1. **Problem stated:** Find the complex roots of $f(x)=-\frac{1}{x-2}-\ln(x-4)$ and its asymptotes. 2. **Formula and domain rules:** A root is found by solving $f(x)=0$. For logarithms, the input must be positive, so $x-4>0$, which means $x>4$. Also, the denominator cannot be zero, so $x\neq 2$. The domain is therefore $x>4$. 3. **Set the function equal to zero:** $$-\frac{1}{x-2}-\ln(x-4)=0$$ 4. **Move one term to the other side:** $$-\frac{1}{x-2}=\ln(x-4)$$ Since the left side is negative for $x>4$, we already know any root must satisfy $\ln(x-4)<0$, so $00$. 6. **Check whether there is an elementary closed-form solution:** This equation mixes a rational function and a logarithm, so it cannot be solved by ordinary algebraic steps. The root must be found numerically. 7. **Numerical solution:** Solving $$-\frac{1}{x-2}-\ln(x-4)=0$$ on the domain $x>4$ gives approximately $$x\approx 4.3669$$ So the real root is approximately $4.3669$. Since the function is only defined for real $x>4$, there are no other real roots from the given expression. 8. **Asymptotes:** The logarithm term has a vertical asymptote where its input approaches $0^+$. That happens at $$x=4$$ So $x=4$ is a vertical asymptote. 9. **Check for other asymptotes:** As $x\to\infty$, $$-\frac{1}{x-2}\to 0$$ and $$\ln(x-4)\to\infty$$ so $$f(x)\to-\infty$$ This means there is no horizontal asymptote and no slant asymptote. 10. **Final answer:** The function has one real root at approximately $x\approx 4.3669$. Its only asymptote is the vertical asymptote $x=4$. There are no horizontal or slant asymptotes.