The function given is $f(x) = \sqrt{x + 4} - 2$. To understand how the graph behaves, let's break it down step by step.

Step-by-Step Analysis:

1. Starting with the base function:

   • The basic form is $\sqrt{x}$, which represents a square root function. Its graph starts at $x = 0$ and increases gradually, always non-negative.

2. Horizontal Shift:

   • The function inside the square root is $x + 4$, so the graph of $\sqrt{x}$ shifts to the left by 4 units. This means the graph starts at $x = -4$ because $\sqrt{x+4}$ is defined when $x + 4 \geq 0$, or $x \geq -4$.

3. Vertical Shift:

   • The $-2$ outside the square root represents a vertical shift downward by 2 units. This shifts the entire graph down by 2 units.

Key Points:

• Domain: The function is defined for $x \geq -4$, as $\sqrt{x+4}$ requires $x + 4 \geq 0$.
• Range: Since the smallest value of $\sqrt{x+4}$ is 0 (at $x = -4$), the smallest value of $f(x)$ will be $0 - 2 = -2$. So, the range is $f(x) \geq -2$.
• Graph Shape: The graph will resemble the basic square root curve, starting at $(-4, -2)$ and increasing as $x$ increases.

To summarize, the graph will start at $(-4, -2)$, rise slowly, and never dip below $-2$.

Would you like further details, or do you have any questions about this?

Here are some follow-up questions:

1. What is the behavior of the graph as $x$ approaches infinity?
2. How does the domain of this function compare to that of the basic square root function $\sqrt{x}$?
3. How does a vertical shift affect the range of a function?
4. Can you calculate the point where the function intersects the x-axis (if it exists)?
5. How would the graph change if the function was $\sqrt{x + 4} + 2$ instead?

Tip: Always analyze transformations (shifts and stretches) to understand how a function's graph moves from its basic form.