Which of the following is a point on the graph as described by the function y=(2x+1)v2−4?(1 point)
(0,−2)
(−1,−3)
(−1,−5)
(1,2)
I think the poster meant y=(2x+1)^2−4
so the graph would include the point (-1,-3)
Ah, thank you for catching that! Yes, if the function is actually y=(2x+1)^2−4, then the graph would include the point (-1,-3).
To determine which point is on the graph of the function y = (2x + 1)² - 4, we need to substitute the x-coordinate of each point into the function and check if it matches the y-coordinate. Let's go through each option:
1. (0,-2):
Plugging x = 0 into the function:
y = (2(0) + 1)² - 4
= (0 + 1)² - 4
= 1² - 4
= 1 - 4
= -3
The y-coordinate of this point is -3, not -2, so this is not a point on the graph.
2. (-1,-3):
Plugging x = -1 into the function:
y = (2(-1) + 1)² - 4
= (-2 + 1)² - 4
= (-1)² - 4
= 1 - 4
= -3
The y-coordinate of this point is -3, which matches, so this is a point on the graph.
3. (-1,-5):
Plugging x = -1 into the function:
y = (2(-1) + 1)² - 4
= (-2 + 1)² - 4
= (-1)² - 4
= 1 - 4
= -3
The y-coordinate of this point is -3, not -5, so this is not a point on the graph.
4. (1,2):
Plugging x = 1 into the function:
y = (2(1) + 1)² - 4
= (2 + 1)² - 4
= 3² - 4
= 9 - 4
= 5
The y-coordinate of this point is 5, not 2, so this is not a point on the graph.
Therefore, the only point that is on the graph of the function y = (2x + 1)² - 4 is (-1, -3).