Vertex Representation: With fractions

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

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

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 $y = (x+8)^2 - 2\frac{1}{4}$. 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 $y = -\left(x - 3\frac{1}{12}\right)^2 - \frac{1}{7}$. 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 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 $y = -\left(x - 2.5\right)^2 + 0.5$, 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}

To find the positive and negative domains of the function $y = \left(x - 9\frac{1}{3}\right)^2 - 4$, we will solve for when $y = 0$ and analyze the intervals:

First, set the function equal to zero:

• $\left(x - 9\frac{1}{3}\right)^2 - 4 = 0$
• $\left(x - 9\frac{1}{3}\right)^2 = 4$
• Take the square root of both sides: $x - 9\frac{1}{3} = \pm 2$

Calculate the roots:

• For $x - 9\frac{1}{3} = 2$, $x = 11\frac{1}{3}$
• For $x - 9\frac{1}{3} = -2$, $x = 7\frac{1}{3}$

The roots are $x = 7\frac{1}{3}$ and $x = 11\frac{1}{3}$. These are the points where the function crosses the x-axis.

To determine the positive and negative domains, consider the following intervals:

• When $x < 7\frac{1}{3}$, the expression $(x - 9\frac{1}{3})^2$ is greater than 4, making $y > 0$, so the function is positive.
• Between $x = 7\frac{1}{3}$ and $x = 11\frac{1}{3}$, the expression is less than 4, making $y < 0$, so the function is negative.
• When $x > 11\frac{1}{3}$, the expression goes back to being greater than 4, making $y > 0$, so the function is positive.

Therefore, the domains are:

• Positive: $x > 11\frac{1}{3}$ or $x < 7\frac{1}{3}$
• Negative: $7\frac{1}{3} < x < 11\frac{1}{3}$

The correct choice based on this analysis is:

x < 0 : 7\frac{1}{3} < x < 11\frac{1}{3}

x > 11\frac{1}{3} or x > 0 : x < 7\frac{1}{3}

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, 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, 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: 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 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 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 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 will follow these steps:

• Step 1: Identify the vertex and the direction in which the parabola opens.

• Step 2: Set the function equal to zero to find critical x-values.

• Step 3: Solve for these x-values to find the specific points where the function changes signs.

• Step 4: Determine which intervals on the x-axis correspond to the function being positive and which are negative.

Now, let's work through each step:
Step 1: The given function is $y = -\left(x-2\right)^2+\frac{1}{6}$. The vertex is at point $(2, \frac{1}{6})$, and since the leading coefficient is negative, the parabola opens downwards.
Step 2: We set the equation equal to zero: $-\left(x-2\right)^2+\frac{1}{6} = 0$.
Step 3: Solving for when the function is zero, we have: $\begin{aligned} -\left(x-2\right)^2 &amp; = -\frac{1}{6}\\ (x-2)^2 &amp; = \frac{1}{6}\\ x-2 &amp; = \pm\sqrt{\frac{1}{6}} \end{aligned}$ This gives us two solutions for x: $x = 2+\sqrt{\frac{1}{6}}$ and $x = 2-\sqrt{\frac{1}{6}}$.
Step 4: We will test intervals determined by these points to see where the function is positive or negative.

• For x < 2-\sqrt{\frac{1}{6}} , plug an x-value less than $2-\sqrt{\frac{1}{6}}$ into the function; the function will be negative because the entire parabola opens downward from the vertex.

• For 2-\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}} , the function is positive.

• For x > 2+\sqrt{\frac{1}{6}} , again, the function returns to being negative.

Therefore, the intervals are:
- Negative domain: x < 0 : x < 2-\sqrt{\frac{1}{6}} and x > 2+\sqrt{\frac{1}{6}}
- Positive domain: x > 0 : 2 -\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}} .

The correct answer is:

x > 2+\sqrt{\frac{1}{6}} or x < 0 : x < 2-\sqrt{\frac{1}{6}}

x > 0 : 2 -\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}}