To determine which expression represents $g(x)$ with a vertical asymptote at $x = 3$, we need to focus on the denominator of the rational function. A vertical asymptote occurs when the denominator equals zero but the numerator does not.

Key Analysis:

1. For $x = 3$ to be a vertical asymptote, the factor $(x - 3)$ must appear in the denominator of $g(x)$.
2. If $(x - 3)$ also appears in the numerator, it would cancel out, creating a hole instead of an asymptote.

Step-by-Step Solution:

Let's analyze the options:

1. Option 1:
   $g(x) = \frac{(x - 3)(3x + 12)}{(x - 2)(x + 4)}$

   • The factor $(x - 3)$ is in the numerator but not in the denominator. This would not create a vertical asymptote at $x = 3$.
     Incorrect.

2. Option 2:
   $g(x) = \frac{(x - 2)(3x + 12)}{(x - 3)(x + 4)}$

   • Here, $(x - 3)$ is in the denominator and not in the numerator.
   • This creates a vertical asymptote at $x = 3$.
     Correct.

3. Option 3:
   $g(x) = \frac{(x + 3)(3x - 12)}{(x + 2)(x - 4)}$

   • The denominator does not contain $(x - 3)$.
     Incorrect.

4. Option 4:
   $g(x) = \frac{(x + 2)(3x - 12)}{(x + 3)(x - 4)}$

   • The denominator does not contain $(x - 3)$.
     Incorrect.

Final Answer:

The correct expression for $g(x)$ is: $g(x) = \frac{(x - 2)(3x + 12)}{(x - 3)(x + 4)}$ This is Option 2.

Do you need further explanation on vertical asymptotes or rational functions? Let me know!

Related Questions:

1. What happens when a factor appears in both the numerator and denominator of a rational function?
2. How do you determine a horizontal asymptote for a rational function?
3. What are the steps to identify holes in a rational function graph?
4. How do you factorize polynomials like $3x + 12$?
5. How can vertical asymptotes be found algebraically?

Tip: Vertical asymptotes occur where the denominator is zero and the numerator is non-zero at that point. Always check for cancellation of factors!