The problem asks which graph corresponds to the equation $y = 2\sqrt{x} + 2$, and it provides a reference graph of $y = \sqrt{x}$. To understand how the transformations work, let's break it down:

1. Starting Function: The graph of $y = \sqrt{x}$ is already given. It is a basic square root function with the following properties:

   • It starts at the origin $(0, 0)$.
   • It increases slowly as $x$ increases.

2. Transformations in $y = 2\sqrt{x} + 2$:

   • Vertical Stretch: The term $2\sqrt{x}$ applies a vertical stretch by a factor of 2. This makes the graph grow twice as fast as $y = \sqrt{x}$. In other words, for the same value of $x$, the $y$-coordinate is doubled.
   • Vertical Shift: The $+2$ at the end shifts the entire graph upward by 2 units.

Step-by-step:

• The original graph of $y = \sqrt{x}$ starts at $(0,0)$.
• The transformation $2\sqrt{x} + 2$ will now start at $(0, 2)$, since the vertical shift moves the graph 2 units upward.
• The vertical stretch by 2 means that for any $x$, the graph's height will be twice as high as the original.

Expected Graph:

• It will start at $(0, 2)$.
• It will have a similar shape to the graph of $y = \sqrt{x}$, but it will rise more steeply due to the vertical stretch.

Would you like further clarification or a detailed graphing example?

Here are 5 questions to explore this concept more:

1. How does changing the coefficient in front of $\sqrt{x}$ affect the graph?
2. What is the domain of the function $y = 2\sqrt{x} + 2$?
3. How would the graph change if the transformation included a horizontal shift?
4. What is the range of the new function?
5. How would you describe the behavior of the function as $x$ approaches infinity?

Tip: Always look at how each transformation (stretch, shift) affects the basic shape of the function.