Graph the piecewise function f(x)={x+5 if x≤−22x+3 if x>−2

Question

𝑓
(
𝑥
)
=
{
𝑥
+
5

if

𝑥
≤
−
2
2
𝑥
+
3

if

𝑥
>
−
2
. Will the graph have any open endpoints? If yes, where?(1 point)
Responses

Yes, at (−2,3)
(
−
2
,
3
)
and (−2,−1)
(
−
2
,
−
1
)
.
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .

Yes, at (−2,−1)
(
−
2
,
−
1
)
.
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .

No, all endpoints are closed.
No, all endpoints are closed.

Yes, at (−2,3)
(
−
2
,
3
)
.

Answer 1

To analyze the piecewise function given by:

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

let's evaluate the function at \( x = -2 \):

1. For \( x \leq -2 \): \[ f(-2) = -2 + 5 = 3 \]

2. For \( x > -2 \), we cannot directly plug in \( x = -2 \) because this part of the piecewise function does not include \( -2 \).

Since the function is defined at \( x = -2 \) as \( 3 \) and the second piece is not defined at \( x = -2 \) (it only applies to \( x > -2 \)), the point \( (-2, 3) \) is included in the graph. However, the value \( f(-2) \) from the second part of the function is not included. Therefore, we can infer that the endpoint at \( x = -2 \) has an open point on the line segment of the function defined by \( 2x + 3 \).

Thus, the graph will have an open endpoint at \((-2, 3)\).

Final Answer:

Yes, at \((-2, 3)\).