1. The problem is to explain how to draw the graph of $f_2(x) = -\sqrt{|x| + 1}$ starting from the graph of $f(x) = \sqrt{x + 1}$.\n\n2. First, understand the graph of $f(x)$. This is the square root function shifted 1 unit left because it is $\sqrt{x + 1}$. It is defined for $x \geq -1$, starting at the point $(-1,0)$ and increasing as $x$ increases.\n\n3. For the function inside $f_2(x)$, note the absolute value $|x|$, so the expression inside the root is symmetric about the y-axis: $|x| + 1$. This means $f_2(x)$ depends on the distance from zero, not direction.\n\n4. The negative sign in front of the square root in $f_2(x) = -\sqrt{|x| + 1}$ flips the graph of the square root function vertically downwards. So, where $f(x)$ is positive and increasing, $f_2(x)$ will be negative and decreasing away from $-\sqrt{1} = -1$.\n\n5. To draw the graph of $f_2(x)$ from $f(x)$, do the following steps:\n a. Reflect the graph of $f(x)$ about the y-axis to incorporate the absolute value (create a symmetric shape around the y-axis).\n b. Apply a vertical reflection by multiplying the $y$-values by $-1$, flipping it below the x-axis.\n\n6. The domain of $f_2$ is all real $x$ because $|x| + 1$ is always $\geq 1$, so $f_2$ is defined everywhere. The graph touches $y = -1$ at $x=0$ and goes downwards as $|x|$ increases.\n\nIn summary: Start with $f(x) = \sqrt{x+1}$ graph, reflect it symmetrically about the y-axis to get $\sqrt{|x|+1}$, then flip it vertically to get $f_2(x) = -\sqrt{|x|+1}$.