\( f(x)=\frac{x^{2}-9}{x-3} \) Choose the correct attribute of the rational function. A. The horizontal asymptote is \( y=3 \). The graph has the oblique asymptote \( \mathrm{y}=\mathrm{x}-3 \). C. The vertical asymptote is \( \mathrm{x}=3 \). D. There is a "hole" in the graph at \( \mathrm{x}=3 \).

Let's analyze the rational function \( f(x) = \frac{x^{2} - 9}{x - 3} \): 1. **Simplify the Function:** \[ f(x) = \frac{x^{2} - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} \] For \( x \neq 3 \), this simplifies to: \[ f(x) = x + 3 \] 2. **Identifying Asymptotes and Holes:** - **Oblique Asymptote:** Since the simplified form is \( f(x) = x + 3 \), the graph has an **oblique asymptote** at \( y = x + 3 \). - **Vertical Asymptote:** Normally, a vertical asymptote occurs where the denominator is zero (i.e., \( x = 3 \)). However, because the factor \( (x - 3) \) cancels out, there's no vertical asymptote at \( x = 3 \). Instead, there's a **"hole"** in the graph at \( x = 3 \). - **Horizontal Asymptote:** Since the degree of the numerator and denominator are the same, there is a horizontal asymptote. However, after simplification, the function behaves like a linear function \( y = x + 3 \), so the oblique asymptote takes precedence, and there is **no horizontal asymptote** at \( y = 3 \). 3. **Evaluating the Options:** - **Option A:** Incorrect. - It incorrectly states that the horizontal asymptote is \( y = 3 \) and the oblique asymptote is \( y = x - 3 \). The correct oblique asymptote is \( y = x + 3 \). - **Option C:** Incorrect. - It incorrectly claims a vertical asymptote at \( x = 3 \). Instead, there's a hole at this point. - **Option D:** Correct. - It correctly identifies that there is a hole in the graph at \( x = 3 \). **Answer:** **D. There is a "hole" in the graph at \( \mathrm{x}=3 \).**