HookEvery energy bill you have ever paid is the equation y = mx + c
When Ofgem announced in August 2022 that the energy price cap for a typical household would rise to around £3,549 a year from that October — before the government stepped in and held it near £2,500 — the headline was a single frightening number. But no household actually pays a single number. An energy bill is built from two separate charges: a standing charge of roughly 40–50p a day that you pay even if you switch everything off and go on holiday, plus a unit rate of around 25–35p for every kilowatt-hour you actually burn. Total cost = unit rate × units used + standing charge. That is not a coincidence of billing; it is the single most important equation in this entire section, dressed in pounds and pence.
Write it in maths and it becomes y = mx + c. The cost y depends on the units used x. The gradient m is the unit rate — how steeply the bill climbs for every extra unit. The intercept c is the standing charge — where the line starts, the bill you owe before you have used a thing. Plot cost against units and you get a straight line that never passes through the origin, because c is not zero. Everything A2.2 asks you to do — plot a line, read its steepness, find where it starts, work out its equation, then do the same trick for curves — is the skill of turning a relationship into a picture and reading the picture back. Foundation papers examine this on both the calculator and non-calculator papers, and it quietly underpins ratio, kinematics and every rate-of-change question you will meet later.
ModelThe coordinate grid — four quadrants, one address system
A coordinate is an address: (x, y) means go along the x-axis first, then up the y-axis. Along the corridor, then up the stairs — always in that order, because (3, 5) and (5, 3) are two completely different points. The origin (0, 0) is where the axes cross, and the two axes carve the plane into four quadrants.
The quadrants are numbered anticlockwise from top-right, and each has its own sign pattern. Top-right: x positive, y positive — (2, 3). Top-left: x negative, y positive — (−2, 3). Bottom-left: both negative — (−2, −3). Bottom-right: x positive, y negative — (2, −3). The classic Foundation error lives entirely in the negatives: a point like (−4, 1) gets plotted at (4, 1) because the minus sign is read as decoration rather than a direction. It is a direction. Negative x means left of the origin; negative y means below it.
Why it matters: once you can place any point in any quadrant confidently, a straight line is just the join-the-dots between points you have plotted, and a quadratic curve is the same thing with more points and a steady hand. Every graph question in this section is built on top of reading and writing coordinates without hesitation, so the two marks it earns directly are the cheapest in the section — and the ten marks it protects downstream are the reason it is worth getting automatic.
Plot A(3, 2), B(−4, 1), C(−2, −3) and D(1, −4), then name the quadrant each sits in. A(3, 2): 3 right, 2 up — top-right, quadrant 1. B(−4, 1): 4 left, 1 up — top-left, quadrant 2. C(−2, −3): 2 left, 3 down — bottom-left, quadrant 3. D(1, −4): 1 right, 4 down — bottom-right, quadrant 4. Now find the midpoint of A and C — a common follow-up: average the x-coordinates and average the y-coordinates. Midpoint = ((3 + −2)/2, (2 + −3)/2) = (0.5, −0.5). Averaging the coordinates works because the midpoint is exactly halfway along in each direction independently.
MechanismPlotting a straight line and reading y = mx + c
To plot a line like y = 2x + 1, build a small table of values. Choose a few x-values, substitute each into the rule, and record the y it produces: at x = 0, y = 1; at x = 1, y = 3; at x = 2, y = 5. Plot (0, 1), (1, 3), (2, 5), lay a ruler across them and draw one clean line right across the grid. The reason three points is the safe minimum, not two, is that the third point is your error-check: if it does not lie on the straight line through the other two, one of your calculations is wrong.
Once a line is drawn, y = mx + c tells you everything about it. The gradient m is the steepness — how many units y rises for every one unit x moves right. You read it as rise ÷ run: pick two points on the line that sit on grid crossings, count the vertical change, count the horizontal change, and divide. A line sloping downhill left-to-right has a negative gradient. The y-intercept c is where the line crosses the y-axis, the value of y when x = 0.
The most punished mistake here is reading the gradient by counting squares instead of using the scale. If each square on the y-axis is worth 2, a rise of 3 squares is a rise of 6, not 3. Gradient is measured in the units printed on the axes, never in squares. The second trap is reading m straight off an equation that is not yet in y = mx + c form: in 2y = 6x + 4 the gradient is not 6. Divide through by 2 first to get y = 3x + 2 — now the gradient is 3.
A line passes through the grid points (0, 3) and (4, 11). Find its equation. Gradient m = rise ÷ run = (11 − 3) ÷ (4 − 0) = 8 ÷ 4 = 2. The line crosses the y-axis at (0, 3), so c = 3. Equation: y = 2x + 3. Check with a third point — at x = 2 the rule gives y = 2(2) + 3 = 7, so (2, 7) must lie on the line, and it does. Now a downhill line through (0, 6) and (3, 0): gradient = (0 − 6) ÷ (3 − 0) = −6 ÷ 3 = −2, intercept 6, so y = −2x + 6. The negative sign is the whole story — miss it and every later value is wrong.
MechanismFinding a line's equation — and spotting parallels
AQA asks two versions of the same job. Given two points, find the gradient with rise ÷ run, then find c by substituting one of the points into y = mx + c and solving. Given one point and a gradient, you already have m, so go straight to substituting the point to find c. Both routes end at the same place: numbers for m and for c, written out as y = mx + c.
Parallel lines have equal gradients — that is the entire rule, and it is examined constantly. y = 3x + 1 and y = 3x − 4 never meet, because they climb at exactly the same rate; only their starting heights differ. So a line parallel to y = 3x + 1 passing through the origin must be y = 3x (same gradient, intercept 0). The reason is geometric: gradient is direction, and two lines pointing the same direction cannot cross. (Perpendicular lines — where the gradients multiply to −1 — are Higher tier only, so on Foundation you are only ever asked about parallel.)
The method mark is earned by showing the gradient calculation as a fraction before you simplify it, and by writing the substitution line when you find c. An unsupported y = 3x + 2 can score full marks if it is right, but scores nothing if the arithmetic slipped, whereas the working keeps most of the marks even with a wrong final digit.
Find the equation of the line through (1, 5) and (3, 11). Gradient = (11 − 5) ÷ (3 − 1) = 6 ÷ 2 = 3. Substitute (1, 5) into y = 3x + c: 5 = 3 × 1 + c, so 5 = 3 + c, giving c = 2. Equation: y = 3x + 2. Verify with the other point: 3 × 3 + 2 = 11 ✓. A follow-up AQA loves: write the equation of the line parallel to this one that passes through (0, −4). Parallel means the same gradient, 3, and it crosses the y-axis at −4, so the answer is simply y = 3x − 4 — no calculation needed once you know parallel means equal gradient.
ModelQuadratics — roots, intercepts and the turning point
A quadratic like y = x² − 2x − 3 draws a smooth U-shaped curve called a parabola (it opens upwards whenever the x² term is positive). Three features carry the marks. The roots are where the curve crosses the x-axis — the x-values that make y = 0. The y-intercept is where it crosses the y-axis, found by setting x = 0. The turning point is the very bottom of the U (a minimum) or the very top of an ∩ (a maximum).
To read roots off a plotted curve, look for where it cuts the x-axis. To deduce them algebraically — which Foundation does expect — set y = 0 and factorise: x² − 2x − 3 = 0 becomes (x − 3)(x + 1) = 0, so x = 3 or x = −1. Those are exactly the x-axis crossings. The y-intercept is instant: at x = 0, y = 0 − 0 − 3 = −3.
The turning point at Foundation is found by symmetry, not by completing the square (that method is Higher only). A parabola is perfectly symmetrical, so its turning point sits exactly halfway between the two roots. Halfway between x = 3 and x = −1 is x = (3 + −1) ÷ 2 = 1; substitute back to get y = 1² − 2 − 3 = −4. Turning point: (1, −4). The classic error is naming the roots as the y-intercept, or vice versa — always tie 'root' to the x-axis (y = 0) and 'y-intercept' to the y-axis (x = 0).
Sketch y = x² + 2x − 8 by finding its key points. Roots: set x² + 2x − 8 = 0 and factorise — two numbers multiplying to −8 and adding to +2 are +4 and −2, so (x + 4)(x − 2) = 0, giving x = −4 and x = 2. Those are the x-axis crossings. y-intercept: at x = 0, y = −8, so the curve cuts the y-axis at (0, −8). Turning point by symmetry: halfway between −4 and 2 is x = (−4 + 2) ÷ 2 = −1, and y = (−1)² + 2(−1) − 8 = 1 − 2 − 8 = −9, so the minimum is (−1, −9). Plot (−4, 0), (2, 0), (0, −8) and (−1, −9), join with a smooth curve, and you have a full-mark sketch — no plotting table required.
CaseThe shape zoo — knowing a graph by its silhouette
AQA expects you to recognise and sketch four families on sight, so that a multiple-choice or matching question is instant. A linear graph (y = mx + c) is a straight line. A quadratic (with an x² term) is a symmetrical U or ∩ parabola. A cubic (with an x³ term, e.g. y = x³) has a distinctive stretched-S shape, rising steeply, flattening near the middle and rising steeply again — and a negative cubic is that S flipped. A reciprocal graph y = 1/x is the odd one out: two separate curves in opposite corners, one in the top-right, one in the bottom-left.
The reciprocal is worth extra attention because its behaviour is examined directly. As x grows huge, 1/x shrinks towards zero but never reaches it, so the curve hugs the x-axis without touching. As x shrinks towards zero, 1/x explodes, so the curve races up the y-axis without touching it. Those lines it approaches but never meets are called asymptotes, and the reason it never touches is arithmetic: you can never divide 1 by a number and get exactly zero, and you can never divide 1 by zero at all. That is why x = 0 is a hole in the graph, not a point on it.
(Exponential graphs — the steepening-forever curves of viral spread and compound growth — and the wave-shaped trigonometric graphs both belong to the Higher tier, so they will not appear on your Foundation paper. Foundation lives entirely in these four shapes.)
DataGraphs of the real world — distance, time and speed
The payoff of the whole section is that a graph's gradient is a rate of change you can name. On a distance–time graph, time runs along the bottom and distance up the side, and the gradient — distance ÷ time — is the speed. A steep line means fast; a shallow line means slow; a horizontal line means the gradient is zero, so distance is not changing — the object has stopped. This is exactly how average-speed cameras on UK motorway roadworks catch you: they measure the distance between two points and the time you took, and the gradient is your speed, whatever it read on your speedometer in between.
Reading these graphs is a matter of translating slopes into sentences. A journey graph that climbs steadily, then goes flat, then climbs again describes travel, a stop, then more travel. A line sloping back down towards zero distance means returning to the start. The units matter: if distance is in metres and time in seconds, the gradient is in metres per second; convert with care if the question mixes kilometres and hours.
Reciprocal relationships show up here too. For a fixed journey, speed and time are inversely proportional — double your speed and the time halves — which plots as the reciprocal curve, not a straight line. Recognising that a real relationship is 'inverse' rather than 'straight-line' is a genuine Foundation skill, and it is the bridge from this section into ratio and proportion.
A cyclist's distance–time graph goes in a straight line from (0 s, 0 m) to (40 s, 800 m), stays flat from 40 s to 60 s, then rises straight from (60 s, 800 m) to (100 s, 1000 m). Describe the journey with numbers. First stage: gradient = 800 ÷ 40 = 20 m/s — a steady 20 metres every second. Middle stage: the line is horizontal, so the gradient is 0 — the cyclist has stopped for 20 seconds. Final stage: gradient = (1000 − 800) ÷ (100 − 60) = 200 ÷ 40 = 5 m/s — moving again, but four times slower than at the start. Three gradients, three plain-English facts: the mark scheme rewards each speed calculation and the interpretation that a flat line means stationary.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Graphs appear on both the non-calculator (Paper 1) and calculator (Papers 2 and 3) papers, and the plotting is almost always marked method-first. For a 'draw the graph of…' question, the marks split predictably: one for a correct table of values, one for plotting the points accurately, one for a single straight line or smooth curve drawn with a ruler or a steady hand right across the given grid. Because a correct table earns its mark even if you then plot a point wrong, always show the table — and always plot a third point as your own error-check, since a point off the line tells you a calculation slipped before the examiner does.
When you find the equation of a line, write the gradient as a fraction (rise over run) before simplifying, and write the substitution line when you solve for c — those are the M marks, and they survive an arithmetic slip that a bare answer would not. For quadratics, state that you are setting y = 0 before factorising for the roots, and remember that at Foundation the turning point comes from the symmetry of the roots, never from completing the square (which is Higher). Reading a gradient as a speed or other rate on a calculator paper still needs the division shown, and the units — m/s, mph, £ per unit — are frequently a mark in themselves. Finally, watch the command words: 'plot' means work out and mark points accurately; 'sketch' means show the shape and key features (roots, intercepts, turning point) without a full table.