Question

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

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

$
click and drag to make a line. click the line to delete it.
click on an endpoint of a line to change it.

Explanation:

Step1: Analyze \( f(x) = 2x + 5 \) for \( x \leq -1 \)

This is a linear function. First, find the endpoint at \( x = -1 \): \( f(-1) = 2(-1) + 5 = 3 \). So the point is \( (-1, 3) \), and since \( x \leq -1 \), this point is included (closed dot). Then, pick another \( x \leq -1 \), e.g., \( x = -2 \): \( f(-2) = 2(-2) + 5 = 1 \), so point \( (-2, 1) \). Draw a line through these points (and extend left) with a closed dot at \( (-1, 3) \).

Step2: Analyze \( f(x) = -2 \) for \( x > 5 \)

This is a horizontal line. For \( x > 5 \), \( y = -2 \). So we start at \( x = 5 \) (but \( x > 5 \), so open dot at \( (5, -2) \)) and draw a horizontal line to the right (e.g., through \( (6, -2) \), \( (7, -2) \), etc.) with an open dot at \( (5, -2) \).

Answer:

• For \( f(x) = 2x + 5 \) (\( x \leq -1 \)): Plot points like \( (-1, 3) \) (closed) and \( (-2, 1) \), draw a line through them (leftward).
• For \( f(x) = -2 \) (\( x > 5 \)): Plot an open dot at \( (5, -2) \) and draw a horizontal line right through \( (6, -2) \), etc.