Gradients and Intercepts of Linear Functions Revision Notes | GCSE AQA Higher Maths

1. Why Gradients and Intercepts Matter

Every straight-line graph tells a story in two numbers: how steep it is, and where it starts. Once you can read those two numbers — from an equation or directly from a graph — you can describe, compare, and construct any linear function the examiner throws at you.

Think about a taxi fare: a fixed charge the moment you get in, plus a rate per kilometre. The fixed charge is the y-intercept; the rate per kilometre is the gradient. That structure — a starting value and a rate of change — appears everywhere in GCSE maths and beyond.

2. The Equation of a Straight Line

The standard form you need to know inside out is $y = mx + c$. Every term earns its place: $m$ controls the steepness and direction of the line, and $c$ pins down exactly where it crosses the y-axis.

Reading $m$ and $c$ from $y = mx + c$ is the most direct algebraic method. For example, in $y = 3x + 7$, the gradient is $3$ and the y-intercept is $7$ — no calculation needed. The challenge comes when the equation is not yet in this form, which we will tackle shortly.

The gradient formula is your tool whenever you are given two coordinates rather than an equation. The order of subtraction must be consistent: if you put $y_2$ on top, you must put $x_2$ on top too. Swapping just one of them flips the sign of your answer.

3. Reading Gradients and Intercepts from a Graph

When you are given a graph rather than an equation, you extract $m$ and $c$ by reading coordinates directly from the line. The y-intercept is straightforward: find where the line crosses the y-axis and read off that $y$-value. The gradient takes one extra step.

To find the gradient graphically, pick two points on the line that sit exactly on grid intersections — this avoids estimation errors. Then calculate the rise (vertical change) and the run (horizontal change) between them, and divide: $m = \frac{\text{rise}}{\text{run}}$.

The graph above shows two lines: $y = 2x + 1$ (steeper, positive slope) and $y = -x + 4$ (gentler, negative slope). Notice that $y = 2x + 1$ crosses the y-axis at $1$ and rises $2$ units for every $1$ unit across. The line $y = -x + 4$ crosses at $4$ and falls $1$ unit for every $1$ unit across — the negative gradient is visible as a downward slope from left to right.

Worked example: Reading Gradient and Intercept from a Graph

A straight line passes through the points $(0, 3)$ and $(4, 11)$ on a grid. Find the gradient and state the equation of the line.

1. The line passes through $(0, 3)$, so $c = 3$. — When $x = 0$, the point is on the y-axis by definition. Read the $y$-value directly. [M1]
2. $(x_1, y_1) = (0, 3)$ and $(x_2, y_2) = (4, 11)$ — Grid intersection points give exact values — no estimation needed.
3. Rise $= y_2 - y_1 = 11 - 3 = 8$ Run $= x_2 - x_1 = 4 - 0 = 4$ — Rise is the vertical change; run is the horizontal change. Keep the subtraction order consistent. [M1]
4. $m = \frac{8}{4} = 2$ — $8 \div 4 = 2$. The line rises $2$ units for every $1$ unit to the right. [A1]
5. $y = 2x + 3$ — Substitute $m = 2$ and $c = 3$ into $y = mx + c$. [A1]

Answer: Gradient $= 2$; y-intercept $= 3$; equation: $y = 2x + 3$

4. Rearranging Equations into $y = mx + c$

Exam questions frequently give you a linear equation that is not yet in $y = mx + c$ form — for instance, $3x + 2y = 10$ or $2y - 6x = 4$. You cannot read off $m$ and $c$ until you rearrange. The target is always to isolate $y$ on the left-hand side.

Worked example: Rearranging into $y = mx + c$

Find the gradient and y-intercept of the line $3x + 2y = 10$.

1. $2y = 10 - 3x$ — Move $3x$ to the right-hand side by subtracting $3x$ from both sides. [M1]
2. $y = \frac{10 - 3x}{2} = 5 - \frac{3}{2}x$ — Divide both sides by $2$ to isolate $y$. Each term on the right is divided separately: $\frac{10}{2} = 5$ and $\frac{3x}{2} = \frac{3}{2}x$. [M1]
3. $y = -\frac{3}{2}x + 5$ — Convention places the $x$-term first. Now $m$ and $c$ are visible directly. [A1]
4. $m = -\frac{3}{2}$, $\quad c = 5$ — Read directly from the rearranged equation. The negative gradient means the line slopes downward left to right. [A1]

Answer: Gradient $= -\frac{3}{2}$; y-intercept $= 5$

Notice that the gradient came out as a fraction — this is completely normal and you should leave it as $-\frac{3}{2}$ rather than converting to a decimal unless the question asks for a decimal. Fraction form is exact and preferred when no calculator is used.

5. Finding the Equation from Two Points or a Point and Gradient

So far you have read $m$ and $c$ from existing equations and graphs. The next skill runs in reverse: you are given information about a line and must construct its equation. There are two common scenarios — given the gradient and one point, or given two points.

In both cases, the strategy is the same: find $m$ first (either it is given, or you calculate it from the two points), then substitute a known point into $y = mx + c$ to find $c$. This two-stage approach keeps the method clean and avoids errors.

Worked example: Finding the Equation Given Two Points

Find the equation of the straight line passing through $(2, 5)$ and $(6, 13)$. Give your answer in the form $y = mx + c$.

1. $m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$ — $13 - 5 = 8$ and $6 - 2 = 4$, so $m = 8 \div 4 = 2$. [M1]
2. $5 = 2(2) + c$ — Use the point $(2, 5)$: substitute $x = 2$, $y = 5$, and $m = 2$. Either point works — choose whichever has simpler numbers. [M1]
3. $5 = 4 + c \Rightarrow c = 5 - 4 = 1$ — $2 \times 2 = 4$, then $c = 5 - 4 = 1$. [A1]
4. $y = 2x + 1$ — Substitute $m = 2$ and $c = 1$ into $y = mx + c$. [A1]
5. When $x = 6$: $y = 2(6) + 1 = 12 + 1 = 13$ ✓ — Substituting the second point confirms the equation is correct. This sanity check costs seconds and saves marks.

Answer: $y = 2x + 1$

6. Interpreting Gradients and Intercepts in Context

GCSE Higher exam questions often embed linear functions in real-world scenarios and ask you to interpret what $m$ and $c$ mean. This is where many students drop marks — they calculate correctly but then describe the numbers without connecting them to the context.

Here is a reliable structure for interpretation questions: state the value, state the units, and state what it represents in the situation. For a graph of cost ($£$) against number of items, a gradient of $£3$ means each additional item costs $£3$. The y-intercept of $£5$ means there is a fixed charge of $£5$ regardless of how many items are bought.

Worked example: Interpreting a Linear Model

A plumber charges according to the formula $C = 40t + 25$, where $C$ is the total charge in pounds and $t$ is the time in hours. Interpret the gradient and the intercept in context.

1. $m = 40$ — The coefficient of $t$ is $40$. This is the gradient of the linear function. [M1]
2. The plumber charges $£40$ per hour. — The gradient represents the rate of change: for each additional hour worked, the total charge increases by $£40$. [A1]
3. $c = 25$ — The constant term is $25$. This is the value of $C$ when $t = 0$. [M1]
4. There is a fixed call-out charge of $£25$. — The y-intercept represents the starting value — the charge before any time is counted. [A1]

Answer: Gradient: the plumber charges $£40$ per hour. Intercept: there is a fixed call-out charge of $£25$.

8. Exam Technique, Common Mistakes, and Mark-Winning Habits

The majority of marks lost on gradient and intercept questions come from a small set of predictable errors. Knowing what they are — and actively checking for them — is the difference between a confident answer and an avoidable slip.

On interpretation questions, examiners award a specific mark for connecting the number to the context. Writing '$m = 40$' alone will not earn that mark. Writing 'the gradient of $40$ means the cost increases by $£40$ for each additional hour' will. The extra sentence is the mark.

Earlier in this lesson, you saw how to read $m$ and $c$ from $y = mx + c$ in one glance. Now you can see why that only works once the equation is fully rearranged — the true-false question above is exactly the trap examiners set to test whether you know the difference.