Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}+9}{\mathrm{x}-9} $$. Find any vertical asymptotes of $$ \mathrm{f}(\mathrm{x}) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, (Type an equation.) B. The function has two vertical asymptotes. The leftmost asymptote is $$ \square $$ and the rightmost asymptote is (Type equations.) There are no vertical asymptotes. Find the intervals where $$ \mathrm{f}(\mathrm{x}) $$ is increasing or decreasing. Select the correct choice below and fill in the answer box(es) to complete your choice. (Type your answer in interval notation. Use a comma to separate answers as needed.) A. The function is increasing on $$ \square $$. It is never decreasing. B. The function is increasing on $$ \square $$. It is decreasing on $$ \square $$.

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**Pertinent Information**

1. **Domain:**  
   The function is  
   $$
   f(x)=\frac{x+9}{x-9}.
   $$  
   The denominator is zero when  
   $$
   x-9=0\quad\Longrightarrow\quad x=9.
   $$  
   Hence, the domain is all real numbers except $$ x=9 $$.

2. **Vertical Asymptote:**  
   Since $$ x=9 $$ makes the denominator zero and does not cancel with the numerator, there is a vertical asymptote at  
   $$
   x=9.
   $$  
   (Choice A is correct: The function has one vertical asymptote, $$ x=9 $$.)

3. **Horizontal Asymptote:**  
   The degrees of the numerator and denominator are equal. Thus, the horizontal asymptote is determined by the ratio of the leading coefficients:
   $$
   y=\frac{1}{1}=1.
   $$

4. **Intercepts:**  
   - **$$ x $$-intercept:** Set $$ f(x)=0 $$:
     $$
     \frac{x+9}{x-9}=0\quad\Longrightarrow\quad x+9=0\quad\Longrightarrow\quad x=-9.
     $$
     So, the $$ x $$-intercept is $$(-9,0)$$.  
   - **$$ y $$-intercept:** Substitute $$ x=0 $$:
     $$
     f(0)=\frac{0+9}{0-9}=\frac{9}{-9}=-1.
     $$
     So, the $$ y $$-intercept is $$(0,-1)$$.

5. **Behavior Near the Vertical Asymptote:**  
   - As $$ x\to9^- $$ (approaching 9 from the left), $$ x-9 $$ is negative and very small, so $$ f(x) \to -\infty $$.  
   - As $$ x\to9^+ $$ (approaching 9 from the right), $$ x-9 $$ is positive and very small, so $$ f(x) \to +\infty $$.

6. **Increasing/Decreasing Intervals:**  
   Compute the derivative using the quotient rule:
   $$
   f(x)=\frac{x+9}{x-9} \quad\Longrightarrow\quad f'(x)=\frac{(1)(x-9)-(x+9)(1)}{(x-9)^2}.
   $$  
   Simplify the numerator:
   $$
   f'(x)=\frac{x-9-x-9}{(x-9)^2}=\frac{-18}{(x-9)^2}.
   $$  
   Since the denominator $$(x-9)^2$$ is always positive (except at $$ x=9 $$, which is not in the domain) and the numerator is $$-18$$, we have:
   $$
   f'(x)<0 \quad \text{for all } x\neq 9.
   $$  
   Thus, the function is strictly decreasing on each interval of its domain. In interval notation, $$ f(x) $$ is decreasing on:
   $$
   (-\infty,9)\cup(9,\infty).
   $$

**Final Answers**

- **Vertical Asymptote:** The function has one vertical asymptote at  
  $$
  x=9.
  $$

- **Intervals of Monotonicity:** The function is decreasing on  
  $$
  (-\infty,9)\cup(9,\infty),
  $$  
  and there are no intervals where the function is increasing.
**Summary:** - **Vertical Asymptote:** $$ x = 9 $$ - **Intervals of Decreasing:** $$ (-\infty, 9) $$ and $$ (9, \infty) $$

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