Positive and Negative Domains: Vertex form of quadratic equation

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 $y = -\left(x + 7\right)^2 - 5$, 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 $y = (x+10)^2 + 2$. 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 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 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 find the positive and negative domains of the function $y = (x+2)^2 + 12$, follow these steps:

• Identify the vertex of the function: The vertex form is $y = (x+2)^2 + 12$, hence the vertex is at $(-2, 12)$.
• Determine the parabola's direction: Given the coefficient of $(x+2)^2$ is positive, the parabola opens upwards.
• Consider the vertex's role: At $x = -2$, the minimum value of $y$ is 12. Since the parabola opens upwards from there, $y \geq 12$ for all $x$.
• Analyze positivity/negativity: Since the minimum $y$-value is 12, the function is always positive for all real $x$, and hence it is not negative for any $x$.

Therefore, the positive domain is all $x$, and there is no negative domain. The final choice is:

$x < 0 :$ 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 $x$

To solve for the positive and negative domains of $y = (x-8)^2 - 1$:

• Identify when $y$ equals zero. Set the equation $(x-8)^2 - 1 = 0$.
• Add 1 to both sides: $(x-8)^2 = 1$.
• Take the square root: $x-8 = \pm 1$.
• Solve for $x$: $x = 8 \pm 1 \rightarrow x = 7$ or $x = 9$.

So, the function $y = (x-8)^2 - 1$ is zero at $x = 7$ and $x = 9$. These points divide the x-axis into intervals:

• Interval 1: $x < 7$, test $x = 6$. $(6-8)^2 - 1 = 3$, which is positive.
• Interval 2: $7 < x < 9$, test $x = 8$. $(8-8)^2 - 1 = -1$, which is negative.
• Interval 3: $x > 9$, test $x = 10$. $(10-8)^2 - 1 = 3$, which is positive.

From this analysis:

The negative domain (where $y < 0$) is $7 < x < 9$.

The positive domain (where $y > 0$) consists of $x < 7$ and $x > 9$.

Therefore, the correct answer from the provided choices is:

x < 0 : 7 < x < 9

x > 9 or x > 0 : x < 7

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 $x$

To solve the problem, we'll find the roots of the quadratic function:

• Step 1: Set $y = 0$ in the equation $-\left(x - 9\right)^2 + 4 = 0$.
• Step 2: Solve for $x$:
  $-\left(x - 9\right)^2 + 4 = 0$ leads to $\left(x - 9\right)^2 = 4$.
  Take the square root: $x - 9 = \pm 2$.
• Step 3: Solve for $x$:
  $x - 9 = 2$ gives $x = 11$.
  $x - 9 = -2$ gives $x = 7$.
• Step 4: Analyze intervals determined by $x = 7$ and $x = 11$:
• For $x < 7$, choose a value (e.g., $x = 0$): $y = -\left(0 - 9\right)^2 + 4 = -81 + 4 = -77$ (negative).
• For $7 < x < 11$, choose a value (e.g., $x = 9$): $y = -\left(9 - 9\right)^2 + 4 = 4$ (positive).
• For $x > 11$, choose a value (e.g., $x = 12$): $y = -\left(12 - 9\right)^2 + 4 = -9 + 4 = -5$ (negative).

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

$x > 11$ or $x < 0 : x < 7$

$x > 0 : 7 < x < 11$

In summary, the correct intervals where the function is positive or negative are identified. The function is positive for $7 < x < 11$ and negative otherwise.

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 $x$

To solve this problem, we'll analyze the function $y = (x-2)^2 + 1$ step-by-step:

• Step 1: Identify the function in its vertex form. The function is already presented as $(x-2)^2 + 1$, with vertex at $(2, 1)$.

• Step 2: Analyze the minimum point. The vertex represents the minimum value of the quadratic because the coefficient of the squared term, 1, is positive, meaning the parabola opens upwards.

• Step 3: Calculate the minimum value. By substituting $x = 2$ into the function, we find $y = (2-2)^2 + 1 = 1$.

• Step 4: Determine the range. Since the minimum value is 1, which is positive, the function never takes negative values. The range of $y$ is $(1, \infty)$.

• Step 5: Establish positive and negative domains: - Negative domain: The function does not have any negative values since it is always at or above 1. - Positive domain: The function is positive for all $x$, because the minimum value itself (1) is positive.

Therefore:

x < 0 : none

x > 0 : all $x$

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

• Step 1: Identify the vertex of the parabola.
  The function $y = -\left(x-3\right)^2-1$ is in vertex form, with vertex $(h, k) = (3, -1)$.
• Step 2: Determine the shape and orientation of the parabola.
  Since $a = -1$ is negative, the parabola opens downwards.
• Step 3: Evaluate the function at the vertex.
  At $x = 3$, $y = -\left(3-3\right)^2 - 1 = -1$.
• Step 4: Analyze the function values as $x$ moves away from the vertex.
  Since the parabola opens downwards, $y$ decreases from the vertex value of $-1$, meaning $y < 0$ for all other $x \neq 3$.

Thus, the function is negative (less than zero) for all $x$. There are no $x$ values for which $y$ is positive (greater than zero).
In conclusion:

• Negative domain: all $x$
• Positive domain: none

Therefore, the solution matches choice 4:

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

To solve this problem, we must determine the intervals where the function $y = -\left(x + 4\right)^2 - 1$ is positive or negative.

Let's follow these steps:

• Step 1: Identify the vertex of the quadratic function. The function is given in vertex form, $-\left(x + 4\right)^2 - 1$, where the vertex is $(-4, -1)$.
• Step 2: Determine the direction of the parabola. Since the coefficient of the squared term is negative (i.e., $-1$), the parabola opens downwards.
• Step 3: Analyze the graph. A downward-opening parabola means that the function reaches its maximum at the vertex, and thus all other points on the parabola are below this maximum value of $-1$.
• Step 4: Determine the sign of $y$. Because the vertex $y$-value is $-1$ and all other points are below it, the function does not take on any positive values. Therefore, the positive domain of $y$ is nonexistent, and the function is entirely in the negative $y$-domain.

Therefore, the positive domain is empty because the parabola of $y = -\left(x + 4\right)^2 - 1$ does not reach any positive $y$-values. Thus, the function is negative for all $x$.

In conclusion, the correct answer is: $x < 0 :$ none and $x > 0 :$ all $x$.

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 find where $y = -\left(x-4\right)^2+1$ is positive or negative.

Step 1: Set $y = 0$ to find the roots.

Solve for $x$:

Take the square root of both sides:

This gives the roots:

Thus, the roots are $x = 5$ and $x = 3$.

Step 2: Analyze intervals defined by roots.

Intervals are $(-\infty, 3)$, $(3, 5)$, and $(5, \infty)$.

Check sign of $y$ in these intervals:

• For $x < 3$: Choose $x = 0$. Then, $y = -\left(0-4\right)^2+1 = -16 + 1 = -15$. So, $y < 0$.
• For $3 < x < 5$: Choose $x = 4$. Then, $y = -\left(4-4\right)^2+1 = 1$. So, $y > 0$.
• For $x > 5$: Choose $x = 6$. Then, $y = -\left(6-4\right)^2+1 = -4 + 1 = -3$. So, $y < 0$.

The function $y = -\left(x-4\right)^2+1$ is positive for $3 < x < 5$ and negative for $x < 3$ or $x > 5$.

Thus, the positive domain is $3 < x < 5$ and the negative domain is $x > 5$ or $x < 3$.

The correct choices align with the intervals found, which are:

$x > 5$ or $x < 0 : x < 3$

$x > 0 : 3 < x < 5$

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 $x$

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 $y = -\left(x-\frac{4}{9}\right)^2 + 1$, 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 $x < 0 : x < -\frac{5}{9}$

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