To create a rational function with the given properties:

1. Vertical asymptote at $x = 2$: The denominator must have a factor of $(x - 2)$.
2. x-intercept at $(-1, 0)$: The numerator must have a factor of $(x + 1)$.
3. Horizontal asymptote at $y = 0$: The degree of the numerator must be less than the degree of the denominator.

A function that satisfies these conditions is:

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

Verifying the conditions:

1. Vertical asymptote at $x = 2$: The denominator becomes $0$ when $x = 2$, so $f(x) \to \infty$ or $f(x) \to -\infty$, creating a vertical asymptote.
2. x-intercept at $(-1, 0)$: The numerator becomes $0$ when $x = -1$, so $f(-1) = 0$, confirming the x-intercept.
3. Horizontal asymptote at $y = 0$: As $x \to \infty$ or $x \to -\infty$, the degree of the numerator (1) is less than the degree of the denominator (2), so $f(x) \to 0$.

Would you like me to explain this in greater detail or graph the function for further insight?

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5 Relative Questions:

1. How does the degree of the numerator and denominator affect horizontal asymptotes?
2. Why does $(x - 2)^2$ result in a vertical asymptote at $x = 2$?
3. Could a function have multiple vertical asymptotes? If so, how would the equation change?
4. What happens if the numerator and denominator have the same degree?
5. Can a rational function have a slant asymptote instead of a horizontal asymptote?

Tip:

Always check the degrees of the numerator and denominator to determine the horizontal asymptote's behavior.