Question

Analyze the graph of the function f(x) to complete the statement.

On a coordinate plane, a curved line, labeled f of x, with a minimum value of (0, negative 3) and a maximum value of (negative 2.4, 17), crosses the x-axis at (negative 3, 0), (negative 1.1, 0), and (0.9, 0), and crosses the y-axis at (0, negative 3).

f(x)<0 over and what other interval?

Answer 1

To determine the intervals where \( f(x) < 0 \), we analyze the x-intercepts where the function crosses the x-axis, as well as the given minimum and maximum values.

The x-intercepts are at:
- \( x = -3 \)
- \( x = -1.1 \)
- \( x = 0.9 \)

This means that \( f(x) \) changes from negative to positive at these points. Specifically:

1. **From \( -\infty \) to \( -3 \)**: Since the function must be negative in this range (as it hasn't crossed the x-axis yet).
2. **Between the x-intercepts**:
- From \( -3 \) to \( -1.1 \) is negative.
- From \( -1.1 \) to \( 0.9 \) the function is positive (as it crosses the x-axis from below).

3. **From \( 0.9 \) to \( +\infty \)**: After crossing at \( 0.9 \), the function remains positive.

The function crosses the y-axis at \( (0, -3) \), meaning it is negative at \( x = 0 \), confirming that it is below the x-axis between \( -3 \) and \( -1.1 \).

### Conclusion:

Thus, \( f(x) < 0 \) over the following intervals:
- \( (-\infty, -3) \)
- \( (-3, -1.1) \)

So the complete statement is:
**\( f(x) < 0 \) over the intervals \( (-\infty, -3) \) and \( (-3, -1.1) \).**