Domain Analysis of y=-1/2x²+5: Finding Valid Input Values

To solve this problem, we'll start by calculating the roots of the function $y = -\frac{1}{2}x^2 + 5$.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -\frac{1}{2}$, $b = 0$, and $c = 5$.

Substituting these values into the quadratic formula gives:
$x = \frac{-0 \pm \sqrt{0^2 - 4(-\frac{1}{2})(5)}}{2(-\frac{1}{2})}$
$x = \frac{\pm \sqrt{10}}{-1}$
$x = \pm \sqrt{10}$.

The roots are $x = \sqrt{10}$ and $x = -\sqrt{10}$. These points divide the x-axis into three intervals: $(-\infty, -\sqrt{10})$, $(-\sqrt{10}, \sqrt{10})$, and $(\sqrt{10}, \infty)$.

Given that the parabola opens downwards (since $a < 0$), the function is positive outside these roots and negative within them.

• The positive domain is $x < -\sqrt{10}$ or $x > \sqrt{10}$.
• The negative domain is $-\sqrt{10} < x < \sqrt{10}$.

Therefore, the positive domain is $-\sqrt{10} < x < \sqrt{10}$. The negative domain is $x > \sqrt{10}$ or $x < -\sqrt{10}$.

This corresponds to choice 1:

$x > 0 : -\sqrt{10} < x < \sqrt{10}$

$x > \sqrt{10}$ or $x < 0 : x < -\sqrt{10}$