To determine the asymptotes of the function $f(x) = \frac{x^2 + 2x - 1}{x^2 + x - 2}$, follow these steps:

Step 1: Identify the domain of $f(x)$

The denominator must not equal zero: $x^2 + x - 2 = 0$ Factoring: $(x + 2)(x - 1) = 0$ Thus, $x \neq -2$ and $x \neq 1$. The domain is $\mathbb{R} \setminus \{-2, 1\}$.

Step 2: Vertical asymptotes

Vertical asymptotes occur when the denominator is zero but the numerator is non-zero at those points.

• At $x = -2$, the numerator $x^2 + 2x - 1 = (-2)^2 + 2(-2) - 1 = 4 - 4 - 1 = -1 \neq 0$. Thus, $x = -2$ is a vertical asymptote.
• At $x = 1$, the numerator $x^2 + 2x - 1 = 1^2 + 2(1) - 1 = 1 + 2 - 1 = 2 \neq 0$. Thus, $x = 1$ is also a vertical asymptote.

Step 3: Horizontal asymptotes

The degree of the numerator and denominator are the same ($\deg = 2$). To find the horizontal asymptote, divide the leading coefficients: $\lim_{x \to \pm\infty} f(x) = \frac{1}{1} = 1$ Thus, the horizontal asymptote is $y = 1$.

Step 4: Summary of asymptotes

The curve has:

1. Vertical asymptotes at $x = -2$ and $x = 1$,
2. A horizontal asymptote at $y = 1$.

Let me know if you'd like a detailed graphing explanation or further clarifications.

Related Questions

1. What happens to the behavior of $f(x)$ near $x = -2$ and $x = 1$?
2. How would you check for oblique asymptotes for higher-degree functions?
3. Why does the horizontal asymptote depend only on the leading terms?
4. How can we graph $f(x)$ using these asymptotes?
5. Are there points where $f(x)$ crosses the horizontal asymptote?

Tip

To confirm your work, always test points near and between the asymptotes to observe the curve's behavior.