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To find the values of \( a \) and \( b \) that make the function \( f(x) \) continuous everywhere, we need to ensure that the function is continuous at the points where the piecewise definition changes, i.e., at \( x = -8 \) and \( x = 4 \).

### Step 1: Continuity at \( x = -8 \)
At \( x = -8 \), the two relevant expressions are:
- \( f(x) = \frac{x^2 - 64}{x + 8} \) for \( x < -8 \)
- \( f(x) = ax^2 - bx - 392 \) for \( -8 \leq x < 4 \)

We want the function to be continuous, so the limit from the left must equal the value from the right at \( x = -8 \).

#### 1.1 Left-hand limit as \( x \to -8^- \):
The expression \( \frac{x^2 - 64}{x + 8} \) simplifies:
\[
f(x) = \frac{(x - 8)(x + 8)}{x + 8} = x - 8 \quad \text{for} \quad x \neq -8
\]
Thus, as \( x \to -8 \), the left-hand limit is:
\[
\lim_{x \to -8^-} f(x) = -8 - 8 = -16
\]

#### 1.2 Right-hand value at \( x = -8 \):
We now substitute \( x = -8 \) into the second piece \( ax^2 - bx - 392 \):
\[
f(-8) = a(-8)^2 - b(-8) - 392 = 64a + 8b - 392
\]

For continuity, the left-hand limit must equal the right-hand value:
\[
64a + 8b - 392 = -16
\]
This simplifies to:
\[
64a + 8b = 376 \quad \text{(Equation 1)}
\]

### Step 2: Continuity at \( x = 4 \)
At \( x = 4 \), the two relevant expressions are:
- \( f(x) = ax^2 - bx - 392 \) for \( -8 \leq x < 4 \)
- \( f(x) = -57x - a + b \) for \( x \geq 4 \)

We want these expressions to be equal at \( x = 4 \).

#### 2.1 Left-hand value at \( x = 4 \):
Substitute \( x = 4 \) into the second piece \( ax^2 - bx - 392 \):
\[
f(4) = a(4)^2 - b(4) - 392 = 16a - 4b - 392
\]

#### 2.2 Right-hand value at \( x = 4 \):
Substitute \( x = 4 \) into the third piece \( -57x - a + b \):
\[
f(4) = -57(4) - a + b = -228 - a + b
\]

For continuity, the left-hand value must equal the right-hand value:
\[
16a - 4b - 392 = -228 - a + b
\]
Simplifying:
\[
16a - 4b + a - b = 164
\]
\[
17a - 5b = 164 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
We now have the system:
1. \( 64a + 8b = 376 \)
2. \( 17a - 5b = 164 \)

Let's solve for \( a \) and \( b \). I'll calculate the values.The values of \( a \) and \( b \) that make the function \( f(x) \) continuous everywhere are:

\[
a = 7 \quad \text{and} \quad b = -9
\]

Would you like further details on this solution or have any other questions?

Here are 5 related questions to expand your understanding:
1. How can you check if a function is differentiable in addition to being continuous?
2. What is the significance of ensuring continuity in piecewise functions?
3. How would you approach this problem if there were more than two points of discontinuity?
4. Can a function be continuous but not differentiable at certain points? Why?
5. How do you handle piecewise functions where one piece involves a trigonometric function?

**Tip:** When solving piecewise function problems, always ensure continuity at the points where the function changes its definition by equating limits from both sides.

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

Solve the problem of finding the values of a and b that ensure the continuity of a piecewise function involving quadratic equations and limits. Learn step-by-step how to apply the continuity condition at different points.

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