Domain Analysis of y = (1/4)x² - 4/5: Finding Valid Inputs

To determine where the function $y = \frac{1}{4}x^2 - \frac{4}{5}$ is positive or negative, we start by finding the roots of the function. These roots occur where

$\frac{1}{4}x^2 - \frac{4}{5} = 0$.

Multiply through by 4 to clear the fraction:

$x^2 = \frac{16}{5}$.

Take the square root of both sides:

$x = \pm \sqrt{\frac{16}{5}}$.

This simplifies to:

$x = \pm \frac{4}{\sqrt{5}}$.

With roots at $x = \frac{4}{\sqrt{5}}$ and $x = -\frac{4}{\sqrt{5}}$, the x-axis is divided into three intervals:

• $x < -\frac{4}{\sqrt{5}}$,
• $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$,
• $x > \frac{4}{\sqrt{5}}$.

Analyze these intervals:

• Choose a test point $x = 0$ in $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$:
• $y(0) = \frac{1}{4}(0)^2 - \frac{4}{5} = -\frac{4}{5} < 0$. Thus, the function is negative in $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$.
• Choose a test point $x = -2$ in $x < -\frac{4}{\sqrt{5}}$:
• $y(-2) = \frac{1}{4}(-2)^2 - \frac{4}{5} = 1 - \frac{4}{5} = \frac{1}{5} > 0$. Thus, the function is positive in $x < -\frac{4}{\sqrt{5}}$.
• Choose a test point $x = 2$ in $x > \frac{4}{\sqrt{5}}$:
• $y(2) = \frac{1}{4}(2)^2 - \frac{4}{5} = 1 - \frac{4}{5} = \frac{1}{5} > 0$. Thus, the function is positive in $x > \frac{4}{\sqrt{5}}$.

Therefore, the negative domain is $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$, and the positive domains are $x < -\frac{4}{\sqrt{5}}$ or $x > \frac{4}{\sqrt{5}}$.

The correct choice based on this analysis is:

Choice 3: $x < 0 : -\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$

and

$x > 0 : x < -\frac{4}{\sqrt{5}}$ or $x > \frac{4}{\sqrt{5}}$