Lesson 17 of 20

Graphing Linear Functions

Transform equations into visual lines on the coordinate plane. Master the table method and the efficient gradient-intercept technique.

45 min 5 MCQ + 3 SAQ MAS-LIN-C-01
y = mx + c
Think First

Before we begin: Without drawing, how many points do you think you need to plot to draw a straight line accurately? Explain why.

Come back to this after you have worked through the lesson.

1
The Big Idea

Every linear equation produces a straight line when graphed on the coordinate plane. Two points are enough to define a unique straight line, but plotting three gives a useful check. The table of values method works for any equation, while the gradient-intercept method is faster when the equation is already in $y = mx + c$ form.

Know
The two main methods for graphing linear functions. What x-intercept and y-intercept mean.
Understand
Why only two points are needed to define a straight line. How gradient and y-intercept determine the position of a line.
Can Do
Graph a linear function using a table of values. Graph using the gradient-intercept method. Identify x and y intercepts.
2
Lesson Objectives
  • Complete a table of values for any linear equation.
  • Plot points accurately and draw a straight line through them.
  • Use the y-intercept and gradient to sketch a line quickly.
  • Find the x-intercept and y-intercept of a linear function.
3
Key Vocabulary
Table of values
A table showing x-values and their corresponding y-values for an equation.
Gradient-intercept method
Graphing using the y-intercept $(0, c)$ and the gradient $m$.
x-intercept
The point where a graph crosses the x-axis ($y = 0$).
y-intercept
The point where a graph crosses the y-axis ($x = 0$).
4
Spot the Trap

Wrong: Plotting only one point for a linear graph. A single point does not determine a unique line.

Right: Always plot at least two points (three is better for checking). Use a ruler to draw the line.

Wrong: Using $\dfrac{\text{run}}{\text{rise}}$ instead of $\dfrac{\text{rise}}{\text{run}}$ when moving from the y-intercept.

Right: From the y-intercept, move run units across and rise units up (or down if negative).

5
Graphing Using a Table of Values
+5 XP

The most reliable method for graphing any linear equation is to calculate points and plot them. Choose at least three x-values (including negative, zero, and positive), substitute each into the equation to find y, plot the points, and draw a straight line through them with a ruler.

Choose x-values. Calculate y. Plot and draw.

x y -1 -1 0 1 1 3
reliable for any equation
Always include x = 0
This gives the y-intercept directly. It is the easiest point to calculate.
Choose easy x-values
Pick numbers that make mental arithmetic simple. Avoid fractions if possible.
Three points for safety
Two points define a line, but three let you spot calculation errors. If they do not line up, recheck.
6
The Gradient-Intercept Method
+5 XP

When the equation is in $y = mx + c$ form, you can graph it quickly without a full table. Plot the y-intercept at $(0, c)$, then use the gradient $m = \dfrac{\text{rise}}{\text{run}}$ to find a second point. Move run units right and rise units up (or down if negative), then draw the line.

Plot $(0, c)$. Move run right, rise up. Draw line.

(0, -2) run 3 rise 2 (3, 0)
fast when in y = mx + c
Start at the y-intercept
$(0, c)$ is always on the y-axis. It is your anchor point.
Move run then rise
Go right by the run amount, then up (or down) by the rise amount. Do not reverse.
Extend both directions
A line continues forever. Draw arrows at both ends or extend beyond your plotted points.
7
Finding Intercepts
+5 XP

The x-intercept is where the line crosses the x-axis ($y = 0$). The y-intercept is where the line crosses the y-axis ($x = 0$). In $y = mx + c$ form, the y-intercept is simply $(0, c)$. To find the x-intercept, substitute $y = 0$ and solve for $x$.

x-intercept: $y = 0$. y-intercept: $x = 0$.

y-int x-int
where the line crosses
y-intercept is easiest
In $y = mx + c$, the y-intercept is $(0, c)$. No calculation needed.
x-intercept: set y = 0
Substitute $y = 0$ into the equation and solve for $x$.
Both intercepts on axes
The x-intercept has y-coordinate 0. The y-intercept has x-coordinate 0.
8
Real-World Linear Relationships
+5 XP

Linear relationships appear everywhere: taxi fares (flat fee plus rate per km), phone bills (base cost plus per-minute charges), water usage (fixed service fee plus per-litre rate). In each case, the y-intercept represents the fixed cost and the gradient represents the variable rate.

y-intercept = fixed cost. gradient = rate.

$50 fixed fee $0.30 per km
C = 0.3d + 50
Fixed vs variable
The y-intercept is what you pay even if you use nothing. The gradient is what you pay per unit.
Break-even point
The x-intercept shows where total cost becomes zero (if it makes sense in context).
Label your axes
In real-world graphs, always label what x and y represent, including units.
Worked Example 1 — Table of Values
Q.Graph $y = 2x + 1$ using a table of values.
Solution
1

Choose x-values: $x = -1, 0, 1, 2$

2

Calculate y: $(-1, -1)$, $(0, 1)$, $(1, 3)$, $(2, 5)$

3

Plot: Mark each point on the coordinate plane.

4

Draw: Connect with a straight line extending in both directions.

Answer: Straight line through $(-1, -1)$, $(0, 1)$, $(1, 3)$, $(2, 5)$ with gradient 2 and y-intercept 1.
Worked Example 2 — Gradient-Intercept Method
Q.Graph $y = -\dfrac{1}{2}x + 3$ using the gradient-intercept method.
Solution
1

y-intercept: $(0, 3)$

2

Gradient: $m = -\dfrac{1}{2}$. From $(0, 3)$, move 2 right and 1 down to $(2, 2)$.

3

Draw: Line through $(0, 3)$ and $(2, 2)$.

4

Check: x-intercept at $y = 0$: $0 = -\frac{1}{2}x + 3$ → $x = 6$. Point $(6, 0)$ should also be on the line.

Answer: Line with y-intercept $(0, 3)$, passing through $(2, 2)$ and $(6, 0)$.
Worked Example 3 — Finding Intercepts
Q.Find the x-intercept and y-intercept of $y = 3x - 6$.
Solution
1

y-intercept: Set $x = 0$: $y = 3(0) - 6 = -6$. So $(0, -6)$.

2

x-intercept: Set $y = 0$: $0 = 3x - 6$ → $3x = 6$ → $x = 2$. So $(2, 0)$.

Answer: y-intercept: $(0, -6)$, x-intercept: $(2, 0)$

Brain Trainer

4 quick-fire drills. Beat the clock.

What is the y-intercept of $y = 3x - 5$?
-5
Which point lies on $y = 2x + 3$?
(2, 7)
Find the x-intercept of $y = 4x - 8$.
2
To graph $y = -\frac{3}{4}x + 2$, from the y-intercept you move:
4 right, 3 down
Practice

5 MCQs and 3 short-answer questions. Target: 80% accuracy.

1.The y-intercept of $y = 3x - 5$ is:
Correct! In $y = mx + c$ form, $c$ is the y-intercept. Here $c = -5$, so the y-intercept is $(0, -5)$.
The y-intercept is the constant term in $y = mx + c$. It is not the coefficient of $x$.
2.Which point lies on the line $y = 2x + 3$?
Correct! Substitute $x = 2$: $y = 2(2) + 3 = 7$. So $(2, 7)$ lies on the line.
Substitute the x-coordinate into the equation and check if you get the y-coordinate. For example, for $(2, 7)$: $y = 2(2) + 3 = 7$.
3.The x-intercept of $y = 4x - 8$ is:
Correct! Set $y = 0$: $0 = 4x - 8$ → $4x = 8$ → $x = 2$. The x-intercept is $(2, 0)$.
The x-intercept occurs where $y = 0$. Substitute $y = 0$ into the equation and solve for $x$.
4.To graph $y = -\dfrac{3}{4}x + 2$ using gradient-intercept, from the y-intercept you should:
Correct! The gradient is $-\dfrac{3}{4}$, so rise = $-3$ (down 3) and run = $4$ (right 4). Move 4 right, then 3 down.
For $m = \dfrac{\text{rise}}{\text{run}} = -\dfrac{3}{4}$, the run is 4 (denominator) and the rise is $-3$ (negative means down). Move run units right first, then rise units up or down.
5.A line with gradient $-2$ passing through $(0, 5)$ has equation:
Correct! The gradient is $m = -2$ and the y-intercept is $c = 5$. So $y = -2x + 5$.
In $y = mx + c$, $m$ is the gradient and $c$ is the y-intercept. The point $(0, 5)$ tells you $c = 5$.
6.Complete a table of values for $y = -2x + 5$ using $x = -2, -1, 0, 1, 2$. Plot the points and describe the pattern you observe.

Your answer:

Correct table of values
Pattern described correctly
Connection to gradient made
Model Answer

| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |

| $y$ | $9$ | $7$ | $5$ | $3$ | $1$ |

The points form a straight line sloping downward from left to right. As $x$ increases by 1, $y$ decreases by 2 (consistent with gradient $-2$).

7.A line passes through $(0, -3)$ and has gradient $\dfrac{2}{5}$. (a) Write the equation of the line. (b) Find the x-intercept. (c) Explain how you would use the gradient-intercept method to sketch this line.

Your answer:

Equation written correctly
x-intercept found correctly
Clear explanation of sketching method
Model Answer

(a) $y = \dfrac{2}{5}x - 3$

(b) $0 = \dfrac{2}{5}x - 3$ → $\dfrac{2}{5}x = 3$ → $x = \dfrac{15}{2} = 7.5$. x-intercept: $(7.5, 0)$

(c) Plot $(0, -3)$. From there, move 5 units right and 2 units up to $(5, -1)$. Draw line through both points.

8.A car rental company charges a $50 flat fee plus $0.30 per kilometre driven. Let $C$ be the total cost and $d$ be the distance in kilometres. (a) Write a linear equation relating $C$ and $d$. (b) Complete a table of values for $d = 0, 50, 100, 200$. (c) Graph the relationship, labelling the axes and the y-intercept. (d) Explain what the gradient and y-intercept represent in this context.

Your answer:

Equation correct (1 mark)
Table correct (1 mark)
Graph description with labels (1 mark)
Gradient and intercept interpreted (2 marks)
Model Answer

(a) $C = 0.3d + 50$ or $C = 50 + 0.3d$

(b) $(0, 50)$, $(50, 65)$, $(100, 80)$, $(200, 110)$

(c) Horizontal axis: distance (km). Vertical axis: cost ($). Plot points and draw straight line. y-intercept at $(0, 50)$.

(d) The gradient ($0.30$) represents the cost per kilometre (30 cents per km). The y-intercept ($50$) represents the fixed base fee regardless of distance.

Review

Consolidate and reflect before moving on.

Stretch Challenge

A line has x-intercept $(4, 0)$ and y-intercept $(0, -6)$. Find its equation in the form $y = mx + c$. Then find another point on this line that has integer coordinates. (Hint: find the gradient first using the two intercepts as points.)

Key Idea

Two methods for graphing linear functions: table of values (works for any equation) and gradient-intercept (fastest when in $y = mx + c$). Two points define a line; three confirm accuracy.

Common Trap

Plotting only one point. Reversing rise and run in the gradient-intercept method. Forgetting that a line extends infinitely in both directions.

Connection

The gradient-intercept method connects directly to the equation form $y = mx + c$. The next lessons will use these skills to find equations from graphs and to explore parallel and perpendicular lines.

Interactive: Line Grapher — enter any linear equation and see the graph, intercepts, and gradient visualised instantly.
Table Master
Gradient Sketcher
Intercept Finder
Daily Challenge

Sketch $y = -\dfrac{2}{3}x + 4$ using the gradient-intercept method. Label the y-intercept, x-intercept, and one other point. Time yourself — can you do it in under 60 seconds?