To find the equation of the line normal to the graph of the function \( f(x) \) at the point \( (2, -1) \), we first need to determine the slope of the tangent line at that point.
Given that \( f'(20) = \frac{1}{2} \), this value represents the slope of the tangent line to the function \( f(x) \) at the point where \( x = 20 \), not at \( x = 2 \). However, for finding the normal line at \( (2, -1) \), we need the slope of the tangent line at that specific point.
If we assume \( f'(2) \) was given or could be calculated, let’s denote it as \( m_t \) (the slope of the tangent line at \( x = 2 \)). The slope of the normal line, \( m_n \), is the negative reciprocal of the slope of the tangent line: \[ m_n = -\frac{1}{m_t} \]
The normal line passes through the point \( (2, -1) \) and has a slope of \( m_n \). We can use the point-slope form of a line to find the equation of the normal line: \[ y - y_1 = m_n (x - x_1) \] where \( (x_1, y_1) = (2, -1) \).
Substituting in the values: \[ y - (-1) = m_n (x - 2) \] \[ y + 1 = m_n (x - 2) \] \[ y = m_n (x - 2) - 1 \]
Without knowing \( f'(2) \), we cannot find a specific value for \( m_n \). To proceed, we need the value of \( f'(2) \).
Assuming that we could obtain \( f'(2) \), you would be able to:
- Calculate \( m_n = -\frac{1}{f'(2)} \).
- Plug this value into the equation of the normal line.
Since you provided \( f'(20) \) but not \( f'(2) \), please check or provide \( f'(2) \) to calculate the exact equation for the normal line.
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