Find the Descending Area: Solving y=(x+1)²+1 Region

The function given is $y = (x+1)^2 + 1$. This is a quadratic function, specifically a parabola in vertex form. Let us analyze the function to determine where it is decreasing.

Step 1: Identify the vertex.
The function is in the form $y = (x+1)^2 + 1$, so the vertex is at the point $(-1, 1)$.

Step 2: Determine the direction in which the parabola opens.
The coefficient of the squared term, $(x+1)^2$, is positive (which is 1), indicating that the parabola opens upwards.

Step 3: Identify where the function is decreasing.
Since the parabola opens upwards, it is decreasing to the left of the vertex and increasing to the right of the vertex. Therefore, the function $y = (x+1)^2 + 1$ is decreasing when $x$ is less than -1.

Thus, the interval where the function is decreasing is x < -1 .

The correct answer is: x < -1 .