Question

Given a function f(x) where f'(2) = 1/2, what is the equation of the line normal to the graph of the function at the point (2, -1)

Answer 1

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:

1. The derivative at that point \( f'(2) = \frac{1}{2} \). This means that the slope of the tangent line at \( x = 2 \) is \( \frac{1}{2} \).

2. 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:

\[
\boxed{y = -2x + 3}
\]