0606/13 May/Jun 2025, Question 9: The Quadratic Formula and Completing the Square

Part (ai) · 3 marks

Rewrite f(x) = −2x² + 9x − 10 in the completed-square form a + b(x + c)², stating constants a, b, c.

1. 1f(x) = −2(x² − (9/2)x) − 10
2. 2= −2[(x − 9/4)² − 81/16] − 10
3. 3= −2(x − 9/4)² + 81/8 − 10
4. 4= 1/8 − 2(x − 9/4)²

Answerf(x) = 1/8 − 2(x − 9/4)²

Part (aii) · 2 marks

Using your completed-square form, decide whether the inverse f⁻¹ exists for 0 ≤ x ≤ 3, justifying briefly.

1. 1The turning point is at (9/4, 1/8)
2. 29/4 lies between 0 and 3
3. 3So f is not one-one on 0 ≤ x ≤ 3
4. 4f⁻¹ does not exist

Answerf⁻¹ does not exist, because the turning point (9/4, 1/8) lies inside the domain 0 ≤ x ≤ 3 (f is not one-one)

Part (bi) · 3 marks

Sketch y = 3ln(5 − 2x) for 0 ≤ x < 2.5, giving exact axis intercepts and the asymptote equation.

1. 1g(0) = 3ln 5, so y-intercept (0, 3ln 5)
2. 2g(x) = 0 when 5 − 2x = 1, x = 2, so x-intercept (2, 0)
3. 3As x → 2.5, 5 − 2x → 0 and g(x) → −∞: asymptote x = 2.5
4. 4g is decreasing throughout

AnswerDecreasing log-type curve for 0 ≤ x < 2.5: starts at (0, 3ln 5) on the y-axis, crosses the x-axis at (2, 0), and tends to −∞ approaching the vertical asymptote x = 2.5

Part (bii) · 2 marks

Find an expression for the inverse function g⁻¹(x).

1. 1y = 3ln(5 − 2x), so y/3 = ln(5 − 2x)
2. 25 − 2x = e^(y/3)
3. 3x = (1/2)(5 − e^(y/3))
4. 4g⁻¹(x) = (1/2)(5 − e^(x/3))

Answerg⁻¹(x) = (1/2)(5 − e^(x/3))

Part (biii) · 2 marks

State the domain and range of g⁻¹, giving both in exact form.

1. 1Domain of g⁻¹ = range of g: g(x) ≤ 3ln 5, so x ≤ 3ln 5
2. 2Range of g⁻¹ = domain of g: 0 ≤ g⁻¹(x) < 2.5

AnswerDomain x ≤ 3ln 5; range 0 ≤ g⁻¹(x) < 2.5

More past-paper questions on the quadratic formula and completing the square

• 0606/22 Feb/Mar 2026 Q1 · 8 marks
• 0606/13 May/Jun 2026 Q1 · 6 marks
• 0606/12 Feb/Mar 2025 Q5 · 8 marks
• 0606/12 Oct/Nov 2025 Q4 · 10 marks
• 0606/23 Oct/Nov 2025 Q3 · 9 marks
• 0580/22 May/Jun 2026 Q21 · 3 marks