Let's analyze the problem presented in the image:

We are given two functions:

1. $f(x) = \sqrt{x}$
2. $g(x) = \sqrt{x} + 5$

You are asked to:

• Use integer values of $x$ from the given set $\{0, 1, 4, 9\}$ to find ordered pairs for both functions.
• Graph the functions in the same rectangular coordinate system.
• Describe how the graph of $g(x)$ is related to the graph of $f(x)$.

Step 1: Calculate the ordered pairs for $f(x)$ and $g(x)$

For $f(x) = \sqrt{x}$:

• $f(0) = \sqrt{0} = 0$
• $f(1) = \sqrt{1} = 1$
• $f(4) = \sqrt{4} = 2$
• $f(9) = \sqrt{9} = 3$

So, the ordered pairs for $f(x)$ are:

• $(0, 0), (1, 1), (4, 2), (9, 3)$

For $g(x) = \sqrt{x} + 5$:

• $g(0) = \sqrt{0} + 5 = 0 + 5 = 5$
• $g(1) = \sqrt{1} + 5 = 1 + 5 = 6$
• $g(4) = \sqrt{4} + 5 = 2 + 5 = 7$
• $g(9) = \sqrt{9} + 5 = 3 + 5 = 8$

So, the ordered pairs for $g(x)$ are:

• $(0, 5), (1, 6), (4, 7), (9, 8)$

Step 2: Graphing the functions

The graph of $g(x) = \sqrt{x} + 5$ is a vertical translation of the graph of $f(x) = \sqrt{x}$. Specifically, the graph of $g(x)$ is shifted 5 units upward from the graph of $f(x)$. This is because adding a constant (in this case, 5) to the function shifts it vertically.

Conclusion:

The graph of $g(x)$ is the same shape as the graph of $f(x)$, but every point on the graph of $g(x)$ is shifted 5 units higher than the corresponding point on the graph of $f(x)$.

Would you like me to graph these functions or provide further details? Here are some additional related questions:

1. What is the domain of both $f(x)$ and $g(x)$?
2. How would the graph change if $g(x) = \sqrt{x} - 5$?
3. Can the functions $f(x)$ and $g(x)$ ever intersect?
4. How would you modify the equation to shift the graph of $f(x)$ 5 units to the right instead?
5. What are some real-world applications of square root functions?

Tip: When graphing, shifts like $+5$ or $-5$ are vertical translations; similarly, terms added or subtracted inside the square root (e.g., $\sqrt{x - h}$) result in horizontal shifts.