AQA GCSE Mathematics 8300: Graphs – 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 section 3.2.2, Graphs (A8 to A16), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Questions and parts marked (Higher) use content from the specification’s “Higher content only” column; everything else suits both tiers. Each question says whether it is non-calculator (Paper 1 style) or calculator (Papers 2 and 3 style).

Learn the content first in the Graphs study guide and the Graphs revision notes. For mixed Topic 2 questions, try the Algebra practice questions. The course hub is AQA GCSE Mathematics and the printable checklist lists every statement.

Questions

1. (non-calculator) A(−3, 2), B(1, 2) and C(1, −4) are three corners of a rectangle ABCD.

(a) Write down the coordinates of D. [1] (b) State which quadrant C is in. [1]

2. (non-calculator) A line has equation y = 4 − 3x.

(a) Complete the table of values for x = −1, 0, 1, 2, 3. [2] (b) Write down the coordinates of the point where the line crosses the y-axis. [1] (c) Show whether the point (5, −11) lies on the line. [1]

3. (non-calculator) Line L has equation 5x + 2y = 14.

(a) Find the gradient and the y-intercept of L. [2] (b) Which of these lines is parallel to L? y = 2.5x + 1, 2y = −5x + 3, y = −5x + 7, 5y = 2x [1]

4. (non-calculator) A line passes through (−4, 5) and (2, 2).

(a) Find the equation of the line in the form y = mx + c. [3] (b) (Higher) Find the equation of the line perpendicular to it that passes through (1, 1). [2]

5. (calculator) A plumber’s total charge, £C, for a job lasting h hours is shown by a straight-line graph through (1, 70) and (4, 145).

(a) Work out the gradient of the line. [2] (b) Find the equation of the line, then explain what the gradient and the intercept mean in this context. [2]

6. (non-calculator) The curve y = x² − 4x − 12 is drawn on a grid.

(a) Find the roots of x² − 4x − 12 = 0 by factorising. [2] (b) Write down the coordinates of the y-intercept. [1] (c) Use the symmetry of the curve to find the turning point. [2] (d) State the line you would draw on the grid to solve x² − 4x − 12 = 9, and give the solutions. [2]

7. (non-calculator) (Higher) f(x) = x² + 8x + 11.

(a) Write f(x) in the form (x + p)² + q. [2] (b) Write down the turning point of y = f(x), and state whether it is a maximum or a minimum, with a reason. [2] (c) Write down the turning point of y = f(x) + 5. [1] (d) Write down the turning point of y = f(x − 4). [1] (e) Write down the turning point of y = −f(x). [1]

8. (non-calculator) Match each description to one of the equations y = x², y = 1/x, y = 2 − x, y = x³.

(a) A straight line with negative gradient crossing the y-axis at 2. [1] (b) Two separate curves in the first and third quadrants that never meet the axes. [1] (c) A curve through the origin, symmetrical about the y-axis, with a minimum at (0, 0). [1] (d) A curve through the origin rising from bottom left to top right, flat at (0, 0). [1]

9. (calculator) (Higher)

(a) Complete a table of values for y = 2ˣ for x = −1, 0, 1, 2, 3, 4. [2] (b) Use your graph of y = 2ˣ to estimate the solution of 2ˣ = 10. [2] (c) For y = sin x with 0° ≤ x ≤ 360°, write down the maximum value of y and the value of x where it occurs. [1] (d) Given sin 30° = 0.5, use the graph of y = sin x to solve sin x = 0.5 for 0° ≤ x ≤ 360°. [2]

10. (calculator) Amira leaves home at 09:00 and cycles 6 km to a park, arriving at 09:20. She stays until 09:45, then cycles home, arriving at 10:15. Her journey is shown on a distance-time graph.

(a) Work out her speed on the way to the park, in km/h. [2] (b) Work out her speed on the way home, in km/h. [2] (c) Describe the section of the graph from 09:20 to 09:45 and say what it shows. [1]

11. (calculator) (Higher) A car’s velocity, v m/s, t seconds after it starts is shown by a smooth curve through these points.

t (s)	0	5	10	15	20
v (m/s)	0	5.25	9	11.25	12

(a) Use four strips of equal width to estimate the distance travelled in the first 20 seconds. [3] (b) The curve bends downwards between these points. Is your answer an overestimate or an underestimate? Give a reason. [1] (c) A tangent drawn to the curve at t = 10 passes through (0, 3) and (20, 15). Estimate the acceleration at t = 10. [2]

12. (non-calculator) (Higher) A circle has equation x² + y² = 45. P is the point (6, 3).

(a) Show that P lies on the circle. [1] (b) Write the radius in the form a√5. [1] (c) Find the equation of the tangent to the circle at P. [4] (d) The tangent meets the x-axis at Q. Find the coordinates of Q. [1]

Answers

1. (a) D has the x-coordinate of A and the y-coordinate of C: D(−3, −4) [1] (b) x positive, y negative: fourth quadrant [1] Examiner insight: Coordinates written the wrong way round, as (−4, −3), score zero even when the point is marked correctly on a diagram.

2. (a) y = 7, 4, 1, −2, −5 [1] for any three correct, all five correct [1] (b) (0, 4) [1] (c) 4 − 3(5) = −11, so yes, it lies on the line [1] Examiner insight: In (c) a bare “yes” earns nothing; the substitution must be shown with a conclusion.

3. (a) 2y = −5x + 14, so y = −2.5x + 7 [1]; gradient −5/2, y-intercept 7 [1] (b) 2y = −5x + 3 gives y = −2.5x + 1.5, so 2y = −5x + 3 [1] Examiner insight: Giving the gradient as 5 or −5 shows the equation was not rearranged and loses both marks in (a).

4. (a) m = (2 − 5) ÷ (2 − (−4)) = −3/6 = −1/2 [1]; 5 = (−1/2)(−4) + c, so c = 3 [1]; y = −½x + 3 [1] (b) Perpendicular gradient = 2 [1]; 1 = 2(1) + c gives c = −1, so y = 2x − 1 [1] Examiner insight: A correct method for c with a wrong gradient from (a) usually earns follow-through credit, so always show the substitution.

5. (a) Gradient = (145 − 70) ÷ (4 − 1) [1] = 25 [1] (b) 70 = 25(1) + c gives c = 45, so C = 25h + 45 [1]. The gradient means £25 for each hour, and the intercept means a fixed £45 charge before any time is worked [1] Examiner insight: The interpretation mark needs both meanings in context with units; “25 is the gradient” restates the maths and scores nothing.

6. (a) (x − 6)(x + 2) = 0 [1]; x = 6 or x = −2 [1] (b) (0, −12) [1] (c) Line of symmetry x = (−2 + 6) ÷ 2 = 2 [1]; y = 4 − 8 − 12 = −16, so (2, −16) [1] (d) Draw y = 9 [1]; it meets the curve at x = −3 and x = 7 [1] Examiner insight: In (d), drawing y = 9 is the method mark; the solutions must match the readings at both intersections, not just one.

7. (a) (x + 4)² − 16 + 11 [1] = (x + 4)² − 5 [1] (b) (−4, −5) [1]; a minimum, because (x + 4)² ≥ 0 so y is never below −5 [1] (c) (−4, 0) [1] (d) (0, −5) [1] (e) (−4, 5) [1] Examiner insight: In (b) a turning point of (4, −5) shows the sign error in reading (x + p)²; follow-through is then allowed in (c) to (e) only if each transformation is applied correctly.

8. (a) y = 2 − x [1] (b) y = 1/x [1] (c) y = x² [1] (d) y = x³ [1] Examiner insight: Each match is a separate mark, so one wrong pairing does not cost the others, but using one equation twice loses a mark.

9. (a) y = 0.5, 1, 2, 4, 8, 16 [1] for at least four correct, all six correct [1] (b) Draw y = 10 to meet the curve [1]; x ≈ 3.3 (accept 3.2 to 3.4) [1] (c) Maximum 1 at x = 90° [1] (d) x = 30° [1]; by symmetry about 90°, x = 30° and x = 150° [1] Examiner insight: An answer read from a graph is accepted within a range around the true value; a trial-and-improvement answer with no line drawn may lose the method mark.

10. (a) 6 km in 20 minutes [1]; 6 × 3 = 18 km/h [1] (b) 6 km in 30 minutes (09:45 to 10:15) [1]; 6 × 2 = 12 km/h [1] (c) A horizontal line: she is stationary at the park [1] Examiner insight: Dividing 6 by 20 and giving 0.3 (km per minute) earns only the method mark when km/h is asked; the time must be converted to hours.

11. (a) Strip width 5 [1]; area = (5/2) × [0 + 2(5.25 + 9 + 11.25) + 12] [1] = 2.5 × 63 = 157.5 m [1] (b) Underestimate: the curve bends downwards, so the tops of the trapezia lie below the curve [1] (c) Gradient = (15 − 3) ÷ (20 − 0) [1] = 0.6 m/s² [1] Examiner insight: An estimate from drawn tangents and trapezia is credited within a tolerance, but units are expected; “157.5” alone may lose the final mark on a units question.

12. (a) 6² + 3² = 36 + 9 = 45, so P is on the circle [1] (b) √45 = √9 × √5 = 3√5 [1] (c) Gradient of OP = 3/6 = 1/2 [1]; tangent gradient = −2 [1]; 3 = −2(6) + c [1]; c = 15, so y = −2x + 15 [1] (d) 0 = −2x + 15, x = 7.5: Q(7.5, 0) [1] Examiner insight: Using the radius gradient 1/2 as the tangent gradient loses the second and final marks in (c), though the substitution mark can still be earned.

Where marks are usually lost

• Reading the gradient from 5x + 2y = 14 without making y the subject.
• Subtracting coordinates in different orders, so the gradient has the wrong sign.
• Keeping the same sign when finding a perpendicular gradient.
• Giving one root, or one intersection, when there are two.
• Writing (4, −5) instead of (−4, −5) as the turning point of (x + 4)² − 5.
• Moving y = f(x − 4) left instead of right.
• Leaving time in minutes when the speed is asked for in km/h.
• Leaving out the over- or underestimate reason, or the units, for areas and gradients.
• Using the radius gradient as the tangent gradient.

Next steps

• Graphs revision notes
• Graphs study guide
• Algebra practice questions
• AQA GCSE Mathematics course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.2 Graphs (A8 to A16).