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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow sections 7.01a to 7.01c of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): coordinates, plotting from tables, and recognising and sketching graphs. Questions 1 to 8 and 12 are Foundation and Higher content. Questions 9 to 11 are (Higher tier only). Learn the methods first from the study guide and the revision notes. The OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics are there for planning. Use squared paper for the plotting questions, and draw sketches freehand.

Questions

1. P(−4, 3), Q(2, 3) and R(2, −1) are three vertices of rectangle PQRS.

(a) Write down the coordinates of S. [1] (b) In which quadrant does P lie? [1] (c) Work out the area of the rectangle. [1]

2. A table of values is made for y = 6 − 4x. The values at x = −2, 0 and 3 are 14, 6 and −6.

(a) Work out y when x = −1, 1 and 2. [2] (b) Plot the graph of y = 6 − 4x for x from −2 to 3. [2] (c) Use your graph to estimate the value of x when y = 0. [1]

3. The equation of a line is 5x + 2y = 20. Work out y when x = 0, 2 and 4, then draw the line. [3]

4. A table of values is made for y = x² − 5x + 4. The values at x = −1, 1 and 6 are 10, 0 and 10.

(a) Work out y when x = 0, 2, 3, 4 and 5. [2] (b) Plot the graph of y = x² − 5x + 4 for x from −1 to 6. [2] (c) Use symmetry to find the coordinates of the turning point. [2] (d) Write down the roots of x² − 5x + 4 = 0 from your graph. [1]

5. A table of values is made for y = x³ + x² − 2x, with y = 0 when x = −2.

(a) Work out y when x = −1, 1 and 2. [2] (b) Plot the graph of y = x³ + x² − 2x for x from −2 to 2, and read off where it crosses the x-axis. [2] (c) Write down the y-intercept. [1]

6. Write down the equation of each graph.

(a) A horizontal line through (0, −4). [1] (b) A vertical line through (5, 0). [1] (c) A straight line through the origin and through (1, −3). [1] (d) A U-shaped curve through the origin, with its lowest point at the origin and passing through (2, 4). [1] (e) A smooth curve through the origin and (1, 1), rising throughout, in quadrants 1 and 3. [1] (f) Two separate branches in quadrants 1 and 3, through (1, 1), never touching either axis. [1]

7. A table of values is made for y = x + 9/x. The values at x = 1 and x = 3 are 10 and 6.

(a) Work out y when x = 2, 4 and 6. [2] (b) Work out y when x = −3. [1] (c) Explain why the graph has no point at x = 0. [1]

8. A curve has equation y = x² − 6x − 16.

(a) Solve x² − 6x − 16 = 0. [2] (b) Write down the y-intercept. [1] (c) Use symmetry to find the turning point. [2] (d) Sketch the curve, labelling the intercepts and the turning point. [1]

9. (Higher tier only) Solve x² − 8x + 5 = 0. Give both solutions to 2 decimal places. [3]

10. (Higher tier only) A curve has equation y = x² + 10x + 14.

(a) Write x² + 10x + 14 in the form (x + p)² + q. [2] (b) Write down the turning point. [1] (c) Find the exact roots of x² + 10x + 14 = 0. [2] (d) Sketch the curve, labelling the turning point and the y-intercept. [2]

11. (Higher tier only) A curve has equation y = 5 − 4x − x².

(a) Write the equation in the form y = a − (x + p)². [2] (b) State the turning point and whether it is a maximum or a minimum. [1] (c) Find the roots of 5 − 4x − x² = 0. [2] (d) Sketch the curve, labelling the turning point and the intercepts. [2]

12. The curve y = x² + bx + c crosses the x-axis at x = −5 and x = 1.

(a) Find the values of b and c. [2] (b) Find the turning point. [2] (c) Sketch the curve, labelling the intercepts and the turning point. [2]

Answers

1. (a) S has the x-value of P and the y-value of R: (−4, −1). [1] (b) x is negative and y is positive: second quadrant. [1] (c) Width 6, height 4, so area 24 square units. [1] Examiner insight: write the coordinates in brackets in (x, y) order. A bare pair of numbers is easy to read as the wrong way round.

2. (a) x = −1: 6 − 4(−1) = 10 [1]. x = 1 gives 2 and x = 2 gives −2. [1] (b) Plot (−2, 14), (−1, 10), (0, 6), (1, 2), (2, −2), (3, −6) [1]. Join with a ruler in one straight line [1]. (c) The line crosses the x-axis at x = 1.5 (reading from the graph). [1] Examiner insight: a plotted point that is a small square out of place often shows an arithmetic slip, so check any point that breaks the straight-line pattern.

3. x = 0 gives y = 10, x = 2 gives y = 5, x = 4 gives y = 0. [1] for the first value, [1] for the other two. The line passes through (0, 10), (2, 5) and (4, 0) and is drawn with a ruler. [1] [3] Examiner insight: when the equation has x and y on the same side, rearrange to y = (20 − 5x)/2 and show this step, so a reader can follow your method.

4. (a) Substitute each value, for example x = 3: 9 − 15 + 4 = −2. [1] The values are 4, −2, −2, 0, 4 for x = 0, 2, 3, 4, 5. [1] (b) All eight points plotted correctly. [1] A smooth parabola through them with no straight segments. [1] (c) The curve is symmetrical, with equal y-values at x = 2 and x = 3, so the turning point has x = 2.5. [1] y = 6.25 − 12.5 + 4 = −2.25, so (2.5, −2.25). [1] (d) The curve crosses the x-axis at x = 1 and x = 4. [1] Examiner insight: the turning point is often between two plotted points, so state both coordinates and not only the x-value.

5. (a) x = −1: −1 + 1 + 2 = 2 [1]. x = 1 gives 0 and x = 2 gives 8. [1] (b) Plot (−2, 0), (−1, 2), (0, 0), (1, 0), (2, 8). [1] Join with one smooth curve that rises, falls, then rises, crossing the x-axis at −2, 0 and 1. [1] (c) At x = 0, y = 0, so the y-intercept is (0, 0). [1] Examiner insight: with so few points, the dip between x = 0 and x = 1 is easy to flatten. Add a point such as x = 0.5 (giving −0.625) to place the turn.

6. (a) y = −4 [1]. (b) x = 5 [1]. (c) y = −3x [1]. (d) y = x² [1]. (e) y = x³ [1]. (f) y = 1/x [1]. [6] Examiner insight: y = −4 and x = 5 are different kinds of line. Horizontal lines are y = a number; vertical lines are x = a number.

7. (a) x = 2: 2 + 9/2 = 6.5 [1]. x = 4 gives 6.25 and x = 6 gives 7.5. [1] (b) −3 + 9/(−3) = −3 − 3 = −6. [1] (c) Putting x = 0 means dividing by 0, which has no value, so no point can be plotted. [1] Examiner insight: in a table with negative x, put the negative number in brackets and divide with care. The sign of 9/(−3) is a common source of error.

8. (a) (x − 8)(x + 2) = 0 [1], so x = 8 or x = −2. [1] (b) At x = 0, y = −16, so the point is (0, −16). [1] (c) The mean of the roots is (8 − 2)/2 = 3 [1]. y = 9 − 18 − 16 = −25, so (3, −25). [1] (d) A U-shaped parabola with labelled axes, crossing at (−2, 0), (8, 0) and (0, −16), with minimum (3, −25). [1] Examiner insight: a sketch that shows the labelled points in the correct order along the x-axis, with the minimum under the midpoint of the roots, is easy to check.

9. x = (8 ± √(64 − 20))/2 [1] = (8 ± √44)/2 = 4 ± √11 [1]. x = 7.32 or x = 0.68 [1] Examiner insight: show the substitution into the formula before simplifying. Keep the exact form until the final rounding.

10. (a) (x + 5)² − 25 + 14 [1] = (x + 5)² − 11. [1] (b) Turning point (−5, −11). [1] (c) (x + 5)² = 11 [1], so x = −5 ± √11. [1] (d) A U-shaped curve with minimum (−5, −11) below the x-axis, crossing the y-axis at (0, 14) [1] and the x-axis twice, at about −8.32 and −1.68. [1] Examiner insight: write the turning point as (−5, −11), not (5, −11): the sign flips because the bracket contains x + 5.

11. (a) 5 − 4x − x² = −(x² + 4x) + 5 = −[(x + 2)² − 4] + 5 [1] = 9 − (x + 2)². [1] (b) Turning point (−2, 9), a maximum. [1] (c) (x + 2)² = 9 [1], so x + 2 = ±3 and x = 1 or x = −5. [1] (d) An upside-down parabola with its maximum at (−2, 9) [1], crossing the x-axis at −5 and 1 and the y-axis at (0, 5). [1] Examiner insight: for a negative x² term, take the minus sign outside the bracket first. It is a common source of wrong signs in the constant.

12. (a) (x + 5)(x − 1) [1] = x² + 4x − 5, so b = 4 and c = −5. [1] (b) Mean of roots (−5 + 1)/2 = −2 [1]. y = 4 − 8 − 5 = −9, so (−2, −9). [1] (c) U-shaped parabola crossing the x-axis at (−5, 0) and (1, 0) and the y-axis at (0, −5) [1], with minimum (−2, −9) below the midpoint of the roots. [1] Examiner insight: working backwards from roots to an equation is checked by substitution: put x = 1 into your answer and confirm it gives 0.

Where marks are usually lost

• Writing coordinates as (y, x).
• Squaring a negative as a negative number when completing a table.
• Joining plotted curve points with straight lines or a pointed bottom.
• Joining the two branches of a reciprocal graph.
• Giving the x-value of a turning point and forgetting the y-value.
• Forgetting the y-intercept, which is the constant term.
• Using (p, q) instead of (−p, q) as the turning point.
• Leaving a decimal when an exact answer was requested.
• Drawing a U-shape for a negative x² coefficient.

Next steps

• Review the revision notes and the study guide.
• Open the OCR GCSE Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

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).