Positive and Negative Domains: Using roots

To determine the positive and negative domains of the function, follow these steps:

• Step 1: Solve $(x+4)^2 = \frac{41}{4}$.
• Step 2: This implies $x+4 = \pm \frac{\sqrt{41}}{2}$.
• Step 3: Solving these gives the roots: $x = \frac{-8+\sqrt{41}}{2}$ and $x = \frac{-8-\sqrt{41}}{2}$.
• Step 4: Divide the real number line using these roots into intervals:
• Interval 1: $x < \frac{-8-\sqrt{41}}{2}$
• Interval 2: $\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$
• Interval 3: $x > \frac{-8+\sqrt{41}}{2}$
• Step 5: Test each interval to see where the function is greater or less than zero, using the sign of $(x+4)^2 - \frac{41}{4}$.

Testing reveals that:

• For Interval 1, the function is positive.
• For Interval 2, the function is negative.
• For Interval 3, the function is positive.

Thus, the negative domain is $\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$ and the positive domains are $x > \frac{-8+\sqrt{41}}{2}$ or $x < \frac{-8-\sqrt{41}}{2}$.

Therefore, the correct answer is:

$x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$

$x > \frac{-8+\sqrt{41}}{2}$ or $x > 0 : x < \frac{-8-\sqrt{41}}{2}$

We begin by finding the roots of the function $y = \frac{1}{6}x^2 - 5$ by setting it to zero:

$\frac{1}{6}x^2 - 5 = 0$

Multiply through by 6 to clear the fraction:

$x^2 - 30 = 0$

Solve for $x^2 = 30$, giving:

$x = \pm \sqrt{30}$

Now, we analyze the intervals determined by these roots:

• For $x \in (-\infty, -\sqrt{30})$, since the parabola opens upwards, $y$ is positive.
• For $x \in (-\sqrt{30}, \sqrt{30})$, the parabola dips below the x-axis, so $y$ is negative.
• For $x \in (\sqrt{30}, \infty)$, the parabola returns above the x-axis, making $y$ positive.

Therefore, the function is negative in the interval $-\sqrt{30} < x < \sqrt{30}$, and positive in the interval $x < -\sqrt{30}$ or $x > \sqrt{30}$.

This gives us the positive and negative domains:

$x < 0 : -\sqrt{30} < x < \sqrt{30}$

$x > \sqrt{30}$ or $x > 0 : x < -\sqrt{30}$

To find the roots of the quadratic equation $\frac{1}{2}x^2 - \frac{4}{9} = 0$, follow these steps:

• Set the equation to zero: $\frac{1}{2}x^2 - \frac{4}{9} = 0$.
• Multiply the entire equation by 9 to clear the fraction: $9 \times \frac{1}{2}x^2 - 4 = 0$, simplifying to $\frac{9}{2}x^2 = 4$.
• Multiply through by 2 to solve for $x^2$: $9x^2 = 8$.
• Divide both sides by 9: $x^2 = \frac{8}{9}$.
• Take the square root of both sides: $x = \pm \frac{\sqrt{8}}{3}$, simplifying $\sqrt{8}$ to $2\sqrt{2}$.
• Thus the roots are $x = \frac{2\sqrt{2}}{3}$ and $x = -\frac{2\sqrt{2}}{3}$.

These roots divide the number line into intervals: $x < -\frac{2\sqrt{2}}{3}$, $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, and $x > \frac{2\sqrt{2}}{3}$.

Evaluate the sign of $y$ in each interval:

• When $x < -\frac{2\sqrt{2}}{3}$, choose a test point and check: $x = -1$, then $y = \frac{1}{2} \times (-1)^2 - \frac{4}{9} = \frac{1}{2} - \frac{4}{9} < 0$.
• When $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 0$, then $y = \frac{1}{2} \times 0^2 - \frac{4}{9} = -\frac{4}{9} < 0$.
• When $x > \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 1$, then $y = \frac{1}{2} \times 1^2 - \frac{4}{9} > 0$.

Therefore, $y$ is positive for $x > \frac{2\sqrt{2}}{3}$ and negative for $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, as well as $x < -\frac{2\sqrt{2}}{3}$.

The positive domain for $y$ is $x > \frac{2\sqrt{2}}{3}$, and the negative domain is $x < -\frac{2\sqrt{2}}{3}$ and $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$.

Thus, the correct answer is:

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

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

To solve the problem of finding the positive and negative domains of the function $y = 5x^2 - \frac{9}{16}$, follow these steps:

• Step 1: Set the function equal to zero to find the critical points: $5x^2 - \frac{9}{16} = 0$.
• Step 2: Solve the equation for $x$:
  $5x^2 = \frac{9}{16}$
  Divide both sides by 5:
  $x^2 = \frac{9}{80}$
  Take the square root of both sides:
  $x = \pm \sqrt{\frac{9}{80}}$.
• Step 3: Simplify the expression further:
  $x = \pm \frac{3}{\sqrt{80}} = \pm \frac{3}{4\sqrt{5}} = \pm \frac{3\sqrt{5}}{20}$.
• Step 4: Identify intervals based on the roots where the function could be positive or negative.

The roots are $x = \frac{3\sqrt{5}}{20}$ and $x = -\frac{3\sqrt{5}}{20}$.
The quadratic opens upwards since the coefficient of $x^2$ is positive. Therefore, $y$ will be negative between the roots, i.e.,
For negative domain: $-\frac{3\sqrt{5}}{20} < x < \frac{3\sqrt{5}}{20}$.
For positive domain: $x < -\frac{3\sqrt{5}}{20}$ or $x > \frac{3\sqrt{5}}{20}$.

Verifying against the choices, the correct answer is:

$x < -\frac{3\sqrt{5}}{20} < x < \frac{3\sqrt{5}}{20}$

$x > \frac{3\sqrt{5}}{20}$ or $x > 0 : x < -\frac{3\sqrt{5}}{20}$

This matches choice 3 in the given options.

To solve this problem, we'll start by calculating the roots of the function $y = -\frac{1}{2}x^2 + 5$.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -\frac{1}{2}$, $b = 0$, and $c = 5$.

Substituting these values into the quadratic formula gives:
$x = \frac{-0 \pm \sqrt{0^2 - 4(-\frac{1}{2})(5)}}{2(-\frac{1}{2})}$
$x = \frac{\pm \sqrt{10}}{-1}$
$x = \pm \sqrt{10}$.

The roots are $x = \sqrt{10}$ and $x = -\sqrt{10}$. These points divide the x-axis into three intervals: $(-\infty, -\sqrt{10})$, $(-\sqrt{10}, \sqrt{10})$, and $(\sqrt{10}, \infty)$.

Given that the parabola opens downwards (since $a < 0$), the function is positive outside these roots and negative within them.

• The positive domain is $x < -\sqrt{10}$ or $x > \sqrt{10}$.
• The negative domain is $-\sqrt{10} < x < \sqrt{10}$.

Therefore, the positive domain is $-\sqrt{10} < x < \sqrt{10}$. The negative domain is $x > \sqrt{10}$ or $x < -\sqrt{10}$.

This corresponds to choice 1:

$x > 0 : -\sqrt{10} < x < \sqrt{10}$

$x > \sqrt{10}$ or $x < 0 : x < -\sqrt{10}$

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 $x > \frac{4}{\sqrt{5}}$

To find the positive and negative domains of the function $y = (x-6)^2 - 3$, follow these steps:

• Step 1: Set $y = 0$ to find the roots of the equation. This gives us $(x-6)^2 - 3 = 0$.
• Step 2: Add 3 to both sides to simplify, resulting in $(x-6)^2 = 3$.
• Step 3: Take the square root of both sides. We have two solutions: $x - 6 = \sqrt{3}$ and $x - 6 = -\sqrt{3}$.
• Step 4: Solve for $x$ from each equation:
• For $x - 6 = \sqrt{3}$, $x = 6 + \sqrt{3}$.
• For $x - 6 = -\sqrt{3}$, $x = 6 - \sqrt{3}$.
• Step 5: Determine the intervals:
• Interval 1: $x < 6 - \sqrt{3}$, insert a test point to determine sign.
• Interval 2: $6 - \sqrt{3} < x < 6 + \sqrt{3}$, insert a test point to determine sign.
• Interval 3: $x > 6 + \sqrt{3}$, insert a test point to determine sign.
• Step 6: Based on the intervals:
• For $x < 6 - \sqrt{3}$ and $x > 6 + \sqrt{3}$, $y > 0$.
• For $6 - \sqrt{3} < x < 6 + \sqrt{3}$, $y < 0$.

The positive domains are: $x < 6 - \sqrt{3}$ or $x > 6 + \sqrt{3}$.

The negative domain is: $6 - \sqrt{3} < x < 6 + \sqrt{3}$.

The correct answer to the problem is:

$x > 6+\sqrt{3}$ or $x < 0 : x < 6-\sqrt{3}$

$x < 0 : 6-\sqrt{3} < x < 6+\sqrt{3}$

To find the positive and negative domains of the function $y = -\left(x - 2.5\right)^2 + 0.5$, we analyze when $y$ is greater than and less than zero.

• Step 1: Solve for the positive domain ($y > 0$).

We need to solve the inequality:
$-\left(x - 2.5\right)^2 + 0.5 > 0$.

Rearrange this to:
$-\left(x - 2.5\right)^2 > -0.5$.

Remove the negative sign by multiplying by $-1$ (which flips the inequality sign):
$\left(x - 2.5\right)^2 < 0.5$.

Taking the square root of both sides gives:
$|x - 2.5| < \sqrt{0.5}$.
This implies:
$-\sqrt{0.5} < x - 2.5 < \sqrt{0.5}$.

Solve for $x$:
$2.5 - \sqrt{0.5} < x < 2.5 + \sqrt{0.5}$.

Step 2: Solve for the negative domain ($y < 0$).

From the inequality:
$-\left(x - 2.5\right)^2 + 0.5 < 0$.

Rearrange to:
$-\left(x - 2.5\right)^2 < -0.5$.

Again, multiply by $-1$:
$\left(x - 2.5\right)^2 > 0.5$.

Taking the square root gives:
$|x - 2.5| > \sqrt{0.5}$.
This implies:
$x - 2.5 < -\sqrt{0.5}$ or $x - 2.5 > \sqrt{0.5}$.

Solving gives:
$x < 2.5 - \sqrt{0.5}$ or $x > 2.5 + \sqrt{0.5}$.

Recall $\sqrt{0.5} = \frac{\sqrt{2}}{2}$, so:
The positive domain is: $2.5 - \frac{\sqrt{2}}{2} < x < 2.5 + \frac{\sqrt{2}}{2}$.
The negative domain is: $x < 2.5 - \frac{\sqrt{2}}{2}$ or $x > 2.5 + \frac{\sqrt{2}}{2}$.

Therefore, the correct answer based on the choices provided is:

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

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

To find the positive and negative domains of the function $y = (x-1)^2 - 2$, we need to determine the points where the function intersects the x-axis, as these will mark changes in sign.

Step 1: Set the function equal to zero to find the roots.
$(x-1)^2 - 2 = 0$

Step 2: Move -2 to the other side and solve:
$(x-1)^2 = 2$

Step 3: Solve for $x$ by taking the square root of both sides:
$x - 1 = \pm \sqrt{2}$

Step 4: Solve for $x$ by isolating it:
$x = 1 \pm \sqrt{2}$

The roots are $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$. These roots divide the x-axis into three parts.

Step 5: Evaluate the function behavior in each interval defined by these roots.

• For $x < 1 - \sqrt{2}$, pick a point such as nearly approaching zero value and test the sign.
• For $1 - \sqrt{2} < x < 1 + \sqrt{2}$, pick a midpoint value and test.
• For $x > 1 + \sqrt{2}$, pick a value greater than root for testing function positivity.

Step 6: Determine where the function is positive and negative:

• Within the interval $[1-\sqrt{2}, 1+\sqrt{2}]$, the function lies below the x-axis and is negative.
• Outside this interval, specifically $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$, the function lies above the x-axis and is positive.

The positive domain is $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$ and the negative domain is $1 - \sqrt{2} < x < 1 + \sqrt{2}$.

Therefore, the solution is:

$x < 0 : 1-\sqrt{2} < x < 1+\sqrt{2}$

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

To solve the problem of finding the positive and negative domains of the function $y = \frac{1}{2}x^2 - 1$, we will follow these steps:

• Step 1: Find the roots of the quadratic equation $\frac{1}{2}x^2 - 1 = 0$ using the quadratic formula.
• Step 2: Determine intervals based on the roots and examine the sign of the function within each interval.
• Step 3: Identify where the function takes positive and negative values.

Step 1: The equation $\frac{1}{2}x^2 - 1 = 0$ can be rewritten as $x^2 = 2$. Solving for $x$ gives $x = \pm \sqrt{2}$.

Step 2: The roots $x = -\sqrt{2}$ and $x = \sqrt{2}$ divide the number line into three intervals:
a) $x < -\sqrt{2}$
b) $-\sqrt{2} < x < \sqrt{2}$
c) $x > \sqrt{2}$

Step 3: Analyze the sign of the function in each interval:

• Interval $x < -\sqrt{2}$: Pick $x = -2$ (any point in the interval). The function $y = \frac{1}{2} x^2 - 1$ becomes $y = 2 - 1 = 1$, which is positive.
• Interval $-\sqrt{2} < x < \sqrt{2}$: Pick $x = 0$. Then $y = \frac{1}{2} \times 0 - 1 = -1$, which is negative.
• Interval $x > \sqrt{2}$: Pick $x = 2$. The function $y = \frac{1}{2} \times 4 - 1 = 1$, which is positive.

Therefore, the positive domain of the function is $x < 0 : x < -\sqrt{2}$ and $x > \sqrt{2}$. The negative domain is $x < 0 : -\sqrt{2} < x < \sqrt{2}$.

To solve this problem, we first determine where the quadratic function $y = -3x^2 + 13$ is equal to zero. Setting $-3x^2 + 13 = 0$ yields:

• Rearrange the equation: $-3x^2 = -13$, leading to $x^2 = \frac{13}{3}$.

• Solve for $x$: $x = \pm \sqrt{\frac{13}{3}}$.

The roots of the equation are $x = \sqrt{\frac{13}{3}}$ and $x = -\sqrt{\frac{13}{3}}$. These roots divide the number line into three intervals: $x < -\sqrt{\frac{13}{3}}$, $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and $x > \sqrt{\frac{13}{3}}$.

Determine the sign of the function in each interval:

• For $x < -\sqrt{\frac{13}{3}}$, select a test point (e.g., $x = -3$), the function value is negative because $-3(-3^2) + 13 = -27 + 13 = -14$.

• For $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 0$), the function value is positive because $-3(0)^2 + 13 = 13$.

• For $x > \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 3$), the function value is negative because $-3(3^2) + 13 = -27 + 13 = -14$.

Therefore, the positive domain of the function is $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and the negative domain is $x < -\sqrt{\frac{13}{3}}$ or $x > \sqrt{\frac{13}{3}}$.

The answer matches choice 3:

$x > \sqrt{\frac{13}{3}}$ or $x < 0 : x < -\sqrt{\frac{13}{3}}$

$x > 0 : -\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$

To solve the problem, we will determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 - 2$ by following these steps:

• Step 1: Find the roots of the quadratic function to identify the intervals.

• Step 2: Analyze the sign of the function in each interval determined by the roots.

Step 1: Finding the Roots
To find the roots, set $y = 0$ in the equation:
$\frac{1}{3}x^2 - 2 = 0$.
Multiply through by 3 to eliminate the fraction:
$x^2 - 6 = 0$.
Rearranging gives:
$x^2 = 6$.
The solutions to this equation are the roots $x = \pm \sqrt{6}$.

Step 2: Analyze Sign in Each Interval
The roots $x = -\sqrt{6}$ and $x = \sqrt{6}$ split the real number line into three intervals: $(-\infty, -\sqrt{6})$, $(-\sqrt{6}, \sqrt{6})$, $(\sqrt{6}, \infty)$.
For each interval, choose a test point and determine the sign of $y$.

- Interval $(-\infty, -\sqrt{6})$: Choose $x = -3$ (since $-3 \lt -\sqrt{6} \approx -2.45$).
$y = \frac{1}{3}(-3)^2 - 2 = \frac{9}{3} - 2 = 3 - 2 = 1 > 0$.
Thus, $y > 0$ in the interval $(-\infty, -\sqrt{6})$.

- Interval $(-\sqrt{6}, \sqrt{6})$: Choose $x = 0$.
$y = \frac{1}{3}(0)^2 - 2 = -2 < 0$.
Thus, $y < 0$ in the interval $(-\sqrt{6}, \sqrt{6})$.

- Interval $(\sqrt{6}, \infty)$: Choose $x = 3$ (since $3 \gt \sqrt{6} \approx 2.45$).
$y = \frac{1}{3}(3)^2 - 2 = \frac{9}{3} - 2 = 3 - 2 = 1 > 0$.
Thus, $y > 0$ in the interval $(\sqrt{6}, \infty)$.

Therefore, the positive domain is given by the intervals $x < -\sqrt{6}$ and $x > \sqrt{6}$, while the negative domain is $-\sqrt{6} < x < \sqrt{6}$.

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

- Positive: $x < -\sqrt{6}$ or $x > \sqrt{6}$

- Negative: $-\sqrt{6} < x < \sqrt{6}$

These match the correct answer choice as follows:

$x < 0 : -\sqrt{6} < x < \sqrt{6}$

$x > \sqrt{6}$ or $x> 0 : x <-\sqrt{6}$

To find the positive and negative domains of the function $y = (x+3)^2 - 5$, we first identify the roots by setting $y = 0$ and solving for $x$.

Let's solve $(x+3)^2 - 5 = 0$:

1. Start with the equation: $(x+3)^2 - 5 = 0$
2. Add 5 to both sides: $(x+3)^2 = 5$
3. Take the square root of both sides: $x+3 = \pm \sqrt{5}$
4. Solve for $x$:
• $x = -3 + \sqrt{5}$
• $x = -3 - \sqrt{5}$

Thus, the roots of the function are $x = -3 + \sqrt{5}$ and $x = -3 - \sqrt{5}$.

Since the parabola opens upwards (the coefficient of $(x+3)^2$ is positive), the function $y$ is:

• Negative between the roots: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive outside these roots: $x < -3 - \sqrt{5}$ and $x > -3 + \sqrt{5}$

Therefore, the positive and negative domains are:

• Negative domain: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive domain: $x < -3 - \sqrt{5}$ or $x > -3 + \sqrt{5}$

Upon reviewing the multiple choice options, the correct answer that corresponds to this solution is:

$x < 0 : -3-\sqrt{5} < x < -3+\sqrt{5}$

$x>-3+\sqrt{5}$ or $x > 0 : x < -3-\sqrt{5}$

To find the positive and negative domains of the quadratic function $y = -6x^2 + 27$, we begin by finding the roots of the equation where $y = 0$.

Step 1: Set the function equal to zero: $-6x^2 + 27 = 0$.

Step 2: Solve for $x^2$: $-6x^2 + 27 = 0 \quad \Rightarrow \quad 6x^2 = 27 \quad \Rightarrow \quad x^2 = \frac{27}{6} \quad \Rightarrow \quad x^2 = \frac{9}{2}$

Step 3: Solve for $x$ by taking the square root: $x = \pm \sqrt{\frac{9}{2}} \quad \Rightarrow \quad x = \pm \frac{3}{\sqrt{2}}$

Step 4: These roots, $x = \frac{3}{\sqrt{2}}$ and $x = -\frac{3}{\sqrt{2}}$, divide the real number line into three intervals: $x < -\frac{3}{\sqrt{2}}$, $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, and $x > \frac{3}{\sqrt{2}}$.

Step 5: Test the sign of the function in each interval:

• For $x < -\frac{3}{\sqrt{2}}$, pick a test point like $x = -2$: $y = -6(-2)^2 + 27 = -24 + 27 = 3$ (positive)
• For $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, pick $x = 0$: $y = -6(0)^2 + 27 = 27$ (positive)
• For $x > \frac{3}{\sqrt{2}}$, pick $x = 2$: $y = -6(2)^2 + 27 = -24 + 27 = 3$ (positive)

The function is negative nowhere as the parabola opens downward (due to negative coefficient of $x^2$), it achieves maximum $y$ at its vertex, and beyond the roots, remains positive.

Therefore, the positive domain, where $y > 0$ is for $x < -\frac{3}{\sqrt{2}}$ and $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$.

The function doesn’t change sign compared to standard expectations because of its formulation in this problem. The negative domain is non-existent.

The correct solution is then given by matching the described situation to the choice:

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

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

To solve this problem, we need to determine when $y = (x+10)^2 - 3$ is greater than and less than zero.

Start by finding the roots of the equation:

Set $y = 0$:

$(x+10)^2 - 3 = 0$

Rearrange the equation to find:

$(x+10)^2 = 3$

Take the square root of both sides:

$x + 10 = \pm\sqrt{3}$

Solving these gives:

• $x = -10 + \sqrt{3}$
• $x = -10 - \sqrt{3}$

These roots divide the number line into three intervals:

• $(-\infty, -10 - \sqrt{3})$
• $(-10 - \sqrt{3}, -10 + \sqrt{3})$
• $(-10 + \sqrt{3}, \infty)$