x- and y-Intercepts
For the equation $2x + 3y = 6$, what happens when $x = 0$? These two points are the quickest way to sketch any straight line — no table needed!
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For the equation $2x + 3y = 6$, what happens when $x = 0$? What happens when $y = 0$? What do these points tell us?
Reveal Answer
When $x = 0$: $3y = 6$, so $y = 2$. This gives the point $(0, 2)$ — the line crosses the y-axis at 2.
When $y = 0$: $2x = 6$, so $x = 3$. This gives the point $(3, 0)$ — the line crosses the x-axis at 3.
These two points, $(0, 2)$ and $(3, 0)$, are called the intercepts. Just plot both and draw the line!
An intercept is a point where a graph crosses one of the coordinate axes.
- x-intercept: where the line meets the x-axis. The coordinates are $(x, 0)$ — the $y$-value is always 0.
- y-intercept: where the line meets the y-axis. The coordinates are $(0, y)$ — the $x$-value is always 0.
Together, the two intercepts give us two points on the line. Since "two points determine a line," we can sketch any straight line quickly without making a table of values.
Method: Substitute $x = 0$ into the equation and solve for $y$. The y-intercept is always the point $(0, y)$.
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1Substitute $x = 0$$2(0) + 3y = 6$
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2Simplify and solve for $y$$3y = 6 \Rightarrow y = 2$
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3Write as a coordinate pointy-intercept: $(0, 2)$
Method: Substitute $y = 0$ into the equation and solve for $x$. The x-intercept is always the point $(x, 0)$.
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1Substitute $y = 0$$2x + 3(0) = 6$
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2Simplify and solve for $x$$2x = 6 \Rightarrow x = 3$
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3Write as a coordinate pointx-intercept: $(3, 0)$
The fastest way to sketch any straight line: find both intercepts, plot them, and draw the line through them.
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1Find the y-intercept (set $x = 0$)$3(0) + 4y = 12 \Rightarrow 4y = 12 \Rightarrow y = 3$y-intercept: $(0, 3)$
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2Find the x-intercept (set $y = 0$)$3x + 4(0) = 12 \Rightarrow 3x = 12 \Rightarrow x = 4$x-intercept: $(4, 0)$
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3Plot both intercepts on the axes
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4Draw a straight line through both interceptsDone! The line $3x + 4y = 12$ passes through $(0, 3)$ and $(4, 0)$.
Linear equations can appear in different forms. Here is how to find intercepts in each:
| Form | Example | y-intercept | x-intercept |
|---|---|---|---|
| $y = mx + c$ | $y = 2x + 5$ | Read directly: $(0, 5)$ | Set $y = 0$: $x = -2.5$, so $(-2.5, 0)$ |
| $ax + by = c$ | $2x + 3y = 6$ | Set $x = 0$: $(0, 2)$ | Set $y = 0$: $(3, 0)$ |
| $y = mx$ | $y = 3x$ | $(0, 0)$ — origin | $(0, 0)$ — both coincide |
Intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ is specially designed so that $a$ is the x-intercept and $b$ is the y-intercept. For example, $\dfrac{x}{3} + \dfrac{y}{2} = 1$ has x-intercept $(3, 0)$ and y-intercept $(0, 2)$.
Horizontal Lines
An equation of the form $y = k$ is a horizontal line. It never crosses the x-axis (unless $k = 0$) but crosses the y-axis at $(0, k)$. Example: $y = 3$ has y-intercept $(0, 3)$ but no x-intercept.
Vertical Lines
An equation of the form $x = k$ is a vertical line. It never crosses the y-axis (unless $k = 0$) but crosses the x-axis at $(k, 0)$. Example: $x = 3$ has x-intercept $(3, 0)$ but no y-intercept.
Wrong: Setting $x = 0$ to find the x-intercept (or $y = 0$ to find the y-intercept).
Right: x-intercept → set $y = 0$; y-intercept → set $x = 0$. Think: "Where does it cross the x-axis?" = where $y = 0$.
Wrong: Writing the y-intercept as $y = 4$ instead of the point $(0, 4)$.
Right: Always write intercepts as coordinate pairs: x-intercept = $(x, 0)$ and y-intercept = $(0, y)$.
Brain Trainer · 10 speed drills
Find the intercepts as quickly as you can. Write the answers as coordinate pairs.
1 Find the y-intercept of $y = 3x + 5$
$(0, 5)$2 Find the x-intercept of $2x + y = 8$
$(4, 0)$3 Find the x-intercept of $y = 2x - 6$
$(3, 0)$4 Find the y-intercept of $3x + 4y = 12$
$(0, 3)$5 Find the x-intercept of $x + 2y = 6$
$(6, 0)$6 Find the y-intercept of $y = -x + 4$
$(0, 4)$7 Find both intercepts of $x + y = 5$
$(5, 0)$ and $(0, 5)$8 Find the y-intercept of $y = 4x$
$(0, 0)$9 Does $y = 2$ have an x-intercept?
No x-intercept10 Find the x-intercept of $\dfrac{x}{4} + \dfrac{y}{3} = 1$
$(4, 0)$
Quick Check · 5 questions
Show Your Working · 3 questions
SAQ 1. Find the x- and y-intercepts of $2x + 5y = 10$, then use them to sketch the line.
Show Solution
y-intercept (set $x = 0$): $5y = 10 \Rightarrow y = 2$. y-intercept: $(0, 2)$
x-intercept (set $y = 0$): $2x = 10 \Rightarrow x = 5$. x-intercept: $(5, 0)$
Sketch: Plot $(0, 2)$ and $(5, 0)$, then draw a straight line through both points.
SAQ 2. Explain why the vertical line $x = 3$ has no y-intercept.
Show Solution
The equation $x = 3$ describes a vertical line through all points where $x = 3$. The y-axis is where $x = 0$. Since the x-coordinate is always 3 and can never be 0, the line never intersects the y-axis — so there is no y-intercept. The line does have an x-intercept at $(3, 0)$.
SAQ 3. Find where the line $y = 2x - 6$ crosses both axes. Show all working.
Show Solution
x-intercept (set $y = 0$): $0 = 2x - 6 \Rightarrow 2x = 6 \Rightarrow x = 3$. x-intercept: $(3, 0)$
y-intercept (set $x = 0$): $y = 2(0) - 6 = -6$. y-intercept: $(0, -6)$
Shortcut check: In $y = 2x - 6$, we have $c = -6$, confirming y-intercept $(0, -6)$.
A line passes through $(4, 0)$ and has y-intercept $(0, -2)$. Find its equation in the form $ax + by = c$.
Show Solution
Using intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ with $a = 4$, $b = -2$: $\dfrac{x}{4} + \dfrac{y}{-2} = 1$. Multiply by 4: $x - 2y = 4$.
Find the x- and y-intercepts of $\dfrac{x}{5} + \dfrac{y}{3} = 1$. What do you notice about the denominators?
Show Solution
x-intercept: set $y = 0$: $\dfrac{x}{5} = 1 \Rightarrow x = 5$. x-intercept: $(5, 0)$.
y-intercept: set $x = 0$: $\dfrac{y}{3} = 1 \Rightarrow y = 3$. y-intercept: $(0, 3)$.
Observation: The denominator under $x$ is the x-intercept, and the denominator under $y$ is the y-intercept — this is why it's called intercept form!
- The y-intercept is where the line crosses the y-axis: set $x = 0$ and solve for $y$.
- The x-intercept is where the line crosses the x-axis: set $y = 0$ and solve for $x$.
- The intercept method: find both intercepts, plot, draw — fastest sketch method.
- For $y = mx + c$: the y-intercept is $(0, c)$ — read it directly.
- Horizontal line $y = k$ ($k \ne 0$): y-intercept $(0, k)$, no x-intercept.
- Vertical line $x = k$ ($k \ne 0$): x-intercept $(k, 0)$, no y-intercept.
Lesson Complete!
You've mastered x- and y-intercepts.