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 intervals where \( f(x) \) is concave upward or downward. Select the correct choice below and fill in the answer box(es) to complete your choice. A. The function is concave upward on \( \square \). It is never concave downward. (Type an exact answer. Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function is concave upward on \( \square \). It is concave downward on \( \square \). (Type exact answers. Type your answers in interval notation. Use commas to separate answers as needed.) C. The function is concave downward on \( \square \). It is never concave upward. (Type an exact answer. Type your answer in interval notation. Use a comma to separate answers as needed.) Find the location of any inflection points of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is an inflection point at \( x=\square \). (Type an integer or a decimal. Use a comma to separate answers as needed.)

To analyze the function \( f(x) = \frac{x^{2}+9x+18}{x^{2}+6x+9} \), we first simplify it. Factoring both the numerator and denominator gives us \( f(x) = \frac{(x+3)(x+6)}{(x+3)^{2}} \). This leads to a removable discontinuity at \( x = -3 \), which we will consider when sketching. The function can be rewritten as \( f(x) = \frac{x+6}{x+3} \) for \( x \neq -3 \). Next, we find the intervals of concavity by calculating the second derivative. 1. The first derivative \( f'(x) \) is found using the quotient rule: \[ f'(x) = \frac{(x+3)(1) - (x+6)(1)}{(x+3)^{2}} = \frac{-3}{(x+3)^{2}}. \] 2. Now, we differentiate \( f'(x) \): \[ f''(x) = \frac{d}{dx}\left(\frac{-3}{(x+3)^2}\right) = \frac{6}{(x+3)^{3}}. \] 3. The sign of \( f''(x) \) determines concavity: - \( f''(x) > 0 \) (concave upward) when \( x > -3 \). - \( f''(x) < 0 \) (concave downward) when \( x < -3 \). Thus, the function is concave upward on \( (-3, \infty) \) and concave downward on \( (-\infty, -3) \). So, the correct choice is: B. The function is concave upward on \( (-3, \infty) \). It is concave downward on \( (-\infty, -3) \). For the inflection point, we note that there is a change in concavity at \( x = -3 \), which is also the removable discontinuity. Therefore, the answer to the inflection point is: A. There is an inflection point at \( x=-3 \).