What are the key features of a linear graph in AQA GCSE Mathematics?

In AQA GCSE Mathematics, a linear graph represents a straight line on a coordinate plane. Key features include the slope (or gradient), which indicates the steepness and direction of the line, and the y-intercept, which is the point where the line crosses the y-axis. The equation of a linear graph is typically in the form y = mx + c, where m is the gradient and c is the y-intercept.

How do you plot a quadratic graph for GCSE Algebra?

To plot a quadratic graph in GCSE Algebra, you start with a quadratic equation in the form y = ax^2 + bx + c. First, identify the vertex, which is the highest or lowest point of the graph. Then, find the axis of symmetry, which is a vertical line that passes through the vertex. Calculate a few points on either side of the vertex by substituting x-values into the equation. Finally, plot these points on a coordinate grid and draw a smooth curve through them to form a parabola.

What is the difference between a linear and a quadratic graph in Mathematics?

In Mathematics, a linear graph is a straight line, represented by the equation y = mx + c, where m is the gradient and c is the y-intercept. It has a constant rate of change. A quadratic graph, on the other hand, is a curve called a parabola, represented by the equation y = ax^2 + bx + c. It has a variable rate of change, with a vertex that is either a maximum or minimum point. The shape of the parabola depends on the coefficient 'a'.

How can you determine the gradient of a line from its graph?

To determine the gradient of a line from its graph, you need to identify two points on the line. The gradient is calculated by taking the difference in the y-coordinates (vertical change) and dividing it by the difference in the x-coordinates (horizontal change) between these two points. This is often expressed as 'rise over run'. A positive gradient indicates an upward slope, while a negative gradient indicates a downward slope.

What is the significance of the y-intercept in a graph?

The y-intercept of a graph is the point where the line or curve crosses the y-axis. It is significant because it represents the value of y when x is zero. In the context of a linear equation y = mx + c, the y-intercept is the constant 'c'. It provides a starting point for plotting the graph and is crucial for understanding the relationship between variables in real-world contexts, such as initial conditions in a problem.

Straight line graph y = 2x + 1

At a glance

Every straight line has equation $y = mx + c$ ; $m$ is the gradient and $c$ is the $y$ -intercept
Gradient $= \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}$
A quadratic graph is a parabola; its roots are where it crosses the $x$ -axis
Know the shapes of cubic, reciprocal ( $y = \frac{1}{x}$ ), exponential ( $y = k^x$ ) and trig graphs
Graph transformations: $f(x) + a$ shifts up; $f(x + a)$ shifts left; $-f(x)$ reflects in $x$ -axis; $f(-x)$ reflects in $y$ -axis
Gradient of a curve at a point = gradient of the tangent drawn at that point
The equation of a circle centred at the origin is $x^2 + y^2 = r^2$ (Higher only)

What you should be able to do

Work with coordinates in all four quadrants and plot straight-line graphs (A8, A9)
Find and interpret the gradient and intercepts of a linear function graphically and algebraically (A10)
Identify and interpret roots, intercepts and turning points of quadratic functions; deduce roots algebraically (A11)
Recognise, sketch and interpret linear, quadratic, cubic, reciprocal, exponential and trigonometric graphs (A12)
Sketch translations and reflections of a given graph using function notation (A13)
Plot and interpret graphs in real contexts including distance-time and velocity-time graphs (A14, A15)
Estimate gradients and areas under non-linear graphs and interpret the results (A15)
Recognise and use the equation of a circle $x^2 + y^2 = r^2$ ; find the equation of a tangent at a given point (A16)

Study notes

Straight-Line Graphs

Plotting, gradient, intercept and the equation y = mx + c — AQA spec A8, A9, A10

Every straight line can be written in the form:

$y = mx + c$

where $m$ is the gradient (steepness) and $c$ is the $y$ -intercept (where the line crosses the $y$ -axis).

Calculating gradient:

$m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$

A positive gradient slopes upward left-to-right; a negative gradient slopes downward.

Finding the equation from two points:

Through $(1, 3)$ and $(3, 7)$ :

$m = \frac{7 - 3}{3 - 1} = \frac{4}{2} = 2$

Substitute into $y = mx + c$ : $3 = 2(1) + c \implies c = 1$ , so $y = 2x + 1$ .

Parallel and perpendicular lines:

Parallel lines have equal gradients.
Perpendicular lines have gradients that are negative reciprocals: $m_1 \times m_2 = -1$ .

Straight line y = 2x + 1 with gradient triangle: a run of 1 produces a rise of 2.

Rearrange to $y = mx + c$ first — never read off $m$ or $c$ from other forms
A horizontal line has $m = 0$ , equation $y = k$ ; a vertical line has undefined gradient, equation $x = k$
Perpendicular to $y = 3x + 1$ has gradient $-\frac{1}{3}$

Common pitfall

When picking two points to calculate gradient, make sure they are clearly on the line and as far apart as possible to minimise reading errors. Always divide rise by run, not run by rise.

Quadratic Graphs

Parabolas, roots, turning points and the quadratic formula — AQA spec A11, A12

The graph of $y = ax^2 + bx + c$ is a parabola.

If $a > 0$ : U-shaped (minimum turning point)
If $a < 0$ : ∩-shaped (maximum turning point)

Key features:

Roots (x-intercepts): where $y = 0$ ; found by factorising or the quadratic formula
$y$ -intercept: set $x = 0$ ; gives $(0, c)$
Turning point (vertex): at $x = -\dfrac{b}{2a}$ ; substitute back for $y$
Line of symmetry: $x = -\dfrac{b}{2a}$

Quadratic formula (when factorising is not straightforward):

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Discriminant $= b^2 - 4ac$ :

$> 0$ : two distinct real roots (crosses $x$ -axis twice)
$= 0$ : one repeated root (touches $x$ -axis)
$< 0$ : no real roots (does not cross $x$ -axis)

Parabola y = x² − 4x + 3 with roots at x = 1 and x = 3, y-intercept (0, 3), turning point (2, −1) and line of symmetry x = 2.

The turning point can also be found by completing the square: $y = a(x+p)^2 + q$ gives vertex $(-p, q)$
A parabola is symmetric about its turning point — roots are equidistant from the line of symmetry
If asked to 'sketch', mark roots, $y$ -intercept and turning point with coordinates

Common pitfall

Do not confuse the $y$ -intercept (set $x = 0$ ) with the roots (set $y = 0$ ). They are different features and tested separately on the AQA paper.

Other Graph Types

Cubic, reciprocal, exponential and trigonometric — AQA spec A12

You must be able to recognise and sketch each of these graph families.

Cubic $y = ax^3 + \ldots$ :

Positive $a$ : rises left-to-right with an S-shape
Has up to 3 roots and up to 2 turning points

Reciprocal $y = \dfrac{k}{x}$ ( $x \neq 0$ ):

Two branches in opposite quadrants (for $k > 0$ : Q1 and Q3)
Asymptotes along both axes — the curve never touches them

Exponential $y = k^x$ (for $k > 1$ ):

Always positive; crosses the $y$ -axis at $(0, 1)$
Increases rapidly; never touches the $x$ -axis (asymptote at $y = 0$ )
For $0 < k < 1$ : decreasing exponential (decay)

Trigonometric graphs (arguments in degrees):

$y = \sin x$ : period 360°, amplitude 1, passes through $(0,0)$
$y = \cos x$ : period 360°, amplitude 1, passes through $(0,1)$
$y = \tan x$ : period 180°, asymptotes at $x = 90°, 270°, \ldots$

Exponential and reciprocal graphs have asymptotes — mark these on sketches
$\sin 0° = 0$ , $\cos 0° = 1$ , $\tan 90°$ is undefined — key anchor values
A cubic with positive leading coefficient goes from bottom-left to top-right

Common pitfall

The reciprocal graph $y = \frac{1}{x}$ never crosses the axes — there are asymptotes at $x = 0$ and $y = 0$ . Drawing it touching either axis will lose marks on a sketch question.

Graph Transformations

Translations and reflections using function notation — AQA spec A13

Transformations describe how a graph moves or flips when the function is modified.

Transformation	Effect on graph
$y = f(x) + a$	Translate by $\begin{pmatrix} 0 \\ a \end{pmatrix}$ (up by $a$ )
$y = f(x) - a$	Translate by $\begin{pmatrix} 0 \\ -a \end{pmatrix}$ (down by $a$ )
$y = f(x + a)$	Translate by $\begin{pmatrix} -a \\ 0 \end{pmatrix}$ (left by $a$ )
$y = f(x - a)$	Translate by $\begin{pmatrix} a \\ 0 \end{pmatrix}$ (right by $a$ )
$y = -f(x)$	Reflect in the $x$ -axis
$y = f(-x)$	Reflect in the $y$ -axis

Example: The graph of $y = \sin x$ is transformed to $y = \sin(x - 30°) + 2$ .

This is a translation of $\begin{pmatrix} 30 \\ 2 \end{pmatrix}$ — 30° to the right and 2 units up.

Horizontal changes inside $f(\ldots)$ are counter-intuitive: $f(x + a)$ moves left
Applying transformations to key points (roots, turning point) is a reliable method
$y = f(2x)$ is a horizontal stretch by factor $\frac{1}{2}$ — outside scope for most GCSE questions but possible at Higher

Common pitfall

$f(x + 3)$ shifts the graph left by 3, not right. The sign inside the bracket is opposite to the direction of shift — this trips up most students.

Real-Life Graphs and Rates of Change

Distance-time, velocity-time, and estimating gradients and areas — AQA spec A14, A15

Graphs are used to represent real-world relationships. You must be able to read off values, find rates of change, and interpret the meaning of gradient and area.

Distance-time graphs:

Gradient = speed
Horizontal line = stationary
Steeper line = faster speed

Velocity-time graphs:

Gradient = acceleration (negative gradient = deceleration)
Area under graph = distance travelled

Estimating gradient of a curve (Higher):

Draw a tangent at the point of interest (a straight line touching the curve at that point without crossing it). Calculate the gradient of the tangent using two points on it.

Estimating area under a curve (Higher):

Use the trapezium rule: divide the area into trapeziums of equal width $h$ :

$A \approx \frac{h}{2}\left[y_0 + 2(y_1 + y_2 + \ldots + y_{n-1}) + y_n\right]$

Or count squares on graph paper as an estimate.

On a distance-time graph, a negative gradient means travelling back toward the start
The area under a velocity-time graph gives displacement, not just distance, if the velocity is negative
When drawing a tangent, use a ruler and extend it to give two widely spaced points for accuracy

Common pitfall

Gradient of a curve is not constant — it changes at every point. You must draw a tangent at the specific point asked, not a chord between two points on the curve.

Circle Equations

Equation of a circle centred at the origin and tangent at a point — AQA spec A16

Equation of a circle with centre $(0, 0)$ and radius $r$ :

$x^2 + y^2 = r^2$

Example: $x^2 + y^2 = 25$ is a circle of radius 5 centred at the origin.

Finding the equation of a tangent at a point (Higher):

A tangent to a circle is perpendicular to the radius at the point of contact.

Method: For the circle $x^2 + y^2 = r^2$ and the point $P(a, b)$ on the circle:

The radius $OP$ has gradient $\dfrac{b}{a}$ .
The tangent is perpendicular, so its gradient is $-\dfrac{a}{b}$ .
Use $y - b = m(x - a)$ with $m = -\dfrac{a}{b}$ .

Example: Find the tangent to $x^2 + y^2 = 25$ at $(3, 4)$ :

Gradient of radius $= \dfrac{4}{3}$ ; tangent gradient $= -\dfrac{3}{4}$

$y - 4 = -\frac{3}{4}(x - 3) \implies 4y - 16 = -3x + 9 \implies 3x + 4y = 25$

Verify a point lies on the circle by substituting: $3^2 + 4^2 = 9 + 16 = 25$ ✓
The tangent equation can be written as $ax + by = r^2$ where $(a, b)$ is the point of contact
The circle $x^2 + y^2 = r^2$ always has centre $(0,0)$ — a circle centred elsewhere has equation $(x-h)^2 + (y-k)^2 = r^2$

Common pitfall

Do not confuse the radius with the diameter when reading off $r$ from the equation. $x^2 + y^2 = 36$ has $r = 6$ (not 36 or $\sqrt{36/2}$ ).

Quick recap

$y = mx + c$ : $m$ = gradient, $c$ = $y$ -intercept; gradient $= \dfrac{y_2 - y_1}{x_2 - x_1}$
Perpendicular gradients satisfy $m_1 \times m_2 = -1$
Quadratic parabola: roots (set $y=0$ ), $y$ -intercept (set $x=0$ ), turning point at $x = -\frac{b}{2a}$
Discriminant $b^2 - 4ac$ : positive → 2 roots; zero → 1 repeated root; negative → no real roots
Know the shape of cubic, reciprocal ( $y = \frac{1}{x}$ ), exponential ( $y = k^x$ ) and trig ( $\sin$ , $\cos$ , $\tan$ ) graphs
Transformations: $f(x)+a$ shifts up; $f(x+a)$ shifts left; $-f(x)$ reflects in $x$ -axis; $f(-x)$ reflects in $y$ -axis
Gradient of a curve = gradient of tangent at that point; area under velocity-time graph = distance
Circle centred at origin: $x^2 + y^2 = r^2$ ; tangent at $(a,b)$ has gradient $-\frac{a}{b}$

Exam tips

When sketching any graph, always label key points: roots, $y$ -intercept, turning point — coordinates, not just positions
For straight-line questions, rearrange to $y = mx + c$ before reading off the gradient
The quadratic formula is given on the AQA formula sheet — you do not need to memorise it, but you must apply it accurately
$f(x + a)$ shifts the graph left — this is the single most common transformation error; draw a table of values if unsure
When drawing a tangent to estimate gradient, extend it well beyond the point of contact so you can choose two widely spaced coordinates
For the circle tangent question, always verify the point lies on the circle first by substituting into $x^2 + y^2 = r^2$

Frequently asked questions

Sources: AQA GCSE Mathematics (8300) Specification (2026 onwards); AQA 8300 Examiners' Reports 2022–2024 · Last reviewed May 2026 · AQA GCSE · Mathematics 8300 Higher

Graphs

At a glance

What you should be able to do

Study notes

Straight-Line Graphs

Quadratic Graphs

Other Graph Types

Graph Transformations

Real-Life Graphs and Rates of Change

Circle Equations

Quick recap

Exam tips

Frequently asked questions

How do I find the gradient between two points?

What is the discriminant and why does it matter?

Why does f(x + 3) shift the graph left, not right?

How do I estimate the area under a curve from a graph?

What does the area under a velocity-time graph represent?

How do I find the equation of a tangent to a circle?