Positive and Negative Domains Practice - Quadratic Function

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 $x < A$

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 $x < A$.

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, let's analyze the graph of the quadratic function around points A and B where it intersects the $x$-axis.

• Step 1: Identify the nature of the quadratic. From the graph, it is clear that the parabola intersects the $x$-axis, suggesting $f(x) = 0$ at these points.
• Step 2: Since the problem indicates points A and B as interceptions, we can conclude the parabola crosses or touches the $x$-axis at these points.
• Step 3: Determine where $f(x) < 0$. Since A and B are roots, the parabola's graph will be below the $x$-axis between A and B if the parabola opens upwards, given by $A < x < B$. If it opens downwards, the parabola would be negative outside A and B. Based on typical quadratic behavior with a vertex below the $x$-axis, the parabola likely opens upwards.

The solution, therefore, is found within the interval between the intercepts on an upward-opening parabola. This conclusion is consistent with the graphical representation of most standard quadratics.

Thus, the values of $x$ where $f(x) < 0$ are precisely in the interval $A < x < B$.

To identify the conditions where $f(x) > 0$, we need to analyze the nature of the quadratic function as represented on the provided graph.

Based on the problem, the graph intersects the x-axis exactly at one point, recognized as point A, the vertex. In a quadratic function $ax^2 + bx + c$, if the vertex intersects at the x-axis and nowhere else, it means the graph is tangent to the x-axis at that vertex.

To determine if the function is positive, examine the orientation: - If $a > 0$, the parabola opens upwards, making it have a minimum at the vertex. - If $a < 0$, the parabola opens downwards, making it have a maximum at the vertex. Given that the problem states the parabola intersects the x-axis only at the vertex, the parabola opens downward. This is inferred from the phrased graph where no areas reach above the x-axis.

Therefore, the function never reaches a value greater than zero, as the parabola is concave down, and the vertex sits on the x-axis.

Conclusively, the range where $f(x) > 0$ is nonexistent given the parameters of the problem.

Therefore, the solution is that there are no such values.