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

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 $x$

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 the problem, we first analyze the quadratic function $y = -\left(x + 10\right)^2 - 4$.

Step 1: Identify the vertex.

The function is in vertex form $y = a(x - h)^2 + k$. Here, $a = -1$, $h = -10$, and $k = -4$. Therefore, the vertex is $(-10, -4)$.

Step 2: Determine the direction of the parabola.

Since $a = -1$, the parabola opens downwards. This means the function can only take on either negative values or zero as it cannot have a maximum (i.e., no positive y-values).

Step 3: Analyze the domain of positivity and negativity.

Because the parabola opens downwards and its vertex is the highest point at $(-10, -4)$, all y-values are negative.

Step 4: Determine intersections with the x-axis.

To check for intersections with the x-axis where y = 0, solve: $-\left(x + 10\right)^2 - 4 = 0$.

Rearranging gives $-\left(x + 10\right)^2 = 4$,

which implies $(x + 10)^2 = -4$. Since this yields an imaginary number when solving, the graph does not intersect the x-axis; thus, it is never zero.

Conclusion:

Since the function is negative for all x-values, the positive domain is effectively non-existent.

Checking the choices provided, plug in our understanding:

• For $x < 0$, the positive domain is none, as the function doesn't achieve positive values.
• For $x > 0$, the negative domain is all $x$, as determined.

Thus, the correct answer is:

$x < 0 :$ none
$x > 0 :$ all $x$

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 solve the problem of finding the positive and negative domains of the function $y = \left(x - 12\frac{1}{2}\right)^2 - 4$, follow these steps:

• Start by setting the quadratic equation to zero:
  $(x - 12\frac{1}{2})^2 - 4 = 0$.

• Add 4 to both sides:
  $(x - 12\frac{1}{2})^2 = 4$.

• Take the square root of both sides to find the x-values where the parabola intersects the x-axis:
  $x - 12\frac{1}{2} = \pm 2$.

Solve for $x$ in both cases:

• For $x - 12\frac{1}{2} = 2$:
  $x = 12\frac{1}{2} + 2 = 14\frac{1}{2}$.

• For $x - 12\frac{1}{2} = -2$:
  $x = 12\frac{1}{2} - 2 = 10\frac{1}{2}$.

Thus, the roots of the quadratic are $x = 10\frac{1}{2}$ and $x = 14\frac{1}{2}$. These points divide the x-axis into three intervals: $x < 10\frac{1}{2}$, $10\frac{1}{2} < x < 14\frac{1}{2}$, and $x > 14\frac{1}{2}$.

Next, solve for where the function is positive or negative in these intervals:

• Interval $x < 10\frac{1}{2}$:
  Choose a test point $x = 0$.
  The function value is $(0 - 12\frac{1}{2})^2 - 4 = (12\frac{1}{2})^2 - 4 = 156.25 - 4 = 152.25$.
  Since 152.25 is positive, $y > 0$ for this interval.

• Interval $10\frac{1}{2} < x < 14\frac{1}{2}$:
  Choose a test point $x = 12$.
  The function value is $(12 - 12\frac{1}{2})^2 - 4 = (0.5)^2 - 4 = 0.25 - 4 = -3.75$.
  Since $-3.75$ is negative, $y < 0$ in this interval.

• Interval $x > 14\frac{1}{2}$:
  Choose a test point $x = 15$.
  The function value is $(15 - 12\frac{1}{2})^2 - 4 = (2.5)^2 - 4 = 6.25 - 4 = 2.25$.
  Since 2.25 is positive, $y > 0$ for this interval.

Thus, the function is negative for $10\frac{1}{2} < x < 14\frac{1}{2}$ and positive for $x < 10\frac{1}{2}$ and $x > 14\frac{1}{2}$.

Therefore, the positive and negative domains are:

Positive domain: $x < 10\frac{1}{2}$ or $x > 14\frac{1}{2}$

Negative domain: $10\frac{1}{2} < x < 14\frac{1}{2}$

The correct answer is choice 4.

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 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 quadratic function $y = (x - 12)^2 + 4$, let's proceed step-by-step:

• Step 1: Identify the structure.
  The function is in vertex form $y = (x - 12)^2 + 4$, which indicates a parabola that opens upwards, with vertex $(12, 4)$.
• Step 2: Determine the minimum value.
  Since the vertex form shows the minimum value at $y = 4$ when $x = 12$, the function never actually reaches negative values.
• Step 3: Analyze positivity.
  Given that the minimum value $y = 4$ when $x = 12$, and the parabola opens upwards, every possible value of $x$ results in $y \geq 4$. Therefore, the function is always positive for all $x$.
• Step 4: Conclusion on domains.
  The function has no negative values for any input. Thus, the negative domain is none, and the positive domain includes all values of $x$. Therefore, we assert the positive domain is: all $x$.

With our analysis complete, we can conclude that the positive and negative domains of the function are:

$x < 0 :$ none

$x > 0 :$ all $x$

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.

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 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 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 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, we need to analyze the function $y = (x-1)^2 + 5$, which is a quadratic in vertex form.

Step 1: Identify the Vertex and Orientation
The function is given as $y = (x-1)^2 + 5$, which is in the form $y = a(x-h)^2 + k$. Here, $h = 1$ and $k = 5$, meaning the vertex of the parabola is at $(1, 5)$. Because $a = 1$ (which is positive), the parabola opens upwards.

Step 2: Determine the Minimum Value of $y$
Since the parabola opens upwards, the minimum value of $y$ occurs at the vertex. At the vertex $(1, 5)$, the value of $y$ is 5.

Step 3: Analyze Positive and Negative Values of $y$
The minimum value of $y$ is 5, which indicates that $y$ is always greater than zero. Thus, for all real values of $x$, $y$ remains positive.

Conclusion:
Since the function $y = (x-1)^2 + 5$ has no negative values and is always positive:

$x < 0 :$ none

$x > 0 :$ all $x$

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

$x < 0 :$ none

$x > 0 :$ all $x$

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

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

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