The image shows a problem asking for the range of the function $y = \sqrt{x^2 + 4x + 2} - 1$. Here's how to find the range of this function:

Step-by-Step Solution:

1. Analyze the Function Inside the Square Root: The function inside the square root is $x^2 + 4x + 2$, which is a quadratic expression. The expression inside the square root must be non-negative, so we start by analyzing this quadratic.

2. Find the Vertex of the Quadratic: The expression $x^2 + 4x + 2$ can be rewritten by completing the square: $x^2 + 4x + 2 = (x^2 + 4x + 4) - 4 + 2 = (x + 2)^2 - 2$ The vertex of this parabola occurs at $x = -2$, and the minimum value of the quadratic function is $(x + 2)^2 - 2$. At $x = -2$, the minimum value is: $(x + 2)^2 - 2 = (0)^2 - 2 = -2$

3. Determine the Minimum Value Under the Square Root: Since the square root function requires non-negative inputs, the expression $x^2 + 4x + 2$ must be greater than or equal to zero. But, the lowest value of $x^2 + 4x + 2$ is -2, so: $\sqrt{x^2 + 4x + 2} \text{ is defined for } x^2 + 4x + 2 \geq 0$

   To ensure this, add 2 to both sides: $x^2 + 4x + 2 + 2 \geq 2 \quad \Rightarrow \quad \sqrt{x^2 + 4x + 2} \geq \sqrt{2}$

4. Range Calculation: Since $\sqrt{x^2 + 4x + 2} \geq \sqrt{2}$, then the function $y = \sqrt{x^2 + 4x + 2} - 1$ will always be at least: $y \geq \sqrt{2} - 1$

   As $x^2 + 4x + 2$ grows, so does the square root, meaning there is no upper bound on $y$.

Conclusion:

The range of the function $y = \sqrt{x^2 + 4x + 2} - 1$ is ( [\