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To determine the domain, range, and graph of the given piecewise function $ f(x) $, let's analyze each part of the function and its domain separately.

The function is given as:
\[ f(x) = \begin{cases} 
(x+2)^2 + 1 & \text{if } x \in (-\infty, -4] \\
\sqrt{x+1} & \text{if } x \in (0, 4) \\
2x - 8 & \text{if } x \in [4, \infty) 
\end{cases} \]

### Domain

The domain of $ f(x) $ is the set of all $ x $ values for which $ f(x) $ is defined. Each piece of the function has its own domain:

1. $ (x+2)^2 + 1 $ is defined for $ x \in (-\infty, -4] $.
2. $ \sqrt{x+1} $ is defined for $ x \in (0, 4) $ (as the square root function requires $ x+1 \geq 0 $).
3. $ 2x - 8 $ is defined for $ x \in [4, \infty) $.

Combining these, the domain of $ f(x) $ is:
\[ \text{Domain} = (-\infty, -4] \cup (0, 4) \cup [4, \infty) \]

### Range

Next, we find the range of each piece of the function:

1. For $ (x+2)^2 + 1 $ when $ x \in (-\infty, -4] $:
   - As $ x $ approaches $ -\infty $, $ (x+2)^2 $ approaches $ \infty $, so $ (x+2)^2 + 1 $ also approaches $ \infty $.
   - At $ x = -4 $, $ f(x) = (-4+2)^2 + 1 = 4 + 1 = 5 $.
   - Therefore, the range for this part is $ [5, \infty) $.

2. For $ \sqrt{x+1} $ when $ x \in (0, 4) $:
   - When $ x = 0^+ $, $ \sqrt{x+1} $ approaches $ \sqrt{1} = 1 $.
   - When $ x = 4^- $, $ \sqrt{x+1} $ approaches $ \sqrt{5} $.
   - Therefore, the range for this part is $ (1, \sqrt{5}) $.

3. For $ 2x - 8 $ when $ x \in [4, \infty) $:
   - When $ x = 4 $, $ 2x - 8 = 0 $.
   - As $ x $ approaches $ \infty $, $ 2x - 8 $ also approaches $ \infty $.
   - Therefore, the range for this part is $ [0, \infty) $.

Combining these, the range of $ f(x) $ is:
\[ \text{Range} = [5, \infty) \cup (1, \sqrt{5}) \cup [0, \infty) = [0, \infty) \]

### Graph

I'll now plot the graph of the function $ f(x) $ based on the defined pieces.

Let's proceed to visualize it using Python.

```python
import matplotlib.pyplot as plt
import numpy as np

# Define the pieces of the function
x1 = np.linspace(-10, -4, 400)
x2 = np.linspace(0, 4, 400)
x3 = np.linspace(4, 10, 400)

y1 = (x1 + 2)**2 + 1
y2 = np.sqrt(x2 + 1)
y3 = 2*x3 - 8

# Plot the function
plt.figure(figsize=(10, 6))

plt.plot(x1, y1, label=r'$(x+2)^2 + 1, x \in (-\infty, -4]$')
plt.plot(x2, y2, label=r'$\sqrt{x+1}, x \in (0, 4)$')
plt.plot(x3, y3, label=r'$2x - 8, x \in [4, \infty)$')

# Highlight the boundary points
plt.scatter([-4], [(4+2)**2 + 1], color='red')  # point at x = -4
plt.scatter([0], [np.sqrt(0+1)], color='red')   # point at x = 0
plt.scatter([4], [np.sqrt(4+1)], color='red')   # point at x = 4, sqrt function


Discover how to find the domain and range of the piecewise function defined by f(x) = (x+2)^2 + 1 for x in (-∞, -4], f(x) = sqrt(x+1) for x in (0, 4), and f(x) = 2x - 8 for x in [4, ∞). This guide includes a step-by-step analysis and graphical representation. Suitable for students in Grades 10-12.

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