The function you provided is:

$f(x) = \frac{x + 2}{x - 2}$

Properties of the Function

1. Domain: The function is defined for all $x$ except where the denominator is zero. The denominator is $x - 2$, so the function is undefined when $x = 2$. Thus, the domain is: $x \in (-\infty, 2) \cup (2, \infty)$

2. Vertical Asymptote: A vertical asymptote occurs when the denominator is zero and the numerator is not zero. Here, the vertical asymptote is at: $x = 2$

3. Horizontal Asymptote: To find the horizontal asymptote, we analyze the behavior of $f(x)$ as $x \to \infty$ or $x \to -\infty$. Since the degrees of the numerator and denominator are the same (both are first-degree polynomials), the horizontal asymptote is the ratio of their leading coefficients. The leading coefficients are both 1, so the horizontal asymptote is: $y = 1$

4. Intercepts:

   • Y-intercept: Set $x = 0$: $f(0) = \frac{0 + 2}{0 - 2} = \frac{2}{-2} = -1$ So, the y-intercept is at $(0, -1)$.

   • X-intercept: Set $f(x) = 0$, which requires the numerator to be zero: $x + 2 = 0 \implies x = -2$ So, the x-intercept is at $(-2, 0)$.

Summary of Key Features

• Domain: $x \in (-\infty, 2) \cup (2, \infty)$
• Vertical Asymptote: $x = 2$
• Horizontal Asymptote: $y = 1$
• Y-intercept: $(0, -1)$
• X-intercept: $(-2, 0)$

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Related Questions:

1. How do you find slant asymptotes for rational functions?
2. What happens to the function $f(x)$ as $x$ approaches the vertical asymptote from both sides?
3. How can we determine the concavity of the function $f(x) = \frac{x+2}{x-2}$?
4. What is the significance of intercepts in analyzing the graph of a function?
5. How would the function $f(x) = \frac{x+2}{x-2}$ change if it were transformed into $f(x) = \frac{x+2}{x-2} + 3$?

Tip: Always analyze both vertical and horizontal asymptotes when dealing with rational functions to understand their end behavior and potential points of discontinuity.