Summarize all pertinent information obtained by applying the graphing strategy and sketch the graph of $$ y=f(x) $$. $$ f(x)=\frac{x^{2}+9 x+18}{x^{2}+6 x+9} $$ Find the location of any local extrema of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. There is a local maximum at $$ x=\square $$. There is no local minimum. (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There is a local maximum at $$ x=\square $$ and there is a local minimum at $$ x=\square $$. (Type integers or decimals. Use a comma to separate answers as needed.) C. There is a local minimum at $$ x=\square $$. There is no local maximum. (Type an integer or a decimal. Use a comma to separate answers as needed.) D. There are no local extrema.

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1. Begin with the given function  
   $$
   f(x)=\frac{x^{2}+9x+18}{x^{2}+6x+9}.
   $$

2. Factor both numerator and denominator. Notice that  
   $$
   x^{2}+9x+18 = (x+3)(x+6)
   $$  
   and  
   $$
   x^{2}+6x+9 = (x+3)^2.
   $$

3. For $$ x\neq -3 $$ the function simplifies to  
   $$
   f(x)=\frac{(x+3)(x+6)}{(x+3)^2} = \frac{x+6}{x+3}.
   $$
   Since $$ x=-3 $$ makes the denominator zero, there is a vertical asymptote there. In the original function, $$ x=-3 $$ gives a $$ \frac{0}{0} $$ indeterminate form, but after cancellation the discontinuity remains nonremovable.

4. Determine the horizontal asymptote. Since the degrees of the numerator and denominator (both are 1 after simplification) are equal, the horizontal asymptote is given by the ratio of the leading coefficients:
   $$
   y=\frac{1}{1}=1.
   $$

5. Compute the derivative of the simplified function using the quotient rule. With  
   $$
   u(x)=x+6 \quad \text{and} \quad v(x)=x+3,
   $$
   we have  
   $$
   u'(x)=1 \quad \text{and} \quad v'(x)=1.
   $$
   Then,  
   $$
   f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{(v(x))^2}=\frac{(1)(x+3)-(x+6)(1)}{(x+3)^2}=\frac{x+3-x-6}{(x+3)^2}=\frac{-3}{(x+3)^2}.
   $$

6. Analyze $$ f'(x)=\frac{-3}{(x+3)^2} $$:  
   - The denominator $$(x+3)^2$$ is always positive for $$ x\neq -3 $$.  
   - Hence, $$ f'(x) $$ is always negative on the domain of $$ f $$.

This means the function is strictly decreasing on each interval of its domain. There is no point where the derivative changes sign, so no local extrema occur.

7. The pertinent information for graphing $$ f(x) $$ includes:  
   - **Domain:** $$ x\neq -3 $$ (vertical asymptote at $$ x=-3 $$).  
   - **Vertical asymptote:** $$ x=-3 $$ with  
     $$
     \lim_{x\to -3^-} f(x)=-\infty \quad \text{and} \quad \lim_{x\to -3^+} f(x)=+\infty.
     $$  
   - **Horizontal asymptote:** $$ y=1 $$ as $$ x\to\pm\infty $$.  
   - **Intercepts:**  
     - $$ x $$-intercept: set $$ f(x)=0 $$ using $$ \frac{x+6}{x+3}=0 $$ yielding $$ x=-6 $$.  
     - $$ y $$-intercept: substitute $$ x=0 $$ to get $$ f(0)=\frac{6}{3}=2 $$.

8. Since $$ f'(x)<0 $$ everywhere in its domain, there is no critical point at which a local minimum or maximum can occur.

The correct choice is:

D. There are no local extrema.
There are no local extrema for the function $$ f(x) = \frac{x^{2}+9x+18}{x^{2}+6x+9} $$.

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