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We are asked to find the equation of the tangent line to the curve \( f(x) = \frac{1}{6 + x^2} \) at the point where \( x = 2 \) and the slope of the tangent line is given as \( -\frac{1}{25} \).

### Steps:
1. **Tangent line equation form**: 
   The general equation of the tangent line at a point \( x = a \) is:
   \[
   y - f(a) = m(x - a)
   \]
   where \( m \) is the slope of the tangent line at \( x = a \).

2. **Given slope**:
   The slope at \( x = 2 \) is provided as \( m = -\frac{1}{25} \).

3. **Find \( f(2) \)**:
   We need the value of the function at \( x = 2 \):
   \[
   f(2) = \frac{1}{6 + 2^2} = \frac{1}{6 + 4} = \frac{1}{10}
   \]

4. **Substitute values into the tangent line equation**:
   Using \( x = 2 \), \( f(2) = \frac{1}{10} \), and \( m = -\frac{1}{25} \), we substitute into the tangent line formula:
   \[
   y - \frac{1}{10} = -\frac{1}{25}(x - 2)
   \]

5. **Simplify the equation**:
   Distribute the slope on the right-hand side:
   \[
   y - \frac{1}{10} = -\frac{1}{25}x + \frac{2}{25}
   \]
   Add \( \frac{1}{10} \) to both sides to solve for \( y \):
   \[
   y = -\frac{1}{25}x + \frac{2}{25} + \frac{1}{10}
   \]
   To combine \( \frac{2}{25} \) and \( \frac{1}{10} \), convert \( \frac{1}{10} \) to have a denominator of 25:
   \[
   \frac{1}{10} = \frac{2.5}{25} = \frac{5}{50}
   ```

For f(x) = 1/(6 + x^2), the slope of the graph of y = f(x) is known to be -1/25 at the point with x-coordinate 2. Find the equation of the tangent line at that point.

Learn how to find the equation of the tangent line to the curve f(x) = 1/(6 + x^2) at the point where x = 2, with a given slope of -1/25. This solution uses calculus, focusing on the derivative and tangent line formula. Suitable for Grades 11-12.

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