WMA11/01 Jun 2022, Question 5: Quadratics from Their Turning Point

Part (a) · 4 marks

Find the quadratic expression f(x) given its maximum turning point (−2, 12) and that the curve crosses the negative x-axis at −5.

1. 1Maximum at (−2, 12): f(x) = 12 − a(x + 2)²
2. 2f(−5) = 0: 12 − a(−3)² = 0
3. 39a = 12, so a = 4/3
4. 4f(x) = 12 − (4/3)(x + 2)² = −(4/3)x² − (16/3)x + 20/3

Answerf(x) = 12 − (4/3)(x + 2)² = −(4/3)x² − (16/3)x + 20/3

Part (b) · 3 marks

Find the equation of l₂, perpendicular to y = (4/5)x and passing through (−5, 0), in the form y = mx + c.

1. 1Gradient of l₁ = 4/5, so gradient of l₂ = −5/4
2. 2y − 0 = −(5/4)(x + 5)
3. 3y = −(5/4)x − 25/4

Answery = −(5/4)x − 25/4

Part (c) · 2 marks

Using the sketch, define the shaded region R with inequalities involving C, l₁ and l₂.

1. 1R is below the curve C: y ≤ −(4/3)x² − (16/3)x + 20/3
2. 2R is above l₁: y ≥ (4/5)x
3. 3R is above l₂: y ≥ −(5/4)x − 25/4

Answery ≥ −(5/4)x − 25/4, y ≥ (4/5)x, y ≤ −(4/3)x² − (16/3)x + 20/3 (strict inequalities also accepted)

More past-paper questions on quadratics from their turning point

• WMA11/01 Jan 2024 Q9 · 10 marks
• WMA11/01 Oct 2024 Q4 · 10 marks
• WMA11/01 Jan 2023 Q8 · 8 marks
• WMA11/01 Jan 2021 Q4 · 8 marks
• WMA11/01 Jun 2021 Q5 · 8 marks