The function $h(x) = \sqrt{x} + 4$ is a transformation of the parent square root function $f(x) = \sqrt{x}$.

Here’s a breakdown of how it is derived from the parent function $f(x) = \sqrt{x}$:

1. Parent function: $f(x) = \sqrt{x}$ is the basic square root function, which starts at $(0, 0)$ and only takes non-negative values of $x$ (i.e., its domain is $[0, \infty)$).

2. Vertical shift: Adding 4 to the square root function, $\sqrt{x} + 4$, shifts the graph vertically upward by 4 units. This means that instead of starting at $(0, 0)$, the graph now starts at $(0, 4)$.

Key Characteristics of $h(x) = \sqrt{x} + 4$:

• Domain: Since the square root function is defined only for non-negative $x$, the domain of $h(x)$ is $[0, \infty)$.
• Range: The minimum value of $\sqrt{x}$ is 0 (when $x = 0$), and since the graph is shifted up by 4 units, the range is $[4, \infty)$.
• Graph behavior: The graph starts at $(0, 4)$ and increases slowly as $x$ increases.

Summary of transformations:

• Vertical shift: The graph of $\sqrt{x}$ is shifted up by 4 units.

Let me know if you'd like a visual representation or more details!

Expansion Questions:

1. What is the domain and range of the parent function $f(x) = \sqrt{x}$?
2. How would the graph change if the function were $h(x) = \sqrt{x} - 4$?
3. What would happen to the graph if there were a horizontal shift involved, e.g., $h(x) = \sqrt{x-2} + 4$?
4. How does the graph of $h(x) = \sqrt{x} + 4$ differ from $h(x) = 2\sqrt{x} + 4$?
5. Can you find the inverse of the function $h(x) = \sqrt{x} + 4$?

Tip:

Always check the domain and range of transformed functions by considering how the parent function behaves under transformations!