To address each part of this problem, we can start by examining the function $f(x) = \log_3(x)$ and its transformation to create $g(x) = \log_3(x + 2)$.

Part 1: Graph of $g(x) = \log_3(x + 2)$

1. Basic Graph: The base function $f(x) = \log_3(x)$ is a logarithmic function with a vertical asymptote at $x = 0$ and is defined for $x > 0$.

2. Transformation: The function $g(x) = \log_3(x + 2)$ is a horizontal shift of $f(x)$ by 2 units to the left. This transformation affects the vertical asymptote, moving it to $x = -2$.

3. Graph:

   • The graph of $g(x) = \log_3(x + 2)$ will have a similar shape to $f(x) = \log_3(x)$, but shifted left by 2 units.
   • The vertical asymptote will be located at $x = -2$.

Part 2: Vertical Asymptote of $g(x)$

The vertical asymptote of $g(x) = \log_3(x + 2)$ is at:

$x = -2$

Part 3: Domain of $g(x) = \log_3(x + 2)$

Since $g(x)$ is defined for values where $x + 2 > 0$, we have:

$x > -2$

Thus, the domain of $g(x)$ in interval notation is:

$(-2, \infty)$

Part 4: Range of $g(x) = \log_3(x + 2)$

The range of a logarithmic function like $g(x) = \log_3(x + 2)$ is all real numbers, as it can take any value from $-\infty$ to $\infty$. Therefore, the range is:

$(-\infty, \infty)$

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Would you like further details on each part or visual assistance with the graph? Here are a few related questions for practice:

1. How would the graph change if the function were $h(x) = \log_3(x - 2)$?
2. What would be the effect of a vertical stretch on $g(x)$?
3. What is the horizontal asymptote of an exponential function with base 3?
4. How does the domain of $\log_b(x + k)$ change with different values of $k$?
5. Can the logarithmic function $\log_b(x)$ ever have a finite range?

Tip: For logarithmic functions, remember that horizontal shifts affect the vertical asymptote, but vertical shifts do not.