Use the gradient-intercept form y = mx + b: read the gradient m and the y-intercept b directly from the equation (rearranging first if needed), write the equation of a line from its graph, and sketch a line from m and b

What this dot point is asking

The gradient-intercept formula is the single most useful way to write a straight line. When a line's equation is in the form

the two numbers you most want are sitting right there in plain sight. $m$ is the gradient (the number multiplying $x$) and $b$ is the $y$-intercept (the constant on its own). NESA bundles three skills around this form, and a question can test any of them. First, read $m$ and $b$ straight from an equation. If the equation is given in a different shape, rearrange it into $y = mx + b$ first. Second, write the equation of a line from its graph: read $b$ off the vertical axis, work out $m$ from a rise/run step or two points, then assemble $y = mx + b$. Third, sketch a line quickly from $m$ and $b$: plot the intercept, step by the gradient to a second point, and rule the line. In a real context the same idea models a cost: $b$ is a fixed starting amount and $m$ is a rate. Interpreting the two numbers in words is part of the dot point too.

The deeper idea that ties it together is that $y = mx + b$ is a recipe for building the line: start at height $b$ on the $y$-axis, then climb (or fall) at the steady rate $m$ for every step to the right. That one sentence explains why $b$ is where the line begins and why $m$ controls how fast it rises. You earn the marks in three ways. Compare the equation to $y = mx + b$ correctly, including the sign of $m$ and reading a missing constant as $b = 0$. Rearrange carefully when the equation is not yet in that form. Lay out a sketch with the intercept marked and a gradient step shown. Marks are lost in predictable ways: reading $b$ as the coefficient of $x$, dropping the minus sign on a negative gradient, forgetting to divide the whole equation through when a coefficient is in front of $y$, and confusing the $y$-intercept with the $x$-intercept.

The answer

Reading the gradient and y-intercept from y = mx + b

When a line is written as $y = mx + b$, you can read both key numbers without any working:

• the gradient $m$ is the coefficient of $x$ (the number it is multiplied by), and
• the $y$-intercept $b$ is the constant term (the number standing alone, with no $x$).

So for $y = 3x - 2$ the gradient is $m = 3$ and the $y$-intercept is $b = -2$. The sign travels with the number: a $-2$ at the end means the line cuts the $y$-axis below the origin at $(0, -2)$, and a negative coefficient of $x$ means a line that slopes downwards. Two cases catch students out. If there is no constant, as in $y = 6x$, the $y$-intercept is $b = 0$: the line passes through the origin. If there is no $x$ term, as in $y = 4$, the gradient is $m = 0$: the line is horizontal at height $4$. Reading these correctly is just a matter of remembering that a missing term means that number is zero.

The order the terms are written in does not change what they mean, but it can disguise them. The equation $y = 8 - x$ has the constant first and the $x$ term second. Reorder it as $y = -x + 8$ and the form is obvious: the coefficient of $x$ is $-1$, so $m = -1$, and the constant is $8$, so $b = 8$. Whenever an equation looks unusual, rewrite it as "($x$ term) $+$ (constant)" before reading off $m$ and $b$.

Rearranging an equation into gradient-intercept form first

An equation only reveals $m$ and $b$ at a glance once it is in the form $y = mx + b$, with $y$ by itself on the left. If it is given any other way, rearrange it first using the same balancing moves you use to solve an equation: whatever you do to one side, do to the other. The goal is always to get $y$ alone.

Take $y - 3x = 4$. Add $3x$ to both sides to free $y$, giving $y = 3x + 4$, so $m = 3$ and $b = 4$. Take $3x + y = 12$. Subtract $3x$ from both sides to get $y = -3x + 12$, so $m = -3$ and $b = 12$. The step that needs the most care is when a number multiplies $y$. For $2y - 8x = 6$, first add $8x$ to both sides to get $2y = 8x + 6$, then divide every term by $2$:

so $m = 4$ and $b = 3$. The classic error is to divide only the $8x$ and forget the $6$ (or vice versa). The line is divided all the way through, so each term on the right must be divided by the same number.

Writing the equation of a line from its graph

Going the other way - from a drawn line to its equation - is the reverse of reading $m$ and $b$. You find the two numbers from the picture, then slot them into $y = mx + b$. There are just two things to read off:

1. Read $b$ from the $y$-axis. Find where the line crosses the vertical axis; that height is $b$. This needs no calculation.
2. Find $m$ from the line. Either count a rise/run triangle between two grid points, or pick two clear points and use $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.

Then write $y = mx + b$ with those values. The line below crosses the $y$-axis at $(0, 1)$, so $b = 1$. Stepping from that point, a run of $1$ across goes with a rise of $2$ up to land on the line again at $(1, 3)$, so $m = \tfrac{2}{1} = 2$. The equation is therefore $y = 2x + 1$. The diagram shows exactly the two things you read: the intercept $b$ on the $y$-axis and the gradient $m$ as a rise-over-run step.

If the line slopes down, the only change is that $m$ is negative. A line cutting the $y$-axis at $(0, 4)$ and dropping to $(3, -2)$ has $b = 4$ and gradient

so its equation is $y = -2x + 4$. Read the intercept, find the gradient with its sign, and write it down.

Sketching a line from m and b

The gradient-intercept form is also the fastest way to draw a line, because $b$ hands you one point for free and $m$ tells you how to reach the next one. A straight line needs only two points, so the method is short:

1. Plot the $y$-intercept $(0, b)$ on the vertical axis.
2. Write the gradient as $\dfrac{\text{rise}}{\text{run}}$. A whole number $m$ is $\dfrac{m}{1}$ (run $1$); a fraction like $\tfrac{1}{2}$ is already rise $1$ over run $2$.
3. Step to a second point. From the intercept, go run units across (to the right) and rise units up. If the gradient is negative, the rise is downwards.
4. Rule the line through the two points, extending it past them with arrows on each end.

The stages below sketch $y = 2x - 3$, where $m = 2$ and $b = -3$. Reading $m = 2$ as $\tfrac{2}{1}$, each step is $1$ across and $2$ up. Plot $(0, -3)$, step to $(1, -1)$, and the line is determined; stepping once more reaches $(2, 1)$ as a free check that the three points line up.

Stage 1, set up the axes

Draw a number plane with enough room for the intercept at $y = -3$ and the line as it climbs. Nothing is plotted yet; this is just the grid you will build on.

Stage 2, plot the $y$-intercept

Read $b = -3$ from the equation and mark the single point $(0, -3)$ on the vertical axis. This is the line's starting point and the one point the equation gives you instantly.

Stage 3, step out using the gradient

Read $m = 2 = \tfrac{2}{1}$. From $(0, -3)$ move a run of $1$ to the right, then a rise of $2$ upwards, landing on $(1, -1)$. That right-angled step is the gradient made visible.

Stage 4, rule the line

Join $(0, -3)$ and $(1, -1)$ with a straight edge and extend the line both ways. Stepping again to $(2, 1)$ shows all three points lie on it, confirming the sketch of $y = 2x - 3$.

A fractional gradient is sketched the same way, you just read the rise and run from the fraction. For $y = \tfrac{1}{2}x + 1$, the intercept is $(0, 1)$ and the gradient $\tfrac{1}{2}$ is a rise of $1$ for a run of $2$, so you step $2$ across and $1$ up to $(2, 2)$, then again to $(4, 3)$. Choosing whole-number steps lands you exactly on grid corners, which keeps the sketch tidy.

Building a real model: cost = fixed amount + rate

The reason $y = mx + b$ matters beyond the classroom is that it models any quantity that starts at a fixed amount and then changes at a steady rate. The most common version in Maths Standard is a cost: a fixed fee plus a charge per unit. The fixed fee is $b$ (the cost before anything happens) and the rate is $m$ (the cost added per unit). The letters are usually changed to fit the situation, but the structure is identical.

Suppose an electrician charges a $60 dollar call-out fee plus$40 dollars per hour. Let $C$ be the total cost in dollars and $h$ the number of hours. The model is

which is $y = mx + b$ with the gradient $m = 40$ and the $y$-intercept $b = 60$. Here $b = 60$ is the fixed call-out fee (the cost for $h = 0$ hours, before any work) and $m = 40$ is the rate of $40 per hour. To predict the cost of any job, substitute the hours: a $3$ hour job costs $C = 60 + 40 \times 3 = 180$ dollars. When a question asks you to "interpret" the gradient or intercept, this is what it wants in words: the intercept is the starting amount with its unit, and the gradient is the rate with its unit (here, dollars per hour). A bare number without the meaning loses the interpretation mark.

How exam questions ask about y = mx + b

The wording points you to which of the three skills to use:

• "State the gradient and the $y$-intercept of $y = \ldots$" means compare to $y = mx + b$ and read off the coefficient of $x$ and the constant, after rearranging if $y$ is not already by itself.
• "Write the equation in the form $y = mx + b$" (or "make $y$ the subject") means rearrange, dividing the whole equation through if a number multiplies $y$.
• "Find the equation of the line" with a graph means read $b$ off the $y$-axis, find $m$ from a step or two points, and write $y = mx + b$.
• "Sketch / draw the line $y = \ldots$" means plot $(0, b)$, step by the gradient to a second point, and rule the line.
• "What does the gradient (or $y$-intercept) represent?" in a context wants words: the gradient is the rate (with units) and the intercept is the fixed starting value (with its unit).
• "Write an equation to model this situation" means identify the fixed amount as $b$ and the per-unit rate as $m$, then write the model with letters that fit the context.

Why m and b are exactly what they are

It is worth seeing why the two letters in $y = mx + b$ land on the gradient and the $y$-intercept, because understanding it removes any need to memorise which is which. Put $x = 0$ into $y = mx + b$: the $x$ term vanishes and you are left with $y = b$. That is precisely the point on the $y$-axis, so $b$ must be the $y$-intercept. Now increase $x$ by $1$, from any value to the next: the term $mx$ grows by exactly $m$, so $y$ goes up by $m$ for every one-unit step across. That is the definition of the gradient, the rise per unit run, so $m$ must be the gradient. This also explains the two special cases at a glance: if $m = 0$ there is no $x$ term, $y$ stays at $b$ for every $x$, and the line is horizontal; and if $b = 0$ the line passes through $(0, 0)$, the origin. The formula is not an arbitrary label, it is the line's own building instructions: start at $b$, climb at rate $m$.