Secondary 3 A-Maths sample questions with answers

Quadratic Functions

1. 1Express x^2 - 8x + 3 in the form (x - h)^2 + k.

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   Answer: (x - 4)^2 - 13

   1. x^2 - 8x + 3 = (x - 4)^2 - 16 + 3
   2. = (x - 4)^2 - 13

2. 2Express 2x^2 + 6x - 1 in the form a(x + p)^2 + q.

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   Answer: 2(x + 3/2)^2 - 11/2

   1. 2x^2 + 6x - 1 = 2(x^2 + 3x) - 1
   2. = 2[(x + 3/2)^2 - 9/4] - 1
   3. = 2(x + 3/2)^2 - 11/2

3. 3Write down the maximum value of y = 7 - 2(x + 1)^2 and the corresponding value of x.

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   Answer: Maximum value 7, when x = -1

   1. (x + 1)^2 ≥ 0, so -2(x + 1)^2 ≤ 0 and y ≤ 7
   2. y = 7 when x + 1 = 0, that is, when x = -1

4. 4Write down the coordinates of the turning point of the curve y = 3(x - 4)^2 + 5, and state whether it is a maximum or a minimum point.

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   Answer: (4, 5), a minimum point

   1. The curve is in the form a(x - h)^2 + k with h = 4 and k = 5, so the turning point is (4, 5).
   2. a = 3 > 0, so it is a minimum point.

5. 5Find the equation of the line of symmetry of the curve y = (x + 7)(x - 3).

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   Answer: x = -2

   1. The x-intercepts are -7 and 3.
   2. The line of symmetry is halfway between them: x = (-7 + 3)/2 = -2

6. 6Show that x^2 + 10x + 30 is positive for all real values of x.

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   Answer: x^2 + 10x + 30 = (x + 5)^2 + 5, which is at least 5

   1. x^2 + 10x + 30 = (x + 5)^2 - 25 + 30 = (x + 5)^2 + 5
   2. Since (x + 5)^2 ≥ 0 for all real x, (x + 5)^2 + 5 ≥ 5 > 0.

7. 7Find the range of values of c for which x^2 - 10x + c is positive for all real values of x.

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   Answer: c > 25

   1. a = 1 > 0, so we need b^2 - 4ac < 0
   2. (-10)^2 - 4(1)(c) < 0, so 100 - 4c < 0
   3. c > 25

8. 8Explain why -3x^2 + 2x - 4 is negative for all real values of x.

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   Answer: a = -3 < 0 and b^2 - 4ac = -44 < 0

   1. a = -3 < 0, so the curve y = -3x^2 + 2x - 4 opens downwards.
   2. b^2 - 4ac = 2^2 - 4(-3)(-4) = 4 - 48 = -44 < 0, so the curve does not meet the x-axis.
   3. So the expression is always negative.

9. 9For the curve y = 2x^2 - 4x - 6, state whether the turning point is a maximum or a minimum point. Find the intercepts on the axes and the equation of the line of symmetry.

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   Answer: A minimum point. The intercepts are (0, -6), (-1, 0) and (3, 0). The line of symmetry is x = 1.

   1. a = 2 > 0, so the curve opens upwards and the turning point is a minimum point.
   2. When x = 0, y = -6. The y-intercept is (0, -6).
   3. When y = 0: 2x^2 - 4x - 6 = 0, so 2(x^2 - 2x - 3) = 0 and 2(x - 3)(x + 1) = 0.
   4. x = 3 or x = -1. The x-intercepts are (-1, 0) and (3, 0).
   5. The line of symmetry is halfway between the x-intercepts: x = (-1 + 3)/2 = 1.

10. 10Express -2x^2 + 12x - 7 in the form a(x - h)^2 + k.

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    Answer: -2(x - 3)^2 + 11

    1. Take -2 out of the first two terms: -2x^2 + 12x - 7 = -2(x^2 - 6x) - 7
    2. Complete the square inside the bracket: x^2 - 6x = (x - 3)^2 - 9
    3. -2[(x - 3)^2 - 9] - 7 = -2(x - 3)^2 + 18 - 7
    4. = -2(x - 3)^2 + 11

11. 11Find the minimum value of y = 3x^2 + 12x + 5 and the corresponding value of x. Hence write down the coordinates of the turning point.

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    Answer: Minimum value -7 when x = -2. The turning point is (-2, -7).

    1. y = 3(x^2 + 4x) + 5 = 3[(x + 2)^2 - 4] + 5 = 3(x + 2)^2 - 7
    2. (x + 2)^2 ≥ 0 for all real x, so 3(x + 2)^2 ≥ 0 and y ≥ -7.
    3. y = -7 when x + 2 = 0, that is, when x = -2.

12. 12Find the maximum value of y = 5 + 8x - 2x^2 and the corresponding value of x.

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    Answer: Maximum value 13 when x = 2

    1. y = -2(x^2 - 4x) + 5
    2. = -2[(x - 2)^2 - 4] + 5 = -2(x - 2)^2 + 13
    3. The maximum value is 13, when x = 2.