0580/41 May/Jun 2024, Question 8: Simultaneous Equations with a Curve
Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 15 marks
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The question, in short
Sketch y = x² + 7x − 18 on axes, marking axis intercepts, then differentiate and locate a turning point, and find where a line meets a parabola to 2 decimal places.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 4 marks
Draw the curve y = x² + 7x − 18 on the given axes, labelling where it crosses both axes.
- 1x² + 7x − 18 = (x + 9)(x − 2), so it meets the x-axis at −9 and 2
- 2When x = 0, y = −18
- 3Draw a U-shaped curve through −9 and 2 on the x-axis and −18 on the y-axis, minimum below the x-axis to the left of the y-axis
AnswerCorrect sketch: U-shaped parabola with roots at x = −9 and x = 2, y-intercept −18, minimum in 3rd quadrant
Part (bi) · 2 marks
Differentiate y = x² − 3x − 28 with respect to x.
- 1Multiply by the power, then take 1 off the power: axⁿ → naxⁿ⁻¹
- 2x² → 2x
- 3−3x → −3
- 4−28 → 0
- 5dy/dx = 2x − 3
Answer2x − 3
Part (bii) · 3 marks
Work out the coordinates of the stationary point on y = x² − 3x − 28.
- 12x − 3 = 0
- 2x = 1.5
- 3y = 1.5² − 3 × 1.5 − 28 = −30.25
- 4Turning point (1.5, −30.25)
Answer(1.5, −30.25)
Part (c) · 6 marks
Solve where y = 5 − 2x meets y = x² − 3x − 28, giving points P and Q to 2 decimal places with full working.
- 1x² − 3x − 28 = 5 − 2x
- 2x² − x − 33 = 0
- 3x = (1 ± √(1 + 132))/2 = (1 ± √133)/2
- 4x = 6.266… or x = −5.266…
- 5y = 5 − 2(6.266) = −7.53, y = 5 − 2(−5.266) = 15.53
- 6(−5.27, 15.53) and (6.27, −7.53)
Answer(−5.27, 15.53 or 15.54) and (6.27, −7.53 or −7.54)
Stuck on a step? Read the free lesson: Simultaneous Equations with a Curve
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Also searched as: 0580/41/M/J/24 · 0580_s24_qp_41 · 0580 s24 41 · question 8.
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