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4MA1/1H Jan 2020, Question 20: Completed Square Form and Turning Points

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 1H (Higher) · Calculator · 4 marks

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The question, in short

Sketch curve C with equation y = 4(x − 1)² − a, a > 4, marking axis intercepts and turning point coordinates in terms of a.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Solution · 4 marks

  1. 1y = 4(x − 1)² − a
  2. 2Find turning point: minimum when (x − 1)² = 0 => x = 1
  3. 3Substitute x = 1: y = 4(0)² − a = −a => turning point (1, −a)
  4. 4Find y-intercept: set x = 0 => y = 4(0 − 1)² − a = 4(1) − a = 4 − a
  5. 5So y-intercept is (0, 4 − a)
  6. 6Find x-intercepts: set y = 0 => 4(x − 1)² − a = 0
  7. 74(x − 1)² = a
  8. 8(x − 1)² = a/4
  9. 9x − 1 = ±(√a)/2
  10. 10x = 1 ± (√a)/2
  11. 11So x-intercepts are (1 + (√a)/2, 0) and (1 − (√a)/2, 0)
  12. 12Since a > 4, √a > 2, so 1 − (√a)/2 < 0 => one intercept on negative x-axis, one on positive
  13. 13Minimum at (1, −a), and a > 4 => y-coordinate of turning point < −4 => in fourth quadrant
  14. 14Sketch parabola opening upwards, vertex in fourth quadrant, crossing y-axis at (0, 4 − a), x-axis at both points found

AnswerSee graph with turning point at (1, −a), x-axis intercepts at (1 + (√a)/2, 0) and (1 − (√a)/2, 0), y-axis intercept at (0, 4 − a)

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