Question

graph the following piece - wise function:
f(x)=\begin{cases}x + 3&xleq0\\-x - 2&0lt xleq2\\x - 1&2lt xend{cases}

Explanation:

Step1: Analyze \(x\leq0\) part

For \(y = x + 3\) when \(x\leq0\), if \(x = 0\), \(y=0 + 3=3\); if \(x=-3\), \(y=-3 + 3 = 0\). Plot points and draw a line with a closed - circle at \(x = 0\) (since \(x\) can be \(0\)).

Step2: Analyze \(0\lt x\leq2\) part

For \(y=-x - 2\) when \(0\lt x\leq2\), if \(x = 0\), \(y=-2\) (open - circle as \(x
eq0\) in this sub - domain), if \(x = 2\), \(y=-2-2=-4\). Plot points and draw a line with a closed - circle at \(x = 2\).

Step3: Analyze \(2\lt x\) part

For \(y=x - 1\) when \(2\lt x\), if \(x = 2\), \(y=2-1 = 1\) (open - circle as \(x
eq2\) in this sub - domain). Then draw a line extending to the right for \(x>2\).

The graph should have a line \(y = x + 3\) for \(x\leq0\), a line \(y=-x - 2\) for \(0\lt x\leq2\) and a line \(y=x - 1\) for \(x>2\) with appropriate open and closed circles at the transition points.

Answer:

The graph of the piece - wise function \(f(x)\) is as described above with correct lines and circle types at \(x = 0\) and \(x = 2\).