Graph the function, not by plotting points, but by starting from the graphs in the figures below. $$ f(x)=\log _{3}(x-3) $$

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We begin with the basic logarithmic function  
$$
g(x)=\log_{3}(x)
$$
which has the following characteristics:
- **Domain:** $$ x>0 $$
- **Vertical asymptote:** $$ x=0 $$
- **x-intercept:** $$ (1,0) $$

The given function is  
$$
f(x)=\log_{3}(x-3).
$$
Notice that the argument of the logarithm, $$ x-3 $$, means that we have a horizontal translation of the graph of $$ g(x) $$.

**Step 1. Horizontal Shift**

The transformation $$ x \to x - 3 $$ translates the basic graph to the right by 3 units. This means:
- The vertical asymptote shifts from $$ x=0 $$ to $$ x=3 $$.
- The x-intercept shifts from $$ (1, 0) $$ to $$ (1+3, 0)=(4, 0) $$.

**Step 2. Domain and Asymptote**

After the shift, the domain of $$ f(x) $$ becomes:
$$
x-3>0 \quad \Longrightarrow \quad x>3.
$$
Thus, the vertical asymptote is now at $$ x=3 $$.

**Step 3. Sketching the Graph Using the Transformed Features**

- Draw the vertical line $$ x=3 $$ as the asymptote.
- Mark the x-intercept at $$ (4,0) $$.
- The graph will pass through the point obtained by shifting another key point. For example, in the graph of $$ g(x) $$, when $$ x = 3 $$ we have $$ g(3)=\log_3 3=1 $$. Shifting right by 3 gives the point $$ (3+3, 1)=(6,1) $$.

Thus, the important characteristics of $$ f(x)=\log_3 (x-3) $$ are:
- **Vertical asymptote:** $$ x=3 $$
- **x-intercept:** $$ (4, 0) $$
- **A known point:** $$ (6, 1) $$
- **Domain:** $$ x>3 $$

Using these transformed features from the original graph of $$ \log_3(x) $$, you can sketch the graph of $$ f(x)=\log_3(x-3) $$ without plotting individual points.
To graph $$ f(x) = \log_{3}(x-3) $$, start with the basic logarithm $$ g(x) = \log_{3}(x) $$. Shift the graph of $$ g(x) $$ to the right by 3 units. This moves the vertical asymptote from $$ x=0 $$ to $$ x=3 $$ and the x-intercept from $$ (1,0) $$ to $$ (4,0) $$. The domain becomes $$ x>3 $$. Key points to plot include the x-intercept at $$ (4,0) $$ and another point like $$ (6,1) $$. Draw the vertical asymptote at $$ x=3 $$ and sketch the curve accordingly.

Graph the function, not by plotting points, but by starting from the graphs in the figures below.

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