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Quiz Chapter 2: Quadratic Functions

10 questions · Form 4 Additional Mathematics Bab 1: Quadratic Functions

Question 1 of 10Score: 0

Given the quadratic curve f(x) = 3x² + bx + 12 touches the x-axis at a single point, find the positive value of b.

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Given the quadratic curve f(x) = 3x² + bx + 12 touches the x-axis at a single point, find the positive value of b.

  1. 12
  2. 6
  3. 144
  4. 24
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Answer: A

Touches x-axis => b² - 4ac = 0 => b² - 4(3)(12) = 0 => b² - 144 = 0 => b² = 144 => b = 12 (positive value).

2. If α and β are roots of 2x² - 6x + 3 = 0, find the value of α + β.

  1. -3
  2. 3
  3. 32
  4. -32
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Answer: B

Sum of roots α + β = -ba = --62 = 62 = 3.

3. Find the range of values of x for which f(x) = 2x² - 8x is negative.

  1. 0 < x < 4
  2. x < 0 or x > 4
  3. -4 < x < 0
  4. x < -4 or x > 0
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Answer: A

Negative means f(x) < 0 => 2x(x - 4) < 0. Critical values are x = 0 and x = 4. Solution: 0 < x < 4.

4. State the y-intercept of the quadratic graph f(x) = -2(x - 1)² + 8.

  1. (0, 8)
  2. (0, 6)
  3. (0, -2)
  4. (0, 10)
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Answer: B

y-intercept occurs when x = 0: f(0) = -2(0 - 1)² + 8 = -2(1) + 8 = 6. Thus, y-intercept is (0, 6).

5. If the quadratic equation x² - px + 9 = 0 has equal roots, find the possible values of p.

  1. p = 6 only
  2. p = -6 only
  3. p = ± 6
  4. p = ± 3
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Answer: C

Equal roots => b² - 4ac = 0 => (-p)² - 4(1)(9) = 0 => p² - 36 = 0 => p² = 36 => p = ±6.

6. What is the axis of symmetry for the quadratic function f(x) = 2x² - 8x + 5?

  1. x = -2
  2. x = 4
  3. x = 2
  4. x = 8
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Answer: C

Axis of symmetry formula: x = -b2a = --82 * 2 = 84 = 2.

7. The quadratic function f(x) = (x - p)² + q has a vertex at (-4, -3). What are the values of p and q?

  1. p = 4, q = -3
  2. p = -4, q = -3
  3. p = -4, q = 3
  4. p = 4, q = 3
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Answer: B

Compare f(x) = (x - p)² + q with vertex (h, k) = (-4, -3): h = p = -4, k = q = -3.

8. Solve the quadratic inequality x² - 4x - 5 < 0.

  1. x < -1 or x > 5
  2. -1 < x < 5
  3. -5 < x < 1
  4. x < -5 or x > 1
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Answer: B

Factorise: (x + 1)(x - 5) < 0. Critical values are x = -1 and x = 5. For '< 0', the solution is between the roots: -1 < x < 5.

9. The quadratic graph f(x) = ax² + bx + c has a maximum point. Which condition MUST be true for a?

  1. a > 0
  2. a = 0
  3. a < 0
  4. a ≥ 1
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Answer: C

A quadratic function has a maximum turning point (∩-shape) if and only if the coefficient of x² is negative, i.e., a < 0.

10. Find the minimum value of the quadratic function f(x) = 2(x + 1)² - 9.

  1. -1
  2. -9
  3. 2
  4. 9
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Answer: B

In vertex form f(x) = a(x - h)² + k with a = 2 > 0, the minimum value is given directly by k, which is -9.

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