On a coordinate plane, a curved line with a maximum value of (negative 1, 4) and minimum values of (negative 1.25, negative 16) and (2.5, negative 3), crosses the x-axis at (negative 2.1, 0), (0.25, 0), (1.75, 0), and (3, 0), and crosses the y-axis at (0, negative 3).

Question

Which intervals show f(x) decreasing? Check all that apply.

[–2.5, –2]
[–2, –1.5]
[–1, 1)
[1.5, 2]
[2, 2.5)
(2.5, 3]

Answer 1

To determine the intervals where the function \( f(x) \) is decreasing, we need to look at the behavior of the function around its critical points (i.e., where the function has maximum and minimum values) as well as where it crosses the x-axis.

Given the details:

1. Maximum value at \((-1, 4)\)
2. Minimum values at \((-1.25, -16)\) and \((2.5, -3)\)
3. X-intercepts at \((-2.1, 0)\), \((0.25, 0)\), \((1.75, 0)\), and \((3, 0)\)
4. Y-intercept at \((0, -3)\)

Let’s analyze the intervals:

- **Interval \([-2.5, -2]\):**
As \( x \) increases from \(-2.5\) to \(-2\), it passes through the x-intercept at \(-2.1\). Before \(-2.1\), the function would be below the x-axis and then it would increase to rise to the maximum value at \(-1\). Therefore, \( f(x) \) is decreasing from \(-2.5\) to \(-2.1\) but not the entire interval \([-2.5, -2]\).

- **Interval \([-2, -1.5]\):**
This interval lies between the x-intercept \(-2.1\) and the y-coordinate where the curve is rising to the maximum at \(-1\). So \( f(x) \) is decreasing in this interval.

- **Interval \([-1, 1)\):**
\(-1\) is a maximum, meaning that from \(-1\) to a value less than \(1\), the function starts decreasing from \(\(-1\)\) as it next crosses at \(0.25\) . So \( f(x) \) is decreasing in this interval.

- **Interval \([1.5, 2]\):**
Here, between \(1.5\) to \(2\) the function value decreases from near zero and negative regions. It reaches minimum between 1.75 intersection and leading critical maximum for observation value.Indicating it's decreasing interval \( f(x) \).

- **Interval \([2, 2.5)\):**
As \( x \) increases from \(2\) to \(2.5\), the function value decreases up to the minimum value at \( (2.5, -3) \).

- **Interval \( (2.5, 3]\):**
Finally, after achieving minimum at \(2.5 \), \(x\) increases from \(2.5\) to \(3\) which showing increasing behavior converging on intersection. \( f(x) \) is rising so not a valid interval where \( f(x) \) is decreasing.

Therefore, the intervals where \( f(x) \) is decreasing are:

- \([-2, -1.5]\)
- \([-1, 1)\)
- \([1.5, 2]\)
- \([2, 2.5)\)

So we check the boxes for these intervals:
- [x] \([-2, -1.5]\)
- [x] \([-1, 1)\)
- [x] \([1.5, 2]\)
- [x] \([2, 2.5)\)