Question

submission
$f(x)=\

$$\begin{cases}(x + 2)^2 \\text{ if } x < -1\\\\|x| + 1 \\text{ if } -1 \\leq x \\leq 1\\\\-\\sqrt{x} \\text{ if } x > 1\\end{cases}$$

$
choose the correct graph of the piecewise function above.

Explanation:

Step1: Analyze $x < -1$ piece

The function is $f(x)=(x+2)^2$, a parabola opening up with vertex at $(-2,0)$. For $x=-1$, $f(-1)=(-1+2)^2=1$, which is an open point (since $x<-1$ excludes $x=-1$).

Step2: Analyze $-1 \leq x \leq 1$ piece

The function is $f(x)=|x|+1$. At $x=-1$, $f(-1)=|-1|+1=2$ (closed point, included). At $x=0$, $f(0)=|0|+1=1$. At $x=1$, $f(1)=|1|+1=2$ (closed point, included). This forms a V-shape between these points.

Step3: Analyze $x > 1$ piece

The function is $f(x)=-\sqrt{x}$, a downward-opening square root curve. At $x=1$, $f(1)=-\sqrt{1}=-1$, which is an open point (since $x>1$ excludes $x=1$).

Step4: Match to correct graph

Compare the key points: closed points at $(-1,2)$, $(0,1)$, $(1,2)$; open points at $(-1,1)$ (from first piece) and $(1,-1)$ (from third piece). The right graph matches these features.

Answer:

The correct graph is the second (right-hand) graph with closed points at $(-1, 2)$, $(0, 1)$, $(1, 2)$, an open point at $(-1, 1)$ and $(1, -1)$, the parabola for $x<-1$, V-shape for $-1\leq x\leq1$, and downward square root for $x>1$.