Let's solve this step by step.

Step 1: Complete the Table for $f(x) = \sqrt{x - 3}$

The function $f(x) = \sqrt{x - 3}$ means that we need to find values of $f(x)$ for various $x$ values by substituting them into the equation. We can only choose values of $x \geq 3$ because the square root requires non-negative values inside.

Let's complete the table for some specific values of $x$:

1. $x = 3$:
   $f(3) = \sqrt{3 - 3} = \sqrt{0} = 0$
2. $x = 4$:
   $f(4) = \sqrt{4 - 3} = \sqrt{1} = 1$
3. $x = 5$:
   $f(5) = \sqrt{5 - 3} = \sqrt{2} \approx 1.41$
4. $x = 6$:
   $f(6) = \sqrt{6 - 3} = \sqrt{3} \approx 1.73$
5. $x = 7$:
   $f(7) = \sqrt{7 - 3} = \sqrt{4} = 2$

So, the table should look like this:

Step 2: Plot the Points on the Graph

Now, let's plot these points on the graph:

• (3, 0)
• (4, 1)
• (5, 1.41)
• (6, 1.73)
• (7, 2)

After plotting, draw a smooth curve that starts from $x = 3$ and follows the square root shape, gradually increasing but at a slower rate as $x$ increases.

Would you like me to assist further with a specific part of this solution?

Follow-up Questions:

1. What are the properties of the function $f(x) = \sqrt{x - 3}$ in terms of domain and range?
2. How does changing the expression inside the square root affect the graph?
3. What would happen to the graph if the function were $f(x) = -\sqrt{x - 3}$?
4. Can you predict the behavior of $f(x) = \sqrt{x - a}$ for any constant $a$?
5. How would we handle graphing if we changed this function to $f(x) = \sqrt{x - 3} + 2$?

Tip:

For square root functions like $f(x) = \sqrt{x - a}$, the domain begins at $x = a$, and the range starts from zero and increases with $x$.