Solve the Quadratic Inequality: When is -4x² - 12 Positive?

Look at the function below:

$y=-4x^2-12$

Determine for which values of $x$ the following is true:

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

The goal is to find the values of $x$ for which $f(x) > 0$ given $y = -4x^2 - 12$. Start by analyzing the equation $y = -4x^2 - 12$.

Since the quadratic term is negative ($a = -4$), the parabola opens downwards. This means the maximum point of the parabola (its vertex) is at the top.

Find the vertex using the formula for the $x$-coordinate of the vertex, $x = -\frac{b}{2a}$. Here, $b = 0$, so the vertex is at $x = 0$.

Calculate $y$-value at the vertex:
$y = -4(0)^2 - 12 = -12$.

This evaluation confirms $y = -12$, which is less than 0 at the vertex.

Since the entire parabola opens downward and the highest point $y$ achieves is still negative ($-12$), the function is never greater than 0 at any point.

No $x$ values satisfy $y > 0$. Therefore, no values of $x$ make the quadratic positive.

Thus, the answer is: No values of $x$.