OxfordAQA International GCSE Mathematics: Graphs of quadratic, cubic, reciprocal and other functions (9260) – 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 cover references A12 and A16 of section 3.2.2 in the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Core content is also assessable at Extension. Questions marked “(Extension only)” use the extra Extension content: exponential and trigonometric graphs, and completing the square for turning points.

Links: study guide, revision notes, course hub, checklist and free diagnostics. For algebra practice, see algebra practice questions.

Questions

1. Describe the graph of each equation: say what shape it is and one feature that identifies it.

(a) y = 7 − x² [1] (b) y = −x³ [1] (c) y = 5/x [1]

2. The curve C has equation y = x² + 6x − 27.

(a) Solve x² + 6x − 27 = 0 to find where C crosses the x-axis. [2] (b) Give the equation of the axis of symmetry. Then find the turning point, stating whether it is a maximum or a minimum. [3] (c) Write down the y-intercept. [1]

3. Here is part of a table for y = x² − 12x + 20.

x	0	2	4	6	8	10	12
y	20	0		−16		0

(a) Find the three missing values. [2] (b) Write down the roots of x² − 12x + 20 = 0 from the table. [1] (c) Use the table to solve x² − 12x + 20 = −12. [1] (d) One solution of x² − 12x + 20 = −7 is x = 3. Use symmetry to find the other solution. [2]

4. A parabola crosses the x-axis at x = −2 and x = 6. It passes through the point (0, 12).

(a) Find the equation of the parabola as an expanded expression in x. [3] (b) Work out where its turning point is. [2]

5. Consider the curve y = x³ − 36x.

(a) Factorise x³ − 36x completely and hence find where the curve crosses the x-axis. [3] (b) When x = 3, y = −81. Use the symmetry of the curve to write down y when x = −3, and explain your answer. [2]

6. The curve y = k/x passes through the point (5, 2).

(a) Find k. [1] (b) Find y when x = −8. [1] (c) State which two quadrants the curve lies in, and explain why it never meets the y-axis. [2]

7. The curve y = x² − 10x + k has its turning point on the x-axis. Find the value of k. [3]

8. (Extension only) The curve y = x² + 8x − 5 has a turning point.

(a) Express x² + 8x − 5 as (x + p)² + q. [2] (b) Give the turning point as a coordinate pair. [1] (c) Find the exact x-coordinates of the points where the curve crosses the x-axis. [2]

9. (Extension only) A curve has equation y = 5 + 6x − x².

(a) Express the equation as y = q − (x − p)², finding p and q. [2] (b) State whether the turning point is a maximum or a minimum, and give its coordinates. [1] (c) State the range of values that y can take. [1]

10. (Extension only) The curve y = kˣ, where k is positive, passes through the point (2, 6.25).

(a) Find k. [2] (b) Find y when x = −1. [1] (c) Find y when x = 3. [1] (d) Write down the coordinates of the point where the curve crosses the y-axis. [1]

11. (Extension only)

(a) Write down the greatest value of sin x and the period of tan x. [2] (b) Find every angle x from 0° to 360° for which sin x = −0.5. [2] (c) Given that cos 40° = 0.766 (to 3 significant figures), write down cos 140° and another angle between 0° and 360° that has a cosine of 0.766. [2]

12. (Extension only) The curve D has equation y = x² + 2x − 48.

(a) Find the x-coordinates of the points where D crosses the x-axis. [2] (b) By completing the square, find the coordinates of the turning point of D. [3] (c) Describe a sketch of D, giving every point where it meets the axes and the turning point. [2] (d) The line y = −45 meets D at two points. Find their x-coordinates. [2]

Answers

1. (a) An upside-down (∩-shaped) parabola with its maximum at (0, 7) [1] (b) An S-shaped cubic through the origin that falls from left to right [1] (c) Two separate branches in the top-right and bottom-left quadrants, not touching either axis [1] [3] Examiner insight: Naming the shape alone is thin; add the feature that separates it from similar graphs, such as the maximum point, the direction or the two branches.

2. (a) (x + 9)(x − 3) = 0 [1] so x = −9 and x = 3 [1] (b) The axis is halfway between the roots: x = (−9 + 3)/2 = −3 [1]. At x = −3, y = 9 − 18 − 27 = −36 [1]. The turning point is (−3, −36), a minimum, because the x² coefficient is positive [1] (c) −27 [1] [6] Examiner insight: A turning point needs both coordinates and a statement of type. Writing only “x = −3” answers half the question.

3. (a) x = 4: 16 − 48 + 20 = −12 [1]; x = 8: 64 − 96 + 20 = −12 and x = 12: 144 − 144 + 20 = 20 [1] (b) x = 2 and x = 10 [1] (c) x = 4 and x = 8 [1] (d) The axis of symmetry is x = 6, because the table has equal values at x = 2 and x = 10 [1]. The other solution is 12 − 3 = 9 [1] [6] Examiner insight: On a “solve using the table” part, show which y-value you used. Giving both x-values matters because a quadratic equation often has two solutions.

4. (a) y = a(x + 2)(x − 6) [1]. Substituting (0, 12): a × 2 × (−6) = 12, so a = −1 [1]. y = −(x² − 4x − 12) = −x² + 4x + 12 [1] (b) The axis is x = (−2 + 6)/2 = 2 [1]. y = −4 + 8 + 12 = 16, so the turning point is (2, 16) [1] [5] Examiner insight: Starting from the factorised form keeps the working short. Check your final equation by testing the point (0, 12) and one root.

5. (a) x(x² − 36) [1] = x(x − 6)(x + 6) [1]. The curve crosses the x-axis at x = 0, x = 6 and x = −6 [1] (b) The curve is rotationally symmetrical about the origin, so (3, −81) maps to (−3, 81) [1]. y = 81 [1] [5] Examiner insight: Stopping at x(x² − 36) is partial factorisation, so take it to linear factors before reading off roots. Give a reason for the symmetry, not only the value.

6. (a) k = xy = 5 × 2 = 10 [1] (b) y = 10/(−8) = −1.25 [1] (c) k is positive, so the curve is in the first and third quadrants [1]. The equation is undefined at x = 0 (you cannot divide by zero), so no point has x = 0 [1] [4] Examiner insight: The explanation for the y-axis should mention that x = 0 is not allowed; “it just goes up forever” does not explain it.

7. A turning point on the x-axis means the curve touches the axis, so the quadratic is a perfect square: x² − 10x + k = (x − a)² [1]. Comparing the x terms, −2a = −10, so a = 5 [1]. Then k = a² = 25 [1] [3] Examiner insight: Check the answer by substituting back: x² − 10x + 25 = (x − 5)² has its minimum at (5, 0), on the axis.

8. (a) (x + 4)² − 16 − 5 [1] = (x + 4)² − 21 [1] (b) (−4, −21) [1] (c) (x + 4)² = 21 [1], so x = −4 + √21 and x = −4 − √21 [1] [5] Examiner insight: The turning point is (−p, q), so the sign of the number in the bracket flips. Leaving roots as exact surds is the neater answer; a decimal alone loses the exact form.

9. (a) 5 + 6x − x² = −(x² − 6x) + 5 = −[(x − 3)² − 9] + 5 [1] = 14 − (x − 3)² [1] (b) A maximum, at (3, 14) [1] (c) y ≤ 14 [1] [4] Examiner insight: Take care with the minus sign when factoring out the negative: the 9 changes sign inside the bracket. Expand your answer to check it returns the original expression.

10. (a) k² = 6.25 [1], so k = 2.5 (k is positive) [1] (b) 2.5⁻¹ = 1/2.5 = 0.4 [1] (c) 2.5³ = 15.625 [1] (d) (0, 1) [1] [5] Examiner insight: Substitute the point into y = kˣ to form an equation first. Remember that kˣ passes through (0, 1) for every positive k, which gives a quick check.

11. (a) Greatest value of sin x is 1 [1]; the period of tan x is 180° [1] (b) The first angle is 210° [1], and the second is 330°. Solutions: x = 210° and x = 330° [1] (c) cos 140° = −0.766 [1], because cos 140° = −cos 40°. Another angle is 360° − 40° = 320° [1] [6] Examiner insight: Within 0° to 360°, sin x = −0.5 has two solutions, so give both. Sketching the wave and drawing the line y = −0.5 shows where the second one comes from.

12. (a) (x − 6)(x + 8) = 0 [1] so x = 6 and x = −8 [1] (b) x² + 2x − 48 = (x + 1)² − 1 − 48 [1] = (x + 1)² − 49 [1]. The turning point is (−1, −49) [1] (c) U-shaped curve crossing the x-axis at (−8, 0) and (6, 0) and the y-axis at (0, −48) [1], with a minimum at (−1, −49), symmetrical about x = −1 [1] (d) x² + 2x − 48 = −45 gives x² + 2x − 3 = 0 [1], so (x + 3)(x − 1) = 0 and x = −3 and x = 1 [1] [9] Examiner insight: A sketch earns credit for the labelled features, so write the coordinates on it. In part (d) the answers must be symmetrical about x = −1, which gives a quick check: −3 and 1 average to −1.

Where marks are usually lost

• Giving the turning point as one number instead of a coordinate pair.
• Getting the sign wrong in a completed square, so (x + 4)² − 21 is read as a turning point at x = 4.
• Factorising only part of an expression, such as stopping at x(x² − 36).
• Stating one solution when the equation or the interval needs two.
• Joining the two branches of y = k/x or letting them reach an axis.
• Drawing y = kˣ below the x-axis or leaving out (0, 1).
• Forgetting that the cosine and sine curves have two solutions in a full turn.
• Reading values off a table without saying which y-value was used.
• Leaving a sketch unlabelled when intercepts or the turning point were asked for.

Next steps

• Read the revision notes.
• Re-read the study guide for any question you could not start.
• Return to the course hub and tick off the checklist.
• Take any of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1 (for exams May/June 2018 onwards), section 3.2.2, references A12 and A16, published by OxfordAQA.