Asked by n
Choose an x-value between two consecutive zeros on the graph. Solve the function at that value to determine if the graph is above or below the x-axis. Plot the point on the same graph that shows the zeros.
zeros are -3,1, and 2
zeros are -3,1, and 2
Answers
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Answered by
GPT-4o mini
To determine whether the graph is above or below the x-axis between two consecutive zeros, we first identify the intervals between the zeros you've given, which are -3, 1, and 2. The intervals are as follows:
1. Between -3 and 1
2. Between 1 and 2
Let's evaluate the function at a point within each of these intervals. Since you didn't provide the specific function, let’s assume an example polynomial function, such as \( f(x) = (x + 3)(x - 1)(x - 2) \). This function will have zeros at -3, 1, and 2.
### Interval 1: Between -3 and 1
- Choose an x-value: \( x = 0 \)
- Evaluate the function:
\[
f(0) = (0 + 3)(0 - 1)(0 - 2) = 3 \cdot (-1) \cdot (-2) = 6
\]
Since \( f(0) = 6 \), the graph is above the x-axis in the interval between -3 and 1.
### Interval 2: Between 1 and 2
- Choose an x-value: \( x = 1.5 \)
- Evaluate the function:
\[
f(1.5) = (1.5 + 3)(1.5 - 1)(1.5 - 2) = 4.5 \cdot 0.5 \cdot (-0.5) = -1.125
\]
Since \( f(1.5) = -1.125 \), the graph is below the x-axis in the interval between 1 and 2.
### Summary of Findings:
- **Between -3 and 1**: Chose \( x = 0 \), \( f(0) = 6 \) (above the x-axis).
- **Between 1 and 2**: Chose \( x = 1.5 \), \( f(1.5) = -1.125 \) (below the x-axis).
### Plotting Points:
- The zeros can be plotted at: (-3, 0), (1, 0), and (2, 0).
- The points evaluated can be plotted:
- (0, 6) for the point above the x-axis
- (1.5, -1.125) for the point below the x-axis
You can draw this on a graph with an x-axis and y-axis, marking the x-values and corresponding function values to visualize the behavior of the function around the zeros.
1. Between -3 and 1
2. Between 1 and 2
Let's evaluate the function at a point within each of these intervals. Since you didn't provide the specific function, let’s assume an example polynomial function, such as \( f(x) = (x + 3)(x - 1)(x - 2) \). This function will have zeros at -3, 1, and 2.
### Interval 1: Between -3 and 1
- Choose an x-value: \( x = 0 \)
- Evaluate the function:
\[
f(0) = (0 + 3)(0 - 1)(0 - 2) = 3 \cdot (-1) \cdot (-2) = 6
\]
Since \( f(0) = 6 \), the graph is above the x-axis in the interval between -3 and 1.
### Interval 2: Between 1 and 2
- Choose an x-value: \( x = 1.5 \)
- Evaluate the function:
\[
f(1.5) = (1.5 + 3)(1.5 - 1)(1.5 - 2) = 4.5 \cdot 0.5 \cdot (-0.5) = -1.125
\]
Since \( f(1.5) = -1.125 \), the graph is below the x-axis in the interval between 1 and 2.
### Summary of Findings:
- **Between -3 and 1**: Chose \( x = 0 \), \( f(0) = 6 \) (above the x-axis).
- **Between 1 and 2**: Chose \( x = 1.5 \), \( f(1.5) = -1.125 \) (below the x-axis).
### Plotting Points:
- The zeros can be plotted at: (-3, 0), (1, 0), and (2, 0).
- The points evaluated can be plotted:
- (0, 6) for the point above the x-axis
- (1.5, -1.125) for the point below the x-axis
You can draw this on a graph with an x-axis and y-axis, marking the x-values and corresponding function values to visualize the behavior of the function around the zeros.
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