Positive and Negative Domains Practice - Quadratic Function

To solve this problem, we need to determine the range of values where the quadratic function $f(x)$ is negative.

Given that the graph of the function does not intersect the $x$-axis, it suggests that all real-valued outputs of the function have the same sign. This occurs because there are no real roots (solutions) to the equation $f(x) = 0$.

We identify that the quadratic function's parabola is opening upwards (concave up) because it does not intersect the $x$-axis, typically implying the entire parabola is either fully below or fully above the axis, without cutting through it.

If the parabola were above the axis, at the vertex (marked A), the function's value would be positive, and all corresponding function values would also be positive along the width of the parabola. Conversely, if it were below the axis and since the graph maintains this position, the entire function would remain negative.

The problem indicates that the parabola does not intersect or touch the $x$-axis, highlighting that $f(x)$ does not reach zero but maintains positivity or negativity uniformly along the span of $x$.

Since the final answer choice deduces that $f(x)$ does not enter a negative domain by naturally coasting along the positive regional track, the suitable conclusion is that the function has no negative domain, so there are no such values.

To solve this problem, we need to determine where the function $f(x)$ is positive. The graph of the parabola intersects the $x$-axis at points $A$ and $B$, indicating these are the roots of the function.

The behavior of the function depends on the direction in which the parabola opens:

• If the parabola opens upwards ($a > 0$), the function is positive between the roots, that is in the interval $(A, B)$.
• Conversely, if the parabola opens downwards ($a < 0$), the function is positive outside of the roots.

In the problem, although the nature (upwards or downwards opening) is not explicitly stated, the most common interpretation for an intersection analysis suggests that the parabola opens upwards ($a > 0$). Thus, the values of $x$ where $f(x) > 0$ are precisely those between the roots $A$ and $B$.

Therefore, the solution to the problem is $A < x < B$.

To solve this problem, we need to determine when $f(x)$ is negative by analyzing the graph provided.

The graph shows a quadratic function shaped as a parabola. Importantly, the parabola intersects the x-axis precisely at point A, which is its vertex. From this, we can deduce two possible scenarios:

1. If the parabola opens upwards (convex), the vertex represents the minimum point. Thus, the y-value at the vertex is greater than any other point on the function, implying there is no region where $f(x) < 0$ since the lowest point is zero.

2. If it were to open downwards, point A would be the maximum, and $f(x)$ could be negative elsewhere, but this contradicts the given information that point A is a vertex on the x-axis, suggesting the opening is upwards.

Since the graph passes through the x-axis only at vertex A and that is the minimum point, the parabola opens upwards. Therefore, the function $f(x)$ never takes negative values as it only touches the x-axis without crossing it.

Thus, the conclusion is that there are no values of $x$ for which $f(x) < 0$.

Hence, the function has no negative domain.

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

• The graph intersects the $x$-axis at points A and B, indicating $f(x) = 0$ at these points.
• The function exhibits a parabolic shape with a vertex located at point C, below the $x$-axis, suggesting that the parabola opens upwards.

For a typical upward opening parabola that intersects the $x$-axis at A and B, the function $f(x)$ is below the $x$-axis (i.e., $f(x) < 0$) outside the interval between A and B.

Therefore, the solution set for which $f(x) < 0$ is $x < A$ or $x > B$. This represents where the parabola lies beneath the $x$-axis.

This corresponds to choice 2: $x > B$ or $x < A$.

The graph of the parabola intersects the X-axis at points A and B. This tells us these are the roots of the quadratic equation, and that $f(x) = 0$ at these points. Given that the shape of the parabola (concave up or down) affects where it is positive or negative:

From the graph:

• If the parabola opens upwards (which it must, if we are finding $f(x) > 0$ outside A and B), it is positive when $x \lt A$ or $x \gt B$, as the parabola dips below the X-axis between A and B.
• If the parabola opens downwards, it would be positive between A and B, however, our task is to identify the actual nature based on a graphical interpretation.

The graph signifies the function is positive outside the interval $A \lt x \lt B$.

Therefore, the intervals where $f(x) > 0$ are:

$x > B$ or $x < A$

The answer choice that corresponds to this interpretation is:

$x > B$ or $x < A$