https://r2.erweima.ai/mathbot/image/1725486251543lmctdari.jpg

Let's analyze and solve the asymptotes for the given functions in the image. The functions provided are:

1. \( f(x) = \frac{x^3 + 1}{x^2} \)
2. \( f(x) = \frac{-3}{x - 3} \)
3. \( f(x) = \frac{x^2 - x - 2}{x^2 - 2x + 1} \)

### 1. \( f(x) = \frac{x^3 + 1}{x^2} \)
**Horizontal Asymptotes:**  
For large values of \( x \), we analyze the degree of the polynomial in the numerator and the denominator. The numerator has degree 3, and the denominator has degree 2.

- When the degree of the numerator is larger than that of the denominator, there is no horizontal asymptote, but there could be an oblique asymptote. To find the oblique asymptote, we divide \( x^3 + 1 \) by \( x^2 \):
  \[
  \frac{x^3 + 1}{x^2} = x + \frac{1}{x^2}
  \]
  So the oblique asymptote is the linear term \( y = x \).

**Vertical Asymptotes:**  
The vertical asymptotes occur where the denominator equals zero, i.e., when \( x^2 = 0 \). Thus, \( x = 0 \) is a vertical asymptote.

### 2. \( f(x) = \frac{-3}{x - 3} \)
**Horizontal Asymptotes:**  
This function is a rational function with a constant numerator and a linear denominator. When \( x \to \infty \), the function tends to 0. So, the horizontal asymptote is \( y = 0 \).

**Vertical Asymptotes:**  
The vertical asymptote occurs where the denominator equals zero, i.e., at \( x = 3 \).

### 3. \( f(x) = \frac{x^2 - x - 2}{x^2 - 2x + 1} \)
**Vertical Asymptotes:**  
First, factor both the numerator and the denominator:
- Numerator: \( x^2 - x - 2 = (x - 2)(x + 1) \)
- Denominator: \( x^2 - 2x + 1 = (x - 1)^2 \)

To determine the vertical asymptote, set the denominator equal to zero:
\[
(x - 1)^2 = 0 \quad \text{so} \quad x = 1
\]
Thus, \( x = 1 \) is a vertical asymptote.

**Horizontal Asymptotes:**  
Compare the degrees of the numerator and the denominator. Both are degree 2, so the horizontal asymptote is the ratio of the leading coefficients. Since both coefficients are 1, the horizontal asymptote is \( y = 1 \).

### Summary of Asymptotes:

1. \( f(x) = \frac{x^3 + 1}{x^2} \):
   - Vertical asymptote: \( x = 0 \)
   - Oblique asymptote: \( y = x \)

2. \( f(x) = \frac{-3}{x - 3} \):
   - Vertical asymptote: \( x = 3 \)
   - Horizontal asymptote: \( y = 0 \)

3. \( f(x) = \frac{x^2 - x - 2}{x^2 - 2x + 1} \):
   - Vertical asymptote: \( x = 1 \)
   - Horizontal asymptote: \( y = 1 \)

If you'd like, I can also sketch the graphs based on these asymptotes. Would you like to see that?

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Here are some related questions:
1. How do you find slant or oblique asymptotes for rational functions?
2. What happens to the behavior of the graph near vertical asymptotes?
3. How can horizontal asymptotes be used to predict the end behavior of a function?
4. What is the difference between horizontal and vertical asymptotes in terms of function limits?
5. How do holes in the graph differ from vertical asymptotes?

**Tip:** Always factor the numerator and denominator when dealing with rational functions to simplify and find potential asymptotes more easily.

Determine la ecuación de las asíntotas de las siguientes funciones, luego realice un esquema de cada una:

1. f(x) = \frac{x^3 + 1}{x^2}
2. f(x) = \frac{-3}{x - 3}
3. f(x) = \frac{x^2 - x - 2}{x^2 - 2x + 1}

Learn how to determine the asymptotes of rational functions with detailed examples. This tutorial covers vertical, horizontal, and oblique asymptotes for different functions, making it ideal for students in Grades 11-12.

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