Question

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

pick -2 as the vlaue

Answer 1

Given the zeros of the function are at \( x = -3 \), \( x = 1 \), and \( x = 2\), let's determine the behavior of the function between these zeros. We can pick \( x = -2 \), which is between the zeros \( -3 \) and \( 1 \).

To determine if the graph is above or below the x-axis at \( x = -2 \), we would need the actual function that has these zeros. However, since we don't have the specific function, we can use the fact that polynomials typically change their signs at each zero.

### Steps to Analyze the Sign

1. **Identify the intervals:**
- From \( -\infty \) to \( -3 \)
- From \( -3 \) to \( 1 \) (choose \( x = -2 \))
- From \( 1 \) to \( 2 \)
- From \( 2 \) to \( +\infty \)

2. **Determine the sign in each interval:**
- In the interval \( (-\infty, -3) \), the function will be positive (above the x-axis) because it hasn't crossed any zeros yet.
- At \( x = -3 \), the function is zero.
- In the interval \( (-3, 1) \) (where \( x = -2 \) is):
- Since the function changes sign at \( -3 \) (from positive to negative), the function will be negative (below the x-axis) between \( -3 \) and \( 1 \).
- At \( x = 1 \), the function is zero.
- In the interval \( (1, 2) \):
- The function changes from negative to positive at \( x = 1\) (so it will be positive in this interval).
- At \( x = 2 \), the function is zero.
- In the interval \( (2, +\infty) \):
- The function will be positive again.

### Conclusion

At \( x = -2 \), the function is below the x-axis. If plotted, you would have:

- Zeros at points: \( (-3, 0) \), \( (1, 0) \), and \( (2, 0) \).
- The point at \( (-2, f(-2)) \) would be plotted below the x-axis.

If you want to visualize this on a graph, you can sketch the x-axis and plot the zeros, with the function dipping below the x-axis between \( -3 \) and \( 1 \). For an accurate graph, you would need the specific function or further points to plot its shape.