OxfordAQA International GCSE Mathematics: Functions, coordinates and straight-line graphs (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 follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, section 3.2.2, statements A9, A10 and A11. Questions 1 to 5 use Core content, which can also be assessed on the Extension papers. Questions 6 to 11 are marked (Extension only). Read the study guide first, or use the revision notes. The course hub and checklist show the full course.

Questions

1. A function machine adds 7 to the input x, then multiplies the result by 3.

(a) Find the output when x = −4. [1] (b) The output is 42. Find the input. [2] (c) Write the output as an expression in x. [1]

2. T(2, −3) and U(−4, −3) are two vertices of an isosceles triangle TUV, with TV = UV. V is in the 2nd quadrant and is 5 units above the line TU.

(a) Find the coordinates of V. [2] (b) Write down the equation of the line of symmetry of the triangle. [1] (c) Work out the area of triangle TUV. [2]

3. A line has equation y = 2x − 3.

(a) Complete a table of values for x = −2, 0 and 3. [2] (b) On squared paper, draw the graph for values of x from −2 to 3. [2] (c) At what point does the line meet the x-axis? [2]

4. A line has equation 5x − 2y = 8.

(a) Rearrange the equation into the form y = mx + c. [2] (b) Write down the gradient and the y-intercept. [1] (c) Find the x-intercept. [1] (d) Which of these lines is parallel to 5x − 2y = 8? y = 2.5x + 1, y = −2.5x + 1, y = 0.4x + 1. [1]

5. Line L₁ has equation y = 3x + 2. Line L₂ passes through (0, −4) and (2, 2). Show that L₁ and L₂ are parallel. [3]

6. (Extension only) f(x) = 4x − 3 and g(x) = x² + 1.

(a) Find f(5) and g(−3). [2] (b) Find fg(x), giving your answer in its simplest form. [2] (c) Solve fg(x) = 37. [2]

7. (Extension only) h(x) = 3 − x², for values of x from −2 to 3 inclusive.

(a) Find the range of h. [3] (b) The function m(x) = 4/(x + 5) has no output for one particular input. Write down that input. [1]

8. (Extension only) f(x) = (5x + 2)/3.

(a) Find f⁻¹(x). [3] (b) Find f⁻¹(4). [1] (c) Solve f(x) = f⁻¹(x). [3]

9. (Extension only) A is the point (−5, −1) and B is the point (3, 5).

(a) Find the equation of line AB, in the form 4y = ax + b. [3] (b) The point (k, 8) lies on line AB. Find k. [2]

10. (Extension only) Line L has equation 5x + 2y = 18.

(a) Find the gradient of L. [2] (b) A second line is at right angles to L and goes through (5, −2). Write its equation in the form y = mx + c. [3] (c) Find the x-coordinate of the point where this perpendicular meets the x-axis. [1]

11. (Extension only) A(1, 2), B(7, 4) and C(6, 7) are three points.

(a) Show that AB is perpendicular to BC. [3] (b) Find the equation of line BC. [2] (c) Find the equation of the line through A that is parallel to BC, and where it crosses the y-axis. [2] (d) ABCD is a rectangle. Find the coordinates of D. [2]

Answers

1. (a) (−4 + 7) × 3 = 3 × 3 = 9 [1] (b) 42 ÷ 3 = 14 [1], then 14 − 7 = 7 [1] (c) 3(x + 7) [1] Examiner insight: Working backwards by undoing each step in reverse order is quick and shows method. If you write 3x + 7 in part (c), the brackets are lost and the order of the machine is not recorded.

2. (a) V lies on the line of symmetry, halfway between x = −4 and x = 2, so its x-coordinate is −1 [1]. Its y-coordinate is −3 + 5 = 2, so V = (−1, 2) [1] (b) The line of symmetry is vertical and passes through V, so it is x = −1 [1] (c) Base TU = 2 − (−4) = 6 [1], so the area is ½ × 6 × 5 = 15 square units [1] Examiner insight: When a coordinate is negative, subtract carefully. Writing 2 − (−4) rather than 2 − 4 shows the method, and a quick sketch of the triangle helps you catch a misplaced vertex.

3. (a) x = −2 gives y = −7 [1]. x = 0 gives y = −3 and x = 3 gives y = 3 [1] (b) Plot (−2, −7), (0, −3) and (3, 3) accurately [1], then draw one straight ruled line through them from x = −2 to x = 3 [1] (c) At the x-axis y = 0, so 2x − 3 = 0 [1], giving x = 1.5. The point is (1.5, 0) [1] Examiner insight: A line drawn freehand or joining points with short segments risks losing credit. Write the coordinate pair with the x-value first, and read it from your own graph as a check on the calculation.

4. (a) 5x − 8 = 2y [1], so y = 2.5x − 4 [1] (b) Gradient 2.5, y-intercept −4 [1] (c) Set y = 0: 5x = 8, so x = 1.6 [1] (d) y = 2.5x + 1 [1] Examiner insight: Dividing every term by 2 is a frequent slip, so check that the −8 also becomes −4. Reading the gradient as 5 from the original equation is a common error.

5. Gradient of L₁ = 3 [1]. Gradient of L₂ = (2 − (−4))/(2 − 0) [1] = 6/2 = 3. The gradients are equal, so the lines are parallel [1] Examiner insight: “Show that” needs the working and a closing statement. Compare the gradients explicitly, and say what equal gradients mean. A bare answer of 3 would not make the argument.

6. (a) f(5) = 4(5) − 3 = 17 [1]. g(−3) = (−3)² + 1 = 10 [1] (b) fg(x) = 4(x² + 1) − 3 [1] = 4x² + 1 [1] (c) 4x² + 1 = 37, so 4x² = 36 and x² = 9 [1], giving x = 3 or x = −3 [1] Examiner insight: Put g inside f, not the other way round. Putting f(x) into g would give a different expression. Give both roots when you take a square root.

7. (a) h(−2) = 3 − 4 = −1, h(0) = 3 and h(3) = 3 − 9 = −6 [1]. The largest value is 3, at x = 0 [1]. The smallest value is −6, so the range is −6 ≤ h(x) ≤ 3 [1] (b) x = −5 [1] Examiner insight: The largest output of h is not at either end of the domain. Checking the turning point is what separates a full answer from the endpoints-only response of −1 to −6.

8. (a) y = (5x + 2)/3, so 3y = 5x + 2 [1], then 5x = 3y − 2 [1], so f⁻¹(x) = (3x − 2)/5 [1] (b) (12 − 2)/5 = 2 [1] (c) (5x + 2)/3 = (3x − 2)/5, so 5(5x + 2) = 3(3x − 2) [1], which gives 25x + 10 = 9x − 6 [1], so 16x = −16 and x = −1 [1] Examiner insight: Test your inverse with part (b): f(2) should equal 4. In part (c), multiply both sides by 15 in one step to clear the two denominators and keep the signs right.

9. (a) Gradient = (5 − (−1))/(3 − (−5)) = 6/8 = 3/4 [1]. y − 5 = (3/4)(x − 3) [1], so 4y − 20 = 3x − 9 and 4y = 3x + 11 [1] (b) 4(8) = 3k + 11 [1], so 3k = 21 and k = 7 [1] Examiner insight: Substitute the other point to check the equation: for A, 4(−1) = −4 and 3(−5) + 11 = −4. When a form is specified, the final line must match it.

10. (a) 2y = −5x + 18 [1], so y = −2.5x + 9 and the gradient is −5/2 [1] (b) The perpendicular gradient is 2/5 [1]. y − (−2) = (2/5)(x − 5) [1], so y = 0.4x − 4 [1] (c) 0.4x − 4 = 0, so x = 4/0.4 = 10 [1] Examiner insight: Turn the gradient upside down and change its sign. Quote the gradient of the original line as a fraction before flipping, so the reciprocal is exact.

11. (a) Gradient AB = (4 − 2)/(7 − 1) = 1/3 [1]. Gradient BC = (7 − 4)/(6 − 7) = −3 [1]. The product is (1/3) × (−3) = −1, so AB is perpendicular to BC and angle ABC is 90° [1] (b) y − 4 = −3(x − 7) [1], so y = −3x + 25 [1] (c) Gradient −3 through A(1, 2): y − 2 = −3(x − 1) [1], so y = −3x + 5, which crosses the y-axis at (0, 5) [1] (d) D = A + C − B = (1 + 6 − 7, 2 + 7 − 4) [1] = (0, 5). Check: gradient CD = (7 − 5)/(6 − 0) = 1/3, the same as AB [1] Examiner insight: A “show that” about perpendicular lines needs both gradients and their product stated. In part (d), notice that the answer lies on your line from part (c); using earlier parts as a check is a good habit.

Where marks are usually lost

• Writing coordinates as (y, x).
• Taking the gradient straight from 5x − 2y = 8 without rearranging.
• Subtracting the coordinates in different orders in the numerator and denominator.
• Missing the turning point when finding the range of a quadratic on a restricted domain.
• Using fg when the question needs gf, or the reverse.
• Giving one root only after solving x² = 9.
• Forgetting to change the sign when finding a perpendicular gradient.
• Drawing a line through only two points, so that a slip is not caught.
• Leaving an equation in a different form from the one the question asks for.

Next steps

• Read the study guide and the revision notes again for any question you missed.
• Visit the course hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.2, statements A9, A10 and A11.