Vertex Representation: Identify the positive and negative domain

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

To find the positive and negative domains of $y=\left(x-8\frac{1}{6}\right)^2-1$, we perform the following steps:

• Step 1: Identify when $y = 0$. Set $\left(x-8\frac{1}{6}\right)^2-1 = 0$, solving gives $\left(x-8\frac{1}{6}\right)^2 = 1$.
• Step 2: Solve for $x$ to get $\left(x-8.1667\right)^2 = 1$. The equation gives $x-8.1667 = \pm 1$ leading to two solutions: $x = 9.1667$ and $x = 7.1667$.
• Step 3: Determine the sign of $y$ in intervals $(-\infty, 7.1667)$, $(7.1667, 9.1667)$, and $(9.1667, \infty)$.
• Step 4: Over the interval $(7.1667, 9.1667)$, $y \lt 0$ because the shifted-square is less than 1, making $(x-8.1667)^2-1 \lt 0$.
• Step 5: Outside this interval, specifically $(-\infty, 7.1667)$ and $(9.1667, \infty)$, $y \gt 0$.

The positive domain is $x \in (-\infty, 7\frac{1}{6}) \cup (9\frac{1}{6}, \infty)$ and the negative domain is $x \in (7\frac{1}{6}, 9\frac{1}{6})$.

Therefore, the solution to the problem is:

$x < 0 : x <7\frac{1}{6} < x < 9\frac{1}{6}$

$x > 9\frac{1}{6}$ or $x > 0 : x <7\frac{1}{6}$

To solve this problem, we must analyze the quadratic function $y = -\left(x - 2\frac{6}{19}\right)^2 - 2$ to determine its positive and negative domains.

• Step 1: Identify the vertex and direction
  The given function is in the form $y = a(x - h)^2 + k$, where $a = -1$, $h = 2\frac{6}{19}$, and $k = -2$. The vertex of the parabola is at $(2\frac{6}{19}, -2)$.
• Step 2: Analyze the direction of the parabola
  Since $a = -1$ (negative), the parabola opens downward. This indicates the vertex is at the maximum point of the parabola.
• Step 3: Determine the function's values
  Since the maximum value of the function (at the vertex) is $y = -2$, and the parabola opens downward, the function cannot be positive anywhere. It is always less than or equal to $-2$, so it's negative for all $x$.
• Step 4: Establish the positive and negative domains
  Since the function is always negative, there are no positive domains. Therefore, the negative domain for the function is all real numbers x \>.

Therefore, the positive and negative domains are:

\( x > 0 : none

$x < 0 :$ all $x$

To solve the problem, we follow these steps:

• Step 1: Recognize that the function $y = -\left(x + \frac{8}{9}\right)^2$ is a quadratic function in vertex form where the leading coefficient $a = -1$.
• Step 2: Determine the direction of the parabola. Since the coefficient $a$ is negative, the parabola opens downwards.
• Step 3: Analyze the function's vertex: The vertex is at $x = -\frac{8}{9}$. At this point, $y = 0$, which is the maximum value.
• Step 4: Consider the behavior of the function on either side of the vertex. Because of the downward opening, $y$ is negative for all $x$ except at the vertex itself.
• Step 5: Conclude the results: The function is negative for all $x$, except $y = 0$ at $x = -\frac{8}{9}$.

Thus, the positive and negative domains are:

$x < 0 : x \ne -\frac{8}{9}$ (negative domain)

$x > 0 :$ none (positive domain)

The correct answer choice is:

$x < 0 : x \ne -\frac{8}{9}$

$x > 0 :$ none

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

• Step 1: Identify the expression squared in the given function.
• Step 2: Determine when the result of this squared expression is zero or greater than zero.
• Step 3: Identify any restriction on $x$ that causes $y$ not to be positive.

Now, let's work through each step:
Step 1: We have $y = \left(x - 3\frac{1}{11}\right)^2$. The expression inside the square is zero when $x = 3\frac{1}{11}$.
Step 2: Since any real number squared is non-negative, $\left(x - 3\frac{1}{11}\right)^2 \geq 0$.
Step 3: The value of $y$ is equal to zero only when $x = 3\frac{1}{11}$; for all other $x$, $y > 0$.

Therefore, the negative domain, where $y < 0$, does not exist in this function, because $y$ can never be negative.
For positive domain, the function is positive for any $x \neq 3\frac{1}{11}$, which includes all $x > 0$ except for $x = 3\frac{1}{11}$.

Conclusively, the positive and negative domains are:

$x < 0 :$ none

$x > 0 : x\ne3\frac{1}{11}$

To find the positive and negative domains of the function, let's first analyze the given function $y = -\left(x + \sqrt{13}\right)^2$.

• Step 1: Analyze the function.
  The function is a downward-opening parabola with vertex at $(-\sqrt{13}, 0)$, because the coefficient of the quadratic term is negative.

• Step 2: Determine the intervals for positive and negative domains.
  A parabola that opens downward from its vertex means the function is negative for all $x \neq -\sqrt{13}$ since there are no values of x that make the function greater than 0 because the vertex is the maximum point.

• Step 3: Consider the function around the vertex.
  The only point where the function equals zero is at the vertex $x = -\sqrt{13}$. Thus, for any other $x$, the function is $y < 0$.

Conclusion:
For $x < 0$, the function satisfies the negative characteristic for all domains except $x = -\sqrt{13}$ where $y = 0$. Thus, the negative domain is $x < 0 : x \neq -\sqrt{13}$.

There are no values of $x$ for which the function becomes positive. Therefore, the positive domain is empty for $x > 0$.

The solution concludes that the positive domain is none, and the negative domain is $x < 0 : x \neq -\sqrt{13}$.

Let's analyze the problem by rewriting the function in its vertex form:

The given function is $y = -\left(x - 11\right)^2$.

Step 1: Identify the vertex and parabola direction.

• The vertex is $(11, 0)$ meaning at $x = 11$, the value of $y$ is zero.
• Since the parabola opens downwards (as the coefficient of $(x - 11)^2$ is negative), the output of the function will always be negative except at the vertex where it is zero.

Step 2: Determine the positive and negative domains of the function.

• The entire real number line minus the vertex point is where the function value is negative.
• The function never achieves positive values, so the positive domain is essentially non-existent. Therefore, all $x \neq 11$ fall into the negative domain.

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

$x < 0 : x\ne11$

$x > 0 :$ none

Hence, the solution is, the function is negative for all values except at $x = 11$, where it is precisely zero.

The correct choice according to our analysis is Choice 2:

$x < 0 : x\ne11$

$x > 0 :$ none

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 solve this problem, we will explore the behavior of the quadratic function $y = (x-6)^2 + 8$.

The function is in vertex form, $y = (x-h)^2 + k$, where the vertex of the parabola is at $(h, k) = (6, 8)$. The parabola opens upwards because the squared term, $(x-6)^2$, has a positive coefficient (which is 1).

Given this upward-opening parabola, the minimum value of $y$ is $8$, which occurs when $x = 6$. As a result, the quadratic expression $y = (x-6)^2 + 8$ will always yield non-negative values, actually, specifically, it will always yield positive values $y \geq 8$ across its entire domain of real numbers. Therefore, there are no negative values for $y$ in the range of this function, as the minimum bound itself is positive.

Thus, the analysis tells us:

• Negative domain $x < 0$: None, meaning no part of the range $y$ becomes negative.
• Positive domain $x > 0$: All $x$, indicating that $y$ is always positive or zero, overriding negative conditions.

Therefore, the solution for the domains is:

$x < 0$: none

$x > 0$: all $x$

To determine the positive and negative domains of the function $y = \left(x + 4\frac{5}{7}\right)^2$, we need to analyze the behavior of the expression inside the square.

The expression $\left(x + 4\frac{5}{7}\right)^2$ represents a perfect square. A perfect square is always greater than or equal to zero.

Consider the following observations:

• The expression is zero when $x = -4\frac{5}{7}$. At this point, $y = 0$.
• For any other value of $x$, $x \neq -4\frac{5}{7}$, the expression $\left(x + 4\frac{5}{7}\right)^2$ is positive.

Given that $\left(x + 4\frac{5}{7}\right)^2$ cannot be negative for any real $x$, the function has no negative domain.

The positive domain of $y$ is all $x$ except when $x = -4\frac{5}{7}$. Hence, the positive domain is the set where $x \neq -4\frac{5}{7}$.

In conclusion, the negative domain is none, and the positive domain is $x \neq -4\frac{5}{7}$.

Therefore, the correct choice is:

$x < 0 :$ none

$x > 0 : x\ne-4\frac{5}{7}$

The given function is $y = -\left(x - \frac{1}{3}\right)^2$. This function is in the vertex form of a quadratic equation, where the vertex is at $x = \frac{1}{3}$. The presence of the negative sign in front of the squared term indicates that the parabola opens downward.

Let's analyze the function domain in terms of where it is positive or negative:

• Since the parabola opens downward, the maximum value at the vertex $x = \frac{1}{3}$ is zero.
• The function value is negative for every $x \neq \frac{1}{3}$.
• For $x < 0$, the function will be negative because the shape of the parabola ensures negativity on either side of the vertex.
• Similarly, for $x > 0$, the function will also be negative.

Thus, for the domain where $f(x) < 0$, we have $x < 0 : x \neq \frac{1}{3}$ (though technically beyond zero these negatives are in the left half, described succinctly under real numbers as $-\infty < x < \frac{1}{3}$), where no exceptions apply for $x > 0$ in theoretical range as the parabola negatively surpasses all specified x-real, non-zero magnitude cuts off considered here by environment instruction formats.

Considering these factors, the function is never positive for any $x$, conforming exactly to the negative streak overlook starting from any negative below boundary tunneled vertex center exclusion applicability.

The correct choice based on the given options is:

• Positive Domain: None
• Negative Domain: $x < 0 : x \neq \frac{1}{3}$

Therefore, the solution is best depicted by choice 3:

$x < 0 : x\ne\frac{1}{3}$

$x > 0 :$ none

Hence, the correct answer matching the function characteristics is:

$x < 0 : x\ne\frac{1}{3}$

$x > 0 :$ none

The given quadratic function is $y = -\left(x-12\right)^2 - 4$. This function is in vertex form $y = a(x-h)^2 + k$, with $a = -1$, $h = 12$, and $k = -4$. Because $a < 0$, the parabola opens downwards.

To find when $y \geq 0$ (positive domain) and $y \leq 0$ (negative domain), we start by identifying where the function is zero, the x-intercepts. Set $y = 0$:

$-\left(x-12\right)^2 - 4 = 0$

Solving for $x$, isolate the squared term:

$-\left(x-12\right)^2 = 4$

$(x-12)^2 = -4$

No real roots exist because $(x-12)^2$ cannot equal a negative number. Thus, the parabola does not intersect the x-axis, meaning it is entirely below it.

Therefore, the function is negative for all $x$. There are no positive values for $y$.

The positive domain $x > 0$ has no points since the graph is always negative; the negative domain is the entire set of real numbers.

Thus, the correct positive and negative domains are:

$x < 0 :$ none

$x > 0 :$ all $x$

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 $x > 2\frac{1}{2} + \frac{\sqrt{2}}{2}$

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

The given quadratic function is $y = -\left(x - 3\frac{1}{12}\right)^2 - \frac{1}{7}$. We start by understanding that the shape of this parabola will open downwards due to the negative sign in front of the square term. To find the roots or x-intercepts, set $y = 0$.

Rewriting the expression for clarity, we have:

$0 = -\left(x - \frac{37}{12}\right)^2 - \frac{1}{7}$

We can solve this by isolating the squared term:

$\left(x - \frac{37}{12}\right)^2 = -\frac{1}{7}$

Since a squared term cannot be negative, it illustrates that there are no real roots. This means the parabola does not cross the x-axis and remains entirely below it, due to the downward opening.

Therefore, the function is negative ($y < 0$) for all x-values. The positive domain is non-existent.

The solution tells us:

• $x < 0 :$ all $x$
• $x > 0 :$ none

In terms of given choices: the correct choice is 3.

The solution to the problem is that the positive and negative domains are:

$x < 0 :$ all $x$

$x > 0 :$ none

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 this problem, let's consider the function $y = -\left(x - \frac{9}{5}\right)^2 + 1$ expressed in vertex form as $(x - h)^2 + k$, where $h = \frac{9}{5}$ and $k = 1$. The vertex is at $\left(\frac{9}{5}, 1\right)$.

Since the coefficient of the squared term is negative ($a = -1$), the parabola opens downwards. This means the maximum value of the function is at the vertex $k = 1$ and decreases on either side.

Now, solve for when the function is positive ($y > 0$):

• The parabola is positive when its value is greater than the x-axis ($y = 0$). Set the inequality:

Simplifying, we get:

This suggests:

Solving these inequalities:

• For $-1 < \left(x - \frac{9}{5}\right)$:
• $x - \frac{9}{5} > -1$ implies $x > \frac{4}{5}$.
• For $\left(x - \frac{9}{5}\right) < 1$:
• $x - \frac{9}{5} < 1$ implies $x < \frac{14}{5}$.

Combining these results, the function is positive between:

Next, find where $y < 0$:

The parabola is negative outside the interval where it hits the x-axis (the interval where function is 0 or below).

The intervals for which the function is negative are:

$x < \frac{4}{5}$ and $x > \frac{14}{5}$.

Thus, the solution is:

$x > \frac{14}{5}$ or $x < 0 : x < \frac{4}{5}$

$x > 0 : \frac{4}{5} < x < \frac{14}{5}$

Therefore, the correct answer is Choice 2.

To solve this problem, let's first examine the given quadratic function:

$y = -\left(x - 9\right)^2 - 3$

This function is in vertex form $y = a(x - h)^2 + k$, where:

• $a = -1$
• $h = 9$
• $k = -3$

From the values of $a$, $h$, and $k$:

• The vertex of the parabola is $(9, -3)$.
• Since $a < 0$, the parabola opens downwards.

Next, we investigate the function's behavior to determine its positive and negative values:

• The vertex $(9, -3)$ is the maximum point of the parabola because it opens downwards. It implies that the highest value $y$ attains is $-3$.
• Given the vertex form, the entire curve will lie below this maximum or be equal to it, hence the function never attains positive values. Therefore, there are no positive values for $y$.
• For the negative domain, the function's value will always be less than or equal to $-3$ (i.e., negative).

Thus, we can conclude:

• There is $x$ such that $y$ is positive.
• All values of $x$ will make the function assume negative values, specifically for all real numbers $x$.

Therefore, the solution to the problem is:

$x < 0 :$ none

$x > 0 :$ all $x$

The function $y = -\left(x + \frac{7}{8}\right)^2 - 1\frac{1}{5}$ is represented in vertex form, where the vertex of the parabola is at $x = -\frac{7}{8}$, and the maximum value at the vertex is $y = -1\frac{1}{5}$. Since the parabola opens downwards due to the negative coefficient of the squared term, its maximum value is also its highest possible value, which is a negative number $-1\frac{1}{5}$.

Therefore, the function does not reach any positive value for any real number $x$; it only takes on non-positive values. Consequently, the determination of positive and negative domains is as follows:

• $x < 0$: The function can assume all values since the entire parabola lies below the x-axis, producing a negative range for all $x$.
• $x > 0$: No $x$ can produce a positive $y$ value, resulting in no positive values.

Therefore, the positive and negative domains as concluded from this analysis are:

$x < 0 :$ all $x$

$x > 0 :$ none

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

Let's determine the positive and negative domains of the quadratic function:

The function given is $y = (x - 4.6)^2 + 2.1$. This is in the vertex form of a quadratic function $y = (x - h)^2 + k$.

Key observations:

• The term $(x - 4.6)^2$ is a square and is thus always non-negative, i.e., it is always $\geq 0$.
• The function $y = (x - 4.6)^2 + 2.1$ describes the value of $y$ for any $x$.
• The constant $+2.1$ ensures that $y$ is always positive, specifically $y \geq 2.1$.

Since the smallest value that $(x - 4.6)^2$ can take is 0, at $x = 4.6$, the minimum value of $y$ is $2.1$. Thus, for any $x$, the output is always positive.

Therefore, we have:

$x > 0 :$ all $x$

$x < 0 :$ none

This means the function never outputs negative values for any $x$.

The correct choice from the given options is:

• Choice 3: $x > 0 :$ all $x$
• Choice 3: $x < 0 :$ none