OCR GCSE Mathematics: Exponential and trigonometric graphs, and transformations of curves (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 OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), statements 7.01b, 7.01d, 7.01e and 7.03a. All of it is Higher tier only, and Foundation tier candidates are not assessed on it. Read the study guide first and keep the revision notes to hand. See also the OCR GCSE Mathematics hub and the printable checklist. Angles are in degrees. Work to 3 significant figures unless told otherwise.

Questions

1. (Higher tier only) The table is for y = 4 × 1.5ˣ.

x | -2 | -1  | 0 | 1 | 2 | 3
y |    | 2.67| 4 |   | 9 |

(a) Work out the missing y-values. [3] (b) Write down the y-intercept and explain why the graph never crosses the x-axis. [2]

2. (Higher tier only) The graph of y = kˣ, where k > 0, passes through (2, 1/9).

(a) Find k. [2] (b) State whether the graph rises or falls from left to right, and give its y-intercept and asymptote. [2]

3. (Higher tier only)

(a) Sketch y = sin x for 0° ≤ x ≤ 360°, marking the coordinates of the maximum, the minimum and the points where it meets the x-axis. [3] (b) How many solutions has sin x = −0.6 for 0° ≤ x ≤ 720°? [2]

4. (Higher tier only)

(a) Sketch y = tan x for 0° ≤ x ≤ 360°. Show the asymptotes and the points where the graph meets the x-axis. [3] (b) State the period of y = tan x. [1]

5. (Higher tier only) Each description fits one of y = sin x, y = cos x or y = tan x. Name the graph. [3]

(i) The graph passes through (0, 1) and its values never exceed 1. (ii) The graph has vertical asymptotes. (iii) The graph passes through (0, 0), has a maximum at (90°, 1) and a minimum at (270°, −1).

6. (Higher tier only) The graph of y = x² is translated 3 units left and 5 units down.

(a) Write down the equation of the new graph. [2] (b) Write down its turning point and its y-intercept. [2]

7. (Higher tier only) The graph of y = x² − 6x has its turning point at (3, −9). It is reflected (a) in the x-axis and (b) in the y-axis. For each, write down the equation of the new graph and its turning point.

(a) Reflection in the x-axis. [2] (b) Reflection in the y-axis. [2]

8. (Higher tier only) The graph of y = sin x is translated 2 units down.

(a) Write down the new equation, its maximum and minimum y-values, and its y-intercept. [4] (b) Describe the translation that maps y = cos x onto y = sin x. [1]

9. (Higher tier only) Start from the graph of y = 2ˣ.

(a) It is translated 3 units down. Give the equation, the y-intercept and the equation of the asymptote. [3] (b) It is reflected in the x-axis. Give the equation and the y-intercept. [2] (c) It is reflected in the y-axis. Give the equation of the new graph, and write it in the form y = kˣ. [1]

10. (Higher tier only) The curve y = 3(x − 1)² − 2 has its turning point at (1, −2) and crosses the y-axis at (0, 1).

(a) The curve is translated 4 units right and 3 units up. Find the new turning point, the new equation, and the new y-intercept. [3] (b) The original curve is reflected in the x-axis. Find the new equation and turning point. [2]

11. (Higher tier only) A car’s value, V pounds, after t years is modelled by V = 18 000 × 0.85ᵗ.

(a) Find V when t = 0 and when t = 3, to the nearest pound. [2] (b) Describe the shape of the graph of V against t, and explain why the model never gives a value of zero. [2] (c) Find the first whole number of years after which the value is below £9000. [3]

12. (Higher tier only) The curve C has equation y = (x + 1)² − 9.

(a) Write down the turning point of C and its y-intercept. [2] (b) Find the x-coordinates of the points where C crosses the x-axis. [2] (c) C is reflected in the x-axis and the result is then translated 3 units right. Find the equation of the final curve, its turning point and its y-intercept. [4]

Answers

1. (a) 1.5⁻² = 1/2.25, so y = 4 × 0.444… = 1.78 at x = −2 [1]. At x = 1, y = 4 × 1.5 = 6 [1]. At x = 3, y = 4 × 3.375 = 13.5 [1] [3] (b) y-intercept (0, 4) [1]. 1.5ˣ is always positive, so y is never zero [1] [2] Examiner insight: Write the power you are evaluating (for example 1.5⁻²) before the decimal, so your working is clear and easy to check.

2. (a) k² = 1/9 [1], so k = 1/3 (k is positive) [1] [2] (b) 0 < k < 1, so the graph falls [1]. The y-intercept is (0, 1) and the asymptote is y = 0 [1] [2] Examiner insight: State the equation before square-rooting. Writing k = ±1/3 and then rejecting the negative value shows you used the condition k > 0.

3. (a) A smooth wave starting at (0, 0) [1], with maximum (90°, 1) and minimum (270°, −1) [1], meeting the x-axis at (0°, 0), (180°, 0) and (360°, 0) [1] [3] (b) The line y = −0.6 lies between 0 and −1, so it cuts each wave twice [1]. Two waves give 4 solutions [1] [2] Examiner insight: A sketch scores best as a smooth curve with the key coordinates written on it. Straight-line segments or an unlabelled wave leave the examiner nothing to check.

4. (a) Vertical asymptotes drawn at x = 90° and x = 270° [1]. The curve passes through (0, 0), (180°, 0) and (360°, 0) [1]. Each branch rises from the left to the asymptote on its right, and the middle branch is continuous through (180°, 0) [1] [3] (b) 180° [1] [1] Examiner insight: Draw asymptotes as dashed lines and keep the curve from touching them. A curve that crosses an asymptote suggests the graph is misunderstood.

5. (i) cos x [1]. (ii) tan x [1]. (iii) The point (0, 0) and the maximum at 90° make it sin x [1] [3] Examiner insight: Give just the name for each description. Where two features are given, check that both fit before choosing.

6. (a) Left 3 means x becomes x + 3 [1]; down 5 gives y = (x + 3)² − 5 [1] [2] (b) Turning point (−3, −5) [1]. When x = 0, y = 9 − 5 = 4, so the y-intercept is (0, 4) [1] [2] Examiner insight: Test the direction of a horizontal move by finding where the bracket equals zero. Here x + 3 = 0 gives x = −3, which is to the left.

7. (a) Change the sign of the whole right-hand side: y = −x² + 6x [1]. Turning point (3, 9) [1] [2] (b) Replace x by −x: (−x)² − 6(−x) gives y = x² + 6x [1]. Turning point (−3, −9) [1] [2] Examiner insight: A reflection in the x-axis changes the sign of every term, not just the first. Check by testing the turning point.

8. (a) y = sin x − 2 [1]. Maximum −1 [1], minimum −3 [1], y-intercept (0, −2) [1] [4] (b) Translation 90° to the right [1] [1] Examiner insight: Name the type of transformation, the size and the direction. “Moves 90” alone would not describe it completely.

9. (a) y = 2ˣ − 3 [1]. y-intercept 1 − 3 = −2, so (0, −2) [1]. Asymptote y = −3 [1] [3] (b) y = −2ˣ [1]. y-intercept (0, −1) [1] [2] (c) y = 2⁻ˣ, which is y = (1/2)ˣ = 0.5ˣ [1] [1] Examiner insight: When a graph is translated vertically, move the asymptote as well as the intercept. Leaving the asymptote at y = 0 is a common slip.

10. (a) Turning point (5, 1) [1]. Replace x by x − 4 and add 3: y = 3(x − 5)² + 1 [1]. When x = 0, y = 3 × 25 + 1 = 76, so (0, 76) [1] [3] (b) y = −3(x − 1)² + 2 [1]. Turning point (1, 2) [1] [2] Examiner insight: Keep the squared bracket together and apply the minus sign to the whole expression. A minus applied only to the 3 would not reflect the curve.

11. (a) t = 0: £18 000 [1]. t = 3: 18 000 × 0.85³ = 11 054.25, so £11 054 [1] [2] (b) It falls steeply at first and then flattens, getting closer to the t-axis [1]. 0.85ᵗ is always positive, so V is never zero [1] [2] (c) Half of 18 000 is 9000. At t = 4, V = 9396.11, still above 9000 [1]. At t = 5, V = 7986.70, below 9000 [1]. So 5 years [1] [3] Examiner insight: Show the values either side of 9000. A bare answer of 5 gives no evidence that 4 was checked and rejected.

12. (a) Turning point (−1, −9) [1]. y-intercept: 1 − 9 = −8, so (0, −8) [1] [2] (b) (x + 1)² = 9 [1], so x + 1 = 3 or −3, giving x = 2 and x = −4 [1] [2] (c) Reflection in the x-axis: y = 9 − (x + 1)² [1]. Moving 3 right replaces x by x − 3: y = 9 − (x − 2)², that is y = −x² + 4x + 5 [1]. Turning point (2, 9) [1]. y-intercept 9 − 4 = 5, so (0, 5) [1] [4] Examiner insight: Do the two transformations in the order stated, and check the final equation against one point: the turning point should end up at (2, 9).

Where marks are usually lost

• Drawing an exponential graph that touches or crosses the x-axis.
• Forgetting that y = kˣ passes through (0, 1), whatever the value of k.
• Moving y = (x + 3)² − 5 to the right instead of the left.
• Not moving the asymptote when an exponential graph is translated.
• Reflecting in the x-axis by changing the sign of only part of the equation.
• Sketching sin x starting at 1 or cos x starting at 0.
• Missing the asymptotes of tan x, or drawing straight lines between points.
• Giving a loose description such as “moves 2” instead of naming the transformation, its size and direction.
• Rounding the answer to a modelling question without showing the values either side of the target.

Next steps

• Re-read the revision notes for the key points.
• Go back to the study guide for worked examples.
• Tick off the checklist and visit the course hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), first assessment 2017, published by OCR: statements 7.01b, 7.01d, 7.01e and 7.03a.