Question 1 – Finding Maximum or Minimum by Completing Square

In this question, we have six quadratic functions. For each, we need to find whether it has a maximum or minimum value, and what that value is. We use the method of completing the square.

The general form of a quadratic is f(x) = ax² + bx + c. If a > 0, the parabola opens upward and has a minimum value. If a < 0, it opens downward and has a maximum value.

To complete the square, we rewrite the function in the form f(x) = a(x – h)² + k. Then the vertex is at (h, k), and the maximum or minimum value is k.

Part (i) f(x) = x² + 6x + 13
Here a = 1 > 0, so it has a minimum.
Complete the square: x² + 6x + 13 = (x² + 6x + 9) + 4 = (x + 3)² + 4.
So the minimum value is 4 at x = –3.

Part (ii) f(x) = x² + 4x
a = 1 > 0, so minimum.
Complete the square: x² + 4x = (x² + 4x + 4) – 4 = (x + 2)² – 4.
Minimum value is –4 at x = –2.

Part (iii) f(x) = –x² + 8x + 13
a = –1 < 0, so maximum.
Factor out –1 first: –(x² – 8x) + 13. Complete square inside: –(x² – 8x + 16) + 13 + 16 = –(x – 4)² + 29.
Maximum value is 29 at x = 4.

Part (iv) f(x) = –x² – 3x – 5
a = –1 < 0, so maximum.
–(x² + 3x) – 5 = –(x² + 3x + 9/4) – 5 + 9/4 = –(x + 3/2)² – 11/4.
Maximum value is –11/4 at x = –3/2.

Part (v) f(x) = 3x² + 6x – 13
a = 3 > 0, so minimum.
Factor out 3: 3(x² + 2x) – 13 = 3(x² + 2x + 1) – 13 – 3 = 3(x + 1)² – 16.
Minimum value is –16 at x = –1.

Part (vi) f(x) = –2x² – x + 21
a = –2 < 0, so maximum.
Factor out –2: –2(x² + (1/2)x) + 21 = –2(x² + (1/2)x + 1/16) + 21 + 1/8 = –2(x + 1/4)² + 169/8.
Maximum value is 169/8 at x = –1/4.

Question 2 – Finding Maximum/Minimum Point by Sketching

In this question, we have six quadratic functions. We need to sketch each graph, find the maximum or minimum point, and determine the domain and range.

For each quadratic, we find the vertex (which is the max or min point), the y-intercept, and x-intercepts if they exist. Then we sketch.

The domain of any quadratic function is always all real numbers.

The range depends on whether it opens upward or downward. If a > 0, range is [minimum value, ∞). If a < 0, range is (–∞, maximum value].

Part (i) f(x) = x² – 4x
a = 1 > 0, so opens upward. Vertex: complete square: (x – 2)² – 4. Vertex (2, –4) is minimum. x-intercepts: x(x – 4) = 0 → x = 0, 4. Domain: all real numbers. Range: [–4, ∞).

Part (ii) f(x) = x² – 5x + 6
a = 1 > 0, opens upward. Vertex: (x – 5/2)² – 1/4. Vertex (2.5, –0.25) is minimum. x-intercepts: (x – 2)(x – 3) = 0 → x = 2, 3. Domain: all real numbers. Range: [–0.25, ∞).

Part (iii) f(x) = –x² + 2x – 8
a = –1 < 0, opens downward. Vertex: –(x – 1)² – 7. Vertex (1, –7) is maximum. x-intercepts: discriminant = 4 – 4(–1)(–8) = 4 – 32 = –28 < 0, so no real x-intercepts. Domain: all real numbers. Range: (–∞, –7].

Part (iv) f(x) = x² – 4x + 4
a = 1 > 0, opens upward. This is (x – 2)². Vertex (2, 0) is minimum. x-intercept: x = 2 (touches x-axis). Domain: all real numbers. Range: [0, ∞).

Part (v) f(x) = x² + 2x – 8.3
a = 1 > 0, opens upward. Complete square: (x + 1)² – 9.3. Vertex (–1, –9.3) is minimum. x-intercepts: solve x² + 2x – 8.3 = 0 using quadratic formula. Domain: all real numbers. Range: [–9.3, ∞).

Part (vi) f(x) = 6 – x – x²
Rewrite as –x² – x + 6. a = –1 < 0, opens downward. Complete square: –(x + 1/2)² + 25/4. Vertex (–0.5, 6.25) is maximum. x-intercepts: –x² – x + 6 = 0 → x² + x – 6 = 0 → (x + 3)(x – 2) = 0 → x = –3, 2. Domain: all real numbers. Range: (–∞, 6.25].

Question 3 – Inverse of Quadratic Functions

In this question, we have six quadratic functions. For each, we need to find the inverse function, and also find the domain and range of both the original and inverse functions.

To find the inverse, we replace f(x) with y, then swap x and y, then solve for y. The domain of the inverse is the range of the original function, and the range of the inverse is the domain of the original function.

However, quadratic functions are not one-to-one over all real numbers. So we restrict the domain to make them one-to-one. The given functions already have restricted domains.

Part (i) f(x) = x² – 3, x ≤ 0
Domain: x ≤ 0. Range: for x ≤ 0, x² ≥ 0, so x² – 3 ≥ –3. Range is [–3, ∞).
Inverse: y = x² – 3 → x = y² – 3 → y² = x + 3 → y = –√(x + 3) (negative because x ≤ 0). So f⁻¹(x) = –√(x + 3). Domain of inverse: [–3, ∞). Range of inverse: (–∞, 0].

Part (ii) f(x) = x² + 6x + 4, x < –3
Complete square: (x + 3)² – 5. Since x < –3, the function is decreasing. Range: as x → –∞, f(x) → ∞; at x = –3, f(–3) = –5. So range is [–5, ∞).
Inverse: y = (x + 3)² – 5 → (x + 3)² = y + 5 → x + 3 = –√(y + 5) (negative because x < –3). So f⁻¹(x) = –3 – √(x + 5). Domain: [–5, ∞). Range: (–∞, –3).

Part (iii) f(x) = 2x² – 8x + 11, x ≥ 2
Complete square: 2(x – 2)² + 3. Since x ≥ 2, range is [3, ∞).
Inverse: y = 2(x – 2)² + 3 → (x – 2)² = (y – 3)/2 → x – 2 = √((y – 3)/2) (positive because x ≥ 2). So f⁻¹(x) = 2 + √((x – 3)/2). Domain: [3, ∞). Range: [2, ∞).

Part (iv) f(x) = 3x² – 2x + 6, x ≥ 5
Complete square: 3(x – 1/3)² + 17/3. For x ≥ 5, the function is increasing. Range: f(5) = 3(25) – 10 + 6 = 75 – 10 + 6 = 71. So range is [71, ∞).
Inverse: y = 3x² – 2x + 6 → 3x² – 2x + (6 – y) = 0. Use quadratic formula: x = [2 ± √(4 – 12(6 – y))]/6 = [2 ± √(12y – 68)]/6 = [1 ± √(3y – 17)]/3. Since x ≥ 5, we take the positive root. So f⁻¹(x) = [1 + √(3x – 17)]/3. Domain: [71, ∞). Range: [5, ∞).

Part (v) f(x) = 2(x – 3)² + 1, x ≥ 3
Domain: x ≥ 3. Range: [1, ∞).
Inverse: y = 2(x – 3)² + 1 → (x – 3)² = (y – 1)/2 → x – 3 = √((y – 1)/2) (positive because x ≥ 3). So f⁻¹(x) = 3 + √((x – 1)/2). Domain: [1, ∞). Range: [3, ∞).

Part (vi) f(x) = –3(x + 4)² – 5, x < –4
Domain: x < –4. Range: as x → –∞, f(x) → –∞; at x = –4, f(–4) = –5. So range is (–∞, –5].
Inverse: y = –3(x + 4)² – 5 → –(y + 5)/3 = (x + 4)² → x + 4 = –√(–(y + 5)/3) (negative because x < –4). So f⁻¹(x) = –4 – √(–(x + 5)/3). Domain: (–∞, –5]. Range: (–∞, –4).

Question 4 – Absolute Value Quadratic Equations and Inequalities

This is a combined question about solving absolute value equations and inequalities involving quadratic expressions. The absolute value |u| means we consider both u and –u.

Part (i) |x² + 1| = 5
Since x² + 1 is always positive, we can drop the absolute value: x² + 1 = 5 → x² = 4 → x = ±2.

Part (ii) |x² + 5x + 4| = 0
The absolute value is zero only when the expression is zero: x² + 5x + 4 = 0 → (x + 1)(x + 4) = 0 → x = –1, –4.

Part (iii) |x² – 6x + 8| = 4
This means x² – 6x + 8 = 4 OR x² – 6x + 8 = –4.
First equation: x² – 6x + 4 = 0 → x = [6 ± √(36 – 16)]/2 = [6 ± √20]/2 = 3 ± √5.
Second equation: x² – 6x + 12 = 0 → discriminant = 36 – 48 = –12 < 0, so no real solutions.
So solutions: x = 3 ± √5.

Part (iv) |3x² – 7x + 2| = |x² – x + 1|
We have two cases: either both expressions are equal, or one is negative of the other.
Case 1: 3x² – 7x + 2 = x² – x + 1 → 2x² – 6x + 1 = 0 → x = [6 ± √(36 – 8)]/4 = [6 ± √28]/4 = [3 ± √7]/2.
Case 2: 3x² – 7x + 2 = –(x² – x + 1) → 3x² – 7x + 2 = –x² + x – 1 → 4x² – 8x + 3 = 0 → x = [8 ± √(64 – 48)]/8 = [8 ± √16]/8 = [8 ± 4]/8 = 12/8 = 3/2 or 4/8 = 1/2.
So solutions: x = (3 ± √7)/2, x = 3/2, x = 1/2.

Part (v) |x² – 4| < 5
This means –5 < x² – 4 < 5. Add 4: –1 < x² < 9. Since x² ≥ 0, the left inequality is automatically true. So x² < 9 → –3 < x < 3.

Part (vi) |x² – 3x + 2| > 4
This means x² – 3x + 2 > 4 OR x² – 3x + 2 < –4.
First: x² – 3x – 2 > 0 → x = [3 ± √(9 + 8)]/2 = [3 ± √17]/2. So x < (3 – √17)/2 or x > (3 + √17)/2.
Second: x² – 3x + 6 < 0 → discriminant = 9 – 24 = –15 < 0, so no real solutions.
So solution: x < (3 – √17)/2 or x > (3 + √17)/2.

Part (vii) |x² – 5x + 6| ≤ x + 2
We need to consider two cases and also ensure x + 2 ≥ 0 (since absolute value is ≥ 0). So x ≥ –2.
Case 1: x² – 5x + 6 ≤ x + 2 → x² – 6x + 4 ≤ 0 → x ∈ [3 – √5, 3 + √5].
Case 2: –(x² – 5x + 6) ≤ x + 2 → –x² + 5x – 6 ≤ x + 2 → –x² + 4x – 8 ≤ 0 → x² – 4x + 8 ≥ 0. Discriminant = 16 – 32 = –16 < 0, so always true.
So combining with x ≥ –2 and case 1: x ∈ [3 – √5, 3 + √5] ∩ [–2, ∞) = [3 – √5, 3 + √5] (since 3 – √5 ≈ 0.76 > –2).

Part (viii) |2x² – 3x – 5| < 4
This means –4 < 2x² – 3x – 5 < 4.
First: 2x² – 3x – 5 > –4 → 2x² – 3x – 1 > 0 → x < (3 – √17)/4 or x > (3 + √17)/4.
Second: 2x² – 3x – 5 < 4 → 2x² – 3x – 9 < 0 → x ∈ ((3 – √81)/4, (3 + √81)/4) = ((3 – 9)/4, (3 + 9)/4) = (–3/2, 3).
Combining both: x ∈ (–3/2, (3 – √17)/4) ∪ ((3 + √17)/4, 3).