Positive and Negative Domains Practice - 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.

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 solve this problem, let's analyze the key characteristics of the parabola:

• Since the parabola does not intersect the $x$-axis, it indicates that it is entirely either above or below the $x$-axis.
• The graph of a parabola $ax^2 + bx + c$ does not intersect the $x$-axis when its discriminant $b^2 - 4ac$ is negative. Thus, it does not have any real roots.
• If the parabola opens upwards, then the function is entirely above the $x$-axis if $a > 0$ and below if $a < 0$.
• Given the problem indicates the parabola never reaches or crosses the $x$-axis and the absence of real roots, a positive opening parabola cannot reach positive territory in when not intersecting the x-axis.

Since the parabola's graph neither touches nor crosses the $x$-axis and isn't stated to be always positive or negative, we conclude:

The function does not have a positive domain.

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, we will look at the behavior of the quadratic function and determine when it is greater than zero:

• Step 1: The intersection point A is the vertex, which means $f(x) = a(x - A)^2 + k$ for some constants $a$ and $k=0$. This implies $f(x)$ changes sign at its vertex.
• Step 2: Determine if the parabola opens upwards or downwards. Since the graph of the function intersects the $x$-axis at the vertex, there are no additional real roots, which indicates either $f(x) \geq 0$ or $f(x) \leq 0$ throughout. As $f(x) > 0$ requires parts of the parabola above the $x$-axis, the parabola must open upwards.
• Step 3: For $f(x) > 0$, the graph being a parabola indicates positive $x$ intervals are outside of the vertex, i.e., $x < A$ and $x > A$.
• Step 4: The answers fitting this description are (b) $x < A$ and (c) $x > A$, which combined correspond to option (d) "Answers (b) + (c) are correct".

Therefore, the correct intervals for $f(x) > 0$ are both $x < A$ and $x > A$, leading to:

Answers (b) + (c) are correct.