Question

graph the following function on the axes provided.
$f(x) = \

$$\begin{cases} 1 & \\text{for} & -5 < x < -2 \\\\ -x - 1 & \\text{for} & -2 < x \\leq 3 \\end{cases}$$

$

Explanation:

Step1: Analyze the first piece

The first part of the piece - wise function is \(f(x)=1\) for \(- 5\lt x\lt - 2\). This is a horizontal line. We can find two points in this interval. Let's take \(x=-4\) (which is in \(-5\lt x\lt - 2\)), then \(f(-4) = 1\). Also, the endpoints: at \(x=-5\) (not included, so we use an open circle) and \(x = - 2\) (not included, open circle) with \(y = 1\).

Step2: Analyze the second piece

The second part is \(f(x)=-x - 1\) for \(-2\lt x\leq3\). This is a linear function. Let's find some points:

• When \(x=-2\) (not included, open circle), \(f(-2)=-(-2)-1=2 - 1=1\).
• When \(x = 0\), \(f(0)=-0 - 1=-1\).
• When \(x = 3\) (included, closed circle), \(f(3)=-3-1=-4\).

Step3: Plot the points and draw the graph

• For the first part (\(y = 1,-5\lt x\lt - 2\)): Draw a horizontal line segment between \(x=-5\) (open circle) and \(x=-2\) (open circle) at \(y = 1\).
• For the second part (\(y=-x - 1,-2\lt x\leq3\)): Plot the points \((-2,1)\) (open circle), \((0,-1)\), \((3,-4)\) (closed circle) and draw a straight line through them.

(Note: Since the question is about graphing, the final answer is the graph constructed as above. If we were to describe the key features: The graph has a horizontal segment at \(y = 1\) from \(x=-5\) (open) to \(x=-2\) (open), and a line with slope \(- 1\) from \(x=-2\) (open) to \(x = 3\) (closed) with equation \(y=-x - 1\))

Answer:

The graph consists of a horizontal line segment \(y = 1\) for \(-5\lt x\lt - 2\) (open circles at \(x=-5\) and \(x=-2\)) and a line segment of \(y=-x - 1\) for \(-2\lt x\leq3\) (open circle at \(x = - 2\), closed circle at \(x = 3\))