Cambridge O Level Mathematics: Graphs of Functions and Sketching Curves — 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 available free from your board.

Related: Graphs of Functions and Sketching Curves revision notes

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Section A

1. State the shape of the graph of (a) y = 3x² − 2, (b) y = 6/x, (c) y = 2ˣ. [3]

2. Find the gradient of the line perpendicular to y = 4x − 1. [2]

3. For y = 3x² − 2, state the y-intercept and state whether the curve has a maximum or a minimum turning point. [2]

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Section B

4. Consider y = x² − 4x − 5.

(a) Find the coordinates where the curve crosses the axes. [4]

(b) Find the coordinates of the turning point. [3]

(c) Sketch the curve, labelling all points found. [2]

5. The graph of y = x² − 3x is drawn. By adding a suitable straight line, solve x² − 4x + 1 = 0 graphically.

(a) Determine the equation of the line that must be added. [3]

(b) Explain how the solutions are read from the graph. [2]

6. A car’s journey is shown on a speed–time graph. It accelerates uniformly from rest to 20 m s⁻¹ in 8 s, travels at constant speed for 12 s, then decelerates uniformly to rest in 5 s.

(a) Calculate the acceleration during the first stage. [2]

(b) Calculate the total distance travelled. [4]

(c) State what the gradient and the area under a speed–time graph each represent. [2]

7. Explain how to estimate the gradient of a curve at a specific point using a tangent. [3]

8. State how many times a cubic graph can cross the x-axis, and explain why this varies. [2]

9. Distinguish between a vertical and a horizontal asymptote, using the reciprocal graph y = a/x + b as your example, and state which type of asymptote an exponential graph has. [3]

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Answers

1. (a) Parabola, opening upward [1]. (b) Hyperbola, with asymptotes at both axes [1]. (c) Exponential growth curve, never touching the x-axis [1].

2. Gradient of given line = 4 [1]. Perpendicular gradient = −1/4 [1].

3. y-intercept: x = 0 gives y = −2, so (0, −2) [1]. The x² coefficient is positive, so the curve has a minimum turning point [1].

4. (a) y-intercept: x = 0 gives y = −5, so (0, −5) [1]. x-intercepts: x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 [1] x = 5 or x = −1 [1], so (5, 0) and (−1, 0) [1].

(b) Complete the square: (x − 2)² − 4 − 5 = (x − 2)² − 9 [1] [1] Turning point (2, −9) [1]. (Or: line of symmetry midway between roots, x = (5 + (−1))/2 = 2.)

(c) Upward parabola through the three intercepts with minimum at (2, −9) [1], smooth curve correctly labelled [1].

5. (a) We need x² − 4x + 1 = 0. Rearrange so one side matches the drawn curve: x² − 3x = x − 1 [1] [1] So add the line y = x − 1 [1].

(b) Read the x-coordinates of the points where the line crosses the curve [1] — these are the solutions of the equation [1]. The answer is the x-value, not the coordinate pair.

6. (a) a = Δv ÷ Δt = 20 ÷ 8 [1] = 2.5 m s⁻² [1].

(b) Area = triangle + rectangle + triangle = (½ × 8 × 20) + (12 × 20) + (½ × 5 × 20) [1] = 80 + 240 + 50 [1] [1] = 370 m [1].

(c) Gradient = acceleration [1]; area under the graph = distance travelled [1].

7. Draw a tangent — a straight line touching the curve at that one point only, locally, without crossing it there [1]. Read off two points on the tangent line [1] and calculate its gradient as change in y over change in x [1].

8. A cubic may cross the x-axis once, twice (with a repeated root), or three times [1] — not every cubic has exactly 3 roots; it depends on the specific cubic [1].

9. A vertical asymptote is a line the curve approaches but never touches, where the function is undefined — for y = a/x + b this occurs at x = 0, where the denominator is zero [1]. A horizontal asymptote is the line the curve approaches as x becomes very large or very negative — for y = a/x + b this is the line y = b [1]. A reciprocal graph has both; an exponential graph y = arˣ + b has a horizontal asymptote only [1] — it never has a vertical one, since arˣ is defined for every real value of x.

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Where marks are usually lost

• Forgetting that a positive x² coefficient gives a minimum, and a negative one a maximum.
• Giving the intersection coordinates instead of just the x-values.
• Forgetting the y-intercept when asked for all axis crossings.
• Joining plotted points with straight segments instead of a smooth curve.
• Confusing gradient with area on a speed–time graph.
• Drawing an inaccurate tangent that crosses the curve rather than just touching it at one point.
• Assuming every cubic has exactly three roots.
• Sketching a curve touching or crossing its own asymptote.