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}