Graph y ≤ (x + 2)^2.
Click on the graph until the correct graph appears.




Answer:
the first one is the correct one
Step-by-step explanation:
a great way to check this is to use demos .com
Answer:
its the first one
Step-by-step explanation:
our vertex is (-2,0)
so replace it in the equation y ≤ (x + 2)^2
so, 0 ≤ (-2 + 2)^2
= 0 ≤ (-4)^2
= 0 ≤ 16
look at that statement, is 0 less than or equal to 16? Yes. Therefore it is correct, so you shade out side of the parabola.
TIP: If the parabola opens up and:
y <, then shade outside.
y >, then shade inside.
Do the opposite if the parabola opens down.