Solving y=(x+2)²-4: Finding X-Axis Intersections

To solve this problem, we need to determine where the parabola $y = (x+2)^2 - 4$ intersects the x-axis. This occurs where $y = 0$.

Step 1: Set the equation equal to zero to find the x-intercepts:
$0 = (x+2)^2 - 4$

Step 2: Simplify the equation:
$(x+2)^2 = 4$

Step 3: Solve for $x$ by taking the square root of both sides:
$x+2 = \pm 2$

Step 4: Solve each equation for $x$:
1. $x+2 = 2$ leads to $x = 0$
2. $x+2 = -2$ leads to $x = -4$

Therefore, the points of intersection are $(-4, 0)$ and $(0, 0)$, where the parabola intersects the x-axis.

The correct answer to the problem is $(-4, 0), (0, 0)$.