To find the minimum point of the function \( f(x) = \frac{1}{2}(x - 11)^2 + 4 \), we observe that this is a quadratic function represented in vertex form, \( a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.
From the function:
\[ f(x) = \frac{1}{2}(x - 11)^2 + 4 \]
we can identify:
- \( h = 11 \)
- \( k = 4 \)
The vertex, which represents the minimum point of the quadratic function (since the coefficient of the squared term, \( \frac{1}{2} \), is positive), is located at the point \( (11, 4) \).
However, the question asks for coordinates relating to the square root function. The square root function that corresponds to the vertex form of the quadratic is often given by the transformation related to the quadratic function.
Nonetheless, none of the choices you've provided match the coordinates of the vertex \( (11, 4) \). If you are looking for points from the options presented:
- (4, -11)
- (4, 11)
- (-4, 11)
- (-4, -11)
None of these correspond to the minimum point of the function \( f(x) = \frac{1}{2}(x - 11)^2 + 4 \) or any relation to the square root function we derived.
Thus, if you meant to find another connection or if there was an error in the options given, please clarify. Otherwise, the answer to the minimum point based on the quadratic function is \( (11, 4) \).
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