Positive and Negative intervals of a Quadratic Function

Based on the given graph characteristics, we conclude that the parabola never intersects the $x$-axis and is therefore entirely above it due to opening upwards. This means the function is always positive for every $x$.

Thus, the correct choice is:

• Choice 3: The domain is always positive.

Therefore, the solution to the problem is the domain is always positive.

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 decide where $f(x) < 0$ for the given parabola, observe the following:

• The parabola does not intersect the x-axis, indicating it is either entirely above or below the x-axis.
• If the parabola were entirely above the x-axis for $f(x) > 0$, it would contradict the question by not giving a valid interval for $f(x) < 0$.
• Therefore, the correct conclusion is that the parabola is entirely below the x-axis, meaning $f(x) < 0$ for all $x$.

Based on the understanding of quadratic functions and their graph behavior, the function does not intersect the x-axis implies it is always negative.

Hence, the domain where $f(x) < 0$ is for all $x$. This leads us to choose:

The domain is always negative.

To determine where the function $f(x) < 0$, it's given that the parabola intersects the X-axis exactly at point A, the vertex, indicating the function has its maximum (if it opens downwards) or minimum (if it opens upwards) at this point.

Since it intersects (not crosses) the X-axis at one point, this must mean the parabola opens downwards, having its vertex at the X-axis. Thus, it tests negative to the left and right of point A, except for the vertex A itself, where $f(x) \geq 0$.

Here's the solution approach:

• Step 1: Determine parabolic orientation (vertex as max or min point).
• Step 2: Verify negative function values f(x) outside vertex.
• Step 3: Solution: If the parabola opens downwards and A is the only intersection, intervals are $x < A$ and $x > A$.

The analysis shows negative regions surrounding the vertex for downwards opening, consistent with options (b) and (c).

Therefore, the solutions are Answers (b) + (c) are correct.

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.