Positive and Negative Domains: Representation of missing b value

Let's solve this step-by-step.

• First, identify the key features of the parabola represented by $y = -5x^2 - 10$. It is a downward-opening parabola because the coefficient of $x^2$, which is -5, is negative.
• Since the parabola opens downwards, the vertex provides the highest point on the graph. The general formula for the vertex $x$-coordinate of the quadratic function $ax^2 + bx + c$ is $-\frac{b}{2a}$. However, as mentioned, $b = 0$, so the vertex is at $x = 0$.
• Substitute $x = 0$ back into the function to find the $y$-coordinate of the vertex:
  $y = -5(0)^2 - 10 = -10$.
• This implies that the entire parabola is situated below the y-axis since the maximum value (vertex) is -10, which is already less than zero.
• Thus, the function $f(x) = -5x^2 - 10$ is always less than zero for all $x$. This is because the vertex, the highest point, is also negative, and it only opens downward from there.

Therefore, the solution to the problem is that $f(x) < 0$ for all values of $x$.

To solve this problem, let's analyze the function $f(x) = -3x^2 - 9$ and determine when it is greater than zero.

The function is a quadratic equation of the form $ax^2 + bx + c$ where $a = -3$, $b = 0$, and $c = -9$.

Our task is to find when $-3x^2 - 9 > 0$.

Step 1: Rewrite the inequality:
$-3x^2 - 9 > 0$

Step 2: Add 9 to both sides to isolate the quadratic term:
$-3x^2 > 9$

Step 3: Divide through by -3 (note that dividing by a negative flips the inequality sign):
$x^2 < -3$

Step 4: Analyze $x^2 < -3$

A square of a real number $x^2$ is always non-negative, meaning $x^2 \geq 0$. Therefore, $x^2 < -3$ is impossible since there are no real values of $x$ such that the square of $x$ is a negative number.

Conclusion: The inequality $f(x) > 0$ has no real solutions. Therefore, no values of $x$ satisfy the inequality.

The correct answer is that no values of $x$ will make $f(x) > 0$.

To solve this problem, we need to analyze the quadratic function given by:

1. Step 1: Find the vertex of the parabola.
The formula for a quadratic function is $y = ax^2 + bx + c$. For our function, $a = -5$, $b = 0$, $c = -10$.
Since $b = 0$, the x-coordinate of the vertex is $x = 0$.

2. Step 2: Calculate the y-coordinate of the vertex.
Substitute $x = 0$ into the function:
$y = -5(0)^2 - 10 = -10$.

3. Step 3: Analyze the parabola.
The vertex is at (0, -10). Since the vertex itself is below the x-axis and the parabola opens downwards (given by $a < 0$), the entire parabola is below the x-axis.

As a result, there are no values of $x$ for which the function $y = -5x^2 - 10$ is greater than 0.

Therefore, the correct answer to the problem is that there are no values of $x$.

Given the quadratic function $f(x) = -3x^2 - 9$, we want to determine when $f(x) < 0$.

First, observe that the function is a downward-opening parabola because the coefficient of $x^2$ is negative ($-3$). This means the parabola opens downwards.

To find when the parabola is below the x-axis ($f(x) < 0$), we should first check whether there are any real roots, since this implies crossing the x-axis.

The function $f(x) = -3x^2 - 9$ is reformulated as:

$0 = -3x^2 - 9$.

To find the roots, rearrange and solve:

$-3x^2 = 9$ or $x^2 = -3$.

Since $x^2 = -3$ yields no real solutions (as no real number squared equals a negative), there are no x-intercepts.

This indicates the parabola does not cross the x-axis and is entirely below it (since it opens downward and has no real roots).

Thus, the function $f(x) < 0$ for all real values of $x$.

Therefore, the condition $f(x) < 0$ is satisfied for all x, meaning the correct choice is:

All x

To determine for which values of $x$ the function $y = 4x^2 + 100$ is greater than 0, we analyze the structure of the quadratic equation:

The function is $y = 4x^2 + 100$, where $4x^2$ is always non-negative for any real number $x$ because squaring any real number gives a non-negative result, and multiplying by a positive constant (4) remains non-negative.

The constant term in the function is $100$, which is positive. Therefore, the smallest value $4x^2$ can take is 0 (when $x = 0$), making the minimum value of the function $y = 0^2*4 + 100 = 100$.

Since $y = 4x^2 + 100$ is always greater than $0$, for all values of $x$, this quadratic function never reaches $0$ or becomes negative.

In conclusion, the function $y = 4x^2 + 100$ is positive for all values of $x$.

Therefore, the solution to the problem is All $x$.

To solve for the values of $x$ where $y = 4x^2 + 100 < 0$, follow these steps:

• Step 1: Analyze the given function $y = 4x^2 + 100$.
• Step 2: Recognize that the quadratic function is in the standard form $y = ax^2 + bx + c$ with $a = 4$, $b = 0$, and $c = 100$.
• Step 3: Note that since the coefficient of $x^2$ (i.e., $a = 4$) is positive, the parabola opens upwards.
• Step 4: The vertex of the parabola, which occurs at the minimum point $x = -\frac{b}{2a} = 0$, gives the minimum function value $y = 4(0)^2 + 100 = 100$.

Since the minimum value of the function is $y = 100$, which is greater than zero, the function never reaches below zero. Hence, there are no $x$ values for which the function $f(x) < 0$.

Therefore, the solution is No x.

To solve this problem, consider the function $y = 2x^2 + 16$. Our objective is to find values of $x$ for which $y < 0$.

The function, $y = 2x^2 + 16$, is a quadratic function in standard form. Here, $a = 2$, $b = 0$, and $c = 16$.

• Step 1: Determine the Vertex
  Since the quadratic is in the form $y = ax^2 + bx + c$, with $a > 0$, it opens upwards. Therefore, its vertex represents the minimum point. The x-coordinate of the vertex is given by $x = \frac{-b}{2a} = \frac{0}{4} = 0$. Substituting $x = 0$ into the function provides $y = 2(0)^2 + 16 = 16$.
• Step 2: Analyze the Discriminant
  The discriminant $D = b^2 - 4ac = 0^2 - 4 \times 2 \times 16 = -128$. Since $D < 0$, the quadratic has no real roots, meaning it does not cross the x-axis. Instead, it remains entirely above the x-axis.

Since the discriminant is negative, the function has no real roots, confirming that the quadratic function never takes a value below zero. The vertex is at $y = 16$, which is above zero, indicating all function values are positive.

Therefore, the function $y = 2x^2 + 16$ is never less than zero, implying that there are no values of $x$ for which $y < 0$.

Hence, the solution is No x.

To solve this problem, let's analyze the quadratic function $y = 2x^2 + 6$:

• The function $y = 2x^2 + 6$ is a parabola opening upwards because the coefficient of $x^2$ is positive ($a = 2$).
• A parabola that opens upwards has its vertex as the lowest point.
• Since the function is in the form $y = ax^2 + c$ with no $x$ term, there are no real roots (our function has no $b$ term, hence no rotations impacting x-intercepts).
• We evaluate $f(x) > 0$: Since the constant term $c = 6$ is positive, this defines a vertical shift of the parabola entirely above the x-axis.
• This implies $y = 2x^2 + 6$ is always greater than zero for any value of $x$.

Therefore, the solution to the problem is all x.

We start by considering the function given: $y = x^2 + 16$. Our task is to find the values of $x$ making $f(x) < 0$, i.e., $x^2 + 16 < 0$.

Let's analyze the expression:

• The square of any real number $x$, denoted as $x^2$, is always non-negative. Thus, $x^2 \geq 0$ for all $x \in \mathbb{R}$.
• Adding 16 to a non-negative number (like $x^2$) results in a number that is at least 16, i.e., $x^2 + 16 \geq 16$.

Thus, the expression $x^2 + 16$ is always at least 16, and there are no $x$ values for which $y < 0$.

The solution is that there are No x values where $f(x) < 0$.

Hence, option 4 is correct: No x.

To solve the problem of determining for which values of $x$ the function $y = x^2 + 16$ is positive, we proceed as follows:

Step 1: Analyze the function $y = x^2 + 16$.
The expression $x^2$ is non-negative (i.e., $x^2 \geq 0$) for all real numbers $x$. Therefore, the smallest value $x^2$ can take is 0.

Step 2: Evaluate the function at its minimum value.
Substituting the minimum value of $x^2$ into the function gives us:

$y = x^2 + 16 = 0 + 16 = 16$.

Step 3: Determine for which $x$ the function is positive.
Since the minimum value of the function is 16, which is greater than 0, the function $y = x^2 + 16$ is greater than 0 for all real numbers $x$.

Thus, the solution to the problem is that the function is positive for all $x$.

The solution to the problem involves finding the values of $x$ where the function $y = -x^2 + 49$ is less than zero. Since it is a downward-opening parabola, its intercepts tell us where the function changes sign.

To start, solve for $f(x) = 0$:

$-x^2 + 49 = 0$

Add $x^2$ to both sides:

$x^2 = 49$

Take the square root of both sides:

$x = \pm 7$

These solutions $x = 7$ and $x = -7$ are the x-intercepts of the parabola. Because the parabola opens downwards, the function is negative outside this interval.

Thus, the function $f(x) < 0$ for the intervals:

• $x > 7$
• $x < -7$

Therefore, the solution to the problem is:

$x > 7$ or $x < -7$

To determine for which values of $x$ the function $f(x) = -x^2 + 49$ is positive, we solve the inequality:

$-x^2 + 49 > 0$

We start by finding the roots of the associated quadratic equation:

$-x^2 + 49 = 0$

This can be rewritten as:

$x^2 = 49$

Solving for $x$, we have:

$x = \pm 7$

The roots are $x = 7$ and $x = -7$. These points are where the parabola intersects the x-axis. Since the parabola opens downwards (as the coefficient of $x^2$ is negative), the function is positive between the roots.

Therefore, the function $f(x) > 0$ over the interval:

$-7 < x < 7$

Thus, the correct choice is:

$-7 < x < 7$

To solve the problem of finding the values of $x$ for which $y = 3x^2 - 27 < 0$, we start by solving the equation $3x^2 - 27 = 0$:

Step 1: Solve the equation $3x^2 - 27 = 0$.

• Rearrange the equation as $3x^2 = 27$.
• Divide both sides by 3 to simplify: $x^2 = 9$.
• Take the square root of both sides: $x = \pm 3$.

The solutions $x = 3$ and $x = -3$ are the roots of the quadratic function. This means the function transitions from negative to non-negative (and vice versa) at these points.

Step 2: Analyze the intervals defined by the roots.

• The potential intervals of interest are $x < -3$, $-3 < x < 3$, and $x > 3$.

Since the quadratic is a parabola opening upwards (coefficient of $x^2$ is positive), the function $y = 3x^2 - 27$ will be negative between the roots.

Therefore, check the interval $-3 < x < 3$:

• When $x = 0$, substitute into the function: $y = 3(0)^2 - 27 = -27$, which is negative.

The function $y = 3x^2 - 27$ is negative in the interval $-3 < x < 3$.

Thus, the values of $x$ for which $f(x) < 0$ are $-3 < x < 3$.

To solve the given problem, follow these steps:

• Step 1: Determine the direction of the parabola by examining the coefficient of $x^2$. Since $a = 3$, the parabola opens upwards.
• Step 2: Analyze the quadratic to find where $y = 3x^2 + 21 < 0$.
• Step 3: Find the vertex to determine the minimum point of the parabola. In this case, since there is no $x$ term (i.e., $b = 0$), the vertex lies on the y-axis, specifically at $x = 0$.
• Step 4: Evaluate the function at the vertex, $y(0) = 3(0)^2 + 21 = 21$, confirming it is always positive and does not drop below zero.
• Step 5: As the discriminant $\Delta = b^2 - 4ac = 0^2 - 4 \times 3 \times 21 = -252$ is negative, there are no real roots, meaning the parabola does not cross the x-axis.

Given the function is always positive and never reaches zero or becomes negative, there are no values of $x$ for which $f(x) < 0$.

Thus, the solution to this problem is No x.

Let's solve the problem by following these steps:

• Step 1: Determine the roots of $f(x) = -2x^2 + 32$.
• Step 2: Find where $f(x)$ changes sign.
• Step 3: Determine intervals where $f(x) < 0$.

Step 1: Solving the equation $-2x^2 + 32 = 0$.

Rearrange to find the roots:

$-2x^2 + 32 = 0$ implies $2x^2 = 32$, and dividing both sides by 2 gives us:

$x^2 = 16$.

Taking the square root on both sides results in:

$x = \pm 4$.

Step 2: Identify the intervals defined by the roots $x = -4$ and $x = 4$.

We have three intervals to test: $(-\infty, -4)$, $(-4, 4)$, and $(4, \infty)$.

Step 3: Analyze the sign of the function in each interval:

• For $x < -4$: Choose $x = -5$ as a test point. Substitute into the function:
  $f(-5) = -2(-5)^2 + 32 = -2(25) + 32 = -50 + 32 = -18$.
  The function is negative.
• For $-4 < x < 4$: Choose $x = 0$ as a test point. Substitute into the function:
  $f(0) = -2(0)^2 + 32 = 32$.
  The function is positive.
• For $x > 4$: Choose $x = 5$ as a test point. Substitute into the function:
  $f(5) = -2(5)^2 + 32 = -2(25) + 32 = -50 + 32 = -18$.
  The function is negative.

Therefore, the function is negative for $x > 4$ or $x < -4$.

The correct choice is: $x > 4$ or $x < -4$.

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

• Step 1: Solve for the roots of the equation $f(x) = -2x^2 + 32 = 0$.
• Step 2: Use the roots to determine intervals and test where the function is positive.

Let's work through each step:

Step 1: Solve the equation for roots:
The equation given is $-2x^2 + 32 = 0$. We can find the roots by isolating $x^2$:

$-2x^2 + 32 = 0$
$-2x^2 = -32$
$x^2 = 16$

Taking the square root of both sides gives $x = \pm 4$. So, the roots are $x = 4$ and $x = -4$.

Step 2: Determine the intervals and test for positivity:

The roots split the real number line into the intervals $(-\infty, -4)$, $(-4, 4)$, and $(4, \infty)$. We test the sign of $f(x)$ within these intervals:

• For $x \in (-\infty, -4)$, choose $x = -5$: $f(-5) = -2(-5)^2 + 32 = -50 + 32 = -18$. So, $f(x) < 0$.
• For $x \in (-4, 4)$, choose $x = 0$: $f(0) = -2(0)^2 + 32 = 32$. So, $f(x) > 0$.
• For $x \in (4, \infty)$, choose $x = 5$: $f(5) = -2(5)^2 + 32 = -50 + 32 = -18$. So, $f(x) < 0$.

Therefore, the function $f(x) = -2x^2 + 32$ is positive only between the roots, i.e., in the interval $-4 < x < 4$.

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

To solve the inequality $2x^2 - 50 < 0$, we follow these steps:

• Set the quadratic equation equal to zero to find the roots: $2x^2 - 50 = 0$.

• Rearrange and solve for $x$:

$\begin{aligned} 2x^2 - 50 &= 0\\ 2x^2 &= 50\\ x^2 &= 25\\ x &= \pm \sqrt{25}\\ x &= \pm 5 \end{aligned}$

These roots, $x = -5$ and $x = 5$, are where the function $y = 2x^2 - 50$ is equal to zero.

We now examine the intervals determined by these roots to find where the function is negative:

• $x < -5$

• $-5 < x < 5$

• $x > 5$

Since the quadratic is an upward opening parabola (coefficient of $x^2$ is positive), it attains its minimum value between its roots and increases outside them.

Testing a point in each interval:

• (For $x = 0$ in the interval $-5 < x < 5$): $y = 2(0)^2 - 50 = -50 < 0$.

• (Other intervals will be positive) such as $x = -6$ or $x = 6$, will have $y > 0$.

Thus, the function is negative in the interval $-5 < x < 5$.

Therefore, the values of $x$ that satisfy $2x^2 - 50 < 0$ are:

$-5 < x < 5$.

To solve the problem of finding the values of $x$ for which $f(x) = -x^2 + 1 > 0$, we'll follow these steps:

• Step 1: Begin with the inequality $-x^2 + 1 > 0$.
• Step 2: Rearrange it to $1 > x^2$.
• Step 3: Recognize this can also be expressed as $x^2 < 1$.
• Step 4: Since $x^2 < 1$, the solutions for $x$ are those values between $-1$ and $1$: $-1 < x < 1$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-1 < x < 1$.

To solve this problem, we need to determine where the quadratic function $y = -x^2 + 1$ is negative.

• First, identify the roots of the equation by setting $y = 0$:
  $-x^2 + 1 = 0$ simplifies to $-x^2 = -1$ or $x^2 = 1$.
• Solving $x^2 = 1$ gives us the roots: $x = 1$ and $x = -1$.
• Now, examine the intervals defined by these roots: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.
• Test each interval:
  • For $x < -1$, choose $x = -2$:
    $y = -(-2)^2 + 1 = -4 + 1 = -3$, which is negative.
  • For $-1 < x < 1$, choose $x = 0$:
    $y = -(0)^2 + 1 = 1$, which is positive.
  • For $x > 1$, choose $x = 2$:
    $y = -(2)^2 + 1 = -4 + 1 = -3$, which is negative.
• Thus, the function is negative for $x < -1$ and $x > 1$.

Therefore, the values of $x$ for which $f(x) < 0$ are $x < -1$ or $x > 1$.

To determine for which values of $x$ the function $f(x) = x^2 - 4$ is less than 0, we need to solve the inequality $x^2 - 4 < 0$.

First, find when the function equals zero by solving $x^2 - 4 = 0$. This gives the roots as $x^2 = 4$, or $x = \pm 2$, specifically $x = 2$ and $x = -2$.

Next, perform a sign analysis of $f(x) = x^2 - 4$ in the intervals defined by these roots: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.

• For $x \in (-\infty, -2)$: Pick a test point, such as $x = -3$. Calculate $f(-3) = (-3)^2 - 4 = 9 - 4 = 5$. Here, the function is positive.
• For $x \in (-2, 2)$: Pick a test point, such as $x = 0$. Calculate $f(0) = 0^2 - 4 = -4$. Here, the function is negative, satisfying $f(x) < 0$.
• For $x \in (2, \infty)$: Pick a test point, such as $x = 3$. Calculate $f(3) = 3^2 - 4 = 9 - 4 = 5$. Here, the function is positive.

Thus, the function $f(x) = x^2 - 4$ is less than 0 for $-2 < x < 2$.

The correct interval representing the solution is $-2 < x < 2$.