Question

practice & problem solving
graph each function.

1. ( y=left{\begin{array}{l}-3, \text { if }-4 leq x<-2 \\ -1, \text { if }-2 leq x<0 \\ 1, \text { if } 0 leq x<2 \\ 3, \text { if } 2 leq x<4end{array}

ight. )

Explanation:

Step1: Analyze each piece of the function

For \(y = - 3\) when \(-4\leqslant x\lt - 2\), it is a horizontal line segment at \(y=-3\) from \(x = - 4\) (closed - end) to \(x=-2\) (open - end).
For \(y=-1\) when \(-2\leqslant x\lt0\), it is a horizontal line segment at \(y = - 1\) from \(x=-2\) (closed - end) to \(x = 0\) (open - end).
For \(y = 1\) when \(0\leqslant x\lt2\), it is a horizontal line segment at \(y = 1\) from \(x = 0\) (closed - end) to \(x=2\) (open - end).
For \(y = 3\) when \(2\leqslant x\lt4\), it is a horizontal line segment at \(y = 3\) from \(x = 2\) (closed - end) to \(x=4\) (open - end).

Step2: Plot the points and draw the line segments

• For \(y=-3\), plot the point \((-4,-3)\) (closed circle) and \((-2,-3)\) (open circle) and connect them with a horizontal line.
• For \(y=-1\), plot the point \((-2,-1)\) (closed circle) and \((0,-1)\) (open circle) and connect them with a horizontal line.
• For \(y = 1\), plot the point \((0,1)\) (closed circle) and \((2,1)\) (open circle) and connect them with a horizontal line.
• For \(y = 3\), plot the point \((2,3)\) (closed circle) and \((4,3)\) (open circle) and connect them with a horizontal line.

Answer:

The graph consists of four horizontal line segments:

• A line segment \(y=-3\) from \(x=-4\) (closed - circle) to \(x = - 2\) (open - circle).
• A line segment \(y=-1\) from \(x=-2\) (closed - circle) to \(x = 0\) (open - circle).
• A line segment \(y = 1\) from \(x = 0\) (closed - circle) to \(x=2\) (open - circle).
• A line segment \(y = 3\) from \(x = 2\) (closed - circle) to \(x=4\) (open - circle).