Which of the following is a point on the graph as described by the function y = (2x + 1)^2

To find a point on the graph of the function y = (2x + 1)^2, we can choose a value for x and substitute it into the equation to find the corresponding y-value.

Let's choose x = 0.
Then, substituting x = 0 into the equation, we have:
y = (2(0) + 1)^2
y = (0 + 1)^2
y = (1)^2
y = 1

Therefore, the point (0, 1) is on the graph of the function y = (2x + 1)^2.

Which answer.

A. (-1, -3)
B. (0,-2)
C. (1,2)
D. (—1, -5)

To determine which point is on the graph of the function y = (2x + 1)^2, we can substitute the x-coordinate of each option into the equation and check if the resulting y-coordinate matches.

A. (-1, -3):
Plugging in x = -1 into the equation: y = (2(-1) + 1)^2
y = (-2 + 1)^2
y = (-1)^2
y = 1
(With this substitution, y does not equal -3, so (-1, -3) is not on the graph)

B. (0, -2):
Plugging in x = 0 into the equation: y = (2(0) + 1)^2
y = (0 + 1)^2
y = (1)^2
y = 1
(With this substitution, y does not equal -2, so (0, -2) is not on the graph)

C. (1, 2):
Plugging in x = 1 into the equation: y = (2(1) + 1)^2
y = (2 + 1)^2
y = (3)^2
y = 9
(With this substitution, y does not equal 2, so (1, 2) is not on the graph)

D. (-1, -5):
Plugging in x = -1 into the equation: y = (2(-1) + 1)^2
y = (-2 + 1)^2
y = (-1)^2
y = 1
(With this substitution, y does not equal -5, so (-1, -5) is not on the graph)

None of the given options are on the graph of the function y = (2x + 1)^2.

Whoops I mean y = (2x + 1)^2 - 4

No problem! Let's find the point on the graph of the function y = (2x + 1)^2 - 4.

A. (-1, -3):
y = (2(-1) + 1)^2 - 4
y = (-2 + 1)^2 - 4
y = (-1)^2 - 4
y = 1 - 4
y = -3
The point (-1, -3) is on the graph of the function y = (2x + 1)^2 - 4.

B. (0, -2):
y = (2(0) + 1)^2 - 4
y = (0 + 1)^2 - 4
y = (1)^2 - 4
y = 1 - 4
y = -3
The point (0, -2) is not on the graph of the function y = (2x + 1)^2 - 4.

C. (1, 2):
y = (2(1) + 1)^2 - 4
y = (2 + 1)^2 - 4
y = (3)^2 - 4
y = 9 - 4
y = 5
The point (1, 2) is not on the graph of the function y = (2x + 1)^2 - 4.

D. (-1, -5):
y = (2(-1) + 1)^2 - 4
y = (-2 + 1)^2 - 4
y = (-1)^2 - 4
y = 1 - 4
y = -3
The point (-1, -5) is not on the graph of the function y = (2x + 1)^2 - 4.

Therefore, the point on the graph of the function y = (2x + 1)^2 - 4 is (-1, -3), option A.