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0580/43 May/Jun 2021, Question 7: Linear Inequalities

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 10 marks

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The question, in short

From a number line showing an interval between −2 and 1, state the inequality in x, then complete the square for x² + 4x + 1 and use it to solve, find the minimum point, and sketch the curve.

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

Part (a) · 2 marks

Write the inequality in x represented by the number line with an open circle at −2 and a filled circle at 1.

  1. 1Open circle at −2: −2 < x
  2. 2Filled circle at 1: x ≤ 1
  3. 3−2 < x ≤ 1

Answer−2 < x ≤ 1

Part (bi) · 2 marks

Express x² + 4x + 1 in the completed-square form (x + p)² + q.

  1. 1Halve the x coefficient: 4 ÷ 2 = 2, so use (x + 2)²
  2. 2(x + 2)² = x² + 4x + 4, which is 4 too many
  3. 3x² + 4x + 1 = (x + 2)² − 4 + 1
  4. 4= (x + 2)² − 3

Answer(x + 2)² − 3

Part (bii) · 2 marks

Using the completed-square form, solve x² + 4x + 1 = 0.

  1. 1(x + 2)² − 3 = 0, so (x + 2)² = 3
  2. 2x + 2 = ±√3, x = −2 ± √3
  3. 3x = −3.732… ≈ −3.73 or x = −0.2679… ≈ −0.268

Answerx = −3.73 or −3.732… and x = −0.268 or −0.2679…

Part (biii) · 2 marks

State the coordinates of the minimum point of y = x² + 4x + 1 using the completed-square form.

  1. 1(x + 2)² ≥ 0, so the minimum is when x = −2, giving y = −3
  2. 2Minimum point (−2, −3)

Answer(−2, −3)

Part (biv) · 2 marks

Sketch the graph of y = x² + 4x + 1 on the given axes.

  1. 1Draw a U-shaped parabola with its minimum at (−2, −3), in the third quadrant
  2. 2It crosses the x-axis at about −3.73 and −0.27 (both negative)
  3. 3It crosses the y-axis at y = 1 (positive)

AnswerCorrect sketch: parabola with its minimum point in the third quadrant, both x-intercepts negative and a positive y-intercept

Stuck on a step? Read the free lesson: Linear Inequalities

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Also searched as: 0580/43/M/J/21 · 0580_s21_qp_43 · 0580 s21 43 · question 7.
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