Find all horizontal and vertical asyinptoles of $$ f(x) $$. $$ f(x)=\frac{4 x^{2}+11 x-3}{x^{2}+3 x-18} $$ Find the horizontal asymptotes. Select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one horizontal asymptote, $$ y=4 $$, (Type an equation.) B. The function has two horizontal asymptotes. The top asymptote is $$ \square $$ and the bottorn asymplote is (Type equations.) Find the vertical asymptotes. Select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, $$ \square $$. (Type an equation.) B. The function has two vertical asymptotes. The leltriost asymptote is $$ \square $$ and the nightmost asymptole is $$ \square $$ (Type equations) C. The function has no vertical asymptotes.

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To analyze the function $$ f(x) = \frac{4x^{2} + 11x - 3}{x^{2} + 3x - 18} $$ for its horizontal and vertical asymptotes, follow these steps:

### **Horizontal Asymptotes**

For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:

1. **Degrees of the Polynomials:**
   - **Numerator:** $$4x^{2} + 11x - 3$$ has a degree of 2.
   - **Denominator:** $$x^{2} + 3x - 18$$ has a degree of 2.

2. **Determining the Horizontal Asymptote:**
   - Since the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients.
   - Leading coefficient of the numerator: 4
   - Leading coefficient of the denominator: 1
   - **Horizontal Asymptote:** $$ y = \frac{4}{1} = 4 $$

**Answer for Horizontal Asymptotes:**
- **Choice A:** The function has one horizontal asymptote, $$ y = 4 $$.

### **Vertical Asymptotes**

Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points.

1. **Finding the Zeros of the Denominator:**
   $$
   x^{2} + 3x - 18 = 0
   $$
   - Using the quadratic formula:
     $$
     x = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm \sqrt{81}}{2} = \frac{-3 \pm 9}{2}
     $$
   - Solutions:
     $$
     x = \frac{6}{2} = 3 \quad \text{and} \quad x = \frac{-12}{2} = -6
     $$

2. **Checking the Numerator at These Points:**
   - **At $$ x = 3 $$:**
     $$
     4(3)^2 + 11(3) - 3 = 36 + 33 - 3 = 66 \neq 0
     $$
   - **At $$ x = -6 $$:**
     $$
     4(-6)^2 + 11(-6) - 3 = 144 - 66 - 3 = 75 \neq 0
     $$
   - Since the numerator is not zero at $$ x = 3 $$ and $$ x = -6 $$, both are vertical asymptotes.

**Answer for Vertical Asymptotes:**
- **Choice B:** The function has two vertical asymptotes.
  - The **leftmost asymptote** is $$ x = -6 $$.
  - The **rightmost asymptote** is $$ x = 3 $$.

### **Summary**

- **Horizontal Asymptote:** $$ y = 4 $$
- **Vertical Asymptotes:** $$ x = -6 $$ and $$ x = 3 $$

**Final Selections:**
- **Horizontal Asymptotes:** **A.** The function has one horizontal asymptote, $$ y = 4 $$.
- **Vertical Asymptotes:** **B.** The function has two vertical asymptotes. The leftmost asymptote is $$ x = -6 $$ and the rightmost asymptote is $$ x = 3 $$.
**Horizontal Asymptote:** $$ y = 4 $$ **Vertical Asymptotes:** $$ x = -6 $$ and $$ x = 3 $$

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