Source: 1976 Euclid Part A Problem 6
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The $y$-intercept of the graph of the function defined by $y=\frac{4(x+3)(x-2)-24}{(x+4)}$ is

$\textbf{(A) } -24 \qquad \textbf{(B) } -12 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } -4 \qquad \textbf{(E) } -48$
1. To find the $y$-intercept of the function \( y = \frac{4(x+3)(x-2) - 24}{x+4} \), we need to set \( x = 0 \) and solve for \( y \).

2. Substitute \( x = 0 \) into the function:
   \[
   y = \frac{4(0+3)(0-2) - 24}{0+4}
   \]

3. Simplify the expression inside the numerator:
   \[
   y = \frac{4 \cdot 3 \cdot (-2) - 24}{4}
   \]

4. Calculate the product and subtraction in the numerator:
   \[
   4 \cdot 3 \cdot (-2) = 4 \cdot (-6) = -24
   \]
   \[
   -24 - 24 = -48
   \]

5. Now, the function simplifies to:
   \[
   y = \frac{-48}{4}
   \]

6. Perform the division:
   \[
   y = -12
   \]

The final answer is \(\boxed{-12}\).