Vertex Representation: Identify the positive and negative domain

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

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

The function given is $y = -\left(x-14\right)^2-6$.

This is a quadratic function in vertex form: $y = a(x-h)^2 + k$ where $a = -1$, $h = 14$, and $k = -6$. The vertex of the function is at $(14, -6)$ and since $a = -1$, the parabola opens downwards.

Step 1: Identify intervals for negative and positive values:
- The vertex at $(14, -6)$ is the maximum point of the parabola.
- For the quadratic to have positive values, $y$ must be greater than 0. Given the vertex and opening direction of the parabola, there are no $x$ values for which $y$ is positive because the parabola is entirely below the x-axis.

Step 2: Analyze $y$ values when $x > 0$ and $x < 0$:
- The parabola is below the x-axis ($y < 0$) for all $x$. Therefore, when checking for $x > 0$, the function remains negative for all positive $x$.

Conclusion: This shows that the function is not positive for any $x$, but is negative for all $x$.

Therefore, the positive and negative domains are as followed:

• $x < 0 :$ none
• $x > 0 :$ all $x$

The correct answer is Choice 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 $( x > -4 + \sqrt{26})$

The function given is $y = -\left(x + 7\right)^2 - 5$, which is a parabola that opens downwards. Let's determine the positive and negative domains:

Firstly, we identify the vertex of the function as $(-7, -5)$. The vertex form tells us that the parabola opens downwards because the coefficient of the squared term is negative ($a = -1$). This indicates that the maximum point of the parabola is at the vertex, and the function decreases on either side of the vertex.

Given the downward opening of the parabola and the maximum value $y = -5$ at $x = -7$, the graph of the parabola lies entirely beneath this maximum point. Thus, the function is always non-positive.

Since the function never crosses the x-axis and is below or equal to the vertex's y-coordinate at all points, we find that:

• No values of $x$ give $y > 0$, so the positive domain is empty.
• All real values of $x$ result in $y \leq 0$, indicating a negative domain for $x$.

Therefore, the solutions are:

$x < 0 :$ none

$x > 0 :$ all $x$

The correct choice is:

Choice 4: $x < 0 :$ none

$x > 0 :$ all $x$

The function given is $y = -\left(x-\frac{4}{9}\right)^2 + 1$, which is a downward-opening parabola because the coefficient of the squared term ($a = -1$) is negative.

The vertex form tells us the vertex of the parabola is at $\left(\frac{4}{9}, 1\right)$.

The function will be zero where $-\left(x-\frac{4}{9}\right)^2 + 1 = 0$. Solving this equation, we set:

Taking the square root of both sides gives:

Thus, $x = \frac{4}{9} + 1 = \frac{13}{9}$ and $x = \frac{4}{9} - 1 = -\frac{5}{9}$.

These are the roots of the quadratic, splitting the domain into three intervals: $(-\infty, -\frac{5}{9})$, $(-\frac{5}{9}, \frac{13}{9})$, and $(\frac{13}{9}, \infty)$.

We need to test the sign of $y$ in each interval:

• For $x < -\frac{5}{9}$, choose a test point like $x = -1$. Substituting into the function yields a negative value, hence the function is negative in this interval.
• For $-\frac{5}{9} < x < \frac{13}{9}$, choose a test point like $x = 0$. Substituting into the function yields a positive value, hence the function is positive in this interval.
• For $x > \frac{13}{9}$, choose a test point like $x = 2$. Substituting into the function yields a negative value, hence the function is negative in this interval.

After analyzing these intervals, the function is positive for $-\frac{5}{9} < x < \frac{13}{9}$ and negative otherwise.

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

$x > \frac{13}{9}$ or $x < 0 : x < -\frac{5}{9}$

$x > 0 : -\frac{5}{9} < x < \frac{13}{9}$

The function given is $y = (x+10)^2 + 2$. This is a quadratic function in vertex form.

The vertex form of a quadratic function is $y = a(x-h)^2 + k$, where the vertex is $(h, k)$. For our function, $h = -10$ and $k = 2$, so the vertex is $(-10, 2)$.

In this case, since the coefficient of $(x+10)^2$ is positive (implicitly 1), the parabola opens upwards. This means the function has a minimum point at the vertex, and $y$ will only increase from that point.

Given that the vertex point has a $y$-value of 2, which is positive, the entire domain yields values of $y$ that are greater than 2. Therefore, $y$ will never be negative.

Now, let's determine the domains:

• For the negative domain, we seek values of $x$ where the function $y$ is negative. Since the minimum $y$-value is 2, no such $x$ satisfies $y \lt 0$. Hence, there is no negative domain.
• Similarly, all values of $x$ yield a positive range of $y$, as $y \geq 2$ for all $x$. Thus, for the positive domain, it is all $x$, as every $y$ value is positive or zero.

Consequently, the specified positive domain is all $x$, and the negative domain is none.

Thus, the correct answer is:

$x < 0 :$ None

$x > 0 :$ All $x$

The function given is $y = (x-4)^2 - 4$, which is a quadratic function in vertex form. This indicates a parabola that opens upwards.

The vertex of this function is $(4, -4)$, indicating the minimum point of the parabola. Since the parabola opens upwards, $y \geq -4$ for all $x$. The parabola crosses the x-axis where $y = 0$, so we solve for these points:

Set $y = 0$:

$(x-4)^2 - 4 = 0$

Add 4 to both sides:

$(x-4)^2 = 4$

Take the square root of both sides:

$x - 4 = \pm 2$

Thus, $x = 6$ or $x = 2$. These are the roots of the equation, indicating where the parabola crosses the x-axis.

For the positive domain (where y > 0 ), analyze the intervals:

• For x < 2 , the parabola is above the x-axis (check any point like $x = 0$ to see y = 12 > 0 ).

• For x > 6 , the parabola is above the x-axis (check any point like $x = 8$ to see y = 12 > 0 ).

Therefore, in terms of the positive domain, the function is positive for x < 2 (or x > 6 ).

For the negative domain (where y < 0 ), analyze the interval between roots:

• For 2 < x < 6 , the parabola is below the x-axis (check any point like $x = 4$ to see y = -4 < 0 ).

Therefore, in terms of the negative domain, the function is negative for 2 < x < 6 .

The solution to the problem is:

x > 6 or x < 0 : x < 2

x < 0 : 2 < x < 6

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

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

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 $x$

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, 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 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}

To determine the positive and negative domains of the function $y=(x+15)^2+6$, we start by analyzing its structure.

The function is given in vertex form, $y=a(x-h)^2+k$, where $a=1$, $h=-15$, and $k=6$. Since a=1 > 0 , the parabola opens upwards.

1. Vertex and Axis of Symmetry:
- Vertex: The vertex of the parabola is at $(-15, 6)$. This indicates the minimum point since the parabola opens upwards.

2. Range of the function:
- As $(x+15)^2$ is always zero or positive, the smallest value for $y$ is when $(x+15)^2=0$, thus $y=6$. Hence, $y \geq 6$.

3. Analyzing the function's values:
- Since the minimum value of $y$ is 6 and it increases as $x$ moves away from -15 in either direction, the function does not achieve any negative values.

4. Conclusion:
- The function is always positive, $y \geq 6$.

Based on this analysis:

Negative domain: The function does not have any negative values, thus, for x < 0 , there are no values where the function is negative.

Positive domain: The entire domain is positive. Therefore, for x > 0 , the function remains positive for all $x$.

Thus, the positive and negative domains are:

x < 0 : None

x > 0 : All $x$

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 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 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 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} .