Respuesta :
Answer:
Step-by-step explanation:
Certainly! Let's analyze the given rational functions to find the one that meets the specified criteria:
1. **Horizontal Asymptote at y = 3**:
- The horizontal asymptote represents the behavior of the function as \(x\) approaches infinity or negative infinity.
- We want the function to approach \(y = 3\) as \(x\) becomes very large or very small.
2. **Vertical Asymptotes at x = 5 and x = –2**:
- Vertical asymptotes occur where the denominator of the rational function becomes zero.
- We need the function to have vertical asymptotes at \(x = 5\) and \(x = -2\).
Let's examine the given options:
1. **\(f(x) = \frac{x^2 - 4}{x - 5}\)**:
- This function has a vertical asymptote at \(x = 5\), but we need another one at \(x = -2\).
- It does not meet the criteria for a horizontal asymptote at \(y = 3\).
2. **\(f(x) = \frac{x^2 - 4}{x + 2}\)**:
- This function has vertical asymptotes at both \(x = 5\) and \(x = -2\).
- However, it does not have a horizontal asymptote at \(y = 3\).
3. **\(f(x) = \frac{3x^2 - 12}{x^2 - 4}\)**:
- This function has vertical asymptotes at both \(x = 5\) and \(x = -2\).
- It also has a horizontal asymptote at \(y = 3\).
Therefore, the correct rational function is \(f(x) = \frac{3x^2 - 12}{x^2 - 4}\). It satisfies all the specified conditions.
