Analyzing Positive and Negative Domains in y = -x² + 1/2x - 3

Find the positive and negative domains of the following function:

$y=-x^2+\frac{1}{2}x-3$

To find the domains where the function is positive and negative, let's follow these steps:

• Identify the quadratic function given: $y = -x^2 + \frac{1}{2}x - 3$.
• Use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, to find the roots.

Substitute $a = -1$, $b = \frac{1}{2}$, and $c = -3$ into the quadratic formula:

$x = \frac{-\frac{1}{2} \pm \sqrt{\left(\frac{1}{2}\right)^2 - 4(-1)(-3)}}{2(-1)}$.

$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} - 12}}{-2}$.

$x = \frac{-\frac{1}{2} \pm \sqrt{-\frac{47}{4}}}{-2}$.

This results in a negative discriminant ($-\frac{47}{4}$), meaning there are no real roots.

Since there are no real roots, the function does not cross the x-axis, and given the parabola opens downwards ($a < 0$), the entire curve lies below the x-axis.

Therefore, the function is negative for all $x$.

This means:
For $x < 0$: the function is negative for all $x$.
For $x > 0$: there are no positive intervals as the function is negative everywhere.

Thus, the solution indicates that the function is always negative, confirming the negative domain spans all real numbers, and the positive domain is nonexistent.

The correct choice aligning with this result is Choice 2: $x < 0$: for all $x$, and $x > 0$: none.