Question

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

$$\begin{cases} x + 4 & \\text{for} & -3 \\leq x \\leq 0 \\\\ -2x + 7 & \\text{for} & 0 < x \\leq 6 \\end{cases}$$

$

Explanation:

Step1: Analyze the first piece \( y = x + 4 \) (for \( -3 \leq x \leq 0 \))

• Find the endpoints:
• When \( x = -3 \), \( y = -3 + 4 = 1 \). So the point is \( (-3, 1) \).
• When \( x = 0 \), \( y = 0 + 4 = 4 \). So the point is \( (0, 4) \).
• This is a linear function with slope \( 1 \), so we can draw a line segment between \( (-3, 1) \) and \( (0, 4) \), including both endpoints.

Step2: Analyze the second piece \( y = -2x + 7 \) (for \( 0 < x \leq 6 \))

• Find the endpoints:
• When \( x = 0 \) (approaching from the right), \( y = -2(0) + 7 = 7 \). But since \( x > 0 \) here, the point \( (0, 7) \) is not included, but we consider the limit as \( x \) approaches \( 0 \) from the right.
• When \( x = 6 \), \( y = -2(6) + 7 = -12 + 7 = -5 \). So the point is \( (6, -5) \), and it is included.
• This is a linear function with slope \( -2 \). We draw a line segment starting just to the right of \( x = 0 \) (from \( (0, 7) \) open circle) to \( (6, -5) \) (closed circle).

Step3: Plot the points and draw the graph

• For the first segment, plot \( (-3, 1) \) and \( (0, 4) \), then draw a line between them.
• For the second segment, plot \( (6, -5) \) and an open circle at \( (0, 7) \), then draw a line from the open circle (approaching \( (0, 7) \) from the right) to \( (6, -5) \).

Answer:

To graph the piece - wise function:

1. For \( y=x + 4\) (\(-3\leq x\leq0\)):

• Plot the points \((-3,1)\) (closed circle) and \((0,4)\) (closed circle) and draw a line segment connecting them.

1. For \(y = - 2x+7\) (\(0\lt x\leq6\)):

• Plot the point \((6, - 5)\) (closed circle) and an open circle at \((0,7)\), then draw a line segment from the open circle (near \((0,7)\) from the right - hand side) to \((6, - 5)\).

(Note: Since the problem asks to graph the function, the above steps describe how to construct the graph. If we were to describe the key points:

• For \(y=x + 4\) (\(-3\leq x\leq0\)): passes through \((-3,1)\) and \((0,4)\).
• For \(y=-2x + 7\) (\(0\lt x\leq6\)): passes through \((6, - 5)\) and has a limit point \((0,7)\) (open circle) as \(x\) approaches \(0\) from the right.)