When the endpoints of the range or the position of the vertex changes depending on the parameter a, it is necessary to consider separate cases.
Example ①
Find the minimum value of y = (x - a)² + 1（0 ≤ x ≤ 2）.
Solution (by cases):
(i) a < 0: The vertex is outside the left end of the range → the minimum occurs at x = 0. The minimum value is a² + 1.
(ii) 0 ≤ a ≤ 2: The vertex is within the range → the minimum occurs at x = a. The minimum value is 1.
(iii) a > 2: The vertex is outside the right end of the range → the minimum occurs at x = 2. The minimum value is (2-a)² + 1.
Find the minimum value of y = (x - a)² + 1（0 ≤ x ≤ 2）.
Solution (by cases):
(i) a < 0: The vertex is outside the left end of the range → the minimum occurs at x = 0. The minimum value is a² + 1.
(ii) 0 ≤ a ≤ 2: The vertex is within the range → the minimum occurs at x = a. The minimum value is 1.
(iii) a > 2: The vertex is outside the right end of the range → the minimum occurs at x = 2. The minimum value is (2-a)² + 1.
Key Point
- There are three cases depending on where the vertex a lies in relation to the range.
- Calculate the values at the endpoints and vertex in each case.
Practice Problems
- Express the minimum value of y = x² - 2ax + 1（0 ≤ x ≤ 1）in terms of a.
Solutions and Explanations
- Solution: y = (x-a)² + 1 - a². The vertex is x = a.
When a < 0, the minimum value is 1（x=0）. When 0 ≤ a ≤ 1, the minimum value is 1-a². When a > 1, the minimum value is 2-2a（x=1）.