The problem asks to find the equation of the tangent line to the graph of the function $y = \frac{2x - 6}{2x + 6}$ at the point $\left( 6, \frac{1}{3} \right)$.

Steps to solve:

1. Find the derivative: We need the derivative $\frac{dy}{dx}$, which represents the slope of the tangent line at any point $x$. This can be found using the quotient rule: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$ where $u(x) = 2x - 6$ and $v(x) = 2x + 6$.

2. Apply the quotient rule:

   • $u(x) = 2x - 6$, so $u'(x) = 2$
   • $v(x) = 2x + 6$, so $v'(x) = 2$

   Using the quotient rule: $\frac{dy}{dx} = \frac{(2x + 6)(2) - (2x - 6)(2)}{(2x + 6)^2}$

3. Simplify:

   = \frac{4x + 12 - (4x - 12)}{(2x + 6)^2} = \frac{4x + 12 - 4x + 12}{(2x + 6)^2} = \frac{24}{(2x + 6)^2}$$ So, $$\frac{dy}{dx} = \frac{24}{(2x + 6)^2}$$.

4. Evaluate the derivative at $x = 6$: $\frac{dy}{dx} \bigg|_{x=6} = \frac{24}{(2(6) + 6)^2} = \frac{24}{(12 + 6)^2} = \frac{24}{18^2} = \frac{24}{324} = \frac{2}{27}$

   Therefore, the slope of the tangent line at $x = 6$ is $m = \frac{2}{27}$.

5. Find the equation of the tangent line: The equation of a line is given by $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point of tangency and $m$ is the slope.

   Here, $(x_1, y_1) = \left( 6, \frac{1}{3} \right)$ and $m = \frac{2}{27}$, so the equation becomes: $y - \frac{1}{3} = \frac{2}{27}(x - 6)$

6. Simplify the equation: $y = \frac{2}{27}(x - 6) + \frac{1}{3}$ Multiply both sides to eliminate fractions if desired.

This is the equation of the tangent line at $\left( 6, \frac{1}{3} \right)$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How is the quotient rule derived?
2. What happens if the denominator in a rational function has higher complexity?
3. How do we interpret the derivative geometrically?
4. Why do we need to find the derivative to determine the tangent line?
5. How does a derivative behave when the function has discontinuities?

Tip: Always simplify the function as much as possible before applying differentiation rules; it reduces errors during calculation.