AQA GCSE Mathematics 8300: Graphs – Revision Notes

These notes condense section 3.2.2, Graphs (A8 to A16), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Basic and additional foundation content is for both tiers; anything from the “Higher content only” column is marked Higher tier only. Graphs can come up on Paper 1 (non-calculator) or on Papers 2 and 3 (calculator).

For full explanations and longer worked examples, use the Graphs study guide. Then try the Graphs practice questions. The Algebra revision notes cover the rest of Topic 2. Course hub: AQA GCSE Mathematics. Statement list: printable checklist. Not sure where to start? Take one of the free 10-minute diagnostics.

Key formulas and facts

Idea	Formula or fact	Tier
Gradient	m = (y₂ − y₁) ÷ (x₂ − x₁)	Both
Straight line	y = mx + c (m gradient, c y-intercept)	Both
Parallel lines	Same gradient	Both
Perpendicular lines	m₁ × m₂ = −1, so m₂ = −1/m₁	Higher tier only
Quadratic line of symmetry	Halfway between the roots	Both
Completed square	y = (x + p)² + q has turning point (−p, q)	Higher tier only
Circle	x² + y² = r², centre (0, 0), radius r	Higher tier only
Tangent to a circle	Perpendicular to the radius at that point	Higher tier only
Distance-time graph	Gradient = speed	Both
Velocity-time graph	Gradient = acceleration; area = distance	Higher tier only (A15)

A8 and A9: coordinates and straight lines

• (x, y): across, then up. Negative x is left of the y-axis; negative y is below the x-axis.
• Plot from a table of values. A point out of line means an arithmetic slip.
• Method: line through two points
  1. m = (difference in y) ÷ (difference in x), same order top and bottom.
  2. Substitute one point into y = mx + c to find c.
  3. Write y = mx + c and check with the other point.
• Method: line through one point with a given gradient: steps 2 and 3 only.
• Perpendicular (Higher tier only): flip the gradient and change its sign. 4 → −1/4; −3/5 → 5/3.

A10: reading and interpreting y = mx + c

• Rearrange to “y = …” first. 3x + 2y = 12 → 2y = −3x + 12 → y = −1.5x + 6.
• From a graph: use two well-separated grid points; check the sign by eye (downhill means negative).
• In context: gradient = rate (per hour, per km, per item); intercept = starting value (fixed charge, initial amount).

A11: quadratic graphs

• Roots: solve ax² + bx + c = 0 (factorising is the both-tier method).
• y-intercept: (0, c).
• Turning point x-value: midway between the roots; substitute to get y.
• Higher tier only: complete the square. x² + 10x + 18 = (x + 5)² − 25 + 18 = (x + 5)² − 7, turning point (−5, −7).
• a > 0 gives a minimum (U shape); a < 0 gives a maximum (∩ shape).

A12: recognise the shape

Equation	Recognise it by
y = mx + c	Straight line
y = ax² + bx + c	Parabola, one turning point
y = x³	S-shape through the origin, flat at (0, 0)
y = 1/x (x ≠ 0)	Two branches, first and third quadrants, never touches the axes
y = kˣ, k > 0 (Higher tier only)	Through (0, 1), always above the x-axis
y = sin x (Higher tier only)	Wave from 0, max 1 at 90°, period 360°
y = cos x (Higher tier only)	Wave from 1, min −1 at 180°, period 360°
y = tan x (Higher tier only)	Repeats every 180°, no value at 90°, 270°, …

A13: transformations (Higher tier only)

Graph	Effect on point (a, b)
y = f(x) + k	(a, b + k)
y = f(x + k)	(a − k, b)
y = −f(x)	(a, −b)
y = f(−x)	(−a, b)

A14 and A15: real contexts, gradients and areas

• Graphical solution (A14): draw the extra line y = (the other side) and read the x-values at the intersections.
• Horizontal section on a distance-time graph = stationary.
• Higher tier only (A15):
  1. Gradient at a point: draw a tangent that touches the curve once; use two far-apart points on it.
  2. Area under a curve: equal-width trapezia, area = (h/2) × [first + 2(middle values) + last].
  3. Say whether it is an over- or underestimate: if the curve bends down, trapezia sit below it (underestimate).
  4. Give units: m/s² for a velocity-time gradient, m for its area.

A16: circles (Higher tier only)

• Radius from x² + y² = r²: square-root the number. x² + y² = 49 has radius 7.
• Tangent at (a, b): radius gradient b/a; tangent gradient −a/b; then find c through (a, b).

Small worked reminders

These use different numbers from the study guide, so you can check the method quickly.

Line with a given gradient through a point. Gradient 2 through (−3, 4):

4 = 2(-3) + c  ->  c = 10     so  y = 2x + 10

Graphical solution. Using the curve y = x² − 6x + 5, solve x² − 6x + 5 = 5:

Draw y = 5. Intersections at x = 0 and x = 6.
Check: x^2 - 6x = 0  ->  x(x - 6) = 0

Trapezium estimate (Higher tier only). Velocities 2, 8 and 11 m/s at t = 0, 3 and 6 s:

Area = (3/2) x [2 + 2(8) + 11] = 1.5 x 29 = 43.5 m

Circle tangent (Higher tier only). x² + y² = 10 at (1, 3):

Radius gradient 3/1 = 3, so tangent gradient -1/3
3 = (-1/3)(1) + c  ->  c = 10/3      so  x + 3y = 10

Trig graph symmetry (Higher tier only). cos x = 0.5 at x = 60°. The cosine graph is symmetrical about x = 180°, so the second solution in 0° to 360° is 360° − 60° = 300°.

Must-know distinctions

• Root vs y-intercept: roots have y = 0; the y-intercept has x = 0.
• Parallel vs perpendicular: same gradient vs negative reciprocal.
• f(x) + a vs f(x + a): the first moves up or down; the second moves left or right, opposite to the sign.
• Chord vs tangent: a chord joins two points on a curve (average rate); a tangent touches at one point (rate at that instant).
• Reciprocal vs exponential: y = 1/x never touches either axis; y = kˣ crosses the y-axis at 1.

Quick self-test

1. Find the gradient of the line through (2, 5) and (6, 17).
2. Find the equation of the line in question 1.
3. Write down the gradient and y-intercept of 3x + 2y = 12.
4. (Higher) Write down the gradient of a line perpendicular to y = 4x − 7.
5. For y = x² − 6x + 5, find the roots and the turning point.
6. (Higher) Write x² + 10x + 18 in the form (x + p)² + q and state the turning point.
7. Name the type of graph: (a) y = 1/x (b) y = x³.
8. (Higher) (3, 1) lies on y = f(x). Write down its image on (a) y = f(x + 2) (b) y = −f(x).
9. (Higher) Write down the radius of x² + y² = 49.
10. (Higher) Write down the coordinates where y = 2ˣ crosses the y-axis, and the value of y when x = 3.
11. A distance-time graph has a straight section from (10 min, 4 km) to (30 min, 12 km). Find the speed in km/h.
12. (Higher) A velocity-time graph is a straight line from (0 s, 0 m/s) to (10 s, 20 m/s). Find the acceleration and the distance travelled.

Answers

1. (17 − 5) ÷ (6 − 2) = 12 ÷ 4 = 3.
2. 5 = 3(2) + c, so c = −1: y = 3x − 1.
3. y = −1.5x + 6: gradient −3/2, y-intercept 6.
4. −1/4.
5. (x − 1)(x − 5) = 0: roots x = 1 and x = 5. Line of symmetry x = 3; y = 9 − 18 + 5 = −4: turning point (3, −4), a minimum.
6. (x + 5)² − 7, turning point (−5, −7).
7. (a) Reciprocal (b) Cubic.
8. (a) (1, 1) (b) (3, −1).
9. 7.
10. (0, 1); y = 2³ = 8.
11. 8 km in 20 minutes = 8 × 3 = 24 km/h.
12. Acceleration = 20 ÷ 10 = 2 m/s²; distance = ½ × 10 × 20 = 100 m.

Where marks are usually lost

• Reading m from an equation such as 3x + 2y = 12 without rearranging first.
• Subtracting coordinates in a different order on the top and bottom of the gradient fraction.
• Writing a perpendicular gradient as the reciprocal with the same sign.
• Plotting a correct table but joining quadratic points with a ruler.
• Stating a turning point from (x + p)² + q as (p, q) instead of (−p, q).
• Translating y = f(x − 2) left instead of right.
• On a reciprocal sketch, letting the curve touch or cross an axis.
• Forgetting units, or giving speed in km per minute when km/h is asked.
• For an area estimate, missing the “under or over” comment, or using unequal strip widths.
• For a circle tangent, using the radius gradient itself instead of its negative reciprocal.

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).