Question

graph the following piecewise function: $f(x)=\begin{cases}-x - 2&xleq - 2\\2x - 2&-2 < x < 1\\-2x + 3&1leq xend{cases}$ show your work here draw

Explanation:

Step1: Analyze first - part of function

For \(y=-x - 2\) when \(x\leq - 2\). Find two points. When \(x=-2\), \(y=-(-2)-2 = 0\). When \(x=-3\), \(y=-(-3)-2=1\). Plot the line segment with a closed - circle at \(x = - 2\) (because \(x=-2\) is included in the domain \(x\leq - 2\)).

Step2: Analyze second - part of function

For \(y = 2x-2\) when \(-2\lt x\lt1\). When \(x=-2\), \(y=2\times(-2)-2=-6\) (but \(x=-2\) is not included, so open - circle). When \(x = 1\), \(y=2\times1-2 = 0\) (open - circle). When \(x = 0\), \(y=-2\). Plot the line segment between \(x=-2\) and \(x = 1\) with open - circles at the endpoints.

Step3: Analyze third - part of function

For \(y=-2x + 3\) when \(x\geq1\). When \(x = 1\), \(y=-2\times1+3 = 1\) (closed - circle). When \(x = 2\), \(y=-2\times2+3=-1\). Plot the line segment starting at \(x = 1\) with a closed - circle (since \(x = 1\) is included in the domain \(x\geq1\)).

Answer:

The graph consists of three line - segments: a line \(y=-x - 2\) for \(x\leq - 2\) with a closed - circle at \(x=-2\), a line \(y = 2x-2\) for \(-2\lt x\lt1\) with open - circles at \(x=-2\) and \(x = 1\), and a line \(y=-2x + 3\) for \(x\geq1\) with a closed - circle at \(x = 1\).