Identify the Vertex in the Equation y=(x+1)²: A Quadratic Guide

The vertex form is $y = (x-h)^2 + k$. When you have $(x+1)$, you need to rewrite it as $(x-(-1))$ to match the form. This shows h = -1, not +1!

The vertex is the turning point of the parabola - either the lowest point (minimum) or highest point (maximum). For $y = (x+1)^2$, (-1,0) is the minimum point since the parabola opens upward.

If there's no constant term after the squared expression, then k = 0. The equation $y = (x+1)^2$ is the same as $y = (x+1)^2 + 0$.

Absolutely! Plot a few points around x = -1. You'll see that (-1,0) is the lowest point, and the parabola is symmetric around the line x = -1.

The vertex would still be (-1,0), but the parabola would open downward instead of upward. The negative sign affects the direction, not the vertex location.