Use \( -\infty \) or \( \infty \) where appropriate to describe the behavior at each zero of the denominator and identify all vertical asymptotes. \[ f(x)=\frac{3 x}{x+6} \] Locate all zeros of the denominator and describe the behavior of the function at those values. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. There are two zeros. The smaller zero is \( x= \) \( \square \) and the larger zero is \( x= \) \( \square \) As \( x \) approaches the smaller zero from the left and right, \( f(x) \) approaches \( \square \) and \( \square \) respectively. As x approaches the larger zero from the left and right, \( f(x) \) approaches \( \square \) and \( \square \) , respectively. B. There is one zero, at \( x= \) \( \square \) As x approaches this zero from the left and right, \( \mathrm{f}(\mathrm{x}) \) approaches \( \square \) and \( \square \) respectively. C. There are no zeros of the denominator. Identify all vertical asymptotes. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. From left to right the asymptotes are \( \square \) , \( \square \). and \( \square \) (Type equations.) B. From left to right the asymptotes are \( \square \) and \( \square \) . (Type equations.) C. The asymptote is \( \square \) (Type an equation.) D. The function has no vertical asymptotes.

Let's analyze the function \( f(x) = \frac{3x}{x + 6} \) to identify the zeros of the denominator, describe the behavior of the function near those points, and determine the vertical asymptotes. ### 1. Zeros of the Denominator and Function Behavior **Denominator Zero:** The denominator \( x + 6 \) is zero when: \[ x + 6 = 0 \implies x = -6 \] So, there is **one zero** of the denominator at \( x = -6 \). **Behavior Near \( x = -6 \):** - **As \( x \) approaches \(-6\) from the **left** (\( x \to -6^- \)): \[ f(x) = \frac{3x}{x + 6} \approx \frac{3(-6)}{\text{small negative}} = \frac{-18}{\text{small negative}} \to +\infty \] - **As \( x \) approaches \(-6\) from the **right** (\( x \to -6^+ \)): \[ f(x) = \frac{3x}{x + 6} \approx \frac{3(-6)}{\text{small positive}} = \frac{-18}{\text{small positive}} \to -\infty \] **Conclusion for Zeros of the Denominator:** - **Option B** is correct. **Filled-in Choice B:** > There is one zero, at \( x = -6 \). As \( x \) approaches this zero from the left and right, \( f(x) \) approaches \( \infty \) and \( -\infty \) respectively. ### 2. Vertical Asymptotes A vertical asymptote occurs at points where the denominator is zero (and the numerator is not zero). Since the only zero of the denominator is at \( x = -6 \), there is a single vertical asymptote at this point. **Conclusion for Vertical Asymptotes:** - **Option C** is correct. **Filled-in Choice C:** > The asymptote is \( x = -6 \). ### Final Answer: **Zeros of the Denominator and Behavior:** B. There is one zero, at \( x = -6 \). As \( x \) approaches this zero from the left and right, \( f(x) \) approaches \( \infty \) and \( -\infty \) respectively. **Vertical Asymptote:** C. The asymptote is \( x = -6 \).