Domain Analysis: Find Valid Inputs for y = -1/2x² + 2.347

Find the positive and negative domains of the following function:

$y=-\frac{1}{2}x^2+2\frac{25}{72}$

The given function is $y = -\frac{1}{2}x^2 + 2\frac{25}{72}$. We need to find when this function is equal to zero to determine the positive and negative domains.

First, identify the coefficients from the function:

• $a = -\frac{1}{2}$, $b = 0$, $c = 2\frac{25}{72}$ which we convert to improper fraction: $\frac{169}{72}$.

We then use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.

Substituting in our values:

$x = \frac{-0 \pm \sqrt{0 - 4\left(-\frac{1}{2}\right)\left(\frac{169}{72}\right)}}{-1}$.

The discriminant calculation is as follows:

$4 \cdot \frac{1}{2} \cdot \frac{169}{72} = \frac{169}{36}$.

So our roots are:

$x = \pm \sqrt{\frac{169}{36}} = \pm \frac{13}{6} = \pm 2\frac{1}{6}$.

The roots are $x = 2\frac{1}{6}$ and $x = -2\frac{1}{6}$. These divide the x-axis into intervals to be tested.

- When $x < -2\frac{1}{6}$, test point $x = -3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x < -2\frac{1}{6}$ gives negative values.

- When $-2\frac{1}{6} < x < 2\frac{1}{6}$, test $x = 0$:
$y = -\frac{1}{2}(0)^2 + \frac{169}{72} > 0$: Positive.
Hence, $-2\frac{1}{6} < x < 2\frac{1}{6}$ gives positive values.

- When $x > 2\frac{1}{6}$, test point $x = 3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x > 2\frac{1}{6}$ gives negative values.

Therefore, the positive domain is $x > 0 : -2\frac{1}{6} < x < 2\frac{1}{6}$ and the negative domain is $x < 0 : x < -2\frac{1}{6}$ or $x > 2\frac{1}{6}$.

In comparing to the provided choices, the correct choice is Choice 2: $x > 2\frac{1}{6}$ or $x < 0 : x < -2\frac{1}{6}$

$x > 0 : - 2\frac{1}{6} < x < 2\frac{1}{6}$