Positive and Negative Domains: Logic and comprehension questions

The problem concerns a downward-opening parabola with a vertex below the x-axis. Let's elaborate on these terms to determine if the given statement is correct:

• Step 1: Identify Quadratic Form
  We consider a quadratic function $f(x) = ax^2 + bx + c$ with $a < 0$ since it opens downwards.
• Step 2: Understand the Vertex's Position
  The vertex form of the quadratic is $f(x) = a(x - h)^2 + k$. Given that the vertex is below the x-axis, the condition $k < 0$ holds. The vertex at coordinate $(h, k)$ is the maximum point of the parabola.
• Step 3: Evaluate the Parabola's Sign
  Since $k < 0$, the vertex $(h, k)$ is the highest point of the parabola lying below the x-axis. Consequently, all other points on the parabola satisfy $f(x) \leq k < 0$.

Therefore, the entire quadratic function is negative for all values of $x$ due to the vertex being the maximum and its y-coordinate being negative. Thus, the statement that such a parabola is always negative is indeed "correct".

Thus, the correct choice is: $\text{Correct}$.

The solution to this problem involves understanding the behavior of parabolas. Given that the parabola opens upwards, indicated by $a > 0$, and the vertex $(h,k)$ is below the x-axis, $k < 0$, here's the detailed explanation:
1. The vertex of the parabola, $(h, k)$, is at the lowest point because the parabola opens upwards.
2. With $k < 0$, the value of $f(h) = k$ is negative. However, as $x$ moves away from the vertex, the function increases since it opens upwards.

Therefore, for large $|x|$, $f(x)$ becomes positive. For instance, at $x = h \pm \sqrt{\frac{-k}{a}}$, the parabola can cross the x-axis and become positive.

Given that the parabola will eventually have positive $y$-values for either very large or very small $x$, the function is not always negative. Hence, the statement that the parabola is always negative is Incorrect.

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

• Step 1: Understand the structure of a quadratic function.
• Step 2: Explore the implications of the vertex location below the x-axis.
• Step 3: Analyze specific conditions where the function might not be always positive.

Step 1:
A parabola $y = ax^2 + bx + c$ opens upwards if $a > 0$.

Step 2:
The vertex form of a quadratic is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. If $k < 0$, the vertex is below the x-axis, making $y$ negative at $x = h$ (the minimum point for an upward-opening parabola).

Step 3:
Since $y$ at the vertex is $k < 0$, it implies the function is negative at least at this point. Thus, the function cannot always be positive, as there exists at least one point where it is non-positive (negative).

Therefore, the assertion that the parabola is always positive is incorrect.

The correct answer is: Incorrect.

To analyze whether a quadratic function with two intersection points with the x-axis has sections where the function values are both positive and negative, consider the following:

• The general form of a quadratic function is $f(x) = ax^2 + bx + c$.
• The parabola intersects the x-axis at points where $f(x) = 0$, leading to the equation having real roots. This occurs when the discriminant $(b^2 - 4ac)$ is positive.
• When a parabola, which describes a quadratic function, has two distinct x-intercepts (roots), it must cross the x-axis twice.
• If $a > 0$, the parabola opens upwards, and if $a < 0$, it opens downwards.
• The curve will be above the x-axis (positive values of $f(x)$) between or outside these intercepts, depending on the sign of $a$.
• Hence, for a parabola with two intersections, there are intervals on the x-axis where the function is both positive and negative, confirming dual domain signs.

Thus, it is correct to conclude that a parabola with two distinct x-axis intersections has both positive and negative function values, satisfying the problem's assertion regarding its range and confirming the correct answer choice is:

Correct

To determine if a parabola with its vertex on the x-axis is always positive except for the vertex point, consider the vertex form of a quadratic function: $y = a(x-h)^2 + k$

• Since the vertex is on the x-axis, we have $k = 0$, so the equation simplifies to $y = a(x-h)^2$.
• The vertex of the parabola is the point $(h, 0)$.
• If $a > 0$, the parabola opens upwards, meaning $y$ values are always non-negative but equal zero at the vertex.
• If $a < 0$, the parabola opens downwards, indicating the vertex is at a maximum point, and all other $y$ values are negative.
• Therefore, the behavior of the parabola (whether it is always positive except at the vertex) is dependent on the sign of $a$, which is not specified in the problem.

Without knowing the sign of $a$, we cannot definitively determine if the parabola is always positive except at the vertex. Thus, the answer is that the outcome "cannot be determined" from the given information.

Therefore, the correct answer choice is Cannot be determined.

To evaluate the statement about the parabola, we need to understand its properties precisely:

Step 1: Given the parabola's vertex is on the x-axis, we can write its equation in the form $y = a(x-h)^2$, where $a > 0$ indicates that it opens upwards.

Step 2: The vertex form of a quadratic function has the vertex $(h, 0)$ with the vertex lying directly on the x-axis. Since $a > 0$, the parabola opens upwards, implying $(h, 0)$ is the minimum point.

Step 3: For points other than the vertex, $(x-h)^2$ is always non-negative. Since $a$ is positive, $y = a(x-h)^2 \geq 0$. Therefore, the function value at the vertex is zero, and for all other $x$, the function value is positive.

Step 4: Analyze if the function can be negative: With $a > 0$, the value $a(x-h)^2$ cannot be negative for any value of $x$.

Conclusion: The given statement that the function is "always negative except for the vertex point" is incorrect since outside the vertex, the function is always non-negative and can be positive. Hence, the statement is incorrect.

Therefore, the correct answer is Incorrect.

To solve this problem, follow these steps:

• Step 1: Recognize that the standard form of a quadratic function is $f(x) = a(x - h)^2 + k$, which represents a parabola.
• Step 2: Since the parabola is upward-opening, the coefficient $a$ is positive, $a > 0$.
• Step 3: Given that the vertex is on the x-axis, the vertex has coordinates $(h, 0)$, which means $k = 0$.
• Step 4: Substitute $k = 0$ into the standard form to get $f(x) = a(x - h)^2$.
• Step 5: Since $a > 0$, the expression $a(x - h)^2$ is always non-negative because $(x - h)^2 \geq 0$ for all $x$.
• Step 6: At the vertex $x = h$, $f(h) = a(h - h)^2 = 0$; at any other point, $f(x) > 0$ since $a(x - h)^2 > 0$.

Therefore, the function value is zero at the vertex and positive everywhere else. This confirms that the statement is correct.

The correct answer to the problem is Correct.

To solve this problem, let's analyze the function:

• Step 1: Given the parabola is in vertex form $y = a(x - h)^2 + k$ where $a > 0$ for an upward opening.
• Step 2: Since the vertex is on the x-axis, $k = 0$. Thus, the equation simplifies to $y = a(x - h)^2$.
• Step 3: Analyze values:
  • At $x = h$, the vertex, $y = a(h - h)^2 = 0$.
  • For $x \neq h$, $y = a(x - h)^2 > 0$ since $a > 0$ and any square is non-negative, but only zero at $h$.
• Step 4: Therefore, the function $y$ is zero only at the vertex $(h, 0)$ and positive for all other $x$.

The statement in the problem says the function is always positive except at the vertex. As we see, the function is indeed zero only at the vertex and positive elsewhere, meaning the statement provided is incorrect in its description if one understands it as implying it should never reach zero, which technically it does only at the vertex.

Therefore, the correct answer is Incorrect.

To solve this problem, let's analyze the given conditions:

• Step 1: We examine the standard form of a downward-opening parabola. The standard vertex form is $y = a(x - h)^2 + k$. If the vertex lies on the x-axis, $k = 0$, so the equation simplifies to $y = a(x - h)^2$.
• Step 2: Since the parabola is bending downwards, it implies that $a < 0$. Therefore, the quadratic term $a(x - h)^2$ is non-positive for all values of $x$, becoming 0 only when $x = h$.
• Step 3: Evaluate the function for $x \neq h$. For any $x \neq h$, $(x - h)^2 > 0$, so $y = a(x - h)^2 < 0$ because $a < 0$.

Conclusively, the function value is always negative for all $x \neq h$, and it is exactly zero at $x = h$ (the vertex). The statement provided corresponds precisely with this behavior.

Therefore, the statement that the function is always negative except at the vertex point is indeed Correct.

To solve this problem, let's analyze the situation:

• Given a quadratic function $y = ax^2 + bx + c$, if it opens upwards, a > 0.

• The vertex form of the parabola is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.

• If the vertex $(h, k)$ is above the x-axis, then k > 0.

First, let's restate what it means for a quadratic to be always positive:
This means the function $y = ax^2 + bx + c$ has no real roots and y > 0 for all $x$.

To ensure this, consider:

• The discriminant $\Delta = b^2 - 4ac$ helps determine if the parabola intersects the x-axis.

• If \Delta < 0, there are no real roots, meaning the parabola doesn't cross the x-axis and stays entirely above it for all $x$.

• Given a > 0 and the vertex is above the x-axis (k > 0), it ensures the function $y = a(x-h)^2 + k$ stays y > 0.

Therefore, with k > 0 and a > 0, $\Delta$ implies no x-intercepts, confirming the parabola is always positive.

The correct answer is: Correct

To solve this problem, let's consider the characteristics of a quadratic function in vertex form:

In vertex form, a quadratic function is written as $y = a(x-h)^2 + k$. Given that the vertex is on the x-axis, the vertex point, $(h, k)$, has $k = 0$. Therefore, the equation becomes $y = a(x-h)^2$.

We need to determine if the function is always negative except for the vertex point. This boils down to the sign of the coefficient $a$:

• If $a < 0$, the parabola opens downward, meaning the function is zero at the vertex and negative elsewhere, satisfying the statement.
• If $a > 0$, the parabola opens upward, meaning the function is zero at the vertex and positive elsewhere, contradicting the statement.
• If $a = 0$, the function is linear and doesn't represent a parabola.

Since no specific information regarding the sign of $a$ is given, we cannot conclusively state that the function is always negative except at the vertex.

Therefore, the solution is: Cannot be determined.

To solve this problem, we'll analyze what is implied if the parabola is smiling (opening upwards) and if its vertex is located above the x-axis:

• Step 1: Identify properties of the quadratic function. A "smiling" parabola in the form $y = a(x-h)^2 + k$ with $a > 0$ means the parabola opens upwards. The vertex $(h, k)$ will determine its intersection with the x-axis.
• Step 2: Evaluate the impact of the vertex's location. If the vertex is above the x-axis, then $k > 0$, indicating the vertex point itself is positive.
• Step 3: Assess when the function is negative. For an upwards-facing parabola, $y$ takes on positive values when above the x-axis and negative values when below. If the vertex is above the x-axis, the quadratic will indeed become positive as $x$ moves away from the vertex, indicating that it cannot always be negative.
• Step 4: Interpretation of statement: Given that the parabola could cross the x-axis in some regions other than the vertex and will achieve positive $y$-values whenever the vertex is above the x-axis, the statement "the parabola is always negative" is incorrect.

Therefore, the given statement is Incorrect.

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

• Step 1: Understand the form of the parabola.
• Step 2: Analyze the nature of values above and below the x-axis.
• Step 3: Determine if the parabola can have positive values.

Now, let's work through each step:

Step 1: Given the parabola is opening downwards, the quadratic formula can be defined as:
$y = a(x-h)^2 + k$, with $a < 0$ and $k > 0$.

Step 2: Because the vertex is above the x-axis ($k > 0$), the vertex itself is positive when considered as a point ($h, k$).

Step 3: A parabola that opens downward will eventually intersect the x-axis, creating two roots unless it remains above the x-axis—which is not generally the case when $k$ is small enough. Therefore, for values of $x$ surrounding the vertex and large enough in magnitude, $y$ can be negative.

Conclusion: The parabola is not always negative as it can be positive near its vertex.

The statement in the problem is thus Incorrect.

To solve this problem, we need to carefully examine the nature of the parabolic function given its attributes:

• The parabola opens downward, which indicates the coefficient $a < 0$ in its standard form $y = ax^2 + bx + c$.
• The vertex is above the x-axis, suggesting that its y-coordinate, given by $c - \frac{b^2}{4a}$, is positive.

However, having the vertex above the x-axis alone does not guarantee that the entire parabola remains above the x-axis. A downward-opening parabola can have parts below the x-axis even if its vertex is above it.

Consider a simple example. Take $y = -x^2 + 6$, which is a downward-opening parabola with vertex $(0, 6)$. While the vertex is above the x-axis, the roots $x = -\sqrt{6}$ and $x = \sqrt{6}$ indicate that it crosses the x-axis, thus having negative values for $x$ in $(-\sqrt{6}, \sqrt{6})$.

Therefore, the statement that the parabola is always positive is Incorrect.

To solve this problem, we need to understand the behavior of a downward-opening parabola with its vertex below the x-axis.

A quadratic function of the form $f(x) = ax^2 + bx + c$ opens downward if $a < 0$. The vertex form is $f(x) = a(x-h)^2 + k$, where the vertex is $(h, k)$. In this problem, the vertex $(h, k)$ is below the x-axis, which means $k < 0$.

For a parabola opening downward with $a < 0$, the function will have values greater than at the vertex as $x$ moves away from $h$. However, since $k < 0$, at the vertex itself, $f(h) = k$ is negative. As $x$ increases significantly away from $h$, the value of $a(x-h)^2$ becomes large and negative, due to $a < 0$, and dominates the function, causing $f(x)$ to also be negative for sufficiently large or small $x$.

Therefore, despite the downward-bending parabola having a vertex below the x-axis, it is incorrect to say the entire function is positive. The parabola will take on negative values when $x$ is sufficiently far from the vertex.

The correct conclusion is that the statement, "If a parabola is bending downwards and its vertex is below the x-axis, then it is always positive," is incorrect.