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 the intervals where $$ \mathrm{f}(\mathrm{x}) $$ is concave upward or downward. 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 concave upward on B. The function is concave downward on is never concave downward. It is never concave upward. C. The function is concave upward on $$ \square $$. It is concave downward on $$ \square $$. Find the iocation of any inflection points of $$ \mathrm{f}(\mathrm{x}) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is an inflection point at $$ \mathrm{x}=\square $$. (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There are no inflection points.

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We begin with the function $$ f(x)=\frac{x+9}{x-9}. $$ ### Graph Summary 1. **Domain:** The function is undefined when the denominator is zero, so $$ x-9=0\quad\Longrightarrow\quad x=9. $$ Thus, the domain is $$(-\infty,9)\cup(9,\infty)$$. 2. **Vertical Asymptote:** At $$ x=9 $$ the function has a vertical asymptote. 3. **Horizontal Asymptote:** Since the degrees of the numerator and denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients: $$ y=\frac{1}{1}=1. $$ 4. **Intercepts:** - **x-intercept:** Set $$ x+9=0 $$ which gives $$ x=-9 $$. - **y-intercept:** Set $$ x=0 $$ which gives $$ f(0)=\frac{0+9}{0-9}=-1. $$ 5. **Behavior at Infinity:** As $$ x\to\pm\infty $$, $$ f(x) \to 1 $$ (consistent with the horizontal asymptote). ### Concavity Analysis 1. **First Derivative:** Using the quotient rule for $$ f(x)=\frac{x+9}{x-9}, $$ let $$ u=x+9 $$ and $$ v=x-9 $$. Then $$ f'(x)=\frac{v\cdot u'-u\cdot v'}{v^2}=\frac{(x-9)(1)-(x+9)(1)}{(x-9)^2}=\frac{x-9-x-9}{(x-9)^2}=\frac{-18}{(x-9)^2}. $$ 2. **Second Derivative:** Write $$ f'(x)=-18(x-9)^{-2}. $$ Then, differentiating using the power rule, we have: $$ f''(x)=-18\cdot (-2)(x-9)^{-3}=\frac{36}{(x-9)^3}. $$ 3. **Intervals of Concavity:** The sign of $$ f''(x)=\frac{36}{(x-9)^3} $$ depends on $$(x-9)^3$$: - For $$ x>9 $$: $$(x-9)^3 > 0$$ so $$ f''(x)>0 $$. The function is **concave upward** on $$ (9,\infty) $$. - For $$ x<9 $$: $$(x-9)^3 < 0$$ so $$ f''(x)<0 $$. The function is **concave downward** on $$ (-\infty,9) $$. ### Inflection Points An inflection point occurs where the concavity changes and the function is continuous. Although the sign of $$ f''(x) $$ changes as $$ x $$ passes through $$ 9 $$, the function has a vertical asymptote at $$ x=9 $$ and is not defined there. There are no other points where $$ f''(x)=0 $$ since the numerator is always $$ 36 $$. Thus, there are **no inflection points**. ### Final Answers - **Concavity:** The function is concave upward on $$ (9, \infty) $$ and concave downward on $$ (-\infty, 9). $$ - **Inflection Points:** There are no inflection points. The function $$ f(x) = \frac{x+9}{x-9} $$ is concave upward on $$ (9, \infty) $$ and concave downward on $$ (-\infty, 9) $$. There are no inflection points.

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