What is the y-intercept of y = 3x + 8?

It is the point (0, 8), not (8, 0). The y-intercept is where the line crosses the y-axis, and every point on the y-axis has x = 0. Substitute x = 0 into y = 3x + 8 to get y = 3 × 0 + 8 = 8, so the line crosses at (0, 8). Writing (8, 0) puts the 8 on the x-axis instead, which is the wrong axis entirely.

How do you find where y = −2x + 6 crosses both axes?

Set each variable to zero in turn. For the y-intercept set x = 0: y = −2 × 0 + 6 = 6, so the line crosses the y-axis at (0, 6). For the x-intercept set y = 0: 0 = −2x + 6, so 2x = 6 and x = 3, giving the point (3, 0). The y-intercept always has x = 0 and the x-intercept always has y = 0.

Why does x = 0 give the y-intercept and y = 0 give the x-intercept?

Because the y-axis is the line where x = 0 and the x-axis is the line where y = 0. To find where a graph crosses the y-axis you need the point on it whose x-coordinate is 0, so you put x = 0 into the equation and solve for y. To find where it crosses the x-axis you need the point whose y-coordinate is 0, so you put y = 0 and solve for x. It feels backwards because each axis is named for the coordinate that varies along it, while the other coordinate is zero.

GCSE Maths Foundation · AQA · Linear graphs

Reading and finding intercepts: the y-intercept of y = 3x + 8 is (0, 8), not (8, 0)

Asked for the y-intercept of $y = 3x + 8$ , students write (8, 0) — the 8 lands on the x-axis. But the y-intercept is where the line crosses the y-axis, and every point on the y-axis has $x = 0$ . So the crossing point is (0, 8), with the 8 as the y-coordinate.

The thirty-second fix: the y-intercept is where x = 0, and the x-intercept is where y = 0. Substitute the zero in and solve for the coordinate left over. For $y = -2x + 6$ , setting $x = 0$ gives $(0,\ 6)$ and setting $y = 0$ gives $(3,\ 0)$ .

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How to spot it in your own work

You put the y-intercept on the x-axis, e.g. writing $(8,\ 0)$ for $y = 3x + 8$ instead of $(0,\ 8)$ .
You wrote a single number, like 8, instead of the coordinate pair $(0,\ 8)$ that names the actual point.
You found an intercept without zeroing the other variable — not setting $x = 0$ for the y-intercept or $y = 0$ for the x-intercept.
You solved for the wrong variable, e.g. setting $y = 0$ but then reporting the answer as a y-coordinate.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (where a line meets the axes):

A straight line has equation

$y = -2x + 6$

Write down the coordinates of the point C where the line crosses the y-axis, and the point D where it crosses the x-axis.

The misconception is to read 6 as a point on its own, or to put it on the wrong axis. The crossings are coordinate pairs, and each is found by setting the other variable to zero.

On the y-axis $x = 0$ , so $y = -2 \times 0 + 6 = 6$ gives C $(0,\ 6)$ . On the x-axis $y = 0$ , so $0 = -2x + 6$ gives $x = 3$ and D $(3,\ 0)$ .

Why students fall for this

The names get crossed. The y-intercept is the point where the graph meets the y-axis — but to be on the y-axis a point must have $x = 0$ , so its only free coordinate is y. Students see the +8 in $y = 3x + 8$ , know 8 is “the intercept”, and attach it to the x-coordinate by reflex, writing $(8,\ 0)$ . That places it on the x-axis, the wrong line entirely.

Substituting fixes the confusion every time. To meet the y-axis, set $x = 0$ : in $y = 3x + 8$ that gives $y = 8$ , so the point is $(0,\ 8)$ . The 8 is the height at which the line crosses, which is exactly a y-coordinate.

The same discipline finds the x-intercept: set $y = 0$ and solve. For $y = -2x + 6$ that is $0 = -2x + 6$ , so $2x = 6$ and $x = 3$ — the point $(3,\ 0)$ . Zero the variable belonging to the axis you are not on, then solve for the one that is left.

The fix: The y-intercept is where x = 0; the x-intercept is where y = 0

For the y-intercept, set x = 0. Every point on the y-axis has $x = 0$ . Substitute it in and solve for y. In $y = 3x + 8$ : $y = 8$ , so the point is $(0,\ 8)$ — never $(8,\ 0)$ .

For the x-intercept, set y = 0. Every point on the x-axis has $y = 0$ . Substitute it in and solve for x. In $y = -2x + 6$ : $0 = -2x + 6$ , so $x = 3$ and the point is $(3,\ 0)$ .

Write a coordinate pair. An intercept is a point, not a loose number — give it as $(0,\ 8)$ or $(3,\ 0)$ , with the zero on the axis you are crossing away from.

Check the zero is in the right slot. The y-intercept has the 0 first $(0,\ y)$ ; the x-intercept has the 0 second $(x,\ 0)$ .

Worked example

Find where $y = -2x + 6$ crosses each axis. The trap is to misplace 6 on the x-axis; the fix is to zero one variable at a time.

y-intercept: set x = 0. On the y-axis the x-coordinate is 0.
$y = -2 \times 0 + 6 = 6 \quad\Longrightarrow\quad C\,(0,\ 6)$
x-intercept: set y = 0. On the x-axis the y-coordinate is 0.
$0 = -2x + 6 \quad\Longrightarrow\quad 2x = 6 \quad\Longrightarrow\quad x = 3$
So the point is D $(3,\ 0)$ .
Put the zeros in the right slots. The y-intercept is $(0,\ 6)$ (zero first); the x-intercept is $(3,\ 0)$ (zero second).
Re-read the y-intercept of $y = 3x + 8$ . Set $x = 0$ : $y = 8$ , so it is $(0,\ 8)$ — not the $(8,\ 0)$ that puts 8 on the wrong axis.

So $y = -2x + 6$ meets the axes at $(0,\ 6)$ and $(3,\ 0)$ , each found by setting the other variable to zero — and the y-intercept always carries the constant as its y-coordinate.

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Common questions

What is the y-intercept of $y = 3x + 8$ ?: It is the point $(0,\ 8)$ , not $(8,\ 0)$ . The y-intercept is where the line crosses the y-axis, and every point on the y-axis has $x = 0$ . Substitute $x = 0$ into $y = 3x + 8$ to get $y = 3 \times 0 + 8 = 8$ , so the line crosses at $(0,\ 8)$ . Writing $(8,\ 0)$ puts the 8 on the x-axis instead, which is the wrong axis entirely.
How do you find where $y = -2x + 6$ crosses both axes?: Set each variable to zero in turn. For the y-intercept set $x = 0$ : $y = -2 \times 0 + 6 = 6$ , so the line crosses the y-axis at $(0,\ 6)$ . For the x-intercept set $y = 0$ : $0 = -2x + 6$ , so $2x = 6$ and $x = 3$ , giving the point $(3,\ 0)$ . The y-intercept always has $x = 0$ and the x-intercept always has $y = 0$ .
Why does $x = 0$ give the y-intercept and $y = 0$ give the x-intercept?: Because the y-axis is the line where $x = 0$ and the x-axis is the line where $y = 0$ . To find where a graph crosses the y-axis you need the point on it whose x-coordinate is 0, so you put $x = 0$ into the equation and solve for y. To find where it crosses the x-axis you need the point whose y-coordinate is 0, so you put $y = 0$ and solve for x. It feels backwards because each axis is named for the coordinate that varies along it, while the other coordinate is zero.

Related misconceptions

Finding the gradient of a straight lineThe neighbouring linear graphs skill: rearrange to y = mx + c before reading the gradient, so 2y = 10x becomes y = 5x with gradient 5, not 10.

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