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WMA11/01 Jun 2023, Question 3: Completed Square Form and Turning Points

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 6 marks

The question, in short

Complete the square on 3x² + 12x + 13 to find integers a, b, c, then sketch y = 3x² + 12x + 13 showing the y intercept and turning point coordinates.

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

Part (a) · 3 marks

Rewrite 3x² + 12x + 13 as a(x + b)² + c and state the integer values of a, b and c.

  1. 1Take out 3 from the x terms: 3(x² + 4x) + 13
  2. 2Complete the square: 3[(x + 2)² − 4] + 13
  3. 3= 3(x + 2)² − 12 + 13 = 3(x + 2)² + 1
  4. 4a = 3, b = 2, c = 1

Answer3(x + 2)² + 1 (a = 3, b = 2, c = 1)

Part (b) · 3 marks

Using your completed square, sketch y = 3x² + 12x + 13, marking the y intercept and the turning point coordinates.

  1. 1a = 3 > 0, so the curve is U-shaped
  2. 2From (a), the minimum point is at (−2, 1), which is above the x-axis, so no x-intercepts
  3. 3When x = 0, y = 13, so the y intercept is (0, 13)
  4. 4Sketch a U shape with minimum at (−2, 1) passing through (0, 13)

AnswerU-shaped parabola with its minimum point at (−2, 1) in the second quadrant, cutting the y-axis at (0, 13) and not meeting the x-axis.

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