Solve the Quadratic Inequality: Finding X When -x²-4 > 0

Given the function:

$y=-x^2-4$

Determine for which values of x the following holds:

$f\left(x\right) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Identify the form and behavior of the given quadratic function.
• Step 2: Analyze the vertical position of the function's vertex.
• Step 3: Determine inequality feasibility for positive $y$.

Step 1: The function given is $y = -x^2 - 4$, which is a parabola opening downwards because of the negative coefficient of $x^2$.

Step 2: The vertex of this parabola occurs at $x = -\frac{b}{2a} = 0$, considering no $x$ term present (i.e., $b = 0$). The vertex value of $y$ is $f(0) = -0^2 - 4 = -4$.

Step 3: The vertex, being the highest point at $(0, -4)$, means the entire parabola remains below the x-axis.

Since the maximum point is negative and the parabola only descends from this vertex, the function $y = -x^2 - 4$ remains less than zero for every $x$. There are no $x$ values making $y$ positive.

Therefore, the solution is No x.