Positive and Negative Domains: Representation of missing b value

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 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 $x$.

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.

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

• Step 1: Analyze the quadratic function given: $y = -x^2 - 4$.
• Step 2: Analyze what $y < 0$ implies for the values of $x$.

Now, let's solve it:

Step 1: Consider the given quadratic function $y = -x^2 - 4$. The function $y = -x^2$ represents a downward-opening parabola due to the negative sign before $x^2$. The entire graph of the parabola, being shifted downward by 4 units with the term $-4$, will be wholly beneath the x-axis since there's no positive vertex or value. The vertex of the parabola is at $(0, -4)$, which is already below zero.

Step 2: Because the quadratic term causes the graph to be a parabola opening downwards, it means that for any $x$, $y = -x^2 - 4 \leq -4$, which is always less than zero. Thus, the inequality $f(x) = -x^2 - 4 < 0$ is satisfied for all real numbers $x$.

Therefore, the solution is that the inequality is true for all $x$.

To determine where the function $f(x) = -x^2 + 9$ is less than 0, we first need to find the points where it equals 0.

We solve the equation:

$-x^2 + 9 = 0$

This can be rearranged to:

$x^2 = 9$

Taking the square root of both sides, we get:

$x = 3$ or $x = -3$

The roots of the equation are $x = 3$ and $x = -3$. These points divide the x-axis into three intervals: $x < -3$, $-3 < x < 3$, and $x > 3$.

Next, we determine the sign of $f(x)$ in each interval:

• For $x < -3$, pick a test point like $x = -4$:
  $f(-4) = -(-4)^2 + 9 = -16 + 9 = -7$. Here, $f(x) < 0$.
• For $-3 < x < 3$, pick a test point like $x = 0$:
  $f(0) = -(0)^2 + 9 = 9$. Here, $f(x) > 0$.
• For $x > 3$, pick a test point like $x = 4$:
  $f(4) = -(4)^2 + 9 = -16 + 9 = -7$. Here, $f(x) < 0$.

Therefore, the function $f(x) = -x^2 + 9$ is negative for $x < -3$ and $x > 3$.

The correct answer is $x > 3$ or $x < -3$.

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 $x < -4$.

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 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 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:

$-5 < x < 5$.

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 this problem, we'll follow these steps:

• Step 1: Solve the quadratic equation to find the roots.
• Step 2: Determine intervals based on these roots.
• Step 3: Test the sign of the function on each interval.

Now, let's work through each step:

Step 1: The given function is $y = 3x^2 - 27$. To find where this function equals zero, solve the equation $3x^2 - 27 = 0$.

Factor the equation:
$3x^2 - 27 = 3(x^2 - 9) = 3(x - 3)(x + 3) = 0$. The solutions are $x = 3$ and $x = -3$.

Step 2: The critical points from step 1 divide the number line into three intervals: $x < -3$, $-3 < x < 3$, and $x > 3$.

Step 3: Test each interval:

• For $x < -3$: Choose a test value like $x = -4$. Plug it into the inequality $3x^2 - 27 > 0$:
  $y = 3(-4)^2 - 27 = 3(16) - 27 = 48 - 27 = 21 > 0$. The function is positive.
• For $-3 < x < 3$: Choose a test value like $x = 0$. Plug it into the inequality:
  $y = 3(0)^2 - 27 = -27 < 0$. The function is negative.
• For $x > 3$: Choose a test value like $x = 4$. Plug it into the inequality:
  $y = 3(4)^2 - 27 = 48 - 27 = 21 > 0$. The function is positive.

We conclude that the function is positive for $x > 3$ or $x < -3$.

Therefore, the solution to the problem is $x > 3$ or $x < -3$.

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:

$-7 < x < 7$

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 find the solution to when $y = x^2 - 4$ is greater than zero, we start by analyzing the inequality:

$x^2 - 4 > 0$

This expression can be factored as:

$(x - 2)(x + 2) > 0$

The roots of the equation $(x - 2)(x + 2) = 0$ are $x = 2$ and $x = -2$. These points divide the real number line into intervals. We need to determine on which intervals the expression is positive. Let's analyze the intervals:

• $x < -2$: Choose $x = -3$, then both factors are negative: $(x - 2) < 0$ and $(x + 2) < 0$, making their product $(x - 2)(x + 2) > 0$.
• $-2 < x < 2$: Choose $x = 0$, then one factor is positive and the other is negative: $(x - 2) < 0$ and $(x + 2) > 0$, making their product negative.
• $x > 2$: Choose $x = 3$, then both factors are positive: $(x - 2) > 0$ and $(x + 2) > 0$, making their product positive.

Therefore, the function $x^2 - 4$ is positive for $x > 2$ or $x < -2$.

Thus, the solution to the inequality $y > 0$ is:

$x > 2$ or $x < -2$.

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

• Step 1: Identify the form and behavior of the given quadratic function.
• Step 2: Analyze the vertical position of the function's vertex.
• Step 3: Determine inequality feasibility for positive $y$.

Step 1: The function given is $y = -x^2 - 4$, which is a parabola opening downwards because of the negative coefficient of $x^2$.

Step 2: The vertex of this parabola occurs at $x = -\frac{b}{2a} = 0$, considering no $x$ term present (i.e., $b = 0$). The vertex value of $y$ is $f(0) = -0^2 - 4 = -4$.

Step 3: The vertex, being the highest point at $(0, -4)$, means the entire parabola remains below the x-axis.

Since the maximum point is negative and the parabola only descends from this vertex, the function $y = -x^2 - 4$ remains less than zero for every $x$. There are no $x$ values making $y$ positive.

Therefore, the solution is No x.

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 this problem, we need to determine where the function $y = 2x^2 - 50$ is greater than zero.

Step 1: Find the roots of the equation $2x^2 - 50 = 0$.

Step 1.1: Solve the equation:

• $2x^2 - 50 = 0$
• $2x^2 = 50$
• $x^2 = 25$
• $x = \pm 5$

The roots are $x = 5$ and $x = -5$. These are the points where the parabola touches the x-axis.

Step 2: Analyze intervals defined by the roots. The $x$-values divide the number line into three intervals: $x < -5$, $-5 < x < 5$, and $x > 5$.

Step 3: Test each interval to find where $2x^2 - 50 > 0$.

• For $x < -5$: Choose a test point, e.g., $x = -6$. Then $y = 2(-6)^2 - 50 = 72 - 50 = 22$. Since 22 is positive, $y > 0$ for $x < -5$.
• For $-5 < x < 5$: Choose a test point, e.g., $x = 0$. Then $y = 2(0)^2 - 50 = -50$. Since -50 is negative, $y < 0$ for $-5 < x < 5$.
• For $x > 5$: Choose a test point, e.g., $x = 6$. Then $y = 2(6)^2 - 50 = 72 - 50 = 22$. Since 22 is positive, $y > 0$ for $x > 5$.

Therefore, the intervals where $y = 2x^2 - 50 > 0$ are $x < -5$ and $x > 5$.

The solution is thus: $x > 5$ or $x < -5$, corresponding to choice 4.

Let's solve this step-by-step:

• The function given is $y = -4x^2 - 12$.
• This quadratic function is of the form $y = ax^2 + bx + c$, where $a = -4$, $b = 0$, and $c = -12$.
• The parabola opens downwards because $a = -4$ is negative.
• The vertex form of the quadratic equation is used to find the maximum point, which in this configuration is the vertex. For this standard form, $x = -\frac{b}{2a} = 0$.
• The vertex is located at $(0, -12)$.
• Calculate the discriminant to identify the x-intercepts (if any):
• The discriminant $\Delta = b^2 - 4ac = 0^2 - 4(-4)(-12) = -192$.
• Since the discriminant is negative, there are no real roots for the quadratic equation. This means the function does not intersect the x-axis and is always below it.

Given that the parabola opens downwards and never meets the x-axis, the function $y = -4x^2 - 12$ is always less than zero for all real $x$.

Therefore, the solution to the problem is that the function is negative for all values of $x$.

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.

The goal is to find the values of $x$ for which $f(x) > 0$ given $y = -4x^2 - 12$. Start by analyzing the equation $y = -4x^2 - 12$.

Since the quadratic term is negative ($a = -4$), the parabola opens downwards. This means the maximum point of the parabola (its vertex) is at the top.

Find the vertex using the formula for the $x$-coordinate of the vertex, $x = -\frac{b}{2a}$. Here, $b = 0$, so the vertex is at $x = 0$.

Calculate $y$-value at the vertex:
$y = -4(0)^2 - 12 = -12$.

This evaluation confirms $y = -12$, which is less than 0 at the vertex.

Since the entire parabola opens downward and the highest point $y$ achieves is still negative ($-12$), the function is never greater than 0 at any point.

No $x$ values satisfy $y > 0$. Therefore, no values of $x$ make the quadratic positive.

Thus, the answer is: No values of $x$.