1. **State the problem:** Find the equation of the tangent line to the function $y=\sqrt{x^2 - 2x}$ at the point where $x=3$. 2. **Recall the formula:** The point-slope form of a line is given by: $$y - y_1 = m(x - x_1)$$ where $m$ is the slope of the tangent line at $x_1$, and $y_1 = f(x_1)$. 3. **Find $y_1$:** Evaluate the function at $x=3$: $$y_1 = \sqrt{3^2 - 2 \times 3} = \sqrt{9 - 6} = \sqrt{3}$$ 4. **Find the derivative $y'$ to get the slope $m$:** Given $y = \sqrt{x^2 - 2x} = (x^2 - 2x)^{1/2}$, using the chain rule: $$y' = \frac{1}{2}(x^2 - 2x)^{-1/2} \times (2x - 2) = \frac{2x - 2}{2\sqrt{x^2 - 2x}} = \frac{x - 1}{\sqrt{x^2 - 2x}}$$ 5. **Evaluate the slope at $x=3$:** $$m = y'(3) = \frac{3 - 1}{\sqrt{3^2 - 2 \times 3}} = \frac{2}{\sqrt{9 - 6}} = \frac{2}{\sqrt{3}}$$ 6. **Write the equation of the tangent line:** $$y - \sqrt{3} = \frac{2}{\sqrt{3}}(x - 3)$$ This is the equation in the form $y - y_1 = m(x - x_1)$. **Final answer:** $$y - \sqrt{3} = \frac{2}{\sqrt{3}}(x - 3)$$