On a coordinate plane, a curved line with a minimum value of (0, 1) and a maximum value of (negative 1.3, 2.2), crosses the x-axis at (negative 2.2, 0) and crosses the y-axis at (0, 1).

Question

Which statement correctly identifies a local minimum of the graphed function?

Over the interval [–3, –2], the local minimum is 0.
Over the interval [–2, –1], the local minimum is 2.2.
Over the interval [–1, 0.5], the local minimum is 1.
Over the interval [0.5, 2], the local minimum is 4.

Answer 1

The key part of the information given is that the curved line has a minimum value of \( (0, 1) \) and a maximum value of \( (-1.3, 2.2) \).

Let's analyze each interval to determine the correct statement:

1. **Over the interval \([-3, -2]\), the local minimum is 0.**

The interval \([-3, -2]\) is to the left of the x-intercept at \((-2.2, 0)\). Therefore, this interval doesn't cover significant points related to local extrema (minimum or maximum) provided in the problem.

2. **Over the interval \([-2, -1]\), the local minimum is 2.2.**

The interval \([-2, -1]\) includes the maximum value at \((-1.3, 2.2)\), not a minimum. Therefore, this statement is not correct.

3. **Over the interval \([-1, 0.5]\), the local minimum is 1.**

The interval \([-1, 0.5]\) includes the minimum value at \( (0, 1) \), thus indicating that over this interval, the local minimum is indeed 1.

4. **Over the interval \([0.5, 2]\), the local minimum is 4.**

This interval doesn't contain any relevant information in the provided data to justify the minimum value as 4.

From the given choices and provided details, the correct statement is:

**Over the interval [-1, 0.5], the local minimum is 1.**