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0580/32 Feb/Mar 2024, Question 5: Quadratic Graphs and the Parabola

Cambridge IGCSE Mathematics 0580 · Paper 32 (Core) · Calculator · 15 marks

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The question, in short

Complete a table for y = −x² + 5x + 7, plot its curve for −1 ≤ x ≤ 6, then answer questions on symmetry, gradients, parallel lines, a line on a grid, and rearranging y = mx + c.

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

Part (ai) · 3 marks

Fill in the missing y-values in the table for y = −x² + 5x + 7 at the given x-values.

  1. 1x = −1: −1 − 5 + 7 = 1
  2. 2x = 0: 7
  3. 3x = 2: −4 + 10 + 7 = 13
  4. 4x = 3: −9 + 15 + 7 = 13
  5. 5x = 5: −25 + 25 + 7 = 7

Answer1, 7, 13, 13, 7

Part (aii) · 4 marks

Plot the curve y = −x² + 5x + 7 on the grid for x from −1 to 6.

  1. 1Plot (−1, 1), (0, 7), (1, 11), (2, 13), (3, 13), (4, 11), (5, 7), (6, 1)
  2. 2Join with a smooth curve, highest point between x = 2 and x = 3

AnswerCompletely correct curve through (−1, 1), (0, 7), (1, 11), (2, 13), (3, 13), (4, 11), (5, 7), (6, 1)

Part (aiiia) · 1 mark

State the equation of the vertical mirror line of the plotted curve.

  1. 1y = 13 at x = 2 and x = 3
  2. 2Line of symmetry halfway: x = 2.5

Answerx = 2.5

Part (aiiib) · 1 mark

Given (−8, −97) and (t, −97) lie on the curve, use symmetry to find t.

  1. 1Points with equal y-values are symmetric about the line of symmetry
  2. 2From the table, y = 11 at x = 1 and x = 4, so the line of symmetry is x = (1 + 4)/2 = 2.5
  3. 3−8 is 2.5 − (−8) = 10.5 to the left of 2.5
  4. 4t = 2.5 + 10.5 = 13

Answer13

Part (b) · 1 mark

State the slope of the straight line y = 9x − 4.

  1. 1y = mx + c, m = 9
  2. 2Gradient = 9

Answer9

Part (c) · 1 mark

Give the equation of any straight line that has the same slope as y = −5x + 19.

  1. 1Parallel lines have the same gradient, so the gradient is −5
  2. 2Keep −5x and change the constant to any number except 19
  3. 3For example y = −5x + 1, i.e. y = −5x + k where k ≠ 19

Answery = −5x + k where k ≠ 19

Part (d) · 2 marks

Work out the equation of the straight line L shown on the grid, in the form y = mx + c.

  1. 1L crosses the y-axis at 2, so c = 2
  2. 2Gradient = −2 ÷ 2 = −1 (from (0, 2) to (2, 0))
  3. 3y = −x + 2

Answery = −x + 2

Part (e) · 2 marks

Rearrange y = mx + c so that x is the subject.

  1. 1y − c = mx
  2. 2x = (y − c)/m

Answerx = (y − c)/m

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Also searched as: 0580/32/F/M/24 · 0580_m24_qp_32 · 0580 m24 32 · question 5.
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