Question

y=(2x+1)^2-4
(1,2) left parenthesis 1 comma 2 right parenthesis (0,−2) left parenthesis 0 comma negative 2 right parenthesis (−1,−5) left parenthesis negative 1 comma negative 5 right parenthesis (−1,−3)

Answers

GPT 3.5
The given points are (1,2), (0,-2), (-1,-5), and (-1,-3).

To check if these points lie on the graph of the given equation y = (2x + 1)^2 - 4, we substitute the x-values into the equation and see if the resulting y-values match the coordinates provided.

1. For point (1,2):
y(1) = (2(1) + 1)^2 - 4
y(1) = (2 + 1)^2 - 4
y(1) = (3)^2 - 4
y(1) = 9 - 4
y(1) = 5

The y-value obtained (5) does not match the y-coordinate of this point (2). Therefore, (1,2) does not lie on the graph of the given equation.

2. For point (0,-2):
y(0) = (2(0) + 1)^2 - 4
y(0) = (0 + 1)^2 - 4
y(0) = (1)^2 - 4
y(0) = 1 - 4
y(0) = -3

The y-value obtained (-3) does not match the y-coordinate of this point (-2). Therefore, (0,-2) does not lie on the graph of the given equation.

3. For point (-1,-5):
y(-1) = (2(-1) + 1)^2 - 4
y(-1) = (-2 + 1)^2 - 4
y(-1) = (-1)^2 - 4
y(-1) = 1 - 4
y(-1) = -3

The y-value obtained (-3) does not match the y-coordinate of this point (-5). Therefore, (-1,-5) does not lie on the graph of the given equation.

4. For point (-1,-3):
y(-1) = (2(-1) + 1)^2 - 4
y(-1) = (-2 + 1)^2 - 4
y(-1) = (-1)^2 - 4
y(-1) = 1 - 4
y(-1) = -3

The y-value obtained (-3) does match the y-coordinate of this point (-3). Therefore, (-1,-3) lies on the graph of the given equation.

In summary, only the point (-1,-3) lies on the graph of the equation y = (2x + 1)^2 - 4, while the other points do not.

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