Cambridge IGCSE Mathematics 0580: 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 from your board or school.

These questions cover 2.10 Graphs of functions and 2.11 Sketching curves in the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (June and November series, and March in India). Questions 1–5 use Core content (C2.10, C2.11), examined on Paper 1 (non-calculator) and Paper 3 (calculator). Questions 6–11 are marked (Extended): they use E2.10 and E2.11 content examined on Paper 2 (non-calculator) and Paper 4 (calculator). Where a question asks you to draw a graph, use graph paper; readings from a graph are accepted within a small tolerance.

They differ from the Algebra and graphs practice questions and the Extended algebra practice. For teaching and worked examples, read the graphs of functions and sketching curves study guide first.

Questions

1. (non-calculator) The table is for y = x² + x − 6.

x   -4   -3   -2   -1    0    1    2    3
y    6    ..   -4   ..   -6   ..    0   ..

(a) Complete the table (each .. is a missing value). [2] (b) Write down the roots of x² + x − 6 = 0. [1] (c) Write down the equation of the line of symmetry of the graph. [1]

2. (calculator) The table is for y = 8/x.

x   -4   -2   -1   -0.5   0.5   1    2    4
y   -2   ..   -8   ..     16    ..   4    ..

(a) Complete the table. [2] (b) Explain why the table has no value for x = 0. [1] (c) The graph of y = 8/x is drawn. Write down the equation of the line you would draw to solve 8/x = 2x, and give the solutions. [2]

3. (non-calculator) Sketch the graph of y = 12 + x − x². Show the coordinates of the points where it crosses the axes and the line of symmetry. [4]

4. (non-calculator) Match each equation to one description.

Equations: A y = 4 − 2x, B y = 6 − x², C y = x² − 3x, D y = 7/x.

Descriptions: P a U-shaped curve through the origin; Q two separate branches in the first and third quadrants; R a straight line with negative gradient; S an n-shaped curve crossing the y-axis at (0, 6). [3]

5. (calculator) The graph of y = x² − 3x − 2 has been drawn for −1 ≤ x ≤ 5.

(a) Find the equation of the straight line you need to draw on the same grid to solve x² − 4x + 1 = 0. [2] (b) Draw the line and give the solutions to 1 decimal place. [2]

6. (Extended, non-calculator) Sketch y = 3/x − 2. Show both asymptotes and the coordinates of any point where the curve crosses an axis. [4]

7. (Extended, non-calculator)

(a) Write x² − 6x + 11 in the form (x + p)² + q. [2] (b) Write down the coordinates of the turning point of y = x² − 6x + 11 and say whether it is a maximum or a minimum. [1] (c) Explain how your answer to (a) shows that x² − 6x + 11 = 0 has no real roots. [1] (d) Sketch y = x² − 6x + 11, showing the y-intercept and the turning point. [2]

8. (Extended, calculator) The mass of a medicine in a patient’s blood, M mg, t hours after a dose, is modelled by M = 200 × 0.8ᵗ.

(a) Complete the table, giving values to the nearest whole number.

t   0     1     2     3     4     5     6
M   200   160   ..    102   ..    ..    52

[2] (b) Draw the graph for 0 ≤ t ≤ 6 and use it to estimate when M = 100. [2] (c) State whether the model shows growth or decay, and describe what happens to M as t keeps increasing. [2]

9. (Extended, calculator) The table is for y = x³ − 4x + 2.

x   -3   -2   -1    0    1    2    3
y   ..    2    ..   2    ..   2    ..

(a) Complete the table. [2] (b) Draw the graph for −3 ≤ x ≤ 3. State how many roots x³ − 4x + 2 = 0 has, and estimate each one to 1 decimal place. [3] (c) Find the equation of the straight line you would draw to solve x³ − 5x + 1 = 0, and state how many solutions this equation has for −3 ≤ x ≤ 3. [3]

10. (Extended, non-calculator)

(a) Factorise x³ − x² − 6x completely. Hence sketch y = x³ − x² − 6x, labelling the points where it crosses the axes. [4] (b) Sketch y = −2x³ + 16, labelling the points where it crosses the axes. [3]

11. (Extended, calculator) The graph of y = x² + 4/x is to be drawn for 0.5 ≤ x ≤ 3.

x   0.5    1    1.5    2    2.5    3
y   8.25   ..   4.92   ..   7.85   10.33

(a) Complete the table. [2] (b) Draw the graph and use it to solve x² + 4/x = 6. [2] (c) Write down the equation of the vertical asymptote of y = x² + 4/x. [1] (d) Use your graph to explain why x² + 4/x = 4 has no solution for x > 0. [1]

Answers

1. (a) x = −3: 9 − 3 − 6 = 0; x = −1: 1 − 1 − 6 = −6 [1]. x = 1: 1 + 1 − 6 = −4; x = 3: 9 + 3 − 6 = 6 [1]. Row: 6, 0, −4, −6, −6, −4, 0, 6. (b) Where y = 0: x = −3 and x = 2 [1]. (c) Halfway between the roots: x = −0.5 [1]. Examiner insight: in (a), (−3)² must be worked as +9; a value that breaks the table’s symmetry usually loses the accuracy mark and then the plotting marks that follow from it.

2. (a) Missing values: x = −2: −4; x = −0.5: −16; x = 1: 8; x = 4: 2. [1] for any two correct, [1] for all four correct. (b) 8 ÷ 0 cannot be calculated, so y = 8/x has no value at x = 0 [1]. (c) Draw y = 2x [1]. The line meets both branches: x = −2 and x = 2 [1] (check: 8/2 = 4 = 2 × 2). Examiner insight: when asked for the line, an equation such as y = 2x is needed; writing “the line 2x” or giving only the solutions does not earn the method mark.

3. y = −(x² − x − 12) = −(x − 4)(x + 3) [1]. Roots (−3, 0) and (4, 0) [1]. The x² coefficient is negative, so an n-shape through (0, 12) [1]. Symmetrical about x = 0.5 [1]. Examiner insight: sketch marks are for the shape and the labelled intercepts; an n-shape drawn as a U loses the shape mark even if every intercept is correct.

4. A → R and D → Q [1]. B → S [1]. C → P [1] (x² − 3x = x(x − 3) passes through the origin). Examiner insight: each mark depends on one feature (sign of x², constant term, a/x form); check that feature rather than guessing.

5. (a) x² − 4x + 1 = 0 becomes x² − 3x − 2 = x − 3 (add x − 3 to both sides) [1]. Draw y = x − 3 [1]. (b) From an accurate graph, x ≈ 0.3 and x ≈ 3.7 [1] [1] (exact values 2 ± √3 = 0.268 and 3.73). Examiner insight: check your line by subtracting it from the curve’s equation; if the result is not the equation you were given, the method mark is lost and both solutions with it.

6. Vertical asymptote x = 0 [1]. Horizontal asymptote y = −2 [1]. Root: 3/x = 2, so x = 1.5, giving (1.5, 0) [1]. Two branches, top-right and bottom-left of the point (0, −2), each approaching both asymptotes without touching them [1]. Examiner insight: asymptotes should be drawn as dashed lines and labelled with their equations; a curve that crosses y = −2 loses the shape mark.

7. (a) x² − 6x + 11 = (x − 3)² − 9 + 11 [1] = (x − 3)² + 2 [1]. (b) (3, 2), a minimum [1]. (c) (x − 3)² ≥ 0, so (x − 3)² + 2 ≥ 2 and can never equal 0 [1]. (d) U-shape crossing the y-axis at (0, 11) [1], lowest point at (3, 2) and entirely above the x-axis [1]. Examiner insight: in (b) the sign of p flips: (x − 3)² gives x = 3, and a turning point written as (−3, 2) earns no credit.

8. (a) 200 × 0.8² = 128, 200 × 0.8⁴ = 81.92 = 82, 200 × 0.8⁵ = 65.536 = 66 [1] for two correct, [1] for all three. (b) Smooth decreasing curve through the plotted points [1]. Read across from M = 100: t ≈ 3.1 hours [1] (accept 3.0 to 3.2). (c) Decay, because the multiplier 0.8 is less than 1 [1]. M keeps decreasing towards 0 but never reaches it; the curve gets closer and closer to the t-axis [1]. Examiner insight: a reading from a graph is credited within a tolerance; show your reading line on the graph so the method can be seen if your value is slightly outside it.

9. (a) x = −3: −27 + 12 + 2 = −13; x = −1: −1 + 4 + 2 = 5 [1]. x = 1: 1 − 4 + 2 = −1; x = 3: 27 − 12 + 2 = 17 [1]. Row: −13, 2, 5, 2, −1, 2, 17. (b) The curve crosses the x-axis three times [1]. Roots x ≈ −2.2 [1], x ≈ 0.5 and x ≈ 1.7 [1] (accurate values −2.21, 0.539, 1.68). (c) x³ − 5x + 1 = 0 becomes x³ − 4x + 2 = x + 1 [1]. Line y = x + 1 [1]. It meets the curve three times, so 3 solutions [1]. Examiner insight: the sign changes in the table (−13 to 2, 2 to −1, −1 to 2) tell you where the three roots are before you draw; a root missing from one of those intervals shows a plotting error.

10. (a) x(x² − x − 6) [1] = x(x − 3)(x + 2) [1]. Roots (−2, 0), (0, 0), (3, 0), and the curve crosses the y-axis at the origin [1]. Positive x³ term: rises from bottom left, up through −2, turns, down through 0, turns, up through 3 [1]. (b) y-intercept (0, 16) [1]. Root: 2x³ = 16, x³ = 8, x = 2, giving (2, 0) [1]. Negative x³ term: the curve falls from top left to bottom right with the cubic’s flattened S-shape, flattest where it crosses the y-axis [1]. Examiner insight: “factorise completely” needs all three factors including x; stopping at x(x² − x − 6) earns only the first mark.

11. (a) x = 1: 1 + 4 = 5 [1]. x = 2: 4 + 2 = 6 [1]. (b) Draw y = 6 [1]. It meets the curve at x ≈ 0.7 and x = 2 [1] (exact values −1 + √3 = 0.732 and 2). (c) x = 0 [1]. (d) The lowest point of the curve for x > 0 is about y = 4.8 (near x = 1.3), so the line y = 4 does not meet the curve [1]. Examiner insight: in (d) the explanation must refer to the graph (the curve stays above y = 4); solving algebraically or saying “it doesn’t work” earns nothing.

Where marks are usually lost

• Squaring negatives wrongly in tables, so one plotted point is out of line with the rest.
• Ruled segments between plotted points on a curve instead of one smooth curve.
• Joining the two branches of a reciprocal graph through x = 0.
• Choosing the wrong line for a graphical solution because the given equation was not rearranged to “curve = line”.
• Giving a y-value when the question asks for the x-value of an intersection.
• Unlabelled intercepts or asymptotes on a sketch.
• (Extended) Sign error in the turning point from completing the square.
• (Extended) An exponential or reciprocal curve drawn crossing its asymptote.

Next steps

• Graphs of functions and sketching curves study guide
• Algebra and graphs revision notes
• Algebra and graphs study guide
• Course hub: Cambridge IGCSE Mathematics
• Printable checklist: 0580 checklist
• Free 10-minute diagnostics: Core and Extended
• Book a free trial class

Official syllabus

Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (Version 3), Cambridge University Press & Assessment: subject content C2.10, C2.11, E2.10 and E2.11.