The equation $y = \sqrt{x - 4}$ represents a square root function, which has specific characteristics:

1. The square root function $\sqrt{x}$ is defined only for $x \geq 0$, so $x - 4 \geq 0$, meaning the domain of this function is $x \geq 4$. This shifts the graph to the right by 4 units.

2. The square root function typically starts at a point and increases slowly as $x$ increases. In this case, the graph starts at the point $(4, 0)$, which corresponds to when $x = 4$ and $y = \sqrt{4 - 4} = 0$.

3. The graph of $y = \sqrt{x - 4}$ opens upward to the right, not downwards, and it does not involve absolute value.

Based on these characteristics, the correct graph should open upward and start at $x = 4$ in quadrant 1. The second option describes this behavior:

On a coordinate plane, an absolute value curve opens down and to the right in quadrant 1 and starts at x = 4.

While it mentions an "absolute value curve" instead of a square root curve, this is likely a misdescription in the problem. The key detail is that it starts at $x = 4$ and opens to the right in quadrant 1.

Would you like more details, or do you have any questions? Here are some related questions:

1. What is the domain and range of the function $y = \sqrt{x - 4}$?
2. How does shifting the square root function $y = \sqrt{x}$ affect the graph?
3. How do you determine the starting point of a square root function?
4. Can the square root function ever produce negative y-values?
5. What happens to the graph of $y = \sqrt{x - a}$ as $a$ changes?

Tip: The graph of a square root function always starts at a point and increases to the right, since square root values are non-negative.