Solving Quadratic Inequality: y = x² + 2x + 2 < 0

To solve this problem, we need to rewrite the quadratic function $y = x^2 + 2x + 2$ into its vertex form. This process allows us to find the vertex and understand the behavior of the function.

First, we complete the square for the quadratic expression. Starting with:
$y = x^2 + 2x + 2$

We take the $x$-terms $x^2 + 2x$ and complete the square as follows:

• Take half of the coefficient of $x$, which is 2, resulting in 1.
• Square this value to get 1.
• Add and subtract this square inside the equation to maintain equality.

Therefore, $x^2 + 2x + 1 - 1 + 2$ can be rewritten as:
$y = (x+1)^2 + 1$

Now, the function is in the form $y = (x+1)^2 + 1$, which shows the vertex at $(-1, 1)$. The vertex is the minimum point because the parabola opens upwards (as the coefficient of $x^2$ is positive).

This vertex indicates that the minimum value of $y$ is 1, which means the function never reaches below zero. As a result, the function never assumes negative values.

Based on this analysis, we conclude that the function has no negative values.

The correct answer is therefore: The function has no negative values.