Standard Representation: With fractions

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 $x$

Therefore, the positive and negative domains of the function are:

x < 0 : 1 < x < 1.5

x > 1.5 or x > 0 : x < 1

To determine the positive and negative domains of $y = 25x^2 + 20x + 4$, we follow these steps:

• Step 1: Calculate the discriminant $\Delta = b^2 - 4ac = 20^2 - 4 \cdot 25 \cdot 4 = 400 - 400 = 0$.
• Step 2: Since the discriminant is zero, the quadratic has one distinct real root, given by $x = \frac{-b}{2a} = \frac{-20}{2 \cdot 25} = -\frac{2}{5}$.
• Step 3: Analyze the sign of $y = 25x^2 + 20x + 4$.

Since the discriminant is zero, the quadratic equation reaches zero only at $x = -\frac{2}{5}$, and is symmetrical around this point. Therefore:

• For $x > 0$, $y$ is positive except where $x = -\frac{2}{5}$.
• For $x < 0$, there are no points where the graph is negative since the parabola opens upwards and does not cross below the x-axis after this root.

Thus, interpreting the domains:

The positive domain for $x > 0$ is all $x$ except $x = -\frac{2}{5}$. For $x < 0$, there is no negative domain because the graph does not descend below the x-axis.

Therefore, the solution to the problem is:

$x > 0 :x\ne\frac{2}{5}$

$x < 0 :$ none

To determine the positive and negative domains of the function $y = -x^2 + 5x + 6$, we first find the roots of the equation by solving:

$-x^2 + 5x + 6 = 0$.

We use the quadratic formula:

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

Here, $a = -1$, $b = 5$, and $c = 6$. Substituting these values, we find:

$x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(6)}}{2(-1)} = \frac{-5 \pm \sqrt{25 + 24}}{-2} = \frac{-5 \pm \sqrt{49}}{-2}$.

$x = \frac{-5 \pm 7}{-2}$.

Solving the two scenarios regarding the $\pm$ gives $x = \frac{2}{-2} = -1$ and $x = \frac{-12}{-2} = 6$.

This means the roots are $x = -1$ and $x = 6$.

We now test the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

- For $x < -1$: pick $x = -2$. Substitute into the function:

$y = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative).

- For $-1 < x < 6$: pick $x = 0$. Substitute:

$y = -(0)^2 + 5(0) + 6 = 6$ (positive).

- For $x > 6$: pick $x = 7$. Substitute:

$y = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative).

Thus, the function is positive in the interval $-1 < x < 6$ and negative in the intervals $x < -1$ and $x > 6$.

Therefore, the solution to the problem is:

x > 0 : -1 < x < 6

x > 6 or x < 0 : x < -1

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 $x > 0 : x < -2$

To determine the positive and negative domains of the quadratic function $y = \frac{1}{2}x^2 + \frac{3}{4}x + \frac{5}{6}$, we start by considering the possibility of real roots using the discriminant.

The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac = \left(\frac{3}{4}\right)^2 - 4\left(\frac{1}{2}\right)\left(\frac{5}{6}\right)$

Calculating gives:

$\Delta = \frac{9}{16} - \frac{20}{12} = \frac{9}{16} - \frac{5}{3} = \frac{9}{16} - \frac{80}{48}$

Convert $\frac{9}{16}$ to a common denominator:

$\Delta = \frac{27}{48} - \frac{80}{48} = -\frac{53}{48}$

The discriminant $\Delta$ is negative, indicating that this quadratic equation has no real roots.

Since the coefficient $a = \frac{1}{2}$ is positive and there are no real roots, the parabola opens upwards and never crosses the x-axis.

This means that the function is always positive for all $x$.

Thus, the positive domain is all $x$, and there is no negative domain.

Therefore, the correct choice is:

$x > 0 :$ for all $x$

$x < 0 :$ none

To determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 + x + 2$, we will follow these general steps:

• Find the roots of the equation $\frac{1}{3}x^2 + x + 2 = 0$.
• Analyze the sign of the quadratic between and beyond these roots.

First, let's identify the roots of the quadratic function:

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = \frac{1}{3}$, $b = 1$, and $c = 2$, the discriminant $\Delta = b^2 - 4ac = 1^2 - 4 \cdot \frac{1}{3} \cdot 2 = 1 - \frac{8}{3} = \frac{-5}{3}$.

Since the discriminant is negative, the quadratic equation $\frac{1}{3}x^2 + x + 2 = 0$ has no real roots.

Given that the coefficient of $x^2$ (i.e., $\frac{1}{3}$) is positive, the parabola opens upwards, meaning the entire function is greater than zero for all real values of $x$.

Therefore, the positive domain is all real numbers, and there is no negative domain.

Therefore, the solution is: $x > 0$: for all $x$; $x < 0$: 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 find the positive and negative domains of the function $y = -3x^2 + 5x - 2$, we follow these steps:

• Step 1: Find the roots of the equation by solving $-3x^2 + 5x - 2 = 0$.
• Step 2: The quadratic equation $ax^2 + bx + c = 0$ has roots $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant, $b^2 - 4ac = 5^2 - 4(-3)(-2) = 25 - 24 = 1$.
• Step 4: Find the roots: $x = \frac{-5 \pm \sqrt{1}}{2(-3)} = \frac{-5 \pm 1}{-6}$.
• Step 5: This gives roots $x = \frac{-4}{-6} = \frac{2}{3}$ and $x = \frac{-6}{-6} = 1$.
• Step 6: Determine intervals: $(-\infty, \frac{2}{3})$, $(\frac{2}{3}, 1)$, $(1, \infty)$.
• Step 7: Test a point from each interval in the original equation to determine sign:
• For $x \in (-\infty, \frac{2}{3})$, choose $x = 0$: $y = -3(0)^2 + 5(0) - 2 = -2$ (negative).
• For $x \in (\frac{2}{3}, 1)$, choose $x = 0.8$: $y = -3(0.8)^2 + 5(0.8) - 2 = -1.28 + 4 - 2 = 0.72$ (positive).
• For $x \in (1, \infty)$, choose $x = 2$: $y = -3(2)^2 + 5(2) - 2 = -12 + 10 - 2 = -4$ (negative).

Therefore, the positive domains of the function are when $\frac{2}{3} < x < 1$, and the negative domains are when $x < \frac{2}{3}$ or $x > 1$.

Thus, the solution to the problem is:

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

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

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 find the positive and negative domains of the function $y = x^2 + 2x + 4.2$, we need to consider the graph of this function and its roots.

First, let's compute the discriminant of the quadratic $y = x^2 + 2x + 4.2$. The discriminant $\Delta$ is given by $b^2 - 4ac$.

Here, $a = 1$, $b = 2$, and $c = 4.2$.

Calculating, we have:

$\Delta = 2^2 - 4 \cdot 1 \cdot 4.2 = 4 - 16.8 = -12.8$.

Since the discriminant is negative, there are no real roots. This means the parabola does not intersect the x-axis.

Next, because $a = 1$ is positive, the parabola opens upwards.

Hence, the entire parabola lies above the x-axis, indicating that the function $y = x^2 + 2x + 4.2$ is positive for all real $x$.

Thus, there is no negative domain for this quadratic since it doesn't dip below the x-axis at any point.

Therefore, the positive and negative domains are:

$x > 0 :$ for all $x$

$x < 0 :$ none

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 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.

To solve for the positive and negative domains of the function $y = 4x^2 - x - 3$, follow these steps:

• Step 1: Identify and apply the quadratic formula to find the roots.
• Step 2: Calculate the discriminant to ensure real roots.
• Step 3: Analyze the sign changes in the resulting intervals.

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

Step 2: Calculate the discriminant: $b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-3) = 1 + 48 = 49$. Since the discriminant is positive, two distinct real roots exist.

Step 3: Calculate the roots using the formula:

$x = \frac{-(-1) \pm \sqrt{49}}{2 \times 4} = \frac{1 \pm 7}{8}$

Thus, the roots $x_1 = \frac{1 + 7}{8} = 1$ and $x_2 = \frac{1 - 7}{8} = -\frac{3}{4}$.

Now we examine the sign of $y = 4x^2 - x - 3$ across the intervals determined by these roots: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

• For $x < -\frac{3}{4}$: Choose $x = -1$. Substitute into the function: $y = 4(-1)^2 - (-1) - 3 = 4 + 1 - 3 = 2$. Positive.
• For $-\frac{3}{4} < x < 1$: Choose $x = 0$. Substitute: $y = 4(0)^2 - (0) - 3 = -3$. Negative.
• For $x > 1$: Choose $x = 2$. Substitute: $y = 4(2)^2 - (2) - 3 = 16 - 2 - 3 = 11$. Positive.

Therefore, the positive domains are $x < -\frac{3}{4}$ and $x > 1$, and the negative domain is $-\frac{3}{4} < x < 1$.

The positive and negative domains are: $x < 0 : -\frac{3}{4} < x < 1$ and $x > 1$ or $x > 0 : x < -\frac{3}{4}$.

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 this problem, let's start by finding the roots of the quadratic equation:

The given function is $y = -4x^2 + x + 3$. We set it to zero to find the roots:

$-4x^2 + x + 3 = 0$

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

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

$x = \frac{-1 \pm \sqrt{1 + 48}}{-8}$

$x = \frac{-1 \pm \sqrt{49}}{-8}$

$x = \frac{-1 \pm 7}{-8}$

The roots are:

• $x_1 = \frac{-1 + 7}{-8} = \frac{6}{-8} = -\frac{3}{4}$
• $x_2 = \frac{-1 - 7}{-8} = \frac{-8}{-8} = 1$

These roots divide the number line into intervals: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

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

Interval $(-\infty, -\frac{3}{4})$: Choose $x = -1$.

• $y = -4(-1)^2 + (-1) + 3 = -4 - 1 + 3 = -2$ (Negative)

Interval $(-\frac{3}{4}, 1)$: Choose $x = 0$.

• $y = -4(0)^2 + 0 + 3 = 3$ (Positive)

Interval $(1, \infty)$: Choose $x = 2$.

• $y = -4(2)^2 + 2 + 3 = -16 + 2 + 3 = -11$ (Negative)

Therefore, the function is positive in the interval $(-\frac{3}{4}, 1)$ and negative in the intervals $(-\infty, -\frac{3}{4})$ and $(1, \infty)$.

Thus, the positive and negative domains of the function are:

$x > 0 : -\frac{3}{4} < x < 1$

$x > 1$ or $x < 0 : x < -\frac{3}{4}$

The correct answer choice corresponds to:

$x > 1$ or $x<0:x<-\frac{3}{4}$

$x > 0 : -\frac{3}{4} < x <1$

To solve this problem, we follow these steps:

• Step 1: Determine if the function intersects the x-axis by finding the roots.
• Step 2: Compute the discriminant of the quadratic equation.
• Step 3: Analyze the parabola's direction and minimum point.

Now, let's work through each step:
Step 1: The given quadratic function is $y = x^2 + \frac{1}{2}x + 5$. The standard quadratic form is $ax^2 + bx + c$ where $a = 1$, $b = \frac{1}{2}$, and $c = 5$.
Step 2: To determine the roots, let's calculate the discriminant, $\Delta = b^2 - 4ac$.
For our function, $\Delta = \left(\frac{1}{2}\right)^2 - 4 \cdot 1 \cdot 5 = \frac{1}{4} - 20 = \frac{1}{4} - 20 = -\frac{79}{4}$.
Since the discriminant is negative, the quadratic has no real roots, indicating that it does not intersect the x-axis. Thus, it does not pass below the x-axis.

Step 3: Since $a = 1$ is positive, the parabola opens upwards. Since there are no real roots, it suggests that the function is always positive.

Therefore, the solution to the problem is that the function is positive for all $x$. There is no $x$ for which the function is negative, since it never crosses the x-axis.

Thus, the solution is:
$x > 0 :$ all $x$
$x < 0 :$ none

This corresponds to choice 4.

To solve this problem, we'll analyze the given quadratic function $y = \frac{1}{2}x^2 + 2x + 3$ and determine where it is positive and where it is negative.

Step 1: Calculate the discriminant to find out if there are real roots.

The quadratic equation has coefficients $a = \frac{1}{2}$, $b = 2$, $c = 3$. The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac = 2^2 - 4(\frac{1}{2})(3) = 4 - 6 = -2$

Since the discriminant is negative ($\Delta < 0$), the quadratic equation does not have real roots; thus, it does not cross the x-axis.

Step 2: Determine the nature of the parabola.

The parabola opens upwards because the leading coefficient $a = \frac{1}{2}$ is positive.

Step 3: Conclude based on the parabola's direction and lack of real roots.

Because the parabola opens upwards and does not intersect the x-axis, the function $y$ is positive for all $x$.

Therefore, the positive domain of the function is for all $x$, and there is no negative domain.

Conclusion:
x > 0 : for all x
x < 0 : none

To solve this problem, we'll determine where the quadratic function $y = -\frac{1}{2}x^2 + x - 1$ is positive and negative.

Step 1: Find the roots of the quadratic equation.
We'll use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -\frac{1}{2}$, $b = 1$, and $c = -1$.

Calculate the discriminant $b^2 - 4ac = 1^2 - 4(-\frac{1}{2})(-1) = 1 - 2 = -1$.
Notice that the discriminant is negative, meaning the quadratic equation has no real roots.

Step 2: Determine the sign of the quadratic function.
Since there are no real roots, the quadratic does not intersect the x-axis. Since $a = -\frac{1}{2}$, which is negative, the parabola opens downwards. Without real roots, it means it is always negative for all values of $x$.

Conclusion: The function is negative for all $x$.

Therefore, the positive and negative domains of the function are:

x < 0 : for all $x$

x > 0 : none

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