0580/32 Oct/Nov 2022, Question 8: Quadratic Graphs and the Parabola

Part (ai) · 3 marks

Fill in the missing y-values in the table for y = x² + 2x − 4 at the given x-values.

1. 1x = −4: 16 − 8 − 4 = 4; x = −2: 4 − 4 − 4 = −4
2. 2x = −1: 1 − 2 − 4 = −5; x = 0: −4
3. 3x = 2: 4 + 4 − 4 = 4

Answer| x | −4 | −2 | −1 | 0 | 2 | | y | 4 | −4 | −5 | −4 | 4 |

Part (aii) · 4 marks

Plot the graph of y = x² + 2x − 4 on the grid for x from −5 to 3.

1. 1Plot (−5, 11), (−4, 4), (−3, −1), (−2, −4), (−1, −5), (0, −4), (1, −1), (2, 4), (3, 11)
2. 2Join with a smooth U-shaped curve, lowest point at (−1, −5)

AnswerCorrect curve

Part (aiii) · 1 mark

Sketch the curve for −5 ≤ x ≤ 3 and state the equation of its line of symmetry.

1. 1Symmetric pairs, e.g. y(−3) = y(1) = −1, around x = −1
2. 2x = −1

Answerx = −1

Part (aiv) · 2 marks

Use your drawn graph to find the solutions of x² + 2x − 4 = 0.

1. 1Read where the curve crosses y = 0 (the x-axis)
2. 2x ≈ −3.2 and x ≈ 1.2

Answerx = −3.3 to −3.1, x = 1.1 to 1.3

Part (b) · 2 marks

Find the equation of the line parallel to y = 5x + 7 that passes through (0, −2).

1. 1Parallel: same gradient 5; through (0, −2): c = −2
2. 2y = 5x − 2

Answery = 5x − 2

More past-paper questions on quadratic graphs and the parabola

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