Solve the Quadratic Inequality: When Does -4x² - 12 Fall Below Zero?

Look at the function below:

$y=-4x^2-12$

Then determine for which values of $x$ the following is true:

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

Let's solve this step-by-step:

• The function given is $y = -4x^2 - 12$.
• This quadratic function is of the form $y = ax^2 + bx + c$, where $a = -4$, $b = 0$, and $c = -12$.
• The parabola opens downwards because $a = -4$ is negative.
• The vertex form of the quadratic equation is used to find the maximum point, which in this configuration is the vertex. For this standard form, $x = -\frac{b}{2a} = 0$.
• The vertex is located at $(0, -12)$.
• Calculate the discriminant to identify the x-intercepts (if any):
• The discriminant $\Delta = b^2 - 4ac = 0^2 - 4(-4)(-12) = -192$.
• Since the discriminant is negative, there are no real roots for the quadratic equation. This means the function does not intersect the x-axis and is always below it.

Given that the parabola opens downwards and never meets the x-axis, the function $y = -4x^2 - 12$ is always less than zero for all real $x$.

Therefore, the solution to the problem is that the function is negative for all values of $x$.