Question

graph the following function on the axes provided. ( f(x) = \begin{cases} 5x + 1 & \text{for } x leq -1 \\ x - 8 & \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: Graph \( f(x) = 5x + 1 \) for \( x \leq -1 \)

First, find the endpoint at \( x = -1 \). Substitute \( x = -1 \) into \( 5x + 1 \): \( f(-1) = 5(-1) + 1 = -5 + 1 = -4 \). So, there is a closed circle at \( (-1, -4) \) (since \( x \leq -1 \) includes \( x = -1 \)). Now, find another point for this line. Let's take \( x = -2 \): \( f(-2) = 5(-2) + 1 = -10 + 1 = -9 \). So, the line passes through \( (-2, -9) \) and \( (-1, -4) \), and we draw it to the left of \( x = -1 \) (since the domain is \( x \leq -1 \)).

Step2: Graph \( f(x) = x - 8 \) for \( x > 5 \)

Find the endpoint at \( x = 5 \) (but it's not included, so open circle). Substitute \( x = 5 \) into \( x - 8 \): \( f(5) = 5 - 8 = -3 \). So, open circle at \( (5, -3) \). Find another point for this line. Let's take \( x = 6 \): \( f(6) = 6 - 8 = -2 \). So, the line passes through \( (6, -2) \) and has an open circle at \( (5, -3) \), and we draw it to the right of \( x = 5 \) (since the domain is \( x > 5 \)).

Answer:

To graph the piecewise function:

• For \( y = 5x + 1 \) (\( x \leq -1 \)): Draw a line with a closed circle at \( (-1, -4) \) and passing through \( (-2, -9) \) (extending left).
• For \( y = x - 8 \) (\( x > 5 \)): Draw a line with an open circle at \( (5, -3) \) and passing through \( (6, -2) \) (extending right).

(Note: Since this is a graphing task, the final answer is the visual graph constructed as described.)