Explore Domains with y=-2/3x²+1/4x-1/5

To find when the function $y = -\frac{2}{3}x^2 + \frac{1}{4}x - \frac{1}{5}$ is positive or negative, we will determine its roots and analyze its sign changes across different intervals of $x$.

**Step 1: Calculate the Roots**

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -\frac{2}{3}$, $b = \frac{1}{4}$, and $c = -\frac{1}{5}$.

Calculate the discriminant:

$b^2 - 4ac = \left(\frac{1}{4}\right)^2 - 4 \left(-\frac{2}{3}\right) \left(-\frac{1}{5}\right) = \frac{1}{16} - \frac{8}{15}$.

Since $\frac{1}{16} = \frac{15}{240}$ and $\frac{8}{15} = \frac{128}{240}$, the discriminant is negative, $\frac{15}{240} - \frac{128}{240} = -\frac{113}{240}$.

The discriminant is negative, indicating no real roots; the parabola does not intersect the x-axis.

**Step 2: Determine the Orientation and Sign**

The coefficient $a = -\frac{2}{3}$ is negative, meaning the quadratic opens downwards.

**Step 3: Analyze the Sign of the Quadratic**

Since the quadratic opens downwards and doesn't intersect the x-axis, it remains negative for all $x$.

Therefore, the negative domain of the function is $x \in (-\infty, \infty)$ and the function has no positive domain.

Consequently:

$x < 0 :$ for all $x$

$x > 0 :$ none

Hence, the correct answer is: Choice 4.