Domain Analysis of y = -1/4x² + 6¼: Finding Valid Inputs

To find the positive and negative domains of the quadratic function $y = -\frac{1}{4}x^2 + 6\frac{1}{4}$, we must determine where the function's output (y-value) is positive and negative.

First, find the roots of the function by solving the equation $-\frac{1}{4}x^2 + 6\frac{1}{4} = 0$.

Set $y = 0$:
$-\frac{1}{4}x^2 + 6\frac{1}{4} = 0$

Multiply through by $-4$ to clear the fraction:
$x^2 = 25$

Taking the square root of both sides, we find the roots:
$x = 5$ and $x = -5$.

These roots divide the x-axis into three intervals: $(- \infty, -5)$, $(-5, 5)$, and $(5, \infty)$.

To determine where the function is positive or negative, test points from each interval in the original equation:

• For $x \in (-\infty, -5)$, choose $x = -6$:
  $y = -\frac{1}{4}(-6)^2 + 6\frac{1}{4} = -\frac{1}{4}(36) + 6.25 = -9 + 6.25 = -2.75$. The function is negative.
• For $x \in (-5, 5)$, choose $x = 0$:
  $y = -\frac{1}{4}(0)^2 + 6\frac{1}{4} = 0 + 6.25 = 6.25$. The function is positive.
• For $x \in (5, \infty)$, choose $x = 6$:
  $y = -\frac{1}{4}(6)^2 + 6\frac{1}{4} = -9 + 6.25 = -2.75$. The function is negative.

Therefore, the function is positive for $x \in (-5, 5)$ and negative for $x \in (-\infty, -5) \cup (5, \infty)$.

In conclusion, the positive domain of the function is $x > 0 : -5 < x < 5$, and the negative domain is $x > 5$ or $x < 0 : x < -5$.