IGCSE Maths 0580 Graphs of Functions Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Algebra and Graphs · 2.10

Every graphs of functions question from 15 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.10. 15 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.

0580 Graphs of Functions past paper questions: Core (Papers 1 and 3)

19 marksCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q18

(a) Complete the table of values for y = x² + x − 3.

x−4−3−2−101234
y9−1−3−3−19

(b) On the grid, draw the graph of y = x² + x − 3 for −4 ≤ x ≤ 4.

(c) Write down the equation of the line of symmetry of the graph.

(d) Use your graph to solve the equation x² + x − 3 = 0.

Show solution
Part (a)
  1. Substitute each x into y = x² + x − 3.
  2. x = −3: (−3)² + (−3) − 3 = 9 − 3 − 3 = 3
  3. x = 2: 2² + 2 − 3 = 4 + 2 − 3 = 3
  4. x = 4: 4² + 4 − 3 = 16 + 4 − 3 = 17
Answer 3 17
Part (b)
  1. Plot all nine points from the table with small crosses.
  2. Join them with one smooth U-shaped curve (do not use a ruler). The lowest point is just below y = −3, between x = −1 and x = 0.
0580/11 May/June 2026 Q18(b) solution – graph of y = x² + x − 3 from x = −4 to 4xy-4-3-2-11234-4-22468101214160
Part (c)
  1. The table has equal y-values at x = −1 and x = 0, so the line of symmetry is halfway between them.
  2. x = −1⁄2
Answer x = −1⁄2
Part (d)
  1. The solutions of x² + x − 3 = 0 are where the curve crosses the x-axis (y = 0).
  2. Read the two x-values: x ≈ −2.3 or x ≈ 1.3
Note: Readings from your graph from −2.4 to −2.2 and 1.2 to 1.4 are accepted.
Answer x ≈ −2.3 or x ≈ 1.3
29 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q13

(a) Complete the table of values for y = 6x − x² − 4.

x0123456
y541−4

(b) Draw the graph of y = 6x − x² − 4 for 0 ≤ x ≤ 6. (c) On the grid, draw the line y = 3. (d) Use your graph to solve 6x − x² − 4 = 3.

Show solution
Part (a)
  1. Substitute each x into y = 6x − x² − 4.
  2. x = 0: 0 − 0 − 4 = −4
  3. x = 1: 6 − 1 − 4 = 1
  4. x = 2: 12 − 4 − 4 = 4
Answer −4 1 4
Part (b)
  1. Plot all seven points from the table with small crosses.
  2. Join them with a smooth ∩-shaped curve. Its highest point is (3, 5).
Part (c)
  1. Draw a horizontal ruled line across the grid at y = 3.
0580/13 May/June 2026 Q13 solution – graph of y = 6x − x² − 4 with the line y = 3xy123456-4-2246y = 3
Part (d)
  1. The equation 6x − x² − 4 = 3 is true where the curve and the line y = 3 meet.
  2. Read the two x-values at the crossing points: x ≈ 1.6 or x ≈ 4.4
Note: Readings from 1.5 to 1.7 and 4.3 to 4.5 are accepted.
Answer x ≈ 1.6 or x ≈ 4.4
310 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q18

(a) Complete the table of values for y = −x² + 3x + 5.

x−2−1012345
y−55−5

(b) On the grid, draw the graph of y = −x² + 3x + 5 for −2 ≤ x ≤ 5.

(c) Write down the coordinates of the highest point on the graph.

(d) Use your graph to solve −x² + 3x + 5 = 0.

Show solution
Part (a)
  1. Substitute each x-value into y = −x² + 3x + 5. Be careful: −x² means −(x²), so it is always negative or zero.
  2. x = −1: −1 − 3 + 5 = 1
  3. x = 0: 0 + 0 + 5 = 5
  4. x = 1: −1 + 3 + 5 = 7
  5. x = 2: −4 + 6 + 5 = 7
  6. x = 4: −16 + 12 + 5 = 1
Answer 1 5 7
Part (b)
  1. Plot every point from the table and join them with a smooth ∩-shaped curve (not straight lines).
  2. The top of the curve is between x = 1 and x = 2, a little above y = 7.
0580/32 May/June 2026 Q18(b) solution – graph of y = −x² + 3x + 5 from x = −2 to 5xy-2-112345-6-5-4-3-2-1123456780
Part (c)
  1. The y-values are symmetrical: 7 at x = 1 and x = 2, so the highest point is half-way, at x = 1.5.
  2. y = −(1.5)² + 3 × 1.5 + 5 = −2.25 + 4.5 + 5 = 7.25.
  3. (1.5, 7.25)
Answer (1.5, 7.25)
Part (d)
  1. −x² + 3x + 5 = 0 means y = 0, so read where the curve crosses the x-axis.
  2. x ≈ −1.2 or x ≈ 4.2
Note: Answers from −1.3 to −1.1 and from 4.1 to 4.3 are accepted.
Answer x ≈ −1.2 or x ≈ 4.2
46 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q16

(a) Complete the table of values for y = 9⁄x.

x−4.5−3−2−1.5−111.5234.5
y−2−3−6−9964.52

(b) On the grid, draw the graph of y = 9⁄x for −4.5 ≤ x ≤ −1 and 1 ≤ x ≤ 4.5.

Show solution
Part (a)
  1. Substitute each x into y = 9⁄x.
  2. x = −2: y = 9 ÷ (−2) = −4.5
  3. x = 3: y = 9 ÷ 3 = 3
Answer −4.5 3
Part (b)
  1. Plot all the points from the table.
  2. Join the points for negative x with one smooth curve and the points for positive x with another smooth curve.
  3. Do not join the two curves across x = 0, because 9 ÷ 0 cannot be worked out.
0580/11 October/November 2025 Q16(b) solution – the two branches of y = 9/xxy-5-4-3-2-112345-9-6-33690
57 marksCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q27

(a) Complete the table of values for y = 12⁄x.

x−4−3−2−11234
y−4−664

(b) Draw the graph of y = 12⁄x for −4 ≤ x ≤ −1 and 1 ≤ x ≤ 4.

(c) Use your graph to write down the solution of the equation 12⁄x = 10.

Show solution
Part (a)
  1. Divide 12 by each x-value.
  2. x = −4: 12 ÷ (−4) = −3
  3. x = −1: 12 ÷ (−1) = −12
  4. x = 1: 12 ÷ 1 = 12
  5. x = 4: 12 ÷ 4 = 3
Answer −3 −12 12 3
Part (b)
  1. Plot all the points. Draw one smooth curve through the negative x-values and another through the positive x-values.
  2. Do not join the two curves: there is no value at x = 0.
0580/32 October/November 2025 Q27(b) solution – the two branches of y = 12/xxy-4-3-2-11234-12-8-448120
Part (c)
  1. Draw across from y = 10 to the curve and read down to the x-axis.
  2. x ≈ 1.2
Note: Answers from 1.1 to 1.3 are accepted. The exact answer is 12 ÷ 10 = 1.2.
Answer x ≈ 1.2
66 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q22

(a) Complete the table of values for y = x² − 4.

x−4−3−2−101234
y120−3−3012

(b) On the grid, draw the graph of y = x² − 4 for −4 ≤ x ≤ 4.

Show solution
Part (a)
  1. Square each x-value, then subtract 4.
  2. x = −3: (−3)² − 4 = 9 − 4 = 5
  3. x = 0: 0² − 4 = −4
  4. x = 3: 3² − 4 = 9 − 4 = 5
Answer 5 −4
Part (b)
  1. Plot all the points from the table.
  2. Join them with a smooth U-shaped curve. Its lowest point is at (0, −4).
0580/33 October/November 2025 Q22(b) solution – graph of y = x² − 4xy-4-3-2-11234-5-4-3-2-11234567891011120
  1. Tip: Do not join the points with straight lines, and do not give the bottom of the curve a flat or pointed shape.
711 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q14

(a) Complete the table of values for y = (x + 3)(x − 2).

x−4−3−2−10123
y6−4−4

(b) On the grid, draw the graph of y = (x + 3)(x − 2) for −4 ≤ x ≤ 3.

0580/12 May/June 2025 Q14 – grid for y = (x + 3)(x − 2), x from −4 to 3, y from −7 to 6xy-4-3-2-1123-7-6-5-4-3-2-11234560

(c) Write down the coordinates of the lowest point of the graph.

(d) Write down the equation of the line of symmetry of the graph.

(e) Use your graph to solve the equation (x + 3)(x − 2) = 3.

Show solution
Part (a)
  1. Substitute each x into y = (x + 3)(x − 2).
  2. x = −3: (0)(−5) = 0. x = −1: (2)(−3) = −6. x = 0: (3)(−2) = −6.
  3. x = 2: (5)(0) = 0. x = 3: (6)(1) = 6.
  4. y = 6, 0, −4, −6, −6, −4, 0, 6
Answer y = 6, 0, −4, −6, −6, −4, 0, 6
Part (b)
  1. Plot all 8 points with small crosses.
  2. Join them with one smooth U-shaped curve (not straight lines). The bottom of the curve is between x = −1 and x = 0, slightly below −6.
  3. Plot the 8 points and join them with a smooth U-shaped curve.
0580/12 May/June 2025 Q14 solution – parabola y = (x + 3)(x − 2) through the plotted pointsxy-4-3-2-1123-7-6-5-4-3-2-11234560
Answer Plot the 8 points and join them with a smooth U-shaped curve.
Part (c)
  1. The y-values are equal at x = −1 and x = 0, so the lowest point is halfway, at x = −0.5.
  2. y = (−0.5 + 3)(−0.5 − 2) = 2.5 × (−2.5) = −6.25.
  3. Lowest point = (−0.5, −6.25)
Answer (−0.5, −6.25)
Part (d)
  1. The line of symmetry is the vertical line through the lowest point.
  2. x = −0.5
Answer x = −0.5
Part (e)
  1. Draw the horizontal line y = 3 across the grid.
  2. Read the x-values where it crosses the curve: x ≈ −3.5 or x ≈ 2.5
Note: Read from your graph; the mark scheme accepts −3.6 to −3.4 and 2.4 to 2.6.
Answer x ≈ −3.5 or x ≈ 2.5

0580 Graphs of Functions past paper questions: Extended (Papers 2 and 4)

84 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q8(c)

P is the point (0, 4) and Q is the point (1, 8). The graph of y = x² + ax + b passes through the points P and Q. This graph also passes through the point (−2, c). Find the value of c.

Show solution
  1. The curve passes through P(0, 4), so substitute x = 0, y = 4: 4 = 0 + 0 + b, so b = 4.
  2. It also passes through Q(1, 8): 8 = 1 + a + 4, so a = 3.
  3. The equation is y = x² + 3x + 4.
  4. Substitute x = −2 to find c: c = (−2)² + 3(−2) + 4 = 4 − 6 + 4 = 2
Answer 2
911 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q8

The table shows some values, correct to 1 decimal place, for y = x³ − 2x² − 2.

x−1−0.500.511.522.53
y−2.6−2−2.4−3.1−27

(a) Complete the table. (b) On the grid, draw the graph of y = x³ − 2x² − 2 for −1 ≤ x ≤ 3.

(c) By drawing a suitable straight line, solve the equation x³ − 2x² − x + 1 = 0.

Show solution
Part (a)
  1. Substitute each missing x-value into y = x³ − 2x² − 2.
  2. x = −1: (−1)³ − 2 × (−1)² − 2 = −1 − 2 − 2 = −5
  3. x = 1: 1 − 2 − 2 = −3
  4. x = 2.5: 15.625 − 12.5 − 2 = 1.125 ≈ 1.1
Answer −5 −3 1.1
Part (b)
  1. Plot all nine points and join them with a smooth curve.
  2. The curve rises to a small peak at (0, −2), dips to about (1.3, −3.2), then climbs steeply to (3, 7).
0580/42 May/June 2026 Q8(b) solution – graph of y = x³ − 2x² − 2 from x = −1 to 3xy-1-0.50.511.522.53-5-4-3-2-112345670
Part (c)
  1. Rearrange the equation so that one side matches the curve. Add x − 3 to both sides: x³ − 2x² − x + 1 + x − 3 = x − 3.
  2. This gives x³ − 2x² − 2 = x − 3. So draw the line y = x − 3, through (0, −3) and (2, −1).
  3. The solutions are the x-values where the line crosses the curve.
0580/42 May/June 2026 Q8(c) solution – the line y = x − 3 crossing the cubic curve three timesxy-1-0.50.511.522.53-5-4-3-2-112345670y = x − 3
  1. x ≈ −0.8, x ≈ 0.55, x ≈ 2.25
Note: Answers from −0.9 to −0.7, 0.5 to 0.6 and 2.2 to 2.35 are accepted.
Answer y = x − 3 x ≈ −0.8, x ≈ 0.55, x ≈ 2.25
1010 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q11

The table shows some values for y = 3 × 2ˣ − 5, correct to 1 decimal place.

x−2−1.5−1−0.500.511.52
y−3.9−2.9−0.813.57

(a) Complete the table. (b) Draw the graph of y = 3 × 2ˣ − 5 for −2 ≤ x ≤ 2.

(c) By drawing a suitable line on the graph, solve the equation 3 × 2ˣ = 7.

(d) Find the equation of the asymptote to the graph of y = 3 × 2ˣ − 5.

Show solution
Part (a)
  1. Substitute each missing x-value into y = 3 × 2ˣ − 5.
  2. x = −2: 2⁻² = 1⁄4 = 0.25, so y = 3 × 0.25 − 5 = −4.25 ≈ −4.3
  3. x = −1: 2⁻¹ = 0.5, so y = 1.5 − 5 = −3.5
  4. x = 0: 2⁰ = 1, so y = 3 − 5 = −2
Answer −4.3 −3.5 −2
Part (b)
  1. Plot all nine points. Draw a smooth curve that is almost flat on the left and rises more and more steeply on the right.
Part (c)
  1. Make the equation match the curve. Subtract 5 from both sides of 3 × 2ˣ = 7: 3 × 2ˣ − 5 = 2.
  2. So draw the line y = 2 and read down from where it crosses the curve.
0580/43 May/June 2026 Q11(b) solution – graph of y = 3 × 2ˣ − 5 with the line y = 2xy-2-1.5-1-0.50.511.52-5-4-3-2-112345670y = 2
  1. x ≈ 1.2
Answer y = 2 x ≈ 1.2
Part (d)
  1. As x becomes very negative, 2ˣ gets closer and closer to 0, but never reaches it.
  2. So y gets closer and closer to 3 × 0 − 5 = −5.
  3. The asymptote is y = −5
Answer y = −5
117 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q22

(a) Expand and simplify (x + 5)(x − 3)².

(b) Sketch the graph of y = (x + 5)(x − 3)², indicating the values where the graph crosses the axes.

Show solution
Part (a)
  1. Expand the squared bracket first: (x − 3)² = (x − 3)(x − 3) = x² − 6x + 9.
  2. Multiply by (x + 5): x(x² − 6x + 9) + 5(x² − 6x + 9).
  3. = x³ − 6x² + 9x + 5x² − 30x + 45.
  4. Collect like terms: −6x² + 5x² = −x² and 9x − 30x = −21x.
  5. x³ − x² − 21x + 45
Answer x³ − x² − 21x + 45
Part (b)
  1. The x³ term is positive, so the curve starts at the bottom left and finishes at the top right.
  2. Set y = 0: x + 5 = 0 gives x = −5, where the curve crosses the x-axis.
  3. (x − 3)² = 0 gives x = 3. A squared factor means the curve touches the axis there and turns back (a minimum).
  4. Set x = 0: y = 5 × (−3)² = 45, where it crosses the y-axis.
  5. Positive cubic crossing the x-axis at −5, touching it at 3, and crossing the y-axis at 45
0580/22 February/March 2026 Q22(b) solution – sketch of y = (x + 5)(x − 3)²−5345xy
Answer Positive cubic crossing the x-axis at −5, touching it at 3, and crossing the y-axis at 45
1210 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q11

The table shows some values for y = x³ − 2x + 3, correct to 2 decimal places where appropriate.

x−2−1.5−1−0.500.511.52
y−143.8832.1327

(a) Complete the table. (b) Draw the graph of y = x³ − 2x + 3 for −2 ≤ x ≤ 2.

(c) By drawing a suitable straight line on the grid, solve the equation x³ − 2.5x + 1 = 0.

Show solution
Part (a)
  1. Substitute each missing x-value into y = x³ − 2x + 3.
  2. x = −1.5: (−1.5)³ − 2 × (−1.5) + 3 = −3.375 + 3 + 3 = 2.625 ≈ 2.63
  3. x = 1.5: 1.5³ − 2 × 1.5 + 3 = 3.375 − 3 + 3 = 3.375 ≈ 3.38
Answer 2.63 3.38
Part (b)
  1. Plot all the points and join them with a smooth S-shaped curve.
Part (c)
  1. Rearrange so that one side matches the curve. Add 0.5x + 2 to both sides of x³ − 2.5x + 1 = 0.
  2. x³ − 2.5x + 1 + 0.5x + 2 = 0.5x + 2, which gives x³ − 2x + 3 = 0.5x + 2.
  3. Draw the line y = 0.5x + 2, through (0, 2) and (2, 3).
  4. Read the x-values where the line crosses the curve.
0580/42 October/November 2025 Q11 solution – graph of y = x³ − 2x + 3 with the line y = 0.5x + 2xy-2-1.5-1-0.50.511.52-2-1123456780y = 0.5x + 2
  1. x ≈ −1.75, x ≈ 0.43, x ≈ 1.32
Note: Answers from −1.85 to −1.7, 0.4 to 0.5 and 1.25 to 1.4 are accepted.
Answer y = 0.5x + 2 x ≈ −1.75, x ≈ 0.43, x ≈ 1.32
133 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q13

The point (5, 1024) lies on the curve y = cˣ, where c is a whole number. Find the y-coordinate of the point on the curve with x-coordinate −2.

Show solution
  1. Substitute the point (5, 1024) into y = cˣ: c⁵ = 1024.
  2. Try whole numbers: 4⁵ = 4 × 4 × 4 × 4 × 4 = 1024, so c = 4.
  3. Now substitute x = −2: y = 4⁻² = 1⁄4² = 1⁄16.
  4. 0.0625
Answer 0.0625
145 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q11

The diagram shows the graph of y = f(x) and the point P(−2, 11). The tangent from P touches the graph of y = f(x) at the point (a, b). The values of a and b are integers.

0580/21 May/June 2025 Q11 – curve y = f(x) for −2 ≤ x ≤ 3 with point P(−2, 11)

(a) By drawing this tangent, find the value of a and the value of b.

(b) Find the equation of the tangent. Give your answer in the form y = mx + c.

Show solution
Part (a)
  1. A tangent touches the curve at one point without crossing it.
  2. Put a ruler on P and turn it until it just touches the curve. Read the touching point.
  3. a = 1, b = 2
0580/21 May/June 2025 Q11 solution – tangent from P(−2, 11) touching the curve at (1, 2)
Answer a = 1, b = 2
Part (b)
  1. The tangent goes through P(−2, 11) and (1, 2).
  2. Gradient = change in y⁄change in x = 2 − 11⁄1 − (−2) = −9⁄3 = −3.
  3. Substitute (1, 2) into y = −3x + c: 2 = −3 + c, so c = 5.
  4. y = 5 − 3x
Answer y = 5 − 3x
156 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q15

The diagram shows the graph of y = 2⁄x − 1.

0580/22 May/June 2025 Q15 – graph of y = 2/x − 1 for −8 ≤ x ≤ 8xy-8-7-6-5-4-3-2-112345678-5-4-3-2-1123450

(a) Write down the coordinates of the point where the graph crosses the x-axis.

(b) Write down the equation of each asymptote.

(c) By drawing a suitable straight line on the grid, solve 2⁄x − x − 1 = 0.

Show solution
Part (a)
  1. The graph crosses the x-axis where y = 0: 2⁄x − 1 = 0.
  2. So 2⁄x = 1, which gives x = 2.
  3. Point = (2, 0)
Answer (2, 0)
Part (b)
  1. You cannot divide by 0, so the graph never reaches x = 0. That is a vertical asymptote.
  2. As x gets very large (or very negative), 2⁄x gets close to 0, so y gets close to −1. That is a horizontal asymptote.
  3. x = 0 and y = −1
Answer x = 0 and y = −1
Part (c)
  1. Rearrange the equation so the left side matches the graph: 2⁄x − x − 1 = 0 becomes 2⁄x − 1 = x.
  2. So draw the line y = x on the grid. The solutions are the x-values where it crosses the curve.
  3. x = −2 or x = 1
0580/22 May/June 2025 Q15 solution – line y = x crossing the curve at x = −2 and x = 1xy-8-7-6-5-4-3-2-112345678-5-4-3-2-1123450y = x
  1. Check: x = 1 gives 2 − 1 − 1 = 0 ✓ and x = −2 gives −1 + 2 − 1 = 0 ✓
Answer x = −2 or x = 1

Question papers used on this page

All graphs of functions questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/11/M/J/260580_s26_qp_11May/June 2026 · Paper 1 (Core)Q18
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q13
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q8(c)
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q18
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q8
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q11
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q22
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q16
0580/32/O/N/250580_w25_qp_32Oct/Nov 2025 · Paper 3 (Core)Q27
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q22
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q11
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q13
0580/12/M/J/250580_s25_qp_12May/June 2025 · Paper 1 (Core)Q14
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q11
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q15

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