Let's break down the problem to find the appropriate form of the function $f(x)$.

Given information:

1. Vertical asymptotes at $x = 2$ and $x = -3$.
2. Horizontal asymptote at $y = 2$.
3. x-intercepts at $x = -4$ and $x = 1$.
4. The graph passes through the point $(-1, 2)$.

Step-by-step solution:

1. Vertical asymptotes:

   • These occur where the denominator of a rational function equals zero.
   • Thus, the factors $(x - 2)$ and $(x + 3)$ should be in the denominator.

2. Horizontal asymptote:

   • The horizontal asymptote is $y = 2$, which suggests the degree of the numerator and denominator must be the same.
   • To have a horizontal asymptote of $y = 2$, the leading coefficient of the numerator must be 2 times that of the denominator.

3. x-intercepts:

   • These occur when the numerator equals zero.
   • The x-intercepts are at $x = -4$ and $x = 1$, so the numerator must include the factors $(x + 4)$ and $(x - 1)$.

4. Equation form:

   • The equation should resemble the third option: $f(x) = \frac{a (x - b)(x - c)}{(x - d)(x - e)}$
   • Substituting the known intercepts and asymptotes: $f(x) = \frac{a (x + 4)(x - 1)}{(x - 2)(x + 3)}$

5. Determine $a$:

   • Since the function passes through the point $(-1, 2)$, substitute $x = -1$ and $f(x) = 2$ to find $a$: $2 = \frac{a (-1 + 4)(-1 - 1)}{(-1 - 2)(-1 + 3)}$ Simplifying: $2 = \frac{a(3)(-2)}{(-3)(2)}$ $2 = \frac{-6a}{-6}$ $2 = a$

Final equation:

$f(x) = \frac{2(x + 4)(x - 1)}{(x - 2)(x + 3)}$

Let me know if you would like more details, and here are 5 related questions:

1. How do you determine horizontal asymptotes in rational functions?
2. What conditions lead to a vertical asymptote in a rational function?
3. How can you find the x-intercepts of a rational function?
4. What happens to the graph of a rational function at points where the numerator equals zero?
5. Why is the horizontal asymptote affected by the degrees of the numerator and denominator?

Tip: Always check for points where the function is undefined to determine vertical asymptotes!