AQA GCSE Mathematics 8300: Graphs – Study Guide

This study guide teaches section 3.2.2, Graphs (specification references A8 to A16), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Graphs can be tested on any of the three papers: Paper 1 is non-calculator, and Papers 2 and 3 allow a calculator. Basic and additional foundation content is for both tiers. Content in the specification’s “Higher content only” column is labelled Higher tier only below.

Use it with the Graphs revision notes and the Graphs practice questions. For the whole of Topic 2, see the Algebra study guide. The course hub is AQA GCSE Mathematics, the printable checklist lists every statement, and the free 10-minute diagnostics show where to start.

What this section covers

Ref	What you must be able to do	Tier
A8	Work with coordinates in all four quadrants	Both
A9	Plot straight-line graphs; use y = mx + c for parallel lines; find a line through two points, or through one point with a given gradient	Both
A9	Use y = mx + c to identify perpendicular lines	Higher tier only
A10	Identify and interpret gradients and intercepts of linear functions, graphically and algebraically	Both
A11	Identify roots, intercepts and turning points of quadratics graphically; deduce roots algebraically	Both
A11	Deduce turning points by completing the square	Higher tier only
A12	Recognise, sketch and interpret linear, quadratic, simple cubic and reciprocal (y = 1/x, x ≠ 0) graphs	Both
A12	Exponential graphs y = kˣ (k > 0); y = sin x, y = cos x, y = tan x for any angle in degrees	Higher tier only
A13	Sketch translations and reflections of a given function	Higher tier only
A14	Plot and interpret graphs, including non-standard functions in real contexts and reciprocal graphs, to find approximate solutions (for example simple kinematics)	Both
A14	The same with exponential graphs	Higher tier only
A15	Calculate or estimate gradients of graphs and areas under graphs, and interpret them	Higher tier only
A16	Equation of a circle centred at the origin; tangent to a circle at a given point	Higher tier only

Coordinates and plotting straight lines (A8, A9)

A point (x, y) is x across and y up. The axes split the plane into four quadrants, so either coordinate can be negative: (−3, 2) is left and up; (2, −3) is right and down.

To plot a straight line, make a table of values, plot the points and join them with a ruler. Three points are enough, but use the full x-range the question gives.

Worked example. Plot y = 3 − 2x for −2 ≤ x ≤ 3.

x  | -2  -1   0   1   2   3
y  |  7   5   3   1  -1  -3

Each step of 1 in x lowers y by 2, so the gradient is −2. The line crosses the y-axis at (0, 3). If one point does not line up with the others, recalculate it before drawing.

Gradient, intercept and y = mx + c (A9, A10)

In y = mx + c, m is the gradient and c is the y-intercept. The equation must start “y =” before you read them off.

Gradient = (change in y) ÷ (change in x). A line sloping down from left to right has a negative gradient. From a graph, pick two points far apart where the line crosses grid corners, then divide.

Parallel lines have the same gradient. Perpendicular lines (Higher tier only) have gradients that multiply to −1, so the perpendicular gradient is the negative reciprocal: 3 becomes −1/3, and −2/5 becomes 5/2.

Worked example (line through two points). Find the equation of the line through (−1, 7) and (3, −1).

m = (-1 - 7) / (3 - (-1)) = -8 / 4 = -2
Substitute (-1, 7) into y = -2x + c:  7 = 2 + c,  so c = 5
Line: y = -2x + 5
Check with (3, -1): -2(3) + 5 = -1   (correct)

Worked example (parallel, then perpendicular). Find the line through (4, 1) that is (a) parallel to y = −2x + 5, (b) perpendicular to it.

(a) m = -2.    1 = -2(4) + c  ->  c = 9.        y = -2x + 9
(b) m = 1/2.   1 = (1/2)(4) + c -> c = -1.      y = (1/2)x - 1   (Higher tier only)

Interpreting gradient and intercept. In context, the gradient is a rate and the intercept is the starting value. A taxi fare is C = 2.5d + 3, where C is the cost in pounds and d is the distance in km. The gradient 2.5 means each extra kilometre costs £2.50; the intercept 3 is a fixed £3 charge before the taxi moves. An 8 km trip costs 2.5 × 8 + 3 = £23. This links to R14 (gradient as a rate of change) in the ratio, proportion and rates of change revision notes.

Quadratic graphs: roots, intercepts, turning points (A11)

The graph of y = ax² + bx + c is a parabola: U-shaped when a > 0 (a minimum), ∩-shaped when a < 0 (a maximum).

• Roots are where the curve crosses the x-axis (y = 0).
• The y-intercept is (0, c).
• The curve is symmetrical about a vertical line through the turning point. The line of symmetry lies halfway between the roots.

Worked example. For y = x² − 2x − 8, find the roots, the y-intercept and the turning point.

Roots:      x^2 - 2x - 8 = 0  ->  (x - 4)(x + 2) = 0  ->  x = 4 or x = -2
Intercept:  x = 0 gives y = -8, so (0, -8)
Symmetry:   halfway between -2 and 4 is x = 1
Turning pt: y = 1 - 2 - 8 = -9, so (1, -9), a minimum

Completing the square (Higher tier only). Write x² + bx + c as (x + b/2)² + k. The turning point is (−b/2, k). This works even when the roots are not whole numbers.

Worked example. Find the turning point of y = x² − 8x + 13.

x^2 - 8x + 13 = (x - 4)^2 - 16 + 13 = (x - 4)^2 - 3
(x - 4)^2 is never negative, so the least value of y is -3 when x = 4
Turning point: (4, -3), a minimum

Recognising and sketching graphs (A12)

Function	Shape and key features	Tier
y = mx + c	Straight line	Both
y = ax² + bx + c	Parabola, one turning point, symmetrical	Both
y = x³ (simple cubic)	Through the origin; rises from bottom left to top right and flattens at (0, 0)	Both
y = 1/x, x ≠ 0	Two separate branches in the first and third quadrants; never touches either axis	Both
y = kˣ, k > 0	Passes through (0, 1); above the x-axis everywhere; rises if k > 1, falls if 0 < k < 1	Higher tier only
y = sin x	Wave between −1 and 1; repeats every 360°; sin 0° = 0, sin 90° = 1	Higher tier only
y = cos x	Same wave shifted; cos 0° = 1, cos 180° = −1	Higher tier only
y = tan x	Repeats every 180°; undefined at 90°, 270°, …; tan 45° = 1	Higher tier only

A sketch must show the correct shape and label key points: intercepts, turning points and, for reciprocal graphs, that the curve approaches but does not meet the axes. For exponential graphs, compare values: at x = 2, 2ˣ = 4 but 3ˣ = 9, so the graph with the larger k is steeper for positive x. For trig graphs, use symmetry: if sin x = 0.5 at 30°, it is also 0.5 at 180° − 30° = 150°. The exact values come from G21.

Translations and reflections (A13, Higher tier only)

For a given graph y = f(x):

New graph	Transformation
y = f(x) + a	Translation by vector (0, a): up a
y = f(x + a)	Translation by vector (−a, 0): left a
y = −f(x)	Reflection in the x-axis
y = f(−x)	Reflection in the y-axis

The change inside the bracket moves the graph the opposite way to the sign.

Worked example. The point (2, 4) lies on y = f(x). Find its image on each graph.

y = f(x - 3):  right 3        -> (5, 4)
y = f(x) + 2:  up 2           -> (2, 6)
y = -f(x):     y changes sign -> (2, -4)
y = f(-x):     x changes sign -> (-2, 4)

Graphs in real contexts and graphical solutions (A14)

Distance-time graphs: the gradient is the speed, and a horizontal section means the object is stationary. A runner whose graph is a straight line from (0, 0) to (40 s, 200 m) runs at 200 ÷ 40 = 5 m/s. Real-context graphs can also be non-standard curves (water depth in a filling vase, temperature over a day) and, on both tiers, reciprocal graphs such as time against speed for a fixed journey. Exponential graphs in contexts such as growth are Higher tier only.

Graphical solutions. To solve an equation from a graph you already have, rearrange so one side is the plotted function, then draw the other side as a line.

Worked example. Using the graph of y = x² − 2x − 8, solve x² − 2x − 8 = −5.

Draw the horizontal line y = -5.
It meets the curve where x = -1 and x = 3.
Check: (-1)^2 - 2(-1) - 8 = -5 and 3^2 - 6 - 8 = -5

Graph readings are approximate, so give answers to the accuracy of the grid.

Gradients and areas of graphs (A15, Higher tier only)

• Gradient of a curve at a point: draw a tangent there and find its gradient. This is the instantaneous rate of change (R15).
• Area under a graph: split it into trapezia (or triangles and rectangles) of equal width and add them.
• On a velocity-time graph, the gradient is acceleration and the area is distance travelled. On a distance-time graph, the gradient is speed. In financial graphs, the gradient is a rate such as pounds per month.

Worked example. A cyclist’s velocity v m/s after t seconds is shown by a smooth curve through these points:

t (s)   | 0   2   4   6   8
v (m/s) | 0   7  12  15  16

(a) Estimate the distance travelled in 8 seconds using four strips.

Area = (2/2) x [0 + 2(7 + 12 + 15) + 16] = 1 x 84 = 84 m

The curve bends downwards, so each trapezium sits below the curve: 84 m is an underestimate.

(b) A tangent drawn at t = 4 passes through (0, 4) and (8, 20). Estimate the acceleration at t = 4.

Gradient = (20 - 4) / (8 - 0) = 16 / 8 = 2 m/s^2

Circles and tangents (A16, Higher tier only)

x² + y² = r² is a circle with centre (0, 0) and radius r. So x² + y² = 20 has radius √20 = 2√5.

A tangent meets the circle at one point and is perpendicular to the radius there.

Worked example. Find the equation of the tangent to x² + y² = 20 at (2, −4).

Check the point: 2^2 + (-4)^2 = 4 + 16 = 20   (on the circle)
Radius gradient: (-4 - 0) / (2 - 0) = -2
Tangent gradient: 1/2  (negative reciprocal)
-4 = (1/2)(2) + c  ->  c = -5
Tangent: y = (1/2)x - 5   (or x - 2y = 10)

Common errors

• Reading the gradient from an equation not yet in the form y = mx + c, such as 2x + 3y = 6 (the gradient is −2/3, not 2).
• Dividing change in x by change in y, or losing a minus sign when one coordinate is negative.
• Using the reciprocal without changing the sign for a perpendicular gradient.
• Joining plotted quadratic points with straight segments, or giving a pointed “V” at the turning point.
• Giving only one root when the curve crosses the x-axis twice.
• Moving y = f(x + 3) to the right instead of the left.
• Drawing the tangent to a curve as a chord that cuts the curve in two places.
• Counting grid squares for an area without converting to the units on the axes.

Where to go next

Fix the facts with the Graphs revision notes, then test yourself with the Graphs practice questions. For other parts of Topic 2, use the Algebra revision notes and Algebra practice questions.

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.2 Graphs (A8 to A16).