Positive and Negative Domains: Using roots

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 determine where the function $y = (x + 5)^2 - 6$ is positive and negative, we start by solving the equation:

$(x + 5)^2 - 6 = 0$

Adding 6 to both sides gives:

$(x + 5)^2 = 6$

Taking the square root of both sides, we obtain two solutions:

$x + 5 = \sqrt{6}$ or $x + 5 = -\sqrt{6}$

Solving these, we get:

$x = -5 + \sqrt{6}$ and $x = -5 - \sqrt{6}$

These roots divide the number line into three intervals: $(- \infty, -5 - \sqrt{6})$, $(-5 - \sqrt{6}, -5 + \sqrt{6})$, and $(-5 + \sqrt{6}, \infty)$.

Next, we determine the sign of the function in each interval:

• For $x < -5 - \sqrt{6}$, choose $x = -10$:

$(x + 5)^2 - 6 = ((-10) + 5)^2 - 6 = (-5)^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

• For $-5 - \sqrt{6} < x < -5 + \sqrt{6}$, choose $x = -5$:

$(x + 5)^2 - 6 = ((-5) + 5)^2 - 6 = 0^2 - 6 = -6 < 0$. Therefore, the function is negative.

• For $x > -5 + \sqrt{6}$, choose $x = 0$:

$(x + 5)^2 - 6 = (0 + 5)^2 - 6 = 5^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

Thus, the function is positive on the intervals $x < -5 - \sqrt{6}$ and $x > -5 + \sqrt{6}$, and negative on the interval $-5 - \sqrt{6} < x < -5 + \sqrt{6}$.

Therefore, the positive domain is $x > -5+\sqrt{6}$ or $x > 0 : x < -5-\sqrt{6}$, and the negative domain is $x < 0 : -5-\sqrt{6} < x < -5+\sqrt{6}$.

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:

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

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

For $x < -10 - \sqrt{3}$: Choose $x = -11$

Then: $y = ((-11 + 10)^2 - 3) = 1 - 3 = -2$

So, $y < 0$ in the interval $(-\infty, -10 - \sqrt{3})$.

For $-10 - \sqrt{3} < x < -10 + \sqrt{3}$: Choose $x = -10$

Then: $y = ((-10 + 10)^2 - 3) = 0 - 3 = -3$

So, $y < 0$ in the interval $(-10 - \sqrt{3}, -10 + \sqrt{3})$.

For $x > -10 + \sqrt{3}$: Choose $x = 0$

Then: $y = ((0 + 10)^2 - 3) = 100 - 3 = 97$

So, $y > 0$ in the interval $(-10 + \sqrt{3}, \infty)$.

Therefore, the positive domain is $x > -10 + \sqrt{3}$ while the negative domain is $x < -10 + \sqrt{3}$.

Using the analysis above and applying it to the choices, the correct response is:

$x < 0 : -10-\sqrt{3} < x < -10+\sqrt{3}$

$x > -10+\sqrt{3}$ or $x > 0 : x < -10-\sqrt{3}$

To solve this problem, follow these steps:

• Step 1: Find the roots of the function. Set $y = 0 = -\left(x-12\right)^2 + 2$.

• Step 2: Rearrange and solve for $x$: $\begin{aligned} -\left(x-12\right)^2 + 2 &= 0 \\ \left(x-12\right)^2 &= 2 \end{aligned}$ Solving gives $x - 12 = \pm \sqrt{2}$, resulting in roots $x = 12 + \sqrt{2}$ and $x = 12 - \sqrt{2}$.

• Step 3: Determine the intervals: $\begin{aligned} x < 12 - \sqrt{2} \\ 12 - \sqrt{2} < x < 12 + \sqrt{2} \\ x > 12 + \sqrt{2} \end{aligned}$Step 4: Test each interval to check the sign of $y$: \begin{itemize}

• For $x < 12 - \sqrt{2}$ and $x > 12 + \sqrt{2}$, $(x-12)^2$ becomes larger than 2, so $y= -[(x-12)^2] + 2$ is negative.

• For $12 - \sqrt{2} < x < 12 + \sqrt{2}$, $(x-12)^2$ is less than 2, so $y = -[(x-12)^2] + 2$ is positive.

Thus, the function is negative for $x > 12 + \sqrt{2}$ or $x < 12 - \sqrt{2}$, and positive for $12 - \sqrt{2} < x < 12 + \sqrt{2}$.

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

$x > 12+\sqrt{2}$ or $x < 0 : x < 12-\sqrt{2}$

$x > 0 : 12-\sqrt{2} < x < 12+\sqrt{2}$

The function given is $y = -\left(x+4\right)^2 + 6$. This is in vertex form $y = a(x-h)^2 + k$ with vertex at $(-4, 6)$.

Step 1: To find the x-values for which the function is positive or negative, set $y = 0$:

$-\left(x+4\right)^2 + 6 = 0$

$\left(x+4\right)^2 = 6$

Step 2: Solve for $x$:

Take the square root of both sides:

$x + 4 = \pm \sqrt{6}$ i.e., $x = -4 \pm \sqrt{6}$

Step 3: Find where the function is positive or negative. The parabola opens downward, so the intervals are:

• Negative domain: $x < -4 - \sqrt{6}$ and $x > -4 + \sqrt{6}$; outside this interval.
• Positive domain: $-4 - \sqrt{6} < x < -4 + \sqrt{6}$; within this interval.

Conclusively:

$x > 0 : -4-\sqrt{26} < x < -4+\sqrt{26}$

$x > -4+\sqrt{26}$ or $x < 0 : x < -4-\sqrt{26}$

Therefore, the solution to this problem is as follows:

For $x > 0$: $-4-\sqrt{26} < x < -4+\sqrt{26}$

For $x < 0$: $x < -4-\sqrt{26}$ and $( x > -4 + \sqrt{26})$

To solve this problem, we follow these essential steps:

• Step 1: Set $y = 0$ to find the roots of the equation.
• Step 2: Analyze the expression $-\left(x+7\right)^2 + 12 = 0$.
• Step 3: Isolate $\left(x+7\right)^2$ by adding $-12$ to both sides, so: $\left(x+7\right)^2 = 12$.
• Step 4: Solve $\left(x+7\right)^2 = 12$ taking the square root of both sides, resulting in $x + 7 = \pm\sqrt{12}$.
• Step 5: Simplify and solve for $x$ to find: $x = -7 \pm 2\sqrt{3}$.

Step 6: Now, determine the positive and negative domains:

• The roots are $x = -7 + 2\sqrt{3}$ and $x = -7 - 2\sqrt{3}$.
• Since the parabola opens downward, the function is positive between the roots $-7 - 2\sqrt{3}$ and $-7 + 2\sqrt{3}$, where $x \gt 0$.
• The function is negative for $x \gt -7 + 2\sqrt{3}$ and $x \lt -7 - 2\sqrt{3}$.

Therefore, the solution to the problem is:
Positive domain: $-7 - 2\sqrt{3} < x < -7 + 2\sqrt{3}$
Negative domain: $x > -7 + 2\sqrt{3}$ or $x < -7 - 2\sqrt{3}$.

As outlined between the choice options, the correct answer is represented under choice 4:

$x > -7+2\sqrt{3}$ or $x < 0 : x < -7-2\sqrt{3}$

$x > 0 : -7-2\sqrt{3} < x < -7+2\sqrt{3}$

To solve this problem, we start by identifying where the given quadratic function is positive and where it is negative.

• Step 1: Convert the vertex form to solve for $y = 0$. The equation is $-\left(x + 10\right)^2 + 2 = 0$.
• Step 2: Rearrange the equation to find roots:
  $(x+10)^2 = 2$.
  Taking the square root, we have $x + 10 = \pm \sqrt{2}$.
• Step 3: Solve for $x$:
  $x = -10 + \sqrt{2}$ and $x = -10 - \sqrt{2}$.
• Step 4: Analyze the intervals:

The roots divide the number line into intervals. We check these intervals for $y > 0$ and $y < 0$.

• For $y > 0$: The function is a downward-opening parabola, hence positive between its roots: $-10-\sqrt{2} < x < -10+\sqrt{2}$.
• For $y < 0$: The function is negative outside the interval where it is positive, giving $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Therefore, for the positive domain $x > 0$, we have the interval $-10-\sqrt{2} < x < -10+\sqrt{2}$. For the negative domain, it is when $x < 0$ such that $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Thus, the correct solution choice is:

$x > 0 : -10-\sqrt{2} < x < -10+\sqrt{2}$

$x > -10+\sqrt{2}$ or $x < 0 : x < -10-\sqrt{2}$

To find the positive and negative domains of the function $y = -\left(x-14\right)^2 + 8$, we'll start by identifying the roots of the quadratic equation.

Step 1: Find the roots of the equation:
To find when the function is zero, set $y = 0$:
$-\left(x-14\right)^2 + 8 = 0$.

Step 2: Solve for $x$:
Rearrange the equation:
$-\left(x-14\right)^2 = -8$
$(x-14)^2 = 8$.

Take the square root on both sides:
$x-14 = \pm\sqrt{8}$.
This simplifies to $x - 14 = \pm 2\sqrt{2}$.

Add 14 to both sides to solve for $x$:
$x = 14 \pm 2\sqrt{2}$.
So, the roots are $x = 14 + 2\sqrt{2}$ and $x = 14 - 2\sqrt{2}$.

Step 3: Analyze intervals between roots and outside:
The roots divide the $x$-axis into three intervals: $x < 14 - 2\sqrt{2}$, $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, and $x > 14 + 2\sqrt{2}$.

- For $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, $y > 0$ because points between roots are above the $x$-axis.
- For $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$, $y < 0$ because points outside of roots are below the $x$-axis.

Conclusion:
The positive domain, where $y > 0$, is $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$.
The negative domain, where $y < 0$, is $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$.

Therefore, the solution is:
Positive domain: $x > 0 : 14-2\sqrt{2} < x < 14+2\sqrt{2}$.
Negative domain: $x > 14+2\sqrt{2}$ or $x < 0 : x < 14-2\sqrt{2}$.

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

The function given is $y = 4x^2 - \frac{49}{100}$, and we need to analyze where it is positive and negative.

First, let's find the roots by setting the function equal to zero:

$4x^2 - \frac{49}{100} = 0$

Solve for $x$:

$4x^2 = \frac{49}{100}$

$x^2 = \frac{49}{400}$

$x = \pm \frac{7}{20}$

We have roots at $x = \frac{7}{20}$ and $x = -\frac{7}{20}$. These roots divide the real line into three intervals: $(-\infty, -\frac{7}{20})$, $(-\frac{7}{20}, \frac{7}{20})$, and $(\frac{7}{20}, \infty)$.

Since the coefficient of $x^2$ is positive (4), the parabola opens upwards, meaning the function is positive outside the interval between the roots and negative within it. Thus:

The function $y = 4x^2 - \frac{49}{100}$ is negative in the interval $-\frac{7}{20} < x < \frac{7}{20}$ and positive in the intervals $x < -\frac{7}{20}$ and $x > \frac{7}{20}$. Therefore:

The positive and negative domains are:

• Positive: $x < -\frac{7}{20}$ or $x > \frac{7}{20}$
• Negative: $-\frac{7}{20} < x < \frac{7}{20}$

Thus, the correct multiple-choice answer is:

$x < 0 : -\frac{7}{20} < x < \frac{7}{20}$

$x > \frac{7}{20}$ or $x > 0 : x < -\frac{7}{20}$

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 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 $x < -\frac{2\sqrt{2}}{3}$.