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To solve for the values of \( a \) and \( b \) that make the function \( f(x) \) continuous everywhere, we need to ensure continuity at the points where the function pieces meet, specifically at \( x = -8 \) and \( x = -1 \).

### Step 1: Continuity at \( x = -8 \)
For \( f(x) \) to be continuous at \( x = -8 \), the left-hand limit (from the first piece) must equal the right-hand limit (from the second piece), and both must equal \( f(-8) \).

1. **Left-hand limit at \( x = -8 \):**
   The function for \( x < -8 \) is:
   \[
   f(x) = \frac{x^2 - 64}{x + 8}
   \]
   Simplifying this expression:
   \[
   f(x) = \frac{(x-8)(x+8)}{x+8} = x - 8 \quad \text{for} \quad x \neq -8
   \]
   So, as \( x \to -8^- \):
   \[
   \lim_{x \to -8^-} f(x) = -8 - 8 = -16
   \]

2. **Right-hand limit at \( x = -8 \):**
   The function for \( -8 \leq x < -1 \) is:
   \[
   f(x) = ax^2 - bx - 448
   \]
   Substituting \( x = -8 \):
   \[
   f(-8) = a(-8)^2 - b(-8) - 448 = 64a + 8b - 448
   \]
   To ensure continuity, we set the left-hand limit equal to the right-hand value:
   \[
   64a + 8b - 448 = -16
   \]
   Simplifying:
   \[
   64a + 8b = 432 \quad \text{(Equation 1)}
   \]

### Step 2: Continuity at \( x = -1 \)
For \( f(x) \) to be continuous at \( x = -1 \), the second piece (for \( -8 \leq x < -1 \)) and the third piece (for \( x \geq -1 \)) must match at \( x = -1 \).

1. **Left-hand limit at \( x = -1 \):**
   Using the second piece \( ax^2 - bx - 448 \), substitute \( x = -1 \):
   \[
   f(-1) = a(-1)^2 - b(-1) - 448 = a + b - 448
   \]

2. **Right-hand limit at \( x = -1 \):**
   The function for \( x \geq -1 \) is \( 436x - a + b \). Substituting \( x = -1 \):
   \[
   f(-1) = 436(-1) - a + b = -436 - a + b
   \]
   Setting the left-hand limit equal to the right-hand limit for continuity:
   \[
   a + b - 448 = -436 - a + b
   \]
   Simplifying:
   \[
   2a = 12 \quad \Rightarrow \quad a = 6 \quad \text{(Equation 2)}
   \]

### Step 3: Solve for \( b \)
Substitute \( a = 6 \) into Equation 1:
\[
64(6) + 8b = 432
\]
\[
384 + 8b = 432
\]
\[
8b = 48 \quad \Rightarrow \quad b = 6
\]

### Final Answer:
The values of \( a \) and \( b \) that make the function continuous everywhere are:
\[
a = 6, \quad b = 6
\]

Would you like further details or have any questions?

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Here are some related questions you might explore:
1. What does it mean for a function to be continuous?
2. How do you solve systems of linear equations in two variables?
3. What are other conditions besides continuity that can apply to functions?
4. How would you check differentiability at the points where the pieces of the function meet?
5. What happens to the function if \( a \) and \( b \) take on different values?

**Tip:** When solving for continuity in piecewise functions, always check the behavior of the function at the points where different expressions are joined.

Find the values of a and b that make f continuous everywhere.

Solve for the values of a and b that make the given piecewise function continuous everywhere. Explore the concepts of continuity, limits, and solving piecewise functions.

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