When a domain is specified, consider separately whether the vertex lies within the domain.
Example ①
Find the maximum and minimum values of y = x² - 4x + 5（1 ≤ x ≤ 4）.
Answer: y = (x-2)² + 1. The vertex x = 2 lies within the domain.
Minimum value: at x = 2, y = 1.
Maximum value: compare the endpoints x = 1（y = 2）and x = 4（y = 5）→ at x = 4, y = 5.
Find the maximum and minimum values of y = x² - 4x + 5（1 ≤ x ≤ 4）.
Answer: y = (x-2)² + 1. The vertex x = 2 lies within the domain.
Minimum value: at x = 2, y = 1.
Maximum value: compare the endpoints x = 1（y = 2）and x = 4（y = 5）→ at x = 4, y = 5.
Key Point
- Vertex within the domain → the vertex's y-value is the minimum（opens upward）or maximum（opens downward）
- Vertex outside the domain → compare the values at both endpoints of the domain
Practice Problems
- Find the maximum and minimum values of y = -x² + 2x + 3（0 ≤ x ≤ 3）.
Answers and Explanations
- Answer: Maximum value 4（x = 1）, minimum value 0（x = 3）. y = -(x-1)² + 4. When x = 3: y = 0.