Pinpointing the Vertex: Analyzing the Quadratic Equation y = (x+1)² - 1

Find the vertex of the parabola

$y=(x+1)^2-1$

The given equation of the parabola is $y = (x+1)^2 - 1$.

This equation is already in the vertex form, $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.

By comparing, we identify:
The expression $(x + 1)$ implies that $h = -1$ (since $x + 1$ is equivalent to $(x - (-1))$).
The constant $-1$ is the $k$ value.

Thus, the vertex $(h, k)$ is $(-1, -1)$.

Therefore, the vertex of the parabola is at the point $(-1,-1)$.