Edexcel International GCSE Maths Graphs Past Paper Questions and Solutions (4MA1 Higher)

Part (a)

1. Midpoint = (mean of the x-coordinates, mean of the y-coordinates) = (^{5 + 13}⁄₂, ^{−4 + 1}⁄₂)

Answer: (9, −1.5)

Part (b)

1. In y = mx + c form this is y = −3x + 2, so the gradient is the number multiplying x.

Answer: −3

Part (c)

1. Substitute x = 100: y = 2 − 3 × 100 = −298
2. −298 is not −302.

Answer: No, because when x = 100, y = −298 (not −302)

1. Work out some points: x = −2 gives y = 7 + 8 = 15; x = 0 gives y = 7; x = 3 gives y = 7 − 12 = −5.
2. The equation is linear, so join the points with a straight line from x = −2 to x = 3.

Answer: Straight line through (−2, 15), (0, 7) and (3, −5)

Part (a)

1. x = −2: 4 + 8 + 3 = 15. x = 1: 1 − 4 + 3 = 0. x = 2: 4 − 8 + 3 = −1. x = 4: 16 − 16 + 3 = 3.

Answer: 15, 0, −1, 3

Part (b)

1. Plot the points and join them with a smooth U-shaped curve (lowest point at (2, −1)).

Answer: Smooth curve through (−2, 15), (−1, 8), (0, 3), (1, 0), (2, −1), (3, 0), (4, 3)

1. The midpoint is the mean of the coordinates.
2. a = ^{−3 + 47}⁄₂ = ⁴⁴⁄₂ = 22
3. ^{11 + b}⁄₂ = −19, so 11 + b = −38 and b = −49

Answer: a = 22, b = −49

1. Read two points on the line where it crosses grid points: (0, 1) and (2, 0).
2. Gradient m = ^{change in y}⁄_{change in x} = ^{0 − 1}⁄_{2 − 0} = −¹⁄₂. The line goes down from left to right, so the gradient is negative.
3. The line crosses the y-axis at 1, so c = 1.

Answer: y = −¹⁄₂x + 1

Part (a)

1. Rearrange into y = mx + c: 2y = −7x + 10, so y = −3.5x + 5

Answer: −3.5

Part (b)

1. On the y-axis x = 0, so y = 5.

Answer: (0, 5)

Part (a)

1. Horizontal change = 7 − (−1) = 8, vertical change = 18 − 3 = 15
2. Pythagoras: distance = √8² + 15² = √64 + 225 = √289

Answer: 17 cm

Part (b)

1. Two circles cross if the distance between their centres is less than the sum of the radii (and more than their difference).
2. Sum of radii = 13 + 6 = 19, which is more than 17. (And 17 is more than 13 − 6 = 7, so one circle is not inside the other.)

Answer: 17 < 13 + 6 = 19, so the circles overlap

1. Pick two clear points on the line: (0, −1) and (2, 2).
2. Gradient = ^{2 − (−1)}⁄_{2 − 0} = ³⁄₂ = 1.5
3. The line crosses the y-axis at −1, so c = −1.

Answer: y = 1.5x − 1

Part (a)

1. Substitute each x value into x³ + 2x + 3. Take care with negative numbers: (−3)³ = −27.
2. x = −3: −27 − 6 + 3 = −30
3. x = 2: 8 + 4 + 3 = 15

Answer: y = −30 when x = −3; y = 15 when x = 2

Part (b)

1. Plot all seven points from the table and join them with a smooth curve (not straight lines).

Answer: Smooth curve through (−3, −30), (−2, −9), (−1, 0), (0, 3), (1, 6), (2, 15), (3, 36)

Part (a)

1. Gradient = ^{−4 − 8}⁄_{9 − 5} = ⁻¹²⁄₄

Answer: −3

Part (b)

1. L has gradient −4. Perpendicular gradients multiply to −1, so take the negative reciprocal.

Answer: ¹⁄₄

1. Gradient = ^{change in y}⁄_{change in x} = ^{−48 − 12}⁄_{19 − (−5)} = ⁻⁶⁰⁄₂₄ = −2.5
2. Use y − y₁ = m(x − x₁) with A(−5, 12): y − 12 = −2.5(x + 5)
3. y − 12 = −2.5x − 12.5, so y = −2.5x − 0.5

Answer: y = −2.5x − 0.5

Part (a)

1. x = 0.5: ²⁄_0.5 = 4 and 5 − ¹⁄_0.5 = 5 − 2 = 3, so y = 4 × 3 = 12
2. x = 2: ²⁄₂ = 1 and 5 − ½ = 4.5, so y = 4.5

Answer: 12 and 4.5

Part (b)

1. Plot the six points and join them with a smooth curve.

Answer: Smooth curve through (0.5, 12), (1, 8), (2, 4.5), (3, 3.1), (4, 2.4), (5, 1.9)

Part (a)

1. x = −2: (−2)³ − 3(−2) + 2 = −8 + 6 + 2 = 0
2. x = 0: 0 − 0 + 2 = 2
3. x = 2: 8 − 6 + 2 = 4

Answer: 0, 2, 4

Part (b)

1. Plot the seven points and join them with a smooth curve.

Answer: Smooth curve through (−2, 0), (−1, 4), (−0.5, 3.4), (0, 2), (1, 0), (1.5, 0.9), (2, 4)

Part (c)

1. Make the left side look like the curve. Divide by 2: x³ − 1.5x + 2 = 0
2. Add and subtract to get x³ − 3x + 2 on the left: x³ − 3x + 2 = −1.5x
3. So draw the line y = −1.5x (through (0, 0), (−2, 3) and (1, −1.5)) and read the x-value where it crosses the curve.

Answer: x = −1.6 (−1.7 also accepted)

Part (a)(i)

1. Draw a tangent at P (a straight line that just touches the curve at x = 2).
2. Pick two points far apart on your tangent, for example (0, 3.6) and (6, 0). Gradient = ^{change in y}⁄_{change in x} = ^{0 − 3.6}⁄_{6 − 0}

Answer: about −0.6 (−0.4 to −0.7 accepted)

Part (a)(ii)

1. The tangent has gradient −0.6 and crosses the y-axis at about 3.6.

Answer: y = −0.6x + 3.6

Part (b)

1. f(x) = k has exactly two solutions where the horizontal line y = k meets the curve exactly twice.
2. y = 3 touches the maximum at x = 0 and meets the curve again at the right-hand end (x = 12). y = −1 touches the minimum at x = 8 and meets the curve again at the left-hand end (x = −4).

Answer: k = 3 and k = −1

Part (i)

1. + 3 outside the function moves the graph 3 units up. The x-coordinate does not change.

Answer: (8, −9)

Part (ii)

1. f(2x) is a stretch parallel to the x-axis with scale factor ½, so the x-coordinate halves and the y-coordinate stays the same.

Answer: (4, −12)

1. The curve is a U-shaped parabola (positive x² coefficient).
2. Turning point: 4(x − 1)² is smallest (0) when x = 1, so the minimum is (1, −a).
3. y-intercept: put x = 0: y = 4(1) − a = 4 − a. Since a > 4, this is negative, so the curve crosses the y-axis below the origin.
4. x-intercepts: 4(x − 1)² = a, so x − 1 = ±^√a⁄₂ and x = 1 ± ^√a⁄₂

Answer: U-shaped curve with minimum (1, −a), y-intercept (0, 4 − a) and x-intercepts (1 ± ^√a⁄₂, 0)

Part (i)

1. Move the curve 6 units down.

Answer: (−3, −2)

Part (ii)

1. f(2x) halves every x-coordinate.

Answer: (−1.5, 4)

1. Gradient of L = ^{3 − 2}⁄_{5 − (−6)} = ¹⁄₁₁
2. M is perpendicular, so its gradient is −11 (the negative reciprocal).
3. Midpoint of AB = (^{−6 + 5}⁄₂, ^{2 + 3}⁄₂) = (−0.5, 2.5)
4. Equation of M: y − 2.5 = −11(x + 0.5)
5. At x = −1: y = 2.5 − 11(−0.5) = 2.5 + 5.5 = 8, so C = (−1, 8)
6. M is the perpendicular bisector of AB, so the triangle’s height is the distance from C to the midpoint: √(−1 + 0.5)² + (8 − 2.5)² = √0.25 + 30.25 = √30.5
7. Base AB = √11² + 1² = √122
8. Area = ½ × √122 × √30.5 = ½ × √3721 = ½ × 61

Answer: 30.5

Part (a)(i)

1. + 5 moves the curve 5 up.

Answer: (−6, 1)

Part (a)(ii)

1. f(3x) divides the x-coordinate by 3.

Answer: (−2, −4)

Part (b)

1. 2g(x) doubles every y-coordinate. g(−1) = 3, g(3) = −1 and g(7) = 3, so the new points are (−1, 6), (3, −2) and (7, 6).

Answer: V-shape from (−1, 6) down to (3, −2) and up to (7, 6)

Part (c)

1. h(x + a) is h(x) moved a units to the left. Its roots are −1 − a and 6 − a.
2. One of these must be 2: −1 − a = 2 gives a = −3; 6 − a = 2 gives a = 4.

Answer: a = −3 or a = 4

Part (i)

1. f(x − 2) moves the curve 2 units right.

Answer: (s + 2, t)

Part (ii)

1. 3f(x) stretches the curve vertically by factor 3, so the y-coordinate is multiplied by 3.

Answer: (s, 3t)

1. The curve goes from −2 up to 2, so the amplitude is 2: a = 2
2. y = 2cos x has its maximum at x = 0 (or 360). This curve has its maximum at x = 225, so cos(x + b) = 1 when x + b = 360
3. 225 + b = 360, so b = 135

Answer: a = 2, b = 135

Part (a)(i)

1. 2f(x) stretches the graph vertically by factor 2: the y-coordinate doubles.

Answer: (3, 10)

Part (a)(ii)

1. f(x) − 7 moves the graph 7 units down.

Answer: (3, −2)

Part (a)(iii)

1. f(−x) reflects the graph in the y-axis: the x-coordinate changes sign.

Answer: (−3, 5)

Part (b)

1. A translation 2 units right replaces x with (x − 2): y = (x − 2)² + 7(x − 2) + 20
2. Expand: x² − 4x + 4 + 7x − 14 + 20
3. Collect like terms.

Answer: y = x² + 3x + 10

Part (a)(i)

1. f(x + 3) moves the curve 3 units left.

Answer: (−12, 15)

Part (a)(ii)

1. ⅓f(x) divides every y-coordinate by 3.

Answer: (−9, 5)

Part (b)

1. The graph goes between −2 and 2, so a = 2.
2. An ordinary cos graph has its maximum at x = 0 (and 360). This graph has its maximum at x = 270, which is the cos graph moved 90 to the left. Moving left by 90 means x + 90.
3. Check: 2cos(0 + 90)° = 2cos 90° = 0 ✓

Answer: a = 2, b = 90

Part (a)

1. f(x − 4) moves the graph 4 units to the right.

Answer: (10, 5)

Part (b)

1. f(3x) squashes the graph horizontally: divide the x-coordinate by 3.

Answer: (2, 5)

Part (i)

1. f(x + 7) is a translation 7 units to the left (the opposite direction to the sign inside the bracket). The y-coordinate does not change.
2. 5 − 7 = −2

Answer: (−2, −4)

Part (ii)

1. f(x) − 6 is a translation 6 units down. The x-coordinate does not change.
2. −4 − 6 = −10

Answer: (5, −10)

Part (a)

1. f(3x) squashes the graph horizontally by a factor of 3: divide the x-coordinate by 3. The y-coordinate stays the same.

Answer: (−2, 9)

Part (b)

1. The point moves from x = 4 to x = 1, which is 3 units left. g(x + 3) moves a graph 3 units left, so a = 3.
2. The point moves from y = −5 to y = −16, which is 11 units down, so b = −11.

Answer: a = 3, b = −11

1. In an isosceles triangle, the line from the apex H to the midpoint of the base is perpendicular to the base. So JK has gradient −¹⁄₂ (perpendicular to gradient 2).
2. Gradient of JK: ^{k − 15}⁄_{6 − j} = −¹⁄₂, so k − 15 = −¹⁄₂(6 − j)
3. Length: (6 − j)² + (k − 15)² = 80. Substitute: (6 − j)² + ¹⁄₄(6 − j)² = 80, so ⁵⁄₄(6 − j)² = 80
4. (6 − j)² = 64, so 6 − j = 8 (it must be positive because j < 0). So j = −2.
5. k − 15 = −¹⁄₂ × 8 = −4, so k = 11
6. Check: M = (2, 13), and the gradient of HM = ^{13 − 1}⁄_{2 − (−4)} = ¹²⁄₆ = 2 ✓

Answer: j = −2, k = 11

Part (a)

1. The maximum value is 2.5 (an ordinary cos graph has maximum 1), so a = 2.5.
2. An ordinary cos graph has its maximum at x = 0. This one has its maximum at x = 60, so it has moved 60 to the right, which means b = −60.

Answer: a = 2.5, b = −60

Part (b)(i)

1. f(2x) halves the x-coordinate.

Answer: (2, 5)

Part (b)(ii)

1. f(x) − 7 moves the curve 7 down.

Answer: (4, −2)

1. Rearrange AC: 2y = −7x + 20, so y = −⁷⁄₂x + 10 and its gradient is −⁷⁄₂
2. The diagonals of a rhombus cross at right angles, so the gradient of BD = ²⁄₇
3. BD passes through M(6, −11): y + 11 = ²⁄₇(x − 6)
4. At the y-axis, x = 0: y + 11 = −¹²⁄₇, so y = −11 − ¹²⁄₇ = −⁸⁹⁄₇

Answer: (0, −⁸⁹⁄₇)

1. M = (^{−1 + 2}⁄₂, ^{5 + 10}⁄₂) = (0.5, 7.5)
2. Gradient of BC = ^{10 − 5}⁄_{2 − (−1)} = ⁵⁄₃
3. AB = AC, so AM is the perpendicular bisector of BC: gradient = −³⁄₅
4. y − 7.5 = −³⁄₅(x − 0.5). Multiply by 10: 10y − 75 = −6x + 3, so 10y + 6x = 78
5. Divide by 2.

Answer: 5y + 3x = 39

Part (i)

1. f(x + 9) moves the curve 9 units left.

Answer: (−4, 7)

Part (ii)

1. + 3 moves it 3 units up.

Answer: (5, 10)

1. 5y − 3x = 6 gives y = ³⁄₅x + ⁶⁄₅, so L has gradient ³⁄₅.
2. Gradient of BD: ^{7 − 1}⁄_{8 − k} = ³⁄₅, so 30 = 24 − 3k and k = −2. So B = (−2, 1).
3. In a kite with AB = AD and CB = CD, the diagonal AC is the perpendicular bisector of BD.
4. Midpoint of BD = (3, 4). Gradient of AC = −⁵⁄₃
5. y − 4 = −⁵⁄₃(x − 3). Multiply by 3: 3y − 12 = −5x + 15

Answer: 5x + 3y = 27

Part (a)

1. Every y-coordinate changes sign and the x-coordinates stay the same.

Answer: Reflection in the x-axis

Part (b)

1. f(x − 1) moves the curve 1 unit right; then 2f(…) doubles every y-coordinate.
2. Key points: minimum (4, −3) → (5, −6); roots (2, 0) → (3, 0) and (6, 0) → (7, 0); ends (1.1, 3) → (2.1, 6) and (7, 3.75) → (8, 7.5), so the ends go off the top of the grid.

Answer: Curve through (3, 0), (5, −6) and (7, 0)

1. The diagonals of a square bisect each other at right angles. So QS passes through the midpoint of PR and is perpendicular to PR.
2. Midpoint of PR = (^{4 + 8}⁄₂, ^{7 + (−5)}⁄₂) = (6, 1)
3. Gradient of PR = ^{−5 − 7}⁄_{8 − 4} = ⁻¹²⁄₄ = −3
4. Perpendicular gradients multiply to −1, so the gradient of QS = ¹⁄₃
5. Line through (6, 1) with gradient ¹⁄₃: y − 1 = ¹⁄₃(x − 6)
6. Multiply by 3: 3y − 3 = x − 6, so 3y = x − 3

Answer: 3y = x − 3