QUESTION IMAGE
Question
graph the following function on the axes provided.
$f(x) = \
$
Step1: Analyze the first piece
The first piece of the function is \( f(x) = 3 \) for \( -4 < x \leq -1 \). This is a horizontal line segment. We need to plot the points:
- At \( x = -1 \), \( f(-1) = 3 \) (closed circle since \( x = -1 \) is included).
- For \( x \) values between \( -4 \) (open circle, since \( x = -4 \) is not included) and \( -1 \), the \( y \)-value is 3. So we draw a horizontal line from \( (-4, 3) \) (open circle) to \( (-1, 3) \) (closed circle).
Step2: Analyze the second piece
The second piece is \( f(x) = x + 3 \) for \( -1 < x < 5 \). This is a linear function. Let's find some key points:
- At \( x = -1 \), \( f(-1) = -1 + 3 = 2 \) (open circle since \( x = -1 \) is not included here).
- At \( x = 5 \), \( f(5) = 5 + 3 = 8 \) (open circle since \( x = 5 \) is not included).
- Let's also find the value at \( x = 0 \): \( f(0) = 0 + 3 = 3 \).
- At \( x = 2 \): \( f(2) = 2 + 3 = 5 \).
- We draw a line segment starting from \( (-1, 2) \) (open circle) with a slope of 1 (since the coefficient of \( x \) is 1) and ending at \( (5, 8) \) (open circle), passing through the points we calculated.
Graph Description:
- For \( -4 < x \leq -1 \): Horizontal line from \( (-4, 3) \) (open) to \( (-1, 3) \) (closed).
- For \( -1 < x < 5 \): Line with slope 1, starting at \( (-1, 2) \) (open) and ending at \( (5, 8) \) (open), passing through points like \( (0, 3) \), \( (2, 5) \) etc.
(Note: Since the problem asks to graph, the above steps describe how to construct the graph. If we were to represent the graph, we would have the two segments as described.)
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The graph consists of a horizontal segment from \( (-4, 3) \) (open circle) to \( (-1, 3) \) (closed circle) and a linear segment from \( (-1, 2) \) (open circle) to \( (5, 8) \) (open circle) with a slope of 1, passing through points like \( (0, 3) \), \( (2, 5) \) etc.