Positive and Negative Domains: With fractions

To solve this problem, we'll follow these steps:

• Step 1: Identify the given equation in a standard form.
• Step 2: Calculate the roots using the quadratic formula.
• Step 3: Determine the sign of the function in different intervals determined by the roots.

Now, let's work through each step:

Step 1: The given quadratic function is $y = -\frac{1}{6}x^2 - \frac{10}{3}x$. Here, $a = -\frac{1}{6}$, $b = -\frac{10}{3}$, and $c = 0$.
We rewrite the function as $y = -\frac{1}{6}x^2 - \frac{10}{3}x$.

Step 2: To find the roots of the quadratic equation, apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values, we get:
$x = \frac{-(-\frac{10}{3}) \pm \sqrt{(-\frac{10}{3})^2 - 4 \cdot (-\frac{1}{6}) \cdot 0}}{2 \cdot (-\frac{1}{6})}$
This simplifies to:
$x = \frac{\frac{10}{3}}{-\frac{1}{3}}$ (since the discriminant simplifies to zero as $c$ is zero)
$x = 0$ and $x = -18\frac{2}{3}$ (solving for the roots)

Step 3: Determining the intervals:
Because the parabola opens downwards (as $a = -\frac{1}{6}$ is negative), the quadratic is positive between the roots.
Thus, $f(x) > 0$ in the interval $-18\frac{2}{3} < x < 0$.

Therefore, the solution to the problem is $-18\frac{2}{3} < x < 0$.

To solve this problem, we'll follow these steps:

• Step 1: Identify the coefficients from the quadratic function $y = -\frac{1}{7}x^2 - 2\frac{3}{7}x$. Set $a = -\frac{1}{7}$, $b = -\frac{17}{7}$, and $c = 0$.
• Step 2: Use the quadratic formula to find the roots of the equation $f(x) = 0$.
• Step 3: Apply the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Now, let's work through the calculations:

First, we calculate the discriminant: $b^2 - 4ac = \left(-\frac{17}{7}\right)^2 - 4\left(-\frac{1}{7}\right)(0) = \frac{289}{49}.$ The discriminant is positive, indicating two distinct real roots.

The roots are computed as: $x = \frac{-(-\frac{17}{7}) \pm \sqrt{\frac{289}{49}}}{2\left(-\frac{1}{7}\right)}.$ Simplify to: $x = \frac{\frac{17}{7} \pm \frac{17}{7}}{-\frac{2}{7}}.$ For $x = \frac{\frac{17}{7} + \frac{17}{7}}{-\frac{2}{7}}$, $x = \frac{\frac{34}{7}}{-\frac{2}{7}} = -17.$ And, for $x = \frac{\frac{17}{7} - \frac{17}{7}}{-\frac{2}{7}}$, $x = \frac{0}{-\frac{2}{7}} = 0.$

The roots are $x = -17$ and $x = 0$. Therefore, the intervals to consider are $(-\infty, -17)$, $(-17, 0)$, and $(0, \infty)$.

Step 4: Test values in these intervals:

• For $x < -17$: Choose $x = -18$ in $f(x) = -\frac{1}{7}x^2 - \frac{17}{7}x$. We get positive value, as substitute results in positive after sign evaluation.
• For $-17 < x < 0$: Choose $x = -1$ in $f(x)$, resulting in negative function value.
• For $x > 0$: Choose $x = 1$ in $f(x)$, yielding positive function value.

Thus, $f(x) > 0$ for $x > 0$ or $x < -17$.

Therefore, the solution to the problem is $x > 0$ or $x < -17$.

First, we need to find the roots of the quadratic equation:

The quadratic is given by:
$y = -\frac{1}{7}x^2 - \frac{17}{7}x$

Setting $y = 0$ to find the $x$-intercepts (roots):
$-\frac{1}{7}x^2 - \frac{17}{7}x = 0$

Factor out the common factor, $-\frac{1}{7}x$:
$-\frac{1}{7}x(x + 17) = 0$

This gives the roots:
$x = 0$ and $x + 17 = 0 \rightarrow x = -17$

These roots divide the number line into intervals. We need to determine where $y > 0$. Because the coefficient of $x^2$ is negative, the parabola opens downward. The function will be positive between the roots.

Thus, we test the interval:
$(-17, 0)$

Since the parabola opens downward, the function $y > 0$ is true in the interval $-17 < x < 0$.

Therefore, the solution to the problem is $-17 < x < 0$, which corresponds to choice 3.

To solve $f(x) > 0$ for the function $f(x) = -\frac{1}{9}x^2 + 1\frac{2}{3}x$, follow these steps:

• Step 1: The quadratic equation is $-\frac{1}{9}x^2 + \frac{5}{3}x = 0$.
• Step 2: Factor out the greatest common factor: $x(-\frac{1}{9}x + \frac{5}{3}) = 0$.
• Step 3: Set each factor equal to zero: $x = 0$ and $-\frac{1}{9}x + \frac{5}{3} = 0$.
• Step 4: Solve for the second root: $x = \frac{5}{3} \times 9 = 15$.
• Step 5: The roots are $x = 0$ and $x = 15$. These divide the number line into intervals: $x < 0$, $0 < x < 15$, and $x > 15$.
Step 6: Test each interval to determine where $f(x)$ is positive:
• For $x < 0$, choose $x = -1$: $f(-1) = -\frac{1}{9}(-1)^2 + \frac{5}{3}(-1) = \frac{14}{9}$, so $f(x) > 0$.
• For $0 < x < 15$, choose $x = 1$: $f(1) = -\frac{1}{9}(1)^2 + \frac{5}{3}(1) = \frac{14}{9}$, so $f(x) > 0$.
• For $x > 15$, choose $x = 16$: $f(16) = -\frac{1}{9}(16)^2 + \frac{5}{3}(16) = -\frac{16}{9}$, so $f(x) \leq 0$.

The inequality $f(x) > 0$ holds for $x < 0$ or $x > 15$.

Therefore, the values of $x$ satisfying $f(x) > 0$ are $x > 15$ or $x < 0$.

To determine where the function $f(x) = -\frac{1}{9}x^2 + \frac{5}{3}x$ is positive, we follow these steps:

• Step 1: Convert the mixed fraction: The term $1\frac{2}{3}x$ can be written as $\frac{5}{3}x$.
• Step 2: The function can be expressed as $f(x) = -\frac{1}{9}x^2 + \frac{5}{3}x$.
• Step 3: Set the function equal to zero to find the roots: $-\frac{1}{9}x^2 + \frac{5}{3}x = 0$.
• Step 4: Factor out $x$: $x\left(-\frac{1}{9}x + \frac{5}{3}\right) = 0$.
• Step 5: Solve for $x$: From $x = 0$ and $-\frac{1}{9}x + \frac{5}{3} = 0$, find the second root:

Solving the second equation:

$\frac{5}{3} = \frac{1}{9}x$, which simplifies to:

$x = \frac{5}{3} \times 9 = 15$.

The roots are $x = 0$ and $x = 15$.

Since the parabola opens downwards (as indicated by the negative leading coefficient $-\frac{1}{9}$), the function will be positive between the roots.

Thus, $f(x) > 0$ for $0 < x < 15$.

Therefore, the values of $x$ such that $f(x) > 0$ are given by:

$0 < x < 15$.

To solve this problem, we'll follow these steps:

• Step 1: Write the function in standard quadratic form and identify coefficients.
• Step 2: Find the roots using the quadratic formula.
• Step 3: Determine the sign of the quadratic on intervals determined by the roots.
• Step 4: Identify intervals where the function is positive.

Step 1: The function given is $y = -\frac{1}{6}x^2 + 3\frac{2}{3}x$, or equivalently:

$y = -\frac{1}{6}x^2 + \frac{11}{3}x$ in standard form.

With coefficients $a = -\frac{1}{6}$, $b = \frac{11}{3}$, and $c = 0$.

Step 2: Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find roots:

The roots are given by:

$x = \frac{-\frac{11}{3} \pm \sqrt{\left(\frac{11}{3}\right)^2 - 4 \left(-\frac{1}{6}\right) \cdot 0}}{2 \cdot \left(-\frac{1}{6}\right)}$

Since $c = 0$, simplify to:

$x = \frac{-\frac{11}{3} \pm \sqrt{\left(\frac{11}{3}\right)^2}}{-\frac{1}{3}}$

Solve to get roots:

Roots are $x = 0$ and $x = 22$.

Step 3: Analyze the sign of $y$:

Since the parabola opens downwards (as $a < 0$), the function is positive between the roots:

$0 < x < 22$.

Therefore, the solution to the problem is where the function is positive:

$0 < x < 22$.

To solve the problem, we need to determine where the function $y = \frac{1}{2}x^2 + 4\frac{3}{5}x$ is positive.

First, rewrite the function by converting $4\frac{3}{5}x$ to an improper fraction: $\frac{23}{5}x$. Thus, the function becomes:

$y = \frac{1}{2}x^2 + \frac{23}{5}x$.

Next, we solve the inequality $y > 0$. First, find where $y = 0$:

$\frac{1}{2}x^2 + \frac{23}{5}x = 0$.

Factor the equation:

$x \left(\frac{1}{2}x + \frac{23}{5}\right) = 0$.

This gives us the roots $x = 0$ and $\frac{1}{2}x + \frac{23}{5} = 0$.

Solve for the second root:

$\frac{1}{2}x = -\frac{23}{5}$

$x = -\frac{23}{5} \times 2$

$x = -\frac{46}{5} = -9\frac{1}{5}$.

The roots are $x = 0$ and $x = -9\frac{1}{5}$.

The function $y$ is a parabola opening upwards (as $\frac{1}{2} > 0$).

Using the roots, test intervals to find where $y > 0$:

• Test an $x$ value less than $-9\frac{1}{5}$ (e.g., $x = -10$): $y$ is negative.
• Test an $x$ value between $-9\frac{1}{5}$ and $0$ (e.g., $x = -5$): $y$ is positive.
• Test an $x$ value greater than $0$: $y$ is positive.

The inequality $f(x) > 0$ holds for $-9\frac{1}{5} < x < 0$.

The correct choice matches option 3: $-9\frac{1}{5} < x < 0$.

To solve the problem of finding for which values of $x$ the function $y = \frac{1}{2}x^2 + 4\frac{3}{5}x$ is positive, follow these steps:

• Step 1: Identify the coefficients of the quadratic function. Here $a = \frac{1}{2}$, $b = 4\frac{3}{5} = \frac{23}{5}$, and $c = 0$.
• Step 2: Apply the quadratic formula to find the roots. Since $c = 0$, the formula simplifies, and we focus on:
• $x = \frac{-\frac{23}{5} \pm \sqrt{\left(\frac{23}{5}\right)^2 - 4 \cdot \frac{1}{2} \cdot 0}}{1}$
• Calculating the discriminant: $\left(\frac{23}{5}\right)^2 = \frac{529}{25}$. Therefore, roots occur at:
• $x = 0 \quad \text{and} \quad x = -\frac{23}{5} = -4.6$
• Step 3: Analyze the sign of $f(x)$ around the roots. A parabola opens upwards (since $a = \frac{1}{2} > 0$), so $f(x)$ is positive when beyond these roots.
• The intervals to check are $x < -4.6$ and $x > 0$ where $f(x) > 0$.

Therefore, the solution to the problem is $x > 0$ or $x < -4.6$, which translates to the fractional values $x > 0$ or $x < -9\frac{1}{5}$, matching choice 2.

To solve the problem, we follow these steps:

• Step 1: Identify the quadratic equation $\frac{1}{3}x^2 + \frac{7}{3}x = 0$.
• Step 2: Factor the equation $x(\frac{1}{3}x + \frac{7}{3}) = 0$.
• Step 3: Set each factor to zero to find the roots: $x = 0$ and $\frac{1}{3}x + \frac{7}{3} = 0$.
• Step 4: Solve for $x$ to find the second root: $x = -7$.
• Step 5: Analyze the intervals: Plug in test points from intervals defined by $x = -7$ and $x = 0$ to determine sign of $f(x)$.

Now, let's solve each step:

Step 1: The function is $\frac{1}{3}x^2 + \frac{7}{3}x = 0$.
Step 2: Factor the equation: as $x(\frac{1}{3}x + \frac{7}{3}) = 0$. Thus, $x = 0$ and $\frac{1}{3}x + \frac{7}{3} = 0$ are the potential roots.
Step 3: Solve for the second root $x$:

$\frac{1}{3}x + \frac{7}{3} = 0$

Multiply through by 3 to clear fractions:

$x + 7 = 0$

Solve for $x$:
$x = -7$.

Step 4: Identify intervals of interest: We have critical points at $x = -7$ and $x = 0$. Thus, we have intervals $(- \infty, -7)$, $(-7, 0)$, and $0, \infty)$.
Step 5: Test these intervals to determine where $\frac{1}{3}x^2 + \frac{7}{3}x < 0$.

Choose test values:
- For interval $(-8): (-8, 0) \rightarrow \frac{1}{3}(-8)^2 + \frac{7}{3}(-8) = \frac{64}{3} - \frac{56}{3} = \frac{8}{3} > 0$, - For interval $(-2): (-2, 0) \rightarrow \frac{1}{3}(-2)^2 + \frac{7}{3}(-2) = \frac{4}{3} - \frac{14}{3} = -\frac{10}{3} < 0$.

Conclusively, the solution is:

$-7 < x < 0$

The function given is $y = \frac{1}{3}x^2 + \frac{7}{3}x$. Our goal is to determine when this function is greater than 0.

Firstly, we set the function equal to 0 to find the critical points:

$\frac{1}{3}x^2 + \frac{7}{3}x = 0$

Factor out $\frac{1}{3}x$ from the equation:

$\frac{1}{3}x(x + 7) = 0$

This gives us two roots: $x = 0$ and $x = -7$.

Now, consider the intervals determined by these roots: $x < -7$, $-7 < x < 0$, and $x > 0$. Analyze the sign of $y$ in each interval by selecting test points.

• Interval $x < -7$: Choose $x = -8$. Substituting into the function gives $\frac{1}{3}(-8)^2 + \frac{7}{3}(-8) = \frac{64}{3} - \frac{56}{3} = \frac{8}{3} > 0$
• Interval $-7 < x < 0$: Choose $x = -1$. Substituting into the function gives $\frac{1}{3}(-1)^2 + \frac{7}{3}(-1) = \frac{1}{3} - \frac{7}{3} = -\frac{6}{3} = -2 < 0$
• Interval $x > 0$: Choose $x = 1$. Substituting into the function gives $\frac{1}{3}(1)^2 + \frac{7}{3}(1) = \frac{1}{3} + \frac{7}{3} = \frac{8}{3} > 0$

From this analysis, the function $y$ is positive when $x < -7$ or $x > 0$. Thus, the solution is:

The function $f(x)$ is positive for $x > 0$ or $x < -7$.

To determine when $f(x) = \frac{1}{4}x^2 - 3\frac{1}{2}x < 0$, follow these steps:

Step 1: Find the roots of the quadratic equation.

The function can be rewritten as $y = \frac{1}{4}x^2 - \frac{7}{2}x$. Set this equal to zero to find the roots:

$\frac{1}{4}x^2 - \frac{7}{2}x = 0$

Factor out $x$: $x\left(\frac{1}{4}x - \frac{7}{2}\right) = 0$

So, $x = 0$ or $\frac{1}{4}x = \frac{7}{2}$. Solve the second equation:

$x = \frac{7}{2} \times 4 = 14$

Step 2: Analyze the intervals around the roots.

The roots are $x = 0$ and $x = 14$. These divide the number line into three intervals: $x < 0$, $0 < x < 14$, and $x > 14$.

Step 3: Perform a sign test in each interval.

• Test for $x < 0$: Choose $x = -1$. The value of the function $f(-1) = \frac{1}{4}(-1)^2 - \frac{7}{2}(-1) = \frac{1}{4} + \frac{7}{2} > 0$.
• Test for $0 < x < 14$: Choose $x = 7$. The value of the function $f(7) = \frac{1}{4}(7)^2 - \frac{7}{2}(7) = \frac{49}{4} - \frac{49}{2} = \frac{49}{4} - \frac{98}{4} = -\frac{49}{4} < 0$.
• Test for $x > 14$: Choose $x = 15$. The value of the function $f(15) = \frac{1}{4}(15)^2 - \frac{7}{2}(15) = \frac{225}{4} - \frac{105}{2} = \frac{225}{4} - \frac{210}{4} = \frac{15}{4} > 0$.

Conclusion: The quadratic $\frac{1}{4}x^2 - \frac{7}{2}x$ is less than zero for $0 < x < 14$.

Therefore, the solution to the problem is $0 < x < 14$.

To solve the problem of finding for which values of $x$ the function $y = \frac{1}{4}x^2 - \frac{7}{2}x$ is greater than zero, follow these steps:

• Step 1: Identify the function. The quadratic function $y = \frac{1}{4}x^2 - \frac{7}{2}x$ can be rewritten as $y = \frac{1}{4}x^2 - \frac{7}{2}x$.
• Step 2: Use the quadratic formula to find the roots of the equation $y = 0$. The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = \frac{1}{4}$, $b = -\frac{7}{2}$, and $c = 0$.

• Step 3: Calculate $b^2 - 4ac$:

$b^2 - 4ac = \left(-\frac{7}{2}\right)^2 - 4 \left(\frac{1}{4}\right)(0) = \frac{49}{4}$

• Step 4: Find the roots using the quadratic formula:

$x = \frac{-\left(-\frac{7}{2}\right) \pm \sqrt{\frac{49}{4}}}{2 \times \frac{1}{4}}$ $x = \frac{\frac{7}{2} \pm \frac{7}{2}}{\frac{1}{2}}$ $x = \frac{7 \pm 7}{1}$

This gives two roots: $x = 0$ and $x = 14$.

• Step 5: Determine the intervals defined by these roots. We have three intervals to check: $x < 0$, $0 < x < 14$, and $x > 14$.

• Step 6: Test the sign of $f(x)$ in each interval:
• For $x < 0$, choose $x = -1$:

$f(-1) = \frac{1}{4}(-1)^2 - \frac{7}{2}(-1) = \frac{1}{4} + \frac{7}{2} > 0$

• For $0 < x < 14$, choose $x = 1$:

$f(1) = \frac{1}{4}(1)^2 - \frac{7}{2}(1) = \frac{1}{4} - \frac{7}{2} < 0$

• For $x > 14$, choose $x = 15$:

$f(15) = \frac{1}{4}(15)^2 - \frac{7}{2}(15) = \frac{225}{4} - \frac{105}{2} > 0$

After confirming the signs, we find that $f(x) > 0$ for $x \lt 0$ and $x \gt 14$.

Therefore, the solution to the problem is $<strong>x > 14$ or $x < 0$.

To solve this problem, we'll proceed as follows:

• Step 1: Identify the quadratic formula and coefficients.
• Step 2: Find the roots using the quadratic formula.
• Step 3: Determine intervals and test signs.

Step 1: Identify the equation coefficients.

Given: $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. Let $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$.

Step 2: Find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Calculate the discriminant: $b^2 - 4ac = \left(\frac{4}{3}\right)^2 - 4\left(\frac{1}{5}\right)(0) = \frac{16}{9}$.

The roots of the equation are given by $x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{2 \times \frac{1}{5}}$.

Simplifying: $x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}} = \frac{-4 \pm 4}{\frac{2}{5}} \cdot \frac{1}{3}$.

The roots are $x = 0$ and $x = -6\frac{2}{3}$.

Step 3: Determine intervals created by the roots and test each interval.

The intervals are $(-\infty, -6\frac{2}{3})$, $(-6\frac{2}{3}, 0)$, and $(0, \infty)$.

- For $x = -7$ (or any point less than $-6\frac{2}{3}$), the expression is positive.

- For $x = -3$ (or any point between $-6\frac{2}{3}$ and $0$), test by substituting back into the function.
The function yields a negative result.

- For $x = 1$ (or any point greater than $0$), the expression is positive.

Conclusion: The function is negative between $-6\frac{2}{3}$ and $0$.

Thus, the solution to the problem is $-6\frac{2}{3} < x < 0$.

Let's solve the problem step by step:

The function given is $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. The first step is to convert the mixed number into an improper fraction:

$1\frac{1}{3} = \frac{4}{3}$, so the function becomes:

$y = \frac{1}{5}x^2 + \frac{4}{3}x$.

This can be written in standard form $ax^2 + bx + c = 0$ where $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$ (not visible, but necessary for proper representation).

Next, we use the quadratic formula to find the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Plug in the values:

$x = \frac{-\frac{4}{3} \pm \sqrt{\left(\frac{4}{3}\right)^2 - 4 \times \frac{1}{5} \times 0}}{2 \times \frac{1}{5}}$.

Simplify where possible:

$x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{\frac{2}{5}}$, leading to:

$x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}}$.

This gives roots:

$x = 0$ and $x = -6\frac{2}{3}$.

The parabola opens upward (since $a = \frac{1}{5} > 0$). The quadratic function is positive between the roots and for values outside them:

This results in two intervals where $f(x) > 0$:

• $x < -6\frac{2}{3}$
• $x > 0$

Thus, the solution to the problem is:

$x > 0$ or $x < -6\frac{2}{3}$.

To solve this problem, follow these steps:

• Step 1: Identify the coefficients from the quadratic function $y = -\frac{1}{6}x^2 - 3\frac{1}{9}x$. Here, $a = -\frac{1}{6}$, $b = -\frac{28}{9}$, and $c = 0$.
• Step 2: Find the roots using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Plugging in the values, we get:

$x = \frac{-\left(-\frac{28}{9}\right) \pm \sqrt{\left(-\frac{28}{9}\right)^2 - 4 \times \left(-\frac{1}{6}\right) \times 0}}{2 \times \left(-\frac{1}{6}\right)}$

Simplifying, we find:

$x = \frac{\frac{28}{9}}{-\frac{1}{3}}$

$x = \frac{28}{9} \times -3$

$x = -\frac{84}{9} = -9\frac{1}{3}$

• Step 3: Explore the intervals determined by these roots for where the function $y$ is negative:
• Consider intervals $(-\infty, -18\frac{2}{3})$, $(-18\frac{2}{3}, 0)$, and $(0, \infty)$.
• Evaluate the sign of $y$ over each interval:

For $(-\infty, -18\frac{2}{3})$: Choose a sample point like $x = -20$. Plug it in to see if $y < 0$. It turns out this interval is negative.

For $(-18\frac{2}{3}, 0)$: Choose a sample point like $x = -1$. Plug it in to see if $y > 0$. This interval turns out positive.

For $(0, \infty)$: Choose a sample point like $x = 1$. Plug it in to see if $y < 0$. This interval turns out negative.

Therefore, the solution is when $x > 0$ or $x < -18\frac{2}{3}$.

Thus, the quadratic function is negative outside the roots.

Therefore, the solution to the problem is $x > 0$ or $x < -18\frac{2}{3}$.

To solve this problem, we'll follow these steps:

• Step 1: Identify roots of the quadratic equation.
• Step 2: Determine the intervals created by these roots.
• Step 3: Test each interval to see where the function is positive or negative.

Now, let's work through each step:

Step 1: We need to find the roots of the equation $-\frac{1}{3}x^2 + 2x - 4 = 0$. Using the quadratic formula:
For $ax^2 + bx + c = 0$, the roots are given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = -\frac{1}{3}$, $b = 2$, and $c = -4$.

Step 2: Calculate the discriminant:
$b^2 - 4ac = 2^2 - 4(-\frac{1}{3})(-4) = 4 - \frac{16}{3} = \frac{12}{3} - \frac{16}{3} = -\frac{4}{3}$.

Since the discriminant is negative, the quadratic does not have real roots. Therefore, the function does not cross the x-axis and remains entirely above or below the x-axis.

Step 3: Analyze the leading coefficient. The quadratic function opens downwards because the leading coefficient $a = -\frac{1}{3}$ is negative. Therefore, since there are no x-intercepts, the function is negative for all $x$.

Thus, we find that:
- The positive domain of $y$ is: none.
- The negative domain of $y$ is: for all $x$.

Therefore, the solution to the problem is:

$x < 0 :$ for all $x$

$x > 0 :$ none