Question

Graph the peicewise function f(x) = { x + 5 if x < -2 & 2x + 3 if x > -2. Will the graph have any open endpoints? If yes, where?

Which anwer is correct?:
1. Yes, at (-2, 3) and (-2, -1).
2. No, all endpoints are closed.
3. Yes, at (-2, 3).
4. Yes, at (-2, -1).

Answer 1

To graph the piecewise function \( f(x) \):

\[
f(x) =
\begin{cases}
x + 5 & \text{if } x < -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]

### Step 1: Determine Each Piece

1. **For \( x < -2 \)**:
\[
f(x) = x + 5
\]
This linear function has a slope of 1 and the y-intercept of 5. However, since we're only interested in values less than \(-2\), we will not include \( (-2, -1) \) in our graph (the endpoint at \( x = -2 \) will not be closed).

If we evaluate it when \( x = -2 \):
\[
f(-2) = -2 + 5 = 3
\]

2. **For \( x > -2 \)**:
\[
f(x) = 2x + 3
\]
This linear function has a slope of 2 and a y-intercept of 3. Similar to the previous piece, we are only considering when \( x \) is greater than \(-2\), so the graph will not include the point at \( x = -2 \).

If we again evaluate when \( x = -2 \):
\[
f(-2) = 2(-2) + 3 = -1
\]

### Step 2: Evaluate the Endpoints

At \( x = -2 \):
- The value of \( f(x) \) from the first piece (if it were included) is 3 (from the function \( x + 5 \)).
- The value of \( f(x) \) from the second piece (for \( x > -2 \)) jumps to -1.

### Conclusion About Open Endpoints

- Since the function does not include the endpoint at \( x = -2 \) for both pieces:
- At \( (-2, 3) \) (from the first piece), this point is open on the left since it only exists for \( x < -2 \).
- At \( (-2, -1) \) (from the second piece), this point is open since it exists for \( x > -2 \).

Thus, there are indeed open endpoints at \( (-2, 3) \) and \( (-2, -1) \).

### Final Answer

The correct choice is:
1. **Yes, at (-2, 3) and (-2, -1)**.