The Equation of a Straight Line Graph

In the equation \( y = mx + c \), the value of \( c \) tells us where the line crosses the y-axis, and \( m \) tells us the gradient. Here, \( m = 3 \) and \( c = 1 \). Substituting \( x = 0 \): \( y = 3(0) + 1 = 1 \), confirming the y-intercept is \( (0, 1) \). Students who mix these up are falling for the “the first number is the intercept” misconception.

The number 3 means “for every 1 unit across, the line goes 3 units up” — it describes the steepness, not where the line starts. You can verify by plotting: at \( x = 0 \), \( y = 1 \); at \( x = 1 \), \( y = 4 \); at \( x = 2 \), \( y = 7 \). The line clearly crosses the y-axis at 1, not 3.

Writing \( y = 4x \) in full gives \( y = 4x + 0 \), so \( c = 0 \) and the y-intercept is \( (0, 0) \). Substituting \( x = 0 \): \( y = 4(0) = 0 \), confirming the line passes through the origin. Having a y-intercept of zero is not the same as having no y-intercept — it simply means the line crosses the y-axis at the origin.

Students often think that because there’s “no number at the end,” the equation is incomplete or the line doesn’t touch the y-axis — this is the “no constant means no y-intercept” misconception. In fact, every non-vertical straight line crosses the y-axis exactly once. Lines like \( y = 4x \), \( y = -2x \), and \( y = x \) all pass through the origin — their y-intercept is 0, not absent.

The gradient can only be read directly when the equation is in the form \( y = mx + c \). Dividing both sides of \( 2y = 6x + 10 \) by 2 gives \( y = 3x + 5 \), so the gradient is 3 (not 6). A common error is to look at the coefficient of \( x \) without first isolating \( y \), which here would give the wrong gradient of 6.

We can verify: when \( x = 0 \), \( y = 5 \); when \( x = 1 \), \( y = 8 \). The change in \( y \) is 3 for a change of 1 in \( x \), confirming the gradient is 3. Students who read the 6 as the gradient have fallen for the “reading the coefficient before rearranging” misconception — you must isolate \( y \) first.

Steepness is determined by the size of the gradient ignoring its sign — that is, its absolute value. The gradient of \( y = -2x + 7 \) is \( -2 \), which has absolute value 2. The gradient of \( y = x + 7 \) is 1, which has absolute value 1. Since \( 2 > 1 \), the first line is steeper.

Students often think a negative gradient means “less steep” because \( -2 < 1 \) — this is the “negative gradient means less steep” misconception. But the negative sign tells us the direction (sloping downwards), not the steepness. A line with gradient \( -2 \) rises or falls 2 units for every 1 unit across, while a line with gradient 1 only rises 1 unit.

The graph crosses the y-axis at 3 units up, so \( c = 3 \). Both equations have this in common. For the gradient, we can pick two clear points on the grid intersections, such as \( (0, 3) \) and \( (2, 2) \). By counting the grid squares, the line goes down 1 unit for every 2 units across.

Rise over run gives \( \frac{-1}{2} \). A gradient of -2 would mean going down 2 units for every 1 unit across, which would be a much steeper line. Students who choose \( y = -2x + 3 \) are confusing the “rise” and the “run”, which is a visual instance of the “rise over run inversion” misconception.

Example: \( y = x \; – \; 3 \)

Another: \( y = 5x \; – \; 3 \)

Creative: \( y = -\frac{1}{2}x \; – \; 3 \) — any gradient works as long as \( c = -3 \). There are infinitely many lines through \( (0, \; -3) \) with different steepnesses.

Trap: \( y = -3x + 1 \). A student might see the \( -3 \) and assume that’s the y-intercept, but here \( -3 \) is the gradient. The y-intercept is actually \( (0, 1) \). The “first number = intercept” misconception leads students to confuse the coefficient of \( x \) with the constant term.

Example: \( y = 2x + 1 \)

Another: \( y = 2x \; – \; 4 \)

Creative: \( y = 2x \) — parallel to the original and passes through the origin. Students might not think of this because there’s no visible “\( + c \)” term.

Trap: \( y = 5x + 2 \). A student might swap the gradient and intercept, thinking that sharing the same numbers somehow makes lines parallel. But parallel lines must have the same gradient — here the gradient is 5, not 2, so this line is not parallel. This exploits the “swapping m and c” misconception.

Example: \( y = \frac{1}{2}x + 3 \)

Another: \( y = \frac{3}{4}x \; – \; 1 \)

Creative: \( y = 0.1x + 100 \) — the gradient can be any value between 0 and 1, including decimals. A gradient of 0.1 means the line is almost flat, rising just 1 unit for every 10 across.

Trap: \( y = \frac{1}{2} \). A student might think this has gradient \( \frac{1}{2} \), but this is actually \( y = 0x + \frac{1}{2} \) — a horizontal line with gradient 0, not \( \frac{1}{2} \). The “confusing the constant with the gradient” misconception catches students who see a fraction and assume it must be the gradient.

Example: \( x = 4 \)

Another: \( x = -2 \)

Creative: \( x = 0 \) — this is the y-axis itself! Students rarely think of the axes as examples of straight lines, but the y-axis is a vertical line with equation \( x = 0 \).

Trap: \( y = 0 \). A student might think a horizontal line has an “undefined” gradient because it “doesn’t go anywhere.” But \( y = 0 \) is a horizontal line with gradient 0 — zero and undefined are very different. A gradient of 0 means perfectly flat; an undefined gradient means perfectly vertical. The “confusing zero gradient with undefined gradient” misconception is very common.

Most straight lines can be written in this form. For example, \( y = 3x + 2 \) and \( y = -x + 7 \) are both in \( y = mx + c \) form. However, vertical lines such as \( x = 5 \) have an undefined gradient, meaning there is no valid number to substitute for \( m \). Because of this, they cannot be written as \( y = mx + c \).

TRUE case: \( y = 2x + 1 \) is already in \( y = mx + c \) form. FALSE case: \( x = 3 \) is a straight line but cannot be written as \( y = mx + c \). Students who assume all lines fit this form have the “every straight line can be written as y = mx + c” misconception — they forget that vertical lines are straight lines too.

Parallel lines have the same gradient, not the same y-intercept. If two different lines share the same y-intercept, they must have different gradients (otherwise they’d be the same line), which means they cross at that y-intercept and go off in different directions — the opposite of parallel.

For example, \( y = 2x + 3 \) and \( y = -x + 3 \) both cross the y-axis at \( (0, 3) \), but one slopes up steeply and the other slopes down — they are definitely not parallel. This targets the “same intercept means parallel” misconception, where students confuse the role of the gradient and the intercept in determining parallelism.

A line with a positive gradient slopes upward from left to right, but it extends infinitely in both directions. No matter how large the y-intercept is, the line will eventually produce negative y-values as \( x \) decreases. For example, \( y = 2x + 10 \) has a positive gradient and \( c = 10 \), but when \( x = -6 \), \( y = 2(-6) + 10 = -2 \), which is negative.

Similarly, \( y = x + 1 \) gives \( y = -1 \) when \( x = -2 \). Because straight lines extend infinitely, a positive gradient guarantees that y-values eventually become negative for sufficiently negative x-values. Students fall for the “positive gradient means everything is positive” misconception, imagining lines only in the first quadrant.

The defining property of a straight line is that its gradient is constant everywhere. If you pick any two points on the line \( y = 3x + 1 \), the gradient between them will always be 3. For example: between \( (0, 1) \) and \( (1, 4) \), gradient \( = \frac{4 \; – \; 1}{1 \; – \; 0} = 3 \). Between \( (5, 16) \) and \( (10, 31) \), gradient \( = \frac{31 \; – \; 16}{10 \; – \; 5} = 3 \).

This is what distinguishes straight lines from curves. On a curve, the gradient changes at every point — but on a straight line, it never does. Students who have started learning about curves sometimes apply the idea of “changing gradient” back to straight lines, or they confuse the fact that y-values change with the idea that the gradient changes.

To find the gradient, the equation must be rearranged into the explicit form \( y = mx + c \). If we rearrange the standard form \( ax + by = c \), we first subtract \( ax \) to get \( by = -ax + c \), and then divide both sides by \( b \) to get \( y = -\frac{a}{b}x + \frac{c}{b} \).

The gradient is therefore always \( -\frac{a}{b} \), not \( a \). For example, in the line \( 2x + 3y = 12 \), the gradient is \( -\frac{2}{3} \), not 2. This targets students who think the number attached to \( x \) is always the gradient, regardless of how the equation is formatted.

The student has used the “rise over run inversion” misconception — they calculated \( \frac{\Delta x}{\Delta y} \) instead of \( \frac{\Delta y}{\Delta x} \). Gradient is always the change in \( y \) divided by the change in \( x \): \( \frac{8 \; – \; 2}{3 \; – \; 1} = \frac{6}{2} = 3 \).

A useful way to remember: gradient measures how much the line goes UP (y-direction) for each step ACROSS (x-direction). The “rise” goes on top, the “run” goes on the bottom. The student’s answer of \( \frac{1}{3} \) describes a very gentle slope, but looking at the points, the line rises 6 units in only 2 units across — that’s a steep line with gradient 3.

The student gets the correct answer of 5, but their method — “gradient = y-intercept ÷ x-intercept” — is not a valid general rule. It happens to work here by coincidence. The x-intercept of \( y = 5x + 10 \) is found by setting \( y = 0 \): \( 0 = 5x + 10 \), so \( x = -2 \) (not 2). Even the student’s x-intercept value is wrong, but the errors cancel out.

To see why this method fails in general, try \( y = 3x + 6 \): the y-intercept is 6, the x-intercept is \( -2 \), and \( 6 \div (-2) = -3 \), which is not the gradient of 3. The correct method is simply to read the coefficient of \( x \) once the equation is in \( y = mx + c \) form. The gradient here is 5 because \( m = 5 \).

The student has fallen for the “ignoring the sign of c” misconception. They correctly identify that the constant term gives the y-intercept, but they ignore the subtraction sign. The equation \( y = 3x \; – \; 7 \) is the same as \( y = 3x + (-7) \), so \( c = -7 \) and the y-intercept is \( (0, -7) \), not \( (0, 7) \).

Substituting \( x = 0 \) confirms: \( y = 3(0) \; – \; 7 = -7 \). The minus sign is part of the value of \( c \). Students often read “\( – \; 7 \)” as “the operation of subtraction” followed by “the number 7” rather than recognising that \( c \) is the signed value \( -7 \). This line crosses the y-axis below the origin, not above it.

The student has made the “y-intercept means y-coordinate goes first” misconception — they’ve confused the order of coordinates. In any coordinate pair, \( x \) always comes first: \( (x, y) \). The y-axis is where \( x = 0 \), so the y-intercept is the point \( (0, 6) \), not \( (6, 0) \).

The point \( (6, 0) \) is actually on the x-axis (where \( y = 0 \)). Substituting into the equation: \( y = 4(6) + 6 = 30 \), which is not 0, confirming that \( (6, 0) \) is not even on this line. The correct answer \( (0, 6) \) can be verified: \( y = 4(0) + 6 = 6 \). ✓

The student calculated the gradient correctly, but they grabbed the y-coordinate from the second point to use as the y-intercept (\( c \)). A y-coordinate is only the y-intercept when its corresponding x-coordinate is 0.

The first point \( (0, 4) \) lies on the y-axis, which explicitly tells them the y-intercept is 4. The correct equation is \( y = 3x + 4 \). You can check this by substituting the second point: if \( x = 2 \), then \( y = 3(2) + 4 = 10 \), which matches perfectly. The student’s flawed equation gives \( y = 3(2) + 10 = 16 \).