Vertex Representation: Identify the positive and negative domain

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 $y = (x+3)^2 - 5$, 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 solve this problem, we need to determine the domains for the given function $y = (x + 2.7)^2 + 0.4$ where $y$ is positive and negative.

The function is a quadratic function in the form $y = (x + h)^2 + k$, representing a parabola opening upwards. The vertex of this parabola is at $x = -2.7$ and $y = 0.4$, meaning this point is the minimum point of the parabola.

The $y$-value of the function at its minimum is $y = 0.4$. Because the parabola opens upwards, it implies that for all $x$, $y \geq 0.4$.

Since the minimum value of $y$ is 0.4, the function never takes negative values; therefore, there is no negative domain.

The positive domain, $x > 0$, can be interpreted as being satisfied by all $x$, since no values make $y$ less than 0. The function's range is therefore always positive, including its minimum value.

Conclusively, the positive domain is all $x$, while the function has no negative domain.

Thus, the final solution is:

$x > 0 :$ all $x$

$x < 0 :$ none

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

To find the positive and negative domains of the function $y = -\left(x - \frac{1}{3}\right)^2 + 4$, we start by determining where the function crosses the x-axis. This happens where $y = 0$.

Set $y = 0$ to get:

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

Solving for $x$:

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

This gives:

Positive root (for $+2$):

$x = \frac{1}{3} + 2 = \frac{7}{3}$

Negative root (for $-2$):

$x = \frac{1}{3} - 2 = -\frac{5}{3}$

The x-intercepts are $x = \frac{7}{3}$ and $x = -\frac{5}{3}$.

Since the quadratic opens downward (as $a = -1$), the graph is above the x-axis between these roots and below outside this interval.

Therefore, the function is positive for:

$x \in \left(-\frac{5}{3}, \frac{7}{3}\right)$

And negative for:

$x > \frac{7}{3}$ or $x < -\frac{5}{3}$

Thus, the positive domain is:

$x > 0 : -\frac{5}{3} < x < \frac{7}{3}$

And the negative domain is:

$x > \frac{7}{3}$ or $x < 0 : x < -\frac{5}{3}$

The correct choice is:

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

The given function is $y = (x+8)^2 - 2\frac{1}{4}$. To find where it is positive or negative, we first find the points where $y = 0$.

Set the function equal to zero:

$(x+8)^2 - \frac{9}{4} = 0$

To handle the fraction, rewrite the equation:

$(x+8)^2 = \frac{9}{4}$

Now, take the square root of both sides:

$x+8 = \pm \frac{3}{2}$

This gives us two solutions for $x$:

$x = -8 + \frac{3}{2}$ and $x = -8 - \frac{3}{2}$

Calculating these values, we have:

$x = -\frac{16}{2} + \frac{3}{2} = -\frac{13}{2} = -6\frac{1}{2}$

$x = -\frac{16}{2} - \frac{3}{2} = -\frac{19}{2} = -9\frac{1}{2}$

Next, test intervals around the roots to determine where the function is positive or negative:

• For $x < -9\frac{1}{2}$, an example point is $x = -10$: $(-10+8)^2 - \frac{9}{4} = 4 - \frac{9}{4}$, which is positive.
• For $-9\frac{1}{2} < x < -6\frac{1}{2}$, an example point is $x = -8$: $(-8+8)^2 - \frac{9}{4} = 0 - \frac{9}{4}$, which is negative.
• For $x > -6\frac{1}{2}$, an example point is $x = -6$: $(-6+8)^2 - \frac{9}{4} = 4 - \frac{9}{4}$, which is positive.

This leads to the conclusion that:

$x < 0 :-9\frac{1}{2} < x < -6\frac{1}{2}$

And for $x > 0$, we have $x > -6\frac{1}{2}$ or $x < 0 : x < -9\frac{1}{2}$

Thus, the correct answer is:

$x < 0 :-9\frac{1}{2} < x < -6\frac{1}{2}$

$x > -6\frac{1}{2}$ or $x > 0 : x < -9\frac{1}{2}$

To solve this problem, we need to determine when the function $y = \left(x + 3\frac{1}{4}\right)^2 + \frac{3}{4}$ is positive or negative across its domain.

The given function is in the vertex form $(x - h)^2 + k$, where $h = -3\frac{1}{4}$ and $k = \frac{3}{4}$. The parabola opens upwards because the coefficient of the squared term is positive.

The vertex of the parabola is at $(-3.25, 0.75)$. This means the minimum value of $y$ is $0.75$, which means $y$ is always greater than zero; the function does not reach zero or negative values.

Therefore, the function is positive for all $x$ and non-negative globally. There are no values of $x$ for which the function is negative.

Thus, for this function, the positive domain is all $x$.

Based on this analysis:

$x < 0 :$ none

$x > 0 :$ for all $x$

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

• Step 1: Recognize that the given quadratic function is in vertex form $y = (x - h)^2 + k$, where $h = 2\frac{1}{9}$ and $k = \frac{5}{6}$.
• Step 2: Identify that the squared term $(x - 2\frac{1}{9})^2$ is always non-negative for any real $x$.
• Step 3: Note that the smallest value that the squared term can obtain is 0, which happens when $x = 2\frac{1}{9}$. Therefore, the smallest value of the whole function $y$ is $\frac{5}{6}$, which is positive. Thus, $y$ is never negative.
• Step 4: Conclude by identifying the domains: $y < 0$ has no solutions and $y > 0$ for all $x$.

After considering the nature of the quadratic function:

Since $y$ cannot be negative, the negative domain is none, which means there are no values where $y < 0$.

On the other hand, for all $x$, $y$ is positive because the minimum value $y$ can take is the constant term $\frac{5}{6}$, which is positive.

Thus, the solution is:

$x < 0 :$ none

$x > 0 :$ for all $x$

To solve this problem, we'll determine the domains where the function is positive or negative:

• Given the function $y = -\left(x + 6\frac{1}{2}\right)^2 - 2\frac{1}{4}$, it's in vertex form, with $a = -1$.
• The vertex is at $\left(x, y\right) = \left(-6\frac{1}{2}, -2\frac{1}{4}\right)$, and since $a < 0$, the parabola opens downward.
• Because the parabola opens downward and has no x-intercepts due to $k < 0$, the function $y = 0$ is never zero nor positive.
• This implies there are no values of $x$ for which $y > 0$. Therefore, the positive domain is nonexistent.
• Thus, the negative domain encompasses all $x$ such that $y \leq 0$, which is the entire set of real numbers.

Since the function's value is never positive, the correct description of positive and negative domains is as follows:
Positive domain: None
Negative domain: For all $x$

Therefore, the solution is:

$x < 0 :$ for all $x$

$x > 0 :$ 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 $y = (x + \sqrt{2})^2$, we begin by noting that this function is a quadratic expression in vertex form. Here, the vertex is located at $x = -\sqrt{2}$, and since the coefficient of the squared term is positive, the parabola opens upwards.

The expression $(x + \sqrt{2})^2$ represents the square of the term $x + \sqrt{2}$. A squared term is always greater than or equal to zero for any real $x$. Therefore, $y$ is never negative for any real $x$.

Since we want to determine where the function is positive, we need $(x + \sqrt{2})^2 > 0$. This inequality holds true for all $x$ except at the point where the expression equals zero, which is $x = -\sqrt{2}$.

Therefore, the positive domain is where $x \neq -\sqrt{2}$, corresponding to $x > 0$ and $x \neq -\sqrt{2}$. However, the condition $x > 0$ is sufficient for this particular problem, indicating the positive domain requirement.

The negative domain does not exist, as the function cannot be negative.

Based on the analysis, the correct answer is:

$x < 0 :$ none

$x > 0 : x \neq -\sqrt{2}$

To solve this problem, we'll investigate the function $y = -(x+7)^2$.

• Step 1: Recognize that the function $y = -(x+7)^2$ is a quadratic function opening downwards due to the negative coefficient.
• Step 2: The vertex form $y = a(x-h)^2 + k$ helps identify that the vertex of this quadratic is $(-7, 0)$.
• Step 3: As $(x+7)^2 \geq 0$ for any real number $x$, $-(x+7)^2 \leq 0$ is always true. Thus, the function is never positive.
• Step 4: Identify when $y = 0$: This occurs only when $x = -7$.
• Step 5: The domain of $x$ where $y < 0$ occurs when $x \ne -7$, covering all other $x$ in the real numbers since $y$ is negative.

Considering the choices provided, the statement matches choice 3:

$x < 0 : x\ne-7$

$x > 0 :$ none

Therefore, the function has no positive domain values, while the negative domain consists of all values except $x = -7$.

To solve the problem of finding the positive and negative domains of the function $y = (x + 2)^2$, we will explore the nature of this function.

Step 1: Given Function and Analysis
The function $y = (x + 2)^2$ is a quadratic function presented in vertex form. The vertex form of a quadratic function is generally given as $y = a(x - h)^2 + k$, but simplifying our function, this translates to $y = 1(x + 2)^2 + 0$, indicating a parabola that opens upwards with vertex at $x = -2$.

Step 2: Zero Point of the Function
The only point where $(x + 2)^2 = 0$ is when $x + 2 = 0$. Solving for $x$, we get $x = -2$. At this point, $y = 0$.

Step 3: Positive and Negative Analysis
Since the parabola opens upwards (as the coefficient 1 in front of $(x+2)^2$ is positive), the function will yield positive $y$ values for all $x$ except at the vertex $x = -2$. Thus, for $x \neq -2$, the function is always positive, and it is never negative.

Step 4: Conclusion
Therefore, since $(x + 2)^2$ can only yield zero at $x = -2$ and positive for every other real $x$, the domains are:
- For $y < 0$ (Negative domain): None
- For $y > 0$ (Positive domain): All $x \neq -2$, including positive numbers only for $x > 0$.

Therefore, the solution to the problem is:
$x < 0 :$ none
$x > 0 : x \ne -2$

To determine the positive and negative domains of the function $y = (x - 5)^2$, follow these steps:

• Step 1: Recognize the function is in vertex form: $y = (x - h)^2 + k$, where the vertex is $(h, k) = (5, 0)$.
• Step 2: Observe the properties of a square. The expression $(x - 5)^2$ is always greater than or equal to zero because a square of any real number cannot be negative.
• Step 3: Analyze when the function is zero. The output $y = 0$ when $x - 5 = 0$, or $x = 5$.
• Step 4: Consider where $y$ is positive. For all $x$ except $x = 5$, the square $(x-5)^2 > 0$.
• Step 5: Determine where $y$ is negative. Since a squared term is never negative, there are no values of $x$ for which $y < 0$.

Based on these steps, the positive domain captures all $x$ except where $x = 5$, and the negative domain is nonexistent because the square is always non-negative.

Therefore, the solution is:

$x < 0 :$ none
$x > 0 : x\ne5$

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 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 $x$

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 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