Asked by Fishy
Which of the following is a point on the graph as described by the function y=(2x+1)2−4
𝑦
=
(
2
𝑥
+
1
)
2
−
4
?(1 point)
Responses
(−1,−5)
(
−
1
,
−
5
)
left parenthesis negative 1 comma negative 5 right parenthesis
(−1,−3)
(
−
1
,
−
3
)
left parenthesis negative 1 comma negative 3 right parenthesis
(0,−2)
(
0
𝑦
=
(
2
𝑥
+
1
)
2
−
4
?(1 point)
Responses
(−1,−5)
(
−
1
,
−
5
)
left parenthesis negative 1 comma negative 5 right parenthesis
(−1,−3)
(
−
1
,
−
3
)
left parenthesis negative 1 comma negative 3 right parenthesis
(0,−2)
(
0
Answers
Answered by
GPT-4o mini
To determine which point is on the graph of the function \( y = (2x + 1)^2 - 4 \), we can substitute each \( x \) value from the given points into the equation and see if we obtain the corresponding \( y \) value.
1. For the point \( (-1, -5) \):
\[
x = -1 \implies y = (2(-1) + 1)^2 - 4 = (-2 + 1)^2 - 4 = (-1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (-1, -5) \) is **not** on the graph.
2. For the point \( (-1, -3) \):
\[
x = -1 \implies y = (2(-1) + 1)^2 - 4 = (-2 + 1)^2 - 4 = (-1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (-1, -3) \) **is** on the graph.
3. For the point \( (0, -2) \):
\[
x = 0 \implies y = (2(0) + 1)^2 - 4 = (0 + 1)^2 - 4 = (1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (0, -2) \) is **not** on the graph.
Therefore, the only point that lies on the graph of the function \( y = (2x + 1)^2 - 4 \) is:
**\((-1, -3)\)**.
1. For the point \( (-1, -5) \):
\[
x = -1 \implies y = (2(-1) + 1)^2 - 4 = (-2 + 1)^2 - 4 = (-1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (-1, -5) \) is **not** on the graph.
2. For the point \( (-1, -3) \):
\[
x = -1 \implies y = (2(-1) + 1)^2 - 4 = (-2 + 1)^2 - 4 = (-1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (-1, -3) \) **is** on the graph.
3. For the point \( (0, -2) \):
\[
x = 0 \implies y = (2(0) + 1)^2 - 4 = (0 + 1)^2 - 4 = (1)^2 - 4 = 1 - 4 = -3
\]
The calculated \( y \) value is \( -3 \), so the point \( (0, -2) \) is **not** on the graph.
Therefore, the only point that lies on the graph of the function \( y = (2x + 1)^2 - 4 \) is:
**\((-1, -3)\)**.
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