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0580/21 May/Jun 2023, Question 20: Quadratic Graphs and the Parabola

Cambridge IGCSE Mathematics 0580 · Paper 21 (Extended) · Calculator · 7 marks

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The question, in short

Table of values for y = 3x² − 2x − 1; complete missing y values, plot the curve for −1 ≤ x ≤ 1.5, then use a straight line to solve 3x² − 4x − 2 = 0.

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

Part (a) · 1 mark

Fill in the two missing y values in the table for x = −0.5 and x = 0.5.

  1. 1x = −0.5: 3 × 0.25 − 2 × (−0.5) − 1 = 0.75 + 1 − 1 = 0.75
  2. 2x = 0.5: 3 × 0.25 − 2 × 0.5 − 1 = 0.75 − 1 − 1 = −1.25

Answer0.75 and −1.25

Part (b) · 3 marks

Plot the graph of y = 3x² − 2x − 1 on the grid for x from −1 to 1.5.

  1. 1Plot (−1, 4), (−0.5, 0.75), (0, −1), (0.5, −1.25), (1, 0), (1.5, 2.75)
  2. 2Join with a smooth U-shaped curve (lowest point near x = 1/3)

AnswerCorrect curve

Part (c) · 3 marks

Draw a suitable straight line on the graph and use it to solve 3x² − 4x − 2 = 0 for −1 ≤ x ≤ 1.5.

  1. 13x² − 4x − 2 = 0 → 3x² − 2x − 1 = 2x + 1
  2. 2Draw the line y = 2x + 1 (through (0, 1) and (1, 3))
  3. 3Read x where the line crosses the curve in −1 ≤ x ≤ 1.5
  4. 4x ≈ −0.39

Answer−0.45 to −0.35

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Also searched as: 0580/21/M/J/23 · 0580_s23_qp_21 · 0580 s23 21 · question 20.
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