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Determine the domain and asymptotes of each rational function.

To analyze the given rational function \( y = \frac{4x^2 - 16}{2x + 9} \), we need to determine its domain and asymptotes.

### 1. **Domain**:
The domain of a rational function is all real numbers except where the denominator equals zero because division by zero is undefined.

The denominator of the function is \( 2x + 9 \).

Set the denominator equal to zero and solve for \( x \):

\[
2x + 9 = 0
\]

\[
2x = -9
\]

\[
x = -\frac{9}{2}
\]

So, the domain is all real numbers except \( x = -\frac{9}{2} \).

**Domain**: \( x \in \mathbb{R}, x \neq -\frac{9}{2} \)

### 2. **Vertical Asymptote**:
Vertical asymptotes occur where the denominator equals zero and the numerator does not simultaneously equal zero.

From the domain calculation, the vertical asymptote is at \( x = -\frac{9}{2} \).

**Vertical Asymptote**: \( x = -\frac{9}{2} \)

### 3. **Horizontal Asymptote**:
To find the horizontal asymptote, we compare the degrees of the numerator and the denominator:

- The degree of the numerator \( 4x^2 - 16 \) is 2.
- The degree of the denominator \( 2x + 9 \) is 1.

Since the degree of the numerator is greater than the degree of the denominator (2 > 1), there is no horizontal asymptote.

**Horizontal Asymptote**: None

### 4. **Oblique Asymptote**:
When the degree of the numerator is one more than the degree of the denominator, there is an oblique (or slant) asymptote. Here, the numerator's degree is 2 and the denominator's degree is 1, so an oblique asymptote exists.

To find the oblique asymptote, perform polynomial long division of the numerator by the denominator:

\[
\frac{4x^2 - 16}{2x + 9}
\]

Performing the division:

- Divide \( 4x^2 \) by \( 2x \) to get \( 2x \).
- Multiply \( 2x \) by \( 2x + 9 \), resulting in \( 4x^2 + 18x \).
- Subtract \( 4x^2 + 18x \) from \( 4x^2 - 16 \), yielding \( -18x - 16 \).
- Divide \( -18x \) by \( 2x \) to get \( -9 \).
- Multiply \( -9 \) by \( 2x + 9 \), resulting in \( -18x - 81 \).
- Subtract \( -18x - 81 \) from \( -18x - 16 \), giving a remainder of \( 65 \).

So, the oblique asymptote is:

\[
y = 2x - 9
\]

**Oblique Asymptote**: \( y = 2x - 9 \)

### Summary:
- **Domain**: \( x \in \mathbb{R}, x \neq -\frac{9}{2} \)
- **Vertical Asymptote**: \( x = -\frac{9}{2} \)
- **Horizontal Asymptote**: None
- **Oblique Asymptote**: \( y = 2x - 9 \)

Would you like a detailed explanation of any part of the solution? Here are some related questions you might consider:

1. What is the significance of the vertical asymptote in graphing the function?
2. How does the oblique asymptote affect the shape of the graph as \( x \) approaches infinity?
3. Can a function have both a horizontal and an oblique asymptote?
4. How would you graph this function, given the asymptotes and the domain?
5. What happens to the function as \( x \) approaches the vertical asymptote?

**Tip**: When dealing with rational functions, always check for common factors between the numerator and denominator to simplify the function if possible, which can reveal removable discontinuities.

Learn how to determine the domain and asymptotes of the rational function y = (4x^2 - 16) / (2x + 9) with detailed step-by-step solutions. This problem covers key concepts in rational functions, including domain restrictions, vertical asymptotes, and the conditions for horizontal and oblique asymptotes. Suitable for high school students.

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