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.
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.
- 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.
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).
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.
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.
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: $x = 0$.
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.
Choose x-values: $x = -1, 0, 1, 2$
Calculate y: $(-1, -1)$, $(0, 1)$, $(1, 3)$, $(2, 5)$
Plot: Mark each point on the coordinate plane.
Draw: Connect with a straight line extending in both directions.
y-intercept: $(0, 3)$
Gradient: $m = -\dfrac{1}{2}$. From $(0, 3)$, move 2 right and 1 down to $(2, 2)$.
Draw: Line through $(0, 3)$ and $(2, 2)$.
Check: x-intercept at $y = 0$: $0 = -\frac{1}{2}x + 3$ → $x = 6$. Point $(6, 0)$ should also be on the line.
y-intercept: Set $x = 0$: $y = 3(0) - 6 = -6$. So $(0, -6)$.
x-intercept: Set $y = 0$: $0 = 3x - 6$ → $3x = 6$ → $x = 2$. So $(2, 0)$.
Brain Trainer
4 quick-fire drills. Beat the clock.
5 MCQs and 3 short-answer questions. Target: 80% accuracy.
Your 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 $-2$).
Your 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.
Your 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.
Consolidate and reflect before moving on.
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.)
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.
Plotting only one point. Reversing rise and run in the gradient-intercept method. Forgetting that a line extends infinitely in both directions.
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.
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?
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