Standard Representation: Identify the positive and negative domain

Let's analyze the function $y = 3x^2 - 6x + 4$ to determine when it is negative.

First, calculate the discriminant $\Delta$:

$\Delta = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$

A negative discriminant ($\Delta < 0$) indicates that there are no real roots, meaning the graph of the function does not intersect the x-axis. Since $a = 3 > 0$, the parabola opens upwards.

This means the vertex of the parabola represents the minimum point, and the entire graph is above the x-axis.

Consequently, the function $y = 3x^2 - 6x + 4$ does not attain any negative values for any real $x$.

The correct interpretation is that the function stays positive, confirming the conclusion:

The function has no negative values.

Let's analyze the function $y = -x^2 + 4x - 5$ and determine the interval where $y$ is negative.

1. **Find the roots using the quadratic formula**:
The function is given by $y = -x^2 + 4x - 5$. The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 4$, and $c = -5$. First, we calculate the discriminant:

$b^2 - 4ac = 4^2 - 4(-1)(-5) = 16 - 20 = -4$

Since the discriminant is negative, the quadratic equation has no real roots, implying that the parabola does not intersect the x-axis. The quadratic formula confirms there are no real solutions, confirming the function does not touch or cross the x-axis.

2. **Analyze the parabola's direction**:
Since $a = -1$, the parabola opens downwards. A downward-opening parabola with no real roots means it lies entirely below the x-axis. Hence, the function $y = -x^2 + 4x - 5$ is negative for all values of $x$.

Therefore, the function is negative for all $x$.

Let's begin by finding the roots of the quadratic function $y = -x^2 + \frac{3}{4}x - 2$. This will help us determine the intervals where the function is positive or negative.

The coefficients of the quadratic are $a = -1$, $b = \frac{3}{4}$, and $c = -2$. Applying the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute in the values:

$x = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4(-1)(-2)}}{2(-1)}$ $x = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - 8}}{-2}$

Calculate inside the square root:

$\frac{9}{16} - 8 = \frac{9}{16} - \frac{128}{16} = \frac{-119}{16}$

The discriminant $b^2 - 4ac$ is negative ($\frac{-119}{16}$), indicating there are no real roots. Therefore, the quadratic function doesn't intersect the x-axis.

Since the parabola is downward (the coefficient of $x^2$ is negative), it is negative for all $x \in \mathbb{R}$.

We conclude that:

• The function $y$ does not have a positive domain for any real $x$.
• The function $y$ is negative for all $x$.

Therefore, the positive and negative domains, based on choices given, are:

$x > 0$: none

$x < 0$: all $x$

The function given is $y = x^2 + 8x + 20$. This is a quadratic function where the coefficient of $x^2$ (which is $a = 1$) is positive, indicating the parabola opens upwards.

Let’s calculate the vertex to find the minimum value of $y$. The vertex of a parabola described by $y = ax^2 + bx + c$ is found at $x = -\frac{b}{2a}$.

Here, $a = 1$, $b = 8$. So the vertex is at:

$x = -\frac{8}{2 \times 1} = -4$

Substitute $x = -4$ into the function $y = x^2 + 8x + 20$ to calculate the minimum value of $y$.

$y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4$

The minimum value of the function is $y = 4$ at $x = -4$.

Given the opening direction of the parabola and the positive minimum value, the function $f(x) = x^2 + 8x + 20$ is always greater than 0.

Thus, the function is positive for all values of $x$.

The function $y = x^2 - 6x + 10$ represents a parabola opening upwards since its leading coefficient $a = 1$ is positive. Our task is to determine when the function is positive.

First, let's find the vertex of this parabola, which occurs at $x = -\frac{b}{2a}$.

Here, $b = -6$ and $a = 1$, so:

\begin{align*} x_{vertex} &= -\frac{-6}{2 \times 1} \\ &= \frac{6}{2} \\ &= 3. \end{align*}

Next, we evaluate the function at this vertex:

\begin{align*} f(3) &= 3^2 - 6 \cdot 3 + 10 \\ &= 9 - 18 + 10 \\ &= 1. \end{align*}

Since $f(3) = 1$, which is greater than zero, we observe that at the vertex the function is indeed positive.

Moreover, because the parabola opens upwards and the vertex value is positive, the entire parabola lies above the x-axis. Consequently, $f(x) > 0$ for all $x \in \mathbb{R}$.

Therefore, the function is positive for all values of $x$.

The given function is $y = x^2 - 4x + 5$. To find where this function is positive, we'll first analyze the properties of this quadratic.

Let's start by completing the square. We have:

$y = x^2 - 4x + 5$

To complete the square, take the coefficient of $x$, which is $-4$, halve it to get $-2$, and then square it to get $4$. Add and subtract this inside the expression:

$y = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1$

Now, the expression is in vertex form $y = (x-2)^2 + 1$, which indicates a parabola with a vertex at $(2, 1)$ and opens upwards. The vertex is the minimum point of the function.

Since the minimum value of $y$ is 1 (when $x = 2$), and the parabola opens upwards, the function is positive for all real $x$, because $(x-2)^2 \geq 0$ for any real number $x$, making $(x-2)^2 + 1 > 0$.

Therefore, the answer is that the function is positive for all values of $x$.

In conclusion, the correct choice is:

The function is positive for all values of $x$.

The problem asks us to determine when the quadratic function $f(x) = x^2 + 4x + 5$ is greater than zero. Here's how we solve it:

Step 1: Analyze the Vertex
The quadratic function $f(x) = x^2 + 4x + 5$ is in the standard form $y = ax^2 + bx + c$, where $a = 1$, $b = 4$, and $c = 5$. Since $a > 0$, the parabola opens upwards, and thus the vertex represents its minimum point.
To find the x-coordinate of the vertex, use the formula $x = -\frac{b}{2a}$:

$x = -\frac{4}{2 \times 1} = -2$

Substitute $x = -2$ back into the function to find the y-coordinate:

$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$

The vertex of the parabola is $(-2, 1)$, which implies the minimum value of the function is 1.

Step 2: Analyze the Discriminant
The discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots:

$\Delta = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$

Since $\Delta < 0$, the quadratic equation has no real roots, meaning it doesn't intersect the x-axis. Therefore, $f(x) > 0$ for all $x$.

Conclusion
Because the vertex is the minimum point and the function does not intersect the x-axis, the function $f(x) = x^2 + 4x + 5$ is positive for all values of $x$.

Therefore, the function is positive for all values of $x$.

The problem asks us to determine where the function $y = x^2 + 10x + 16$ is less than zero.

• Step 1: Find the roots of the equation $x^2 + 10x + 16 = 0$ using the quadratic formula.
• Step 2: Calculate the discriminant $\Delta = b^2 - 4ac = 10^2 - 4 \times 1 \times 16 = 100 - 64 = 36$.
• Step 3: Find the roots using $x = \frac{{-b \pm \sqrt{\Delta}}}{2a} = \frac{{-10 \pm \sqrt{36}}}{2}$.
• Step 4: Calculate the roots: $x = \frac{{-10 + 6}}{2} = -2$ $x = \frac{{-10 - 6}}{2} = -8$

The roots are $x = -2$ and $x = -8$. These roots will divide the number line into intervals.

• Step 5: Analyze the sign of the quadratic on the intervals $-\infty, -8$, $-8, -2$, and $-2, \infty$.
• Step 6: Choose test points: - For $x < -8$, choose $x = -9$, - For $-8 < x < -2$, choose $x = -5$, - For $x > -2$, choose $x = 0$.

Test each interval:

• For $x = -9$, $y = (-9)^2 + 10 \times (-9) + 16 = 81 - 90 + 16 = 7$ (positive).
• For $x = -5$, $y = (-5)^2 + 10 \times (-5) + 16 = 25 - 50 + 16 = -9$ (negative).
• For $x = 0$, $y = (0)^2 + 10 \times 0 + 16 = 16$ (positive).

The function is negative between $x = -8$ and $x = -2$. Therefore, the solution to $f(x) < 0$ is $-8 < x < -2$.

Therefore, the correct answer is $-8 < x < -2$.

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

x < 0 : 1 < x < 1.5

x > 1.5 or x > 0 : x < 1

The task is to determine where the function $y=3x^2+18x+27$ is positive and negative. A quadratic function is upward-facing if the coefficient of $x^2$ is positive. Here, $a = 3 > 0$, indicating an upward parabola.

First, find the roots using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculate the discriminant:

$b^2 - 4ac = 18^2 - 4 \times 3 \times 27 = 324 - 324 = 0$

The discriminant is 0, implying one repeated root at:

$x = \frac{-18}{2 \times 3} = -3$

The vertex at $x = -3$ means the parabola touches the x-axis without crossing it, and there are no intervals where $y < 0$.

Since the parabola is always above the x-axis except at this point, for $x > -3$ and $x < -3$, $y > 0$ except at $x = -3$ where $y = 0$.

Therefore, the function is positive for $x \ne -3$.

The positive domain where $y > 0$ is:

$x > 0 : x \ne -3$

There is no negative domain (where $y < 0$).

$x < 0 :$ none

The correct choice is option 4:

\($x > 0 : x \ne -3$\) and \($x < 0 :$ none\).

To determine for which values of $x$ the function $f(x) = -3x^2 + 6x - 9 < 0$, follow these steps:

• Step 1: Identify the coefficient information. The function is $-3x^2 + 6x - 9$ with $a = -3$, $b = 6$, $c = -9$.
• Step 2: Determine the orientation of the parabola. The leading coefficient $a = -3$ is negative, so the parabola opens downward.
• Step 3: Calculate the vertex using the formula $x = -\frac{b}{2a} = -\frac{6}{2 \times (-3)} = 1$. The vertex is at $x = 1$.
• Step 4: Find the function's value at the vertex: $f(1) = -3(1)^2 + 6(1) - 9 = -3 + 6 - 9 = -6$. The function is negative at the vertex.
• Step 5: Calculate the discriminant $\Delta = b^2 - 4ac = 6^2 - 4(-3)(-9) = 36 - 108 = -72$. The discriminant is negative, indicating no real roots.

Since the quadratic opens downward and does not cross or touch the x-axis, it remains entirely below the x-axis for all values of $x$. Therefore, the function is negative for all $x$.

Thus, the solution is: The function is negative for all $x$.

To determine for which values of $x$ the function $y = 2x^2 - 4x + 5$ is positive, we will analyze its characteristics.

Step 1: Determine the direction of the parabola.
The given quadratic function $y = 2x^2 - 4x + 5$ has a leading coefficient $a = 2$, which is positive. Therefore, the parabola opens upwards.

Step 2: Check for real roots.
To identify where the function might be zero, calculate the discriminant $\Delta = b^2 - 4ac$.
Here, $a = 2$, $b = -4$, $c = 5$.
The discriminant $\Delta = (-4)^2 - 4 \cdot 2 \cdot 5 = 16 - 40 = -24$.
Since the discriminant is negative, the quadratic has no real roots, meaning it doesn't intersect the x-axis.

Step 3: Analyze positivity over the entire domain.
Since the parabola opens upwards and has no real roots, the function does not touch or cross the x-axis. Therefore, $y = 2x^2 - 4x + 5$ is always positive.

Conclusion.
The function $y = 2x^2 - 4x + 5$ is positive for all values of $x$.

Therefore, the solution to the problem is The function is positive for all values of $x$.

To determine the positive and negative domains of $y = 25x^2 + 20x + 4$, we follow these steps:

• Step 1: Calculate the discriminant $\Delta = b^2 - 4ac = 20^2 - 4 \cdot 25 \cdot 4 = 400 - 400 = 0$.
• Step 2: Since the discriminant is zero, the quadratic has one distinct real root, given by $x = \frac{-b}{2a} = \frac{-20}{2 \cdot 25} = -\frac{2}{5}$.
• Step 3: Analyze the sign of $y = 25x^2 + 20x + 4$.

Since the discriminant is zero, the quadratic equation reaches zero only at $x = -\frac{2}{5}$, and is symmetrical around this point. Therefore:

• For $x > 0$, $y$ is positive except where $x = -\frac{2}{5}$.
• For $x < 0$, there are no points where the graph is negative since the parabola opens upwards and does not cross below the x-axis after this root.

Thus, interpreting the domains:

The positive domain for $x > 0$ is all $x$ except $x = -\frac{2}{5}$. For $x < 0$, there is no negative domain because the graph does not descend below the x-axis.

Therefore, the solution to the problem is:

$x > 0 :x\ne\frac{2}{5}$

$x < 0 :$ none

To determine the positive and negative domains of the function $y = -x^2 + 5x + 6$, we first find the roots of the equation by solving:

$-x^2 + 5x + 6 = 0$.

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = -1$, $b = 5$, and $c = 6$. Substituting these values, we find:

$x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(6)}}{2(-1)} = \frac{-5 \pm \sqrt{25 + 24}}{-2} = \frac{-5 \pm \sqrt{49}}{-2}$.

$x = \frac{-5 \pm 7}{-2}$.

Solving the two scenarios regarding the $\pm$ gives $x = \frac{2}{-2} = -1$ and $x = \frac{-12}{-2} = 6$.

This means the roots are $x = -1$ and $x = 6$.

We now test the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

- For $x < -1$: pick $x = -2$. Substitute into the function:

$y = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative).

- For $-1 < x < 6$: pick $x = 0$. Substitute:

$y = -(0)^2 + 5(0) + 6 = 6$ (positive).

- For $x > 6$: pick $x = 7$. Substitute:

$y = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative).

Thus, the function is positive in the interval $-1 < x < 6$ and negative in the intervals $x < -1$ and $x > 6$.

Therefore, the solution to the problem is:

x > 0 : -1 < x < 6

x > 6 or x < 0 : x < -1

To determine the positive and negative domains of the function $y = x^2 - 8x + 16$, follow these steps:

• Step 1: Re-express the quadratic in standard form. Notice that $x^2 - 8x + 16$ can be rewritten as $(x-4)^2$ because it is a perfect square trinomial. This simplifies identification of intercepts.
• Step 2: Determine the vertex and the direction of the parabola. Since $(x - 4)^2$ is a perfect square, it has a minimum at $x = 4$ when $y = 0$. Thus, the vertex is (4, 0), and the parabola opens upwards.
• Step 3: Identify intervals on the x-axis where the function is positive or zero. Since it opens upwards, $y = (x-4)^2 \geq 0$ for all $x$ and is zero exactly at $x = 4$.
• Step 4: Recognize that the function can only be non-positive or zero, never negative, given its parabolic nature with an upward opening.

Therefore, the positive domain of the function is $x > 0$ excluding $x = 4$, and there is no negative domain, consistent with the solution: $x > 0 : x \ne 4$; $x < 0 :$ none.

This conclusion leads us to verify with choice options. The correct option matches this solution, indicating the positive interval where the function remains non-negative and is bounded by $x = 4$ as a minimum.

To determine the positive and negative domains of the quadratic function $y = 3x^2 + 7x + 2$, we will first find its roots using the quadratic formula.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For this function, $a = 3$, $b = 7$, and $c = 2$.

First, calculate the discriminant $\Delta$:

$\Delta = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25$.

Since the discriminant is positive, the function has two distinct real roots.

Now, calculate the roots:

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

$x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}$.

This results in:

• Root 1: $x = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3}$
• Root 2: $x = \frac{-7 - 5}{6} = \frac{-12}{6} = -2$

The roots are $x = -2$ and $x = -\frac{1}{3}$, dividing the x-axis into three intervals: $x < -2$, $-2 < x < -\frac{1}{3}$, and $x > -\frac{1}{3}$.

For $x \to -\infty$, the quadratic $y = 3x^2 + 7x + 2$ is positive, as the leading coefficient $a = 3$ is positive, indicating the parabola opens upwards.

Evaluate the sign of $y$ within the intervals:

• For $x < -2$, pick $x = -3$: $y = 3(-3)^2 + 7(-3) + 2 = 27 - 21 + 2 = 8$. Hence, $y > 0$.
• For $-2 < x < -\frac{1}{3}$, pick $x = -1$: $y = 3(-1)^2 + 7(-1) + 2 = 3 - 7 + 2 = -2$. Hence, $y < 0$.
• For $x > -\frac{1}{3}$, pick $x = 0$: $y = 3(0)^2 + 7(0) + 2 = 2$. Hence, $y > 0$.

Therefore, the positive domain of the function is $x < -2$ and $x > -\frac{1}{3}$, and the negative domain is $-2 < x < -\frac{1}{3}$.

Thus, the solution matches the given correct answer:

$x < 0 : -2 < x < -\frac{1}{3}$

$x > -\frac{1}{3}$ or $x > 0 : x < -2$

To determine the positive and negative domains of the quadratic function $y = \frac{1}{2}x^2 + \frac{3}{4}x + \frac{5}{6}$, we start by considering the possibility of real roots using the discriminant.

The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac = \left(\frac{3}{4}\right)^2 - 4\left(\frac{1}{2}\right)\left(\frac{5}{6}\right)$

Calculating gives:

$\Delta = \frac{9}{16} - \frac{20}{12} = \frac{9}{16} - \frac{5}{3} = \frac{9}{16} - \frac{80}{48}$

Convert $\frac{9}{16}$ to a common denominator:

$\Delta = \frac{27}{48} - \frac{80}{48} = -\frac{53}{48}$

The discriminant $\Delta$ is negative, indicating that this quadratic equation has no real roots.

Since the coefficient $a = \frac{1}{2}$ is positive and there are no real roots, the parabola opens upwards and never crosses the x-axis.

This means that the function is always positive for all $x$.

Thus, the positive domain is all $x$, and there is no negative domain.

Therefore, the correct choice is:

$x > 0 :$ for all $x$

$x < 0 :$ none

To determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 + x + 2$, we will follow these general steps:

• Find the roots of the equation $\frac{1}{3}x^2 + x + 2 = 0$.
• Analyze the sign of the quadratic between and beyond these roots.

First, let's identify the roots of the quadratic function:

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = \frac{1}{3}$, $b = 1$, and $c = 2$, the discriminant $\Delta = b^2 - 4ac = 1^2 - 4 \cdot \frac{1}{3} \cdot 2 = 1 - \frac{8}{3} = \frac{-5}{3}$.

Since the discriminant is negative, the quadratic equation $\frac{1}{3}x^2 + x + 2 = 0$ has no real roots.

Given that the coefficient of $x^2$ (i.e., $\frac{1}{3}$) is positive, the parabola opens upwards, meaning the entire function is greater than zero for all real values of $x$.

Therefore, the positive domain is all real numbers, and there is no negative domain.

Therefore, the solution is: $x > 0$: for all $x$; $x < 0$: none.

To determine the positive and negative domains of the quadratic function $y = x^2 - 22x + 121$, we start by finding its roots. The roots will help us identify intervals where the function is positive or negative.

First, apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The given quadratic equation is $x^2 - 22x + 121$. Here, $a = 1$, $b = -22$, and $c = 121$. Calculate the discriminant:

$\Delta = b^2 - 4ac = (-22)^2 - 4 \times 1 \times 121 = 484 - 484 = 0$

The discriminant is zero, indicating a repeated root. Applying the quadratic formula gives:

$x = \frac{-(-22) \pm \sqrt{0}}{2 \times 1} = \frac{22 \pm 0}{2} = 11$

This means there is only one root, $x = 11$. A quadratic with a double root at $x = 11$ indicates the function touches the x-axis at $x = 11$ and opens upwards (since $a > 0$). This implies that the function is non-negative for all $x$. Therefore, the function does not have a negative domain.

As the quadratic is positive for $x \ne 11$, the positive domain is all $x$ except when $x = 11$.

Therefore, the positive domain is $x > 0 : x \ne 11$.

The negative domain, where the function would take negative values, is nonexistent as the parabola never crosses beneath the x-axis.

The solution to the problem is:

$x > 0 : x \ne 11$

$x < 0 :$ none

To determine the values of $x$ for which the quadratic function $y = -x^2 - 6x - 8$ is greater than 0, we will first find the roots of the quadratic equation where it equals zero.

We apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = -1$, $b = -6$, and $c = -8$ into the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-8)}}{2(-1)}$

Simplifying inside the square root and the rest of the expression:

$x = \frac{6 \pm \sqrt{36 - 32}}{-2}$ $x = \frac{6 \pm \sqrt{4}}{-2}$

Since $\sqrt{4} = 2$, the equation becomes:

$x = \frac{6 \pm 2}{-2}$

This gives us two potential solutions:

- $x = \frac{8}{-2} = -4$

- $x = \frac{4}{-2} = -2$

The roots divide the x-axis into three intervals: $x < -4$, $-4 < x < -2$, and $x > -2$.

To find where the function is positive, choose test points from these intervals:

• For $x < -4$ (e.g., $x = -5$): $f(-5) = -(-5)^2 - 6(-5) - 8 = -25 + 30 - 8 = -3$
• For $-4 < x < -2$ (e.g., $x = -3$): $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$
• For $x > -2$ (e.g., $x = 0$): $f(0) = -(0)^2 - 6(0) - 8 = -8$

From this, the function is positive on the interval $-4 < x < -2$.

Therefore, the solution to the problem is $-4 < x < -2$.