OCR GCSE Mathematics: Plotting and sketching linear, quadratic, cubic and reciprocal graphs (J560) – Revision Notes

These revision notes follow the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), sections 7.01a to 7.01c. For full explanations and worked examples, read the study guide. Then test yourself with the practice questions. Links: OCR GCSE Mathematics hub, printable checklist and 10-minute diagnostics.

Everything here is assessed at both tiers except solving with the quadratic formula or by completing the square, and sketching with the turning point from completing the square, which are Higher tier only.

Plot or sketch?

	Plot	Sketch
How	Calculate points, mark them on squared paper, join them	Freehand, no calculated table needed
Must show	Accurate points, ruler or smooth curve	Labelled axes, correct quadrants, correct shape and behaviour at the ends, correct contact with the axes
Labels	Axes scaled	Intercepts and turning points only when the question asks

7.01a Coordinates

• (x, y): across first, then up. Negative x is left, negative y is down.
• Quadrants run anticlockwise from top right: 1st (+, +), 2nd (−, +), 3rd (−, −), 4th (+, −).
• On the x-axis, y = 0. On the y-axis, x = 0.
• Fourth vertex of a rectangle or square: match the x-value of one neighbour and the y-value of the other.

Reminder. Rectangle with vertices (−5, 1), (2, 1) and (2, −3): the fourth is (−5, −3), in the third quadrant.

7.01b Plotting from a table

Method in steps:

1. Substitute each x into the equation. Use brackets for negatives.
2. Plot the crosses carefully, reading the scale on each axis.
3. Join with a ruler (linear) or one smooth curve (anything else).
4. If a point sits off the pattern, recheck its arithmetic.

Equation type	Table pattern	Shape
Linear, such as y = 1 − 2x	y changes by the same amount each step	Straight line
Quadratic, such as y = x² − 2x − 6	y-values symmetrical about the turning point	U or upside-down U
Cubic, such as y = x³ − 8x	Rises, falls, rises (or the reverse)	Smooth S-like curve with two turns
Reciprocal, such as y = x + 4/x	No value at x = 0	Two separate branches

For an equation like 3x + 2y = 12, rearrange to y = (12 − 3x)/2, or find the intercepts: put x = 0 for the y-intercept (0, 6) and y = 0 for the x-intercept (4, 0).

Reminder values (all checked):

Equation	Values
y = 1 − 2x	x = −2, 0, 3 give y = 5, 1, −5
y = x² − 2x − 6	x = −2, 1, 4 give y = 2, −7, 2
y = x³ − 8x	x = −3, 0, 2 give y = −3, 0, −8
y = x + 4/x	x = −4, −1, 2 give y = −5, −5, 4

7.01c Recognise and sketch

Equation	Graph
y = c	Horizontal line through (0, c)
x = c	Vertical line through (c, 0)
y = mx	Line through the origin; up if m > 0, down if m < 0
y = x²	U-shape, minimum (0, 0)
y = −x²	Upside-down U, maximum (0, 0)
y = x³	Rising S through (0, 0); quadrants 1 and 3
y = 1/x	Two branches in quadrants 1 and 3; touches neither axis

Must-know distinctions:

• y = x² against y = x³. Both pass through the origin. The square is symmetrical about the y-axis and never goes below it. The cube goes through quadrant 3 as well as quadrant 1.
• y = x³ against y = 1/x. The cube is one unbroken curve through the origin. The reciprocal has a break at x = 0 and never reaches either axis.
• Horizontal against vertical. y = 2 is horizontal. x = 1 is vertical.

Quadratics: intercepts, roots and the turning point

For y = ax² + bx + c:

Feature	How
Shape	a > 0: U-shape with a minimum. a < 0: upside down with a maximum
y-intercept	Put x = 0. The answer is (0, c)
Roots (x-intercepts)	Solve ax² + bx + c = 0 algebraically
Turning point x-value	The mean of the two roots, because the curve is symmetrical
Turning point y-value	Substitute that x into the equation

Method for roots.

1. Rearrange so that one side is 0.
2. Factorise, if you can, and set each bracket to 0.
3. (Higher tier only) If it will not factorise, use the quadratic formula (or complete the square) and leave surds exact until the end.
4. If the part under the root is negative, there are no roots.

Worked reminder. y = x² + 2x − 24.

(x + 6)(x − 4) = 0, roots −6 and 4
Turning point x = (−6 + 4)/2 = −1, y = 1 − 2 − 24 = −25, so (−1, −25)
y-intercept (0, −24)

Worked reminder (Higher tier only). x² − 6x + 3 = 0 gives x = 3 ± √6, which is about 5.45 and 0.55.

Worked reminder (upside-down). y = 8 + 2x − x² has roots 4 and −2, a maximum at (1, 9) and y-intercept (0, 8).

Completing the square (Higher tier only)

If y = (x + p)² + q, the turning point is (−p, q). To get there:

1. Halve the coefficient of x to get p.
2. Write (x + p)² and subtract p² to compensate.
3. Add the constant that is left.

x² + 6x + 7 = (x + 3)² − 9 + 7 = (x + 3)² − 2
Turning point (−3, −2), minimum
Roots: (x + 3)² = 2, x = −3 ± √2

For a negative x², take out the minus sign first: −x² + 4x − 7 = −(x − 2)² − 3. The maximum is (2, −3). Since the maximum is below the x-axis, there are no roots.

Quick test for roots: a minimum below the x-axis, or a maximum above it, means two roots. A minimum above the x-axis, or a maximum below it, means none.

Sketch checklist

Run through this before you move on from any sketch question:

1. Are both axes labelled, with x and y written at the ends?
2. Is the curve in the right quadrants for its equation?
3. Does it show the right behaviour at each end, such as rising on the right for a positive cube or levelling towards an axis for a reciprocal?
4. Does it meet the axes only where it should?
5. Have you labelled every feature the question asked for, using coordinates and not only numbers on the axes?

Sketch method in steps

For a quadratic that factorises:

1. Decide the shape from the sign of the x² term.
2. Find the y-intercept by putting x = 0.
3. Factorise and write down the roots.
4. Take the mean of the roots for the turning point’s x-value, then substitute to get y.
5. Draw the smooth parabola through all four features and label them.

For a cubic or reciprocal such as y = x³ or y = 1/x, you do not need a table. Recall the shape, check the quadrants, and show that the curve never crosses an axis where it should not.

Quick self-test

1. Which quadrant contains (−6, 4)?
2. Three vertices of a rectangle are (−2, 3), (5, 3) and (5, −1). Write down the fourth.
3. Complete the table for y = 3x + 2 at x = −2, 0 and 4.
4. Find y when x = −3 on y = x² + 2x − 8.
5. Where does the line 2x + 5y = 20 cross each axis?
6. Which curve has two separate branches: y = x², y = x³ or y = 1/x?
7. Write down the equation of the vertical line through (−3, 0).
8. Solve x² − 3x − 10 = 0.
9. Find the turning point of y = x² − 3x − 10 using symmetry.
10. State whether y = 6 − x² has a minimum or a maximum, and give its y-intercept.
11. (Higher tier only) Write x² + 12x + 30 in the form (x + p)² + q.
12. (Higher tier only) Give the turning point of y = x² + 12x + 30.

Answers

1. Second quadrant (negative x, positive y).
2. (−2, −1).
3. y = −4, 2, 14.
4. (−3)² + 2(−3) − 8 = 9 − 6 − 8 = −5.
5. The y-axis at (0, 4) and the x-axis at (10, 0).
6. y = 1/x.
7. x = −3.
8. (x − 5)(x + 2) = 0, so x = 5 or x = −2.
9. Mean of roots: x = (5 − 2)/2 = 1.5. Then y = 2.25 − 4.5 − 10 = −12.25. The point is (1.5, −12.25).
10. Maximum, because the x² coefficient is negative. The y-intercept is (0, 6).
11. (x + 6)² − 6.
12. (−6, −6).

Where marks are usually lost

• Reading a coordinate pair as (y, x).
• Squaring a negative number with the wrong sign when filling in a table.
• Plotting accurately, then joining a curve with short straight segments or a flat bottom.
• Joining the two branches of a reciprocal graph or crossing an axis.
• Leaving out the labelled intercepts or turning point that a sketch question asked for.
• Giving only the x-value of a turning point.
• Writing the turning point as (p, q) instead of (−p, q).
• Using a decimal approximation when the question asks for an exact root.
• Drawing a curve that crosses the x-axis at points that disagree with the roots you found.
• Choosing a U-shape for a quadratic with a negative x² coefficient.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, section 7.01 Graphs of equations and functions (7.01a to 7.01c).