Positive and Negative Domains - Examples, Exercises and Solutions

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 the problem of finding all $x$ values where $f(x) < 0$, we analyze the graph provided:

The graph of the function $f(x)$ shows it is below the x-axis in the interval from $x = -10$ to $x = -2$. Between these points, $f(x)$ is negative because the complete span of the graph resides beneath the x-axis between these points.

Steps to validate this are:

• Recognize the x-intercepts, which occur at $x = -10$ and $x = -2$, where the curve crosses the x-axis.
• The graph stays below the x-axis between these intercepts, indicating the function is negative.

Thus, the correct interval where $f(x) < 0$ is $-10 < x < -2$.

Therefore, the solution to the problem is $-10 < x < -2$.

To solve the given problem using the graph, we need to determine the intervals along the x-axis where the quadratic function $f(x)$ is positive, based on its x-intercepts $x = -11$ and $x = -1$ as shown on the graph.

• Step 1: Identify the x-intercepts from the graph: $x = -11$ and $x = -1$.
• Step 2: Interpret the graph of the quadratic function. Since it is a parabola opening upwards and touches the x-axis at $x = -11$ and $x = -1$, these are points where the quadratic changes sign.
• Step 3: Determine the intervals: The graph is above the x-axis (positive) between the x-intercepts because the parabola is opening upwards. Therefore, the function is positive for $-11 < x < -1$.

The conclusion is that the quadratic function $f(x)$ is greater than zero in the interval $-11 < x < -1$.

Therefore, the correct answer is $\mathbf{-11 < x < -1}$.

To determine where the function $f(x)$ is less than 0, observe the graphical representation:

• The roots are located at $x = -11$, $x = -6$, and $x = -1$. These are the x-values where the function intersects the x-axis.
• Considering the general behavior of quadratic functions, the function is negative between the outer roots unless it passes through below x-axis at multiple roots due to shape.

The given graph suggests the function dips below the x-axis between $x = -11$ and $x = -1$, passing through $x = -6$.

After analyzing the intervals:

• The interval to the left: $x < -11$
• The interval to the right: $x > -1$

Therefore, values of $x$ for which the function $f(x)$ is less than 0 are $x > -1$ or $x < -11$.

The correct choice is: $x > -1$ or $x < -11$

In this problem, we are tasked with determining the values of $x$ for which the function $f(x)$ is positive. We have been provided a graphical representation of the function, and we will use this graph to find our solution.

Based on the graph, we observe the following behavior of the function $f(x)$:

• The function intersects the x-axis at $x = -4$. This indicates a potential root or turning point where the function transitions from positive to negative or vice versa.
• From the graph, it appears that the function is above the x-axis on both sides of $x = -4$, except exactly at $x = -4$, where it touches the x-axis.

Hence, the function $f(x)$ is positive for $x > -4$ and for $x < -4$. Note that exactly at $x = -4$, the function is zero, not positive.

Therefore, the solution is: $x > -4$ or $x < -4$.

In conclusion, the function $f(x)$ is positive for these values of $x$, except the point where it touches the x-axis.

The corresponding choice given the problem's options is:

x > -4 or x < -4

Let's analyze the graph to determine where $f(x) < 0$.

The process to follow is:

• Identify the x-axis intersections (roots) where $f(x) = 0$.
• Notice where the graph dips below the x-axis, indicating $f(x) < 0$.
• The graph crosses and only touches the x-axis at $x = -4$.
• The graph lies below the x-axis both to the left and right of $x = -4$.

From this analysis, the function $f(x)$ is negative for all $x$ except at $x = -4$, where it touches but doesn’t dip below the x-axis.

Therefore, the solution is that the function is negative for $x > -4$ or $x < -4$.

Based on the graph provided, we can see the entire function lies below the x-axis. Thus, there is no interval where $f(x) > 0$.

To solve this problem, here's what we observed:

• Visual inspection of the graph reveals that it never crosses the x-axis from below.
• Consequently, the function remains non-positive for all x-values visible, indicating it's non-positive overall within the range observable.

Therefore, the function has no domain where it is positive. Therefore, the solution is:

The function has no domain where it is positive

To solve the problem of determining where $f(x) < 0$:

• Step 1: Identify the x-intercepts of the graph, which are $x = 1$ and $x = 7$.
• Step 2: Examine the section of the graph between these intercepts. Since the graph dips below the x-axis between these values, $f(x) < 0$ in that interval.

Therefore, the function is negative between the roots, i.e., $1 < x < 7$.

Thus, the solution to the problem is: $1 < x < 7$.

To solve for when f(x) < 0 on the graph, we follow these steps:

• Step 1: Locate the x-intercepts, where the curve intersects the x-axis. These intercepts are $x = 2$ and $x = 8$.

• Step 2: Analyze the sections determined by these intercepts. The graph is below the x-axis to the left of $x = 2$ and to the right of $x = 8$.

By visually inspecting the graph, it is evident that:

• The function $f(x)$ is below the x-axis (i.e., negative) for x < 2 and x > 8 .

Therefore, the solution to the problem is that the graph of the function is negative for x > 8 or x < 2 .

The problem is asking us to identify for which $x$ values $f(x) > 0$ based on the graph provided, which seems to depict a quadratic function. Let's go step-by-step:

First, we need to determine the points where the function intersects the x-axis, which are the roots of the function. The graph shows these intersections at $x = -2$ and $x = 2$. These are the points where the function is equal to zero, $f(x) = 0$.

Next, we observe the overall shape of the graph to understand where $f(x) > 0$ (i.e., where the graph is above the x-axis). Typically for a quadratic function, which is a parabola, the parabola will be above the x-axis outside the roots if it opens upwards, and between the roots if it opens downwards, given that $a(x - x_1)(x - x_2) = 0$ with root analysis on $a > 0$.

In the provided graph, the parabola appears to open upwards. Therefore, the function $f(x)$ is positive when $x$ is less than the smaller root, $-2$, or greater than the larger root, $2$. This is a typical behavior for a quadratic function which opens upwards, where it takes negative values inside the range of its roots and positive values outside.

Conclusively, $f(x) > 0$ for the intervals where $x < -2$ or $x > 2$.

Therefore, the solution to the problem is $x > 2$ or $x < -2$.

To solve the problem of finding for which $x$ values the function $f(x) < 0$, we proceed as follows:

First, we observe the provided graph of the function. Our goal is to identify the intervals on the $x$-axis where the curve of the function is below the line $y = 0$ (the x-axis). These intervals represent where the function $f(x)$ takes negative values.

Upon examining the graph, we notice that:

• The curve descends below the x-axis starting once it crosses $x = -2$.
• It continues below the x-axis until it reaches $x = 2$.
• Therefore, the function is negative between these two points.

Based on the graph, the interval where $f(x) < 0$ is from $x = -2$ to $x = 2$. Thus, the correct mathematical statement for the values of $x$ where $f(x) < 0$ is $-2 < x < 2$.

The correct choice from the options given is \(\text{$-2 < x < 2$}\).

Therefore, the solution to the problem is $-2 < x < 2$.

First, we examine the provided graph of the quadratic function $f(x)$. The graph clearly shows the x-intercepts (where the function crosses the x-axis) at $x = -3$ and $x = 3$.

Since the quadratic function appears to be a standard parabola opening upwards, the portion of the graph between these two x-intercepts will be above the x-axis, which means that $f(x) > 0$ in this interval.

The intervals to the left of $x = -3$ and to the right of $x = 3$ will be where the graph lies below the x-axis, meaning $f(x) < 0$ in those regions.

Thus, the graph shows that the function $f(x)$ is positive between $x = -3$ and $x = 3$. Therefore, the solution to the problem is:

$-3 < x < 3$