On a coordinate plane, a curved line with a minimum value of (5.1, negative 7) and a maximum value of (0, 25), crosses the x-axis at (negative 3.4, 0), (3.9, 0), and (6, 0), and crosses the y-axis at (0, 25).

Question

Which statement is true about the local minimum of the graphed function?

Over the interval [–4, –2], the local minimum is 0.
Over the interval [–2, –1], the local minimum is 25.
Over the interval [–1, 4], the local minimum is 0.
Over the interval [4, 7], the local minimum is -7.

Answer 1

To determine the true statement about the local minimum of the graphed function, let's analyze the given information and check each interval:

1. **Over the interval \([-4, -2]\), the local minimum is 0.**
- The function crosses the x-axis at \((-3.4, 0)\). In this interval, there is no information available to suggest there's a lower value than 0 since the provided information is sparse between \([-4, -2]\). This interval could plausibly have a minimum of 0 at \((-3.4, 0)\).

2. **Over the interval \([-2, -1]\), the local minimum is 25.**
- The function has a maximum value of \((0, 25)\), which is outside the interval \([-2, -1]\). There's no crossing or mentioned value near \((0, 25)\) within this interval, so this statement is almost certainly incorrect.

3. **Over the interval \([-1, 4]\), the local minimum is 0.**
- In the interval \([-1, 4]\), the function crosses the x-axis at \((3.9, 0)\) and suggests being 0 on the x-axis. This could mean the local minimum in this range is indeed 0, likely at \((3.9, 0)\).

4. **Over the interval \([4, 7]\), the local minimum is -7.**
- In the interval \([4, 7]\), the function has a minimum value at \((5.1, -7)\). This directly meets the interval condition and minimum value given previously.

Based on the analysis, the correct statement is:
**Over the interval \([4, 7]\), the local minimum is -7.**