Solve the Quadratic Inequality for y=x²+4x+5 When f(x) < 0

To solve the problem, we need to determine the conditions under which the quadratic function $y = x^2 + 4x + 5$ satisfies $y < 0$.

We start by analyzing the discriminant of the quadratic equation. The standard form of a quadratic equation is:

$ax^2 + bx + c$, where here $a = 1$, $b = 4$, and $c = 5$.

The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$.

Calculating the discriminant, we have:

$\Delta = 4^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4$.

Since the discriminant is negative ($\Delta < 0$), the quadratic function has no real roots. This means the parabola does not intersect the x-axis and opens upwards (since $a = 1 > 0$).

Next, we find the vertex of the parabola to determine its minimum point. The vertex $(h, k)$ of a parabola given by $ax^2 + bx + c$ is:

$h = -\frac{b}{2a} = -\frac{4}{2 \times 1} = -2$.

$k = f(h) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$.

Thus, the vertex is at $(-2, 1)$, and since the vertex is the minimum point of the upward-opening parabola and its value ($k$) is positive, the parabola $y = x^2 + 4x + 5$ is always above the x-axis.

Therefore, the function is never negative, and the solution is:

The function has no negative values.