Standard Representation: Identify the positive and negative domain

To solve this problem, we'll find the intervals where the given quadratic function $y = -2x^2 + 7x - 3$ is greater than zero (positive) and less than zero (negative).

Step 1: Find the roots of the quadratic function.

The general form of the quadratic equation is $ax^2 + bx + c = 0$. Here, $a = -2$, $b = 7$, and $c = -3$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the roots.

First, calculate the discriminant:

$b^2 - 4ac = 7^2 - 4 \times (-2) \times (-3) = 49 - 24 = 25$.

Thus, the roots are:

$x = \frac{-7 \pm \sqrt{25}}{2 \times (-2)} = \frac{-7 \pm 5}{-4}$.

Calculating for the two roots:

• $x_1 = \frac{-7 + 5}{-4} = \frac{-2}{-4} = \frac{1}{2}$
• $x_2 = \frac{-7 - 5}{-4} = \frac{-12}{-4} = 3$

The roots are $x = \frac{1}{2}$ and $x = 3$.

Step 2: Determine the sign of the function in each interval.

The function is defined as:

$(-\infty, \frac{1}{2}), \left(\frac{1}{2}, 3\right), (3, \infty)$.

Test each interval to determine where the function is positive or negative:

• For $x < \frac{1}{2}$, choose $x = 0$:
  $y = -2(0)^2 + 7(0) - 3 = -3$ (negative)
• For $\frac{1}{2} < x < 3$, choose $x = 1$:
  $y = -2(1)^2 + 7(1) - 3 = 2$ (positive)
• For $x > 3$, choose $x = 4$:
  $y = -2(4)^2 + 7(4) - 3 = -3$ (negative)

Conclusion: The positive domain is $\frac{1}{2} < x < 3$, and the negative domain is $x < \frac{1}{2}$ or $x > 3$.

Therefore, the correct option is:

$x > 0 :\frac{1}{2} < x < 3$

$x > 3$ or $x < 0 : x <\frac{1}{2}$

Let's begin by finding the roots of the quadratic function $y = -x^2 + \frac{3}{4}x - 2$. This will help us determine the intervals where the function is positive or negative.

The coefficients of the quadratic are $a = -1$, $b = \frac{3}{4}$, and $c = -2$. Applying the quadratic formula:

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

Substitute in the values:

$x = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4(-1)(-2)}}{2(-1)}$ $x = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - 8}}{-2}$

Calculate inside the square root:

$\frac{9}{16} - 8 = \frac{9}{16} - \frac{128}{16} = \frac{-119}{16}$

The discriminant $b^2 - 4ac$ is negative ($\frac{-119}{16}$), indicating there are no real roots. Therefore, the quadratic function doesn't intersect the x-axis.

Since the parabola is downward (the coefficient of $x^2$ is negative), it is negative for all $x \in \mathbb{R}$.

We conclude that:

• The function $y$ does not have a positive domain for any real $x$.
• The function $y$ is negative for all $x$.

Therefore, the positive and negative domains, based on choices given, are:

$x > 0$: none

$x < 0$: all $x$

To find the positive and negative domains of the function $y = -x^2 + \frac{3}{2}x - \frac{21}{4}$, we first determine the roots of the equation:

Set $y = 0$, giving us:

$-x^2 + \frac{3}{2}x - \frac{21}{4} = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{3}{2}$, and $c = -\frac{21}{4}$, we calculate:

• First, compute the discriminant: $b^2 - 4ac = \left(\frac{3}{2}\right)^2 - 4(-1)\left(-\frac{21}{4}\right) = \frac{9}{4} - \frac{21}{1} = \frac{9}{4} - \frac{84}{4} = -\frac{75}{4}$.
• Since the discriminant is negative, the quadratic has no real roots.

This implies that the parabola does not intersect the $x$-axis and since the quadratic coefficient $a = -1$ is negative, the parabola opens downwards.

Thus, the function is always negative for all $x$. Therefore, the positive domain is empty, and the negative domain is the entire set of real numbers.

Conclusion: The solution to the problem is as follows:

$x < 0$: for all $x$

$x > 0$: none

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

• Step 1: Identify the roots of the quadratic function.
• Step 2: Use the roots to determine intervals.
• Step 3: Test each interval to determine if the function is positive or negative.

Now, let's work through each step:
Step 1: The quadratic function is $y = -\frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}$. We apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots, where $a = -\frac{1}{3}$, $b = \frac{2}{3}$, and $c = -\frac{1}{3}$.

Calculating the discriminant $b^2 - 4ac$:

$\left(\frac{2}{3}\right)^2 - 4\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right) = \frac{4}{9} - \frac{4}{9} = 0$.

The discriminant is zero, indicating a repeated root at:

$x = \frac{-\frac{2}{3}}{2\left(-\frac{1}{3}\right)} = 1$.

Step 2: The repeated root is $x = 1$. For $x < 1$ and $x > 1$, evaluate the sign of the function.

Step 3: Testing intervals:

• For $x = 0$ (as a test point for $x < 1$), substitute into the original function:

$y = -\frac{1}{3}(0)^2 + \frac{2}{3}(0) - \frac{1}{3} = -\frac{1}{3}$. The function is negative.

• For $x > 1$, any positive $x$ will substitute into the function and confirm it remains negative, as the parabola opens downwards and cannot turn positive again.

Therefore, the solution to the problem is:

$x < 0 : x \ne 1$, and $x > 0$: none.

To find the positive and negative domains of the function $y = -\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4}$, we must determine where the function is above or below the x-axis.

Step 1: Find the roots of the quadratic equation. This requires solving:

$-\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4} = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = -\frac{1}{2}$, $b = \frac{1}{3}$, and $c = -\frac{1}{4}$, we calculate:

• Discriminant: $b^2 - 4ac = \left(\frac{1}{3}\right)^2 - 4\left(-\frac{1}{2}\right)\left(-\frac{1}{4}\right) = \frac{1}{9} - \frac{1}{2} = \frac{1}{9} - \frac{4}{8} = \frac{1}{9} - \frac{9}{18} = \frac{1 - 9}{18} = -\frac{8}{18} = -\frac{4}{9}$

The discriminant is negative, indicating no real roots.

Step 2: Analyze the parabola's orientation. Because the leading term is negative, the parabola opens downwards. With no x-intercepts, this implies the entire graph is below the x-axis.

Therefore, the function is negative for all x-values. In the context of positive and negative domains:

$x > 0 :$ none, as the function doesn't cross the x-axis in positive domain.

$x < 0 :$ all $x$, as the function is always negative.

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

To solve the given problem, we will perform the following steps:

• Calculate the roots of the quadratic equation $y = -2x^2 + 3x + 2$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Given $a = -2$, $b = 3$, and $c = 2$, use the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(2)}}{2(-2)} = \frac{-3 \pm \sqrt{9 + 16}}{-4} = \frac{-3 \pm \sqrt{25}}{-4}$

• The roots are $x = \frac{-3 + 5}{-4} = -\frac{1}{2}$ and $x = \frac{-3 - 5}{-4} = 2$.
• The roots $-\frac{1}{2}$ and $2$ divide the x-axis into three intervals: $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 2)$, and $(2, \infty)$.
• Test a point from each interval in the function to determine the sign of the function in those intervals:

Choose $x = -1$ for interval $(- \infty, -\frac{1}{2})$:
Substitute into the function: $y = -2(-1)^2 + 3(-1) + 2 = -2 - 3 + 2 = -3$ (negative).

Choose $x = 0$ for interval $(- \frac{1}{2}, 2)$:
Substitute into the function: $y = -2(0)^2 + 3(0) + 2 = 2$ (positive).

Choose $x = 3$ for interval $(2, \infty)$:
Substitute into the function: $y = -2(3)^2 + 3(3) + 2 = -18 + 9 + 2 = -7$ (negative).

Therefore, the positive domain where the function is positive is $-\frac{1}{2} < x < 2$, and the negative domains are $x < -\frac{1}{2}$ or $x > 2$.

The solution to the problem is:

$x > 0 : -\frac{1}{2} < x < 2$

$x > 2$ or $x < 0 : x < -\frac{1}{2}$

To solve for the positive and negative domains of the function $y = 2x^2 - 4x + 2$, we follow these steps:

• Step 1: Find the roots of the equation using the quadratic formula. For $ax^2 + bx + c = 0$, the roots are found by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 2: For our specific function, identify: $a = 2$, $b = -4$, $c = 2$.

Let's calculate the discriminant: $b^2 - 4ac = (-4)^2 - 4 \times 2 \times 2 = 16 - 16 = 0$.
Since the discriminant is zero, we have a repeated root, which means the graph touches the x-axis at one point.

The root is found as follows:
$x = \frac{-(-4) \pm \sqrt{0}}{2 \times 2} = \frac{4}{4} = 1$.

This root $x = 1$ represents a vertex-touching parabola, with no intervals of $x$ such that $y < 0$.

To find the positive domain ($y > 0$), we note the parabola $y = 2x^2 - 4x + 2$ opens upwards (since $a = 2 > 0$) and only touches the x-axis at one point $(1, 0)$. Thus, the positive domain is $x \ne 1$.

The function does not take any negative values, since it opens upwards and only touches the x-axis.

Therefore, the positive domain is $x > 0 :x \ne 1$ and the negative domain is $x < 0 :$ none.

In conclusion, the solution to the problem is:

$x > 0 :x \ne 1$

$x < 0 :$ none

To solve this problem, follow these steps:

• Step 1: Factor the quadratic function or recognize it as a perfect square trinomial.
• Step 2: Identify the vertex of the parabola.
• Step 3: Analyze the intervals between roots and the surrounding areas to determine positivity or negativity.

Now, work through these steps:

Step 1: Recognize the quadratic function $y = x^2 + 10x + 25$ as a perfect square trinomial. It can be rewritten as $y = (x + 5)^2$.

Step 2: The vertex of the parabola, which also represents its minimum point since the parabola opens upwards, occurs at $x = -5$.

Step 3: Since $(x + 5)^2 \geq 0$ for all real $x$ (because a square is always non-negative), the function is non-negative everywhere.

Therefore, the function is never negative, and since it equals zero at $x = -5$, it is positive for all $x \neq -5$.

Therefore, the function's positive domain is $x > 0 : x \neq -5$. For $x < 0$, the function is not negative, hence there is no such domain.

The solution to the problem is:

$x > 0 : x \neq -5$

$x < 0 :$ none

To find the domains where the function is positive and negative, let's follow these steps:

• Identify the quadratic function given: $y = -x^2 + \frac{1}{2}x - 3$.
• Use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, to find the roots.

Substitute $a = -1$, $b = \frac{1}{2}$, and $c = -3$ into the quadratic formula:

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

$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} - 12}}{-2}$.

$x = \frac{-\frac{1}{2} \pm \sqrt{-\frac{47}{4}}}{-2}$.

This results in a negative discriminant ($-\frac{47}{4}$), meaning there are no real roots.

Since there are no real roots, the function does not cross the x-axis, and given the parabola opens downwards ($a < 0$), the entire curve lies below the x-axis.

Therefore, the function is negative for all $x$.

This means:
For $x < 0$: the function is negative for all $x$.
For $x > 0$: there are no positive intervals as the function is negative everywhere.

Thus, the solution indicates that the function is always negative, confirming the negative domain spans all real numbers, and the positive domain is nonexistent.

The correct choice aligning with this result is Choice 2: $x < 0$: for all $x$, and $x > 0$: none.

To solve this problem, let's find the roots of the quadratic function by using the quadratic formula. The function is given as:

$y = -x^2 - 6x - 9$

The quadratic formula is:

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

Given $a = -1$, $b = -6$, and $c = -9$, let's calculate the discriminant:

$\Delta = b^2 - 4ac = (-6)^2 - 4(-1)(-9) = 36 - 36 = 0$

Since the discriminant is zero, there is exactly one real root, which is also the vertex of the parabola. Let's find this root:

$x = \frac{-(-6) \pm \sqrt{0}}{2(-1)} = \frac{6}{-2} = -3$

So, the vertex of the parabola is at $x = -3$. The quadratic function opens downwards ($a < 0$), which means the function will be zero at $x = -3$, negative for all other values of $x$. There is no positive domain because the parabola does not go above the x-axis.

Therefore, the negative domain of the function is:

$x < 0 : x \ne -3$

and there is no positive domain.

This matches choice 3.

Therefore, the solution to the problem is:

$x < 0 : x\ne-3$$x > 0 :$ none

To find the positive and negative domains of the quadratic function $y = -x^2 - 4x - 4$, we begin by solving for its roots.

Step 1: Calculate the discriminant.
The function is in standard form: $y = ax^2 + bx + c$ with $a = -1$, $b = -4$, $c = -4$.
The discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(-1)(-4) = 16 - 16 = 0$.

Step 2: Find the roots using the quadratic formula.
The quadratic formula is $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.
Since $\Delta = 0$, we have a double root at $x = \frac{-(-4)}{2(-1)} = -2$. Thus, the root is $x = -2$.

Step 3: Determine the sign of $y$ without further roots.
For a quadratic $ax^2 + bx + c$ with $a < 0$, the parabola opens downward. Thus, it will only be positive between the roots if distinct or negative if the root is unique, which, in this case, is at $x = -2$.
$y < 0$ for $x < -2$ and $x > -2$. Since the vertex $(x = -2)$ coincides with the root, this implies $y = 0$ only at $x = -2$.

Step 4: Determine positive and negative domains.
Since the parabola does not exist for positive or zero intervals beyond the vertex and the only root, we conclude:
- For $x < 0$ (all valid $x$ where function changes), $y < 0$ except at $x = -2$.
- For $x > 0$, the function does not cross the x-axis, and $y$ remains negative.

Thus, the final answer is as follows:
$x < 0 : x\ne-2$ and for $x > 0 :$ none.

To determine the positive and negative domains of the quadratic function $y = 3x^2 + 7x + 2$, we will first find its roots using the quadratic formula.

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

For this function, $a = 3$, $b = 7$, and $c = 2$.

First, calculate the discriminant $\Delta$:

$\Delta = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25$.

Since the discriminant is positive, the function has two distinct real roots.

Now, calculate the roots:

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

$x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}$.

This results in:

• Root 1: $x = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3}$
• Root 2: $x = \frac{-7 - 5}{6} = \frac{-12}{6} = -2$

The roots are $x = -2$ and $x = -\frac{1}{3}$, dividing the x-axis into three intervals: $x < -2$, $-2 < x < -\frac{1}{3}$, and $x > -\frac{1}{3}$.

For $x \to -\infty$, the quadratic $y = 3x^2 + 7x + 2$ is positive, as the leading coefficient $a = 3$ is positive, indicating the parabola opens upwards.

Evaluate the sign of $y$ within the intervals:

• For $x < -2$, pick $x = -3$: $y = 3(-3)^2 + 7(-3) + 2 = 27 - 21 + 2 = 8$. Hence, $y > 0$.
• For $-2 < x < -\frac{1}{3}$, pick $x = -1$: $y = 3(-1)^2 + 7(-1) + 2 = 3 - 7 + 2 = -2$. Hence, $y < 0$.
• For $x > -\frac{1}{3}$, pick $x = 0$: $y = 3(0)^2 + 7(0) + 2 = 2$. Hence, $y > 0$.

Therefore, the positive domain of the function is $x < -2$ and $x > -\frac{1}{3}$, and the negative domain is $-2 < x < -\frac{1}{3}$.

Thus, the solution matches the given correct answer:

$x < 0 : -2 < x < -\frac{1}{3}$

$x > -\frac{1}{3}$ or $x > 0 : x < -2$

Let's analyze the function $y = 3x^2 - 6x + 4$ to determine when it is negative.

First, calculate the discriminant $\Delta$:

$\Delta = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$

A negative discriminant ($\Delta < 0$) indicates that there are no real roots, meaning the graph of the function does not intersect the x-axis. Since $a = 3 > 0$, the parabola opens upwards.

This means the vertex of the parabola represents the minimum point, and the entire graph is above the x-axis.

Consequently, the function $y = 3x^2 - 6x + 4$ does not attain any negative values for any real $x$.

The correct interpretation is that the function stays positive, confirming the conclusion:

The function has no negative values.

To determine the positive and negative domains of the quadratic function $y = x^2 - 22x + 121$, we start by finding its roots. The roots will help us identify intervals where the function is positive or negative.

First, apply the quadratic formula:

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

The given quadratic equation is $x^2 - 22x + 121$. Here, $a = 1$, $b = -22$, and $c = 121$. Calculate the discriminant:

$\Delta = b^2 - 4ac = (-22)^2 - 4 \times 1 \times 121 = 484 - 484 = 0$

The discriminant is zero, indicating a repeated root. Applying the quadratic formula gives:

$x = \frac{-(-22) \pm \sqrt{0}}{2 \times 1} = \frac{22 \pm 0}{2} = 11$

This means there is only one root, $x = 11$. A quadratic with a double root at $x = 11$ indicates the function touches the x-axis at $x = 11$ and opens upwards (since $a > 0$). This implies that the function is non-negative for all $x$. Therefore, the function does not have a negative domain.

As the quadratic is positive for $x \ne 11$, the positive domain is all $x$ except when $x = 11$.

Therefore, the positive domain is $x > 0 : x \ne 11$.

The negative domain, where the function would take negative values, is nonexistent as the parabola never crosses beneath the x-axis.

The solution to the problem is:

$x > 0 : x \ne 11$

$x < 0 :$ none

To determine where the function $y = x^2 + 5x + 4$ is greater than zero, we first find its roots by setting $x^2 + 5x + 4 = 0$.

Step 1: Factor the quadratic equation.

The expression $x^2 + 5x + 4$ can be factored as $(x + 1)(x + 4) = 0$.

Step 2: Solve for the roots.

Setting each factor to zero gives the roots as follows:
$x + 1 = 0 \Rightarrow x = -1$
$x + 4 = 0 \Rightarrow x = -4$

Step 3: Determine the sign of the quadratic on the intervals defined by the roots.

• Interval 1: $x < -4$. Pick $x = -5$, then $(x + 1)(x + 4) = (-5 + 1)(-5 + 4) = (-4)(-1) = 4 > 0.$
• Interval 2: $-4 < x < -1$. Pick $x = -3$, then $(x + 1)(x + 4) = (-3 + 1)(-3 + 4) = (-2)(1) = -2 < 0.$
• Interval 3: $x > -1$. Pick $x = 0$, then $(x + 1)(x + 4) = (0 + 1)(0 + 4) = 1 \times 4 = 4 > 0.$

Conclusion: The function $y = x^2 + 5x + 4$ is positive when $x < -4$ or $x > -1$.

Thus, the solution is $x > -1$ or $x < -4$.

To solve for the positive and negative domains of the quadratic function $y=3x^2+12x+12$, we will follow these steps:

• Step 1: Calculate the roots using the quadratic formula.
• Step 2: Analyze the sign of the function between and beyond these roots.

Step 1: Find the roots of the quadratic function using the quadratic formula:

The given quadratic function is $y = 3x^2 + 12x + 12$. Identifying coefficients, we have $a = 3$, $b = 12$, and $c = 12$.

Using the quadratic formula:

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

Substitute the values of $a$, $b$, and $c$:

$x = \frac{-12 \pm \sqrt{12^2 - 4 \times 3 \times 12}}{2 \times 3}$ $x = \frac{-12 \pm \sqrt{144 - 144}}{6}$ $x = \frac{-12 \pm \sqrt{0}}{6}$ $x = \frac{-12}{6}$ $x = -2$

The quadratic function has a single root at $x = -2$, meaning it is a perfect square trinomial, and there is only one point where the function equals zero.

Step 2: Analyze the sign of the quadratic:

• Because the parabola opens upwards (since $a = 3 > 0$), it implies the function is positive for all $x$ except at the vertex $x = -2$.
• The function $y$ reaches its minimum value (vertex) at $x = -2$, where $y = 0$.
• Therefore, $y = 3x^2 + 12x + 12 > 0$ for all $x \neq -2$.

Therefore, the function is positive for all $x$ but not for $x = -2$, where it is zero. It never reaches negativity.

To summarize, the positive domain (where $y > 0$) is $x \neq -2$ and the negative domain (where $y < 0$) does not exist.

In terms of the choices given, the correct answer is:

$x > 0 :x\ne-2$

$x < 0 :$ none

To solve this problem, we need to analyze the function $y = \frac{1}{3}x^2 + \frac{1}{2}x + \frac{2}{3}$ to determine where it is positive or negative. This function is quadratic, so it is a parabola. Let us find the domain of positive and negative values.

First, let's determine where the function is zero by finding its roots. This involves using the quadratic formula:

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

Here, we have $a = \frac{1}{3}$, $b = \frac{1}{2}$, and $c = \frac{2}{3}$. The discriminant is calculated as follows:

$b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{1}{4} - \frac{8}{9}$

Calculating gives:

$\frac{1}{4} \approx 0.25$ and $\frac{8}{9} \approx 0.888$ which results in:

$0.25 - 0.888 = -0.638$

Since the discriminant is negative, there are no real roots. As the parabola opens upwards (since $a = \frac{1}{3} > 0$), the function never crosses the x-axis. Therefore, the function remains positive for all real values of $x$ and is never negative.

Thus, the positive domain consists of all real numbers, while there is no negative domain:

$x > 0$: \) for all $x$

$x < 0$: \) none