Find the Domain of y=-2x²-4x-2: Analyzing Negative Quadratic Functions

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

• Step 1: Determine the vertex of the quadratic.
• Step 2: Calculate the roots of the function using the quadratic formula.
• Step 3: Analyze the sign of the quadratic in intervals determined by the roots.

Now, let's work through each step:
Step 1: Since $y = -2x^2 - 4x - 2$, the vertex occurs at $x = -\frac{b}{2a} = -\frac{-4}{2(-2)} = -1$.
Step 2: The roots of the quadratic function are given by solving $-2x^2 - 4x - 2 = 0$. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-2)(-2)}}{2(-2)}$ $x = \frac{4 \pm \sqrt{16 - 16}}{-4} = \frac{4 \pm 0}{-4} = -1$ The solution indicates a repeated root at $x = -1$, implying the parabola just touches the x-axis only at this point without crossing it.
Step 3: The parabola opens downward (as the leading coefficient, $-2$, is negative). This implies the function is negative for all $x \neq -1$, and it is zero exactly at $x = -1$.

Therefore, the positive domain does not exist, and the function is negative for all other $x$. The domain where $y < 0$ is $x < 0$ where $x \neq -1$, as for all $x > 0$, $-2x^2 - 4x - 2$ is strictly decreasing.
Checking with the choices provided, this matches choice 3.

Therefore, the solution is:

$x < 0 : x\ne-1$

$x > 0 :$ none