To find the equation of the normal line to the graph of the function \( f(x) \) at the point \( (2, -1) \), we start with the information given:
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The derivative at that point \( f'(2) = \frac{1}{2} \). This means that the slope of the tangent line at \( x = 2 \) is \( \frac{1}{2} \).
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The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope \( m_{\text{normal}} \) of the normal line is:
\[ m_{\text{normal}} = -\frac{1}{\frac{1}{2}} = -2 \]
Now that we have the slope of the normal line, we can use the point-slope form of the equation of a line, which is:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point of tangency, and \( m \) is the slope. Plugging in our point \( (2, -1) \) and the slope \( -2 \):
\[ y - (-1) = -2(x - 2) \]
This simplifies to:
\[ y + 1 = -2(x - 2) \]
\[ y + 1 = -2x + 4 \]
\[ y = -2x + 4 - 1 \]
\[ y = -2x + 3 \]
Thus, the equation of the line normal to the graph of the function at the point \( (2, -1) \) is:
\[ \boxed{y = -2x + 3} \]