B

Step 1: Given the function \( f(x) = x^2 - 6x + 8 \).
Step 2: This is a quadratic function, and we can find its critical points by taking the derivative.
\( f'(x) = 2x - 6 \).
Step 3: Set the derivative equal to zero to find critical points:
\( 2x - 6 = 0 \) \( \Rightarrow x = 3 \)
Step 4: Next, we evaluate the function at the endpoints of the interval \( [2, 4] \) and the critical point \( x = 3 \):
\( f(2) = 2^2 - 6(2) + 8 = 4 - 12 + 8 = 0 \)
\( f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1 \)
\( f(4) = 4^2 - 6(4) + 8 = 16 - 24 + 8 = 0 \)
Step 5: Therefore, the values of the function are:
\( f(2) = 0, \; f(3) = -1, \; f(4) = 0 \).
Step 6: The function vanishes (i.e., is equal to zero) at \( x = 2 \) and \( x = 4 \).
Hence, the function is minimum at \( x = 3 \), which does not vanish.
Thus, the values of \( x \) for which \( f(x) \) vanishes is at the set \( \{2, 4\} \).
The answer is \( x = 2 \) or \( x = 4 \). Therefore, the midpoint of the interval is required. Thus, the answer is \( B \) (since it only mentions the points where it vanishes).
Therefore, B.