Let's evaluate each point to see if it lies on the graph as described by the function y = (2x + 1)^2 - 4:
1. Point (-1, -5):
y = (2(-1) + 1)^2 - 4
y = (-1)^2 - 4
y = 1 - 4
y = -3
Therefore, (-1, -5) is not a point on the graph of the function.
2. Point (-1, -3):
y = (2(-1) + 1)^2 - 4
y = (-1)^2 - 4
y = 1 - 4
y = -3
Therefore, (-1, -3) is a point on the graph of the function.
3. Point (1, 2):
y = (2(1) + 1)^2 - 4
y = (2 + 1)^2 - 4
y = 3^2 - 4
y = 9 - 4
y = 5
Therefore, (1, 2) is not a point on the graph of the function.
4. Point (0, -2):
y = (2(0) + 1)^2 - 4
y = (0 + 1)^2 - 4
y = 1^2 - 4
y = 1 - 4
y = -3
Therefore, (0, -2) is not a point on the graph of the function.
So, the only point that lies on the graph of the function y = (2x + 1)^2 - 4 is (-1, -3).
which of the following is a point on the graph as described by the function y = ( 2x + 1) ^2 - 4?
(-1,-5)
(-1,-3)
(1,2)
(0,-2)
1 answer