______ → y = ______ . Point: ( ______ , ______ )

Step 2 — x-intercept: set y = 0.

2x + 3( ____ ) = 6 → 2x = ______ → x = ______ . Point: ( ______ , ______ )

Step 3 — Plot both intercepts and join with a straight line.

Mark ( ______ , ______ ) on the y-axis and ( ______ , ______ ) on the x-axis, then draw.

Step 4 — Check with another point. At x = ____ (try a value between the intercepts), y should sit on the line. Pick one and verify.

Stuck? Revisit lesson § Card 4 (Intercept Method) — best for standard form ax + by = c.

3. You do — independent practice

For each, choose the fastest method, list your two points, and describe the sketch. (You don't have to draw a perfect graph — state the two points clearly and identify the type of line.)

Foundation — gradient-intercept form

3.1 Sketch y = x + 2. State the y-intercept and one other point.    1 mark

3.2 Sketch y = 3x − 2. State the y-intercept and one other point.    1 mark

3.3 Sketch y = −x + 4. State the y-intercept and one other point.    1 mark

3.4 Sketch y = −½x + 3 starting from the y-intercept. (Use m = −1/2: run 2 right, rise 1 down.)    1 mark

Standard — intercept method

3.5 Sketch x + y = 5 using the intercept method. State both intercepts.    2 marks

3.6 Sketch 3x + 4y = 12 using the intercept method. State both intercepts.    2 marks

Extension — special lines

3.7 Sketch y = −3 and describe the line in one sentence. Then sketch x = 2 and describe it in one sentence.    2 marks

3.8 Sketch y = 2x using the two-point method (pick x = 0 and x = 2). Why is this both a line through the origin AND a special case?    2 marks

Stuck on 3.7? y = c (a number) is horizontal. x = c is vertical. Vertical lines aren't even functions of x!

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (2x + 3y = 6)

Step 1: 2(0) + 3y = 6 → 3y = 6 → y = 2. Point: (0, 2).
Step 2: 2x + 3(0) = 6 → 2x = 6 → x = 3. Point: (3, 0).
Step 3: plot (0, 2) and (3, 0); draw line through both.
Step 4: e.g. at x = 1.5: 2(1.5) + 3y = 6 → 3 + 3y = 6 → 3y = 3 → y = 1. Point (1.5, 1) sits exactly on the line ✓.

3.1 — y = x + 2

y-intercept (0, 2). Use m = 1: run 1, rise 1 → second point (1, 3). Line goes uphill through both points.

3.2 — y = 3x − 2

y-intercept (0, −2). Use m = 3: run 1, rise 3 → second point (1, 1). Steep uphill line.

3.3 — y = −x + 4

y-intercept (0, 4). Use m = −1: run 1, rise −1 → second point (1, 3). Gentle downhill line; x-intercept (4, 0).

3.4 — y = −½x + 3

y-intercept (0, 3). Use m = −1/2: run 2 right, rise 1 down → second point (2, 2). x-intercept (6, 0).

3.5 — x + y = 5

y-int (x = 0): y = 5 → (0, 5). x-int (y = 0): x = 5 → (5, 0). Plot both and join. Downhill line of gradient −1.

3.6 — 3x + 4y = 12

y-int: 4y = 12 → y = 3 → (0, 3). x-int: 3x = 12 → x = 4 → (4, 0). Join both; the line falls gently from upper-left to lower-right.

3.7 — y = −3 and x = 2

y = −3: horizontal line (gradient 0) crossing the y-axis at (0, −3) — every point has y = −3.
x = 2: vertical line crossing the x-axis at (2, 0) — every point has x = 2. (Vertical lines are NOT in y = mx + c form because their gradient is undefined.)

3.8 — y = 2x via two points

At x = 0: y = 0 → (0, 0). At x = 2: y = 4 → (2, 4). Line passes through the origin with gradient 2. Special case: c = 0, so the y-intercept and x-intercept are both the origin (0, 0).