Domain Analysis of y = -1/2x² + 18: Finding Valid Inputs

Find the positive and negative domains of the function:

$y=-\frac{1}{2}x^2+18$

To find the positive and negative domains of the function $y = -\frac{1}{2}x^2 + 18$, we'll follow these steps:

• Step 1: Set the function equal to zero to find roots using the quadratic equation.
• Step 2: Solve for $x$ using the quadratic formula.
• Step 3: Identify intervals determined by the roots.
• Step 4: Test intervals to determine where the function is positive or negative.

Step 1: Set the equation to zero:
$-\frac{1}{2}x^2 + 18 = 0$

Step 2: Solve for $x$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, $a = -\frac{1}{2}$, $b = 0$, and $c = 18$.

Substitute the values into the quadratic formula:
$x = \frac{-0 \pm \sqrt{0^2 - 4(-\frac{1}{2})(18)}}{2(-\frac{1}{2})}$
$x = \frac{\pm \sqrt{36}}{-1}$
$x = \frac{\pm 6}{-1}$

So, the roots are $x = -6$ and $x = 6$.

Step 3: The roots divide the x-axis into intervals: $(-\infty, -6)$, $(-6, 6)$, and $(6, \infty)$.

Step 4: Test points in each interval to determine the sign of $y$:
For $(-\infty, -6)$ and $(6, \infty)$, choose test points like $x = -7$ and $x = 7$.
Both yield negative values for $y$ since the parabola opens downwards.
For $(-6, 6)$, test with $x = 0$:
$y = -\frac{1}{2}(0)^2 + 18 = 18$ (positive).

Thus, the function is positive for $-6 < x < 6$ and negative for $x > 6$ or $x < -6$.

The correct answer is the interval: $x > 6$ or $x < 0 : x < -6$

The function is positive for \( x > 0 : -6 < x < 6