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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Graphs

Transformations of Graphs — Worked Solutions

Full worked solutions to all 12 questions (32 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 5–6)
  1. (3 marks)The point (3, 4) lies on the curve y = f(x). Write down the coordinates of the image of this point on
    (a) y = f(x) + 2 (b) y = f(x − 1) (c) y = −f(x).
    Solution
    1. (a) Up 2: (3, 6).
    2. (b) Right 1: (4, 4).
    3. (c) Reflect in the x-axis: (3, −4).
    (a) (3, 6) (b) (4, 4) (c) (3, −4)
  2. (2 marks)Describe fully the single transformation that maps y = x² onto y = (x − 4)² + 1.
    Solution
    1. (x − 4) inside: right 4. +1 outside: up 1.
    Translation by the vector (4, 1)
  3. (2 marks)The graph of y = sin x is transformed to y = 3 sin x.
    (a) Describe the transformation. (b) Write down the maximum value of y = 3 sin x.
    Solution
    1. (a) Stretch parallel to the y-axis, scale factor 3.
    2. (b) Maximum of sin x is 1, so 3 × 1 = 3.
    (a) vertical stretch, sf 3 (b) 3
  4. (2 marks)The graph of y = cos x is transformed to y = cos 2x. Describe the transformation, and state the period of y = cos 2x.
    Solution
    1. f(2x): stretch parallel to the x-axis, scale factor ½.
    2. The period of cos x is 360°, so the period of cos 2x is 180°.
    Horizontal stretch, sf ½; period 180°
Section B — Core (grades 6–7)
  1. (4 marks)The curve y = f(x) has a turning point at (2, −5). Write down the turning point of
    (a) y = f(x + 3) (b) y = f(x) − 4 (c) y = 2f(x) (d) y = f(−x).
    Solution
    1. (a) Left 3: (−1, −5).
    2. (b) Down 4: (2, −9).
    3. (c) y × 2: (2, −10).
    4. (d) x changes sign: (−2, −5).
    (a) (−1, −5) (b) (2, −9) (c) (2, −10) (d) (−2, −5)
  2. (3 marks)f(x) = x² − 4x. Show that f(x + 2) = x² − 4.
    Solution
    1. f(x + 2) = (x + 2)² − 4(x + 2).
    2. = x² + 4x + 4 − 4x − 8 = x² − 4.
    x² − 4
  3. (3 marks)The maximum point of y = sin x for 0 ≤ x ≤ 360 is (90, 1). Find the maximum point of y = sin(x − 60) + 2.
    Solution
    1. (x − 60): right 60, so the x-coordinate becomes 90 + 60 = 150.
    2. +2: up 2, so the y-coordinate becomes 1 + 2 = 3.
    (150, 3)
  4. (2 marks)The diagram shows the curve y = x² − 2x and the tangent to the curve at (3, 3). The tangent also passes through (4, 7). Find the gradient of the curve at x = 3.
    xy-112345-2-1123456789(3, 3)(4, 7)
    Solution
    1. The gradient of the curve at a point equals the gradient of the tangent there.
    2. Gradient = (7 − 3) ÷ (4 − 3) = 4.
    4
  5. (3 marks)The graph of y = x² − 2x − 3 has been drawn. Find the equation of the straight line you would draw to solve x² − 3x − 4 = 0, and write down the solutions.
    Solution
    1. x² − 3x − 4 = 0 → x² − 2x − 3 = x + 1.
    2. Draw y = x + 1. It meets the curve at x = −1 and x = 4.
    y = x + 1; x = −1, x = 4
Section C — Stretch (grades 8–9)
  1. (2 marks)The graph of y = f(x) passes through (0, 3) and (4, 0). Find the coordinates of the images of these points on y = 2f(x − 1).
    Solution
    1. (x − 1): right 1. 2f: y-values doubled.
    2. (0, 3) → (1, 6); (4, 0) → (5, 0).
    (1, 6) and (5, 0)
  2. (3 marks)f(x) = x². Show that the graph of y = f(2x) is the same as the graph of y = 4f(x), and describe both as single transformations of y = f(x).
    Solution
    1. f(2x) = (2x)² = 4x², and 4f(x) = 4x², so they are the same graph.
    2. y = f(2x): stretch parallel to the x-axis, scale factor ½.
    3. y = 4f(x): stretch parallel to the y-axis, scale factor 4.
    Both equal 4x²
  3. (3 marks)The graph of y = a cos x + b, where a > 0, has a maximum value of 5 and a minimum value of 1. Find a and b.
    Solution
    1. cos x ranges from −1 to 1, so a cos x + b ranges from b − a to b + a.
    2. b + a = 5 and b − a = 1.
    3. Adding: 2b = 6 → b = 3, so a = 2.
    a = 2, b = 3
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