If for `-2 le x lt 3`, range of `y = x^(2)-4x-9` is
`[a,b], a, b in I` , then `a + 5b =`

Explanation

To solve the problem, we need to find the range of the function \( y = x^2 - 4x - 9 \) for the interval \( -2 \leq x < 3 \) and then calculate \( a + 5b \) where the range is given as \([a, b]\).

### Step-by-step Solution:

1. **Identify the Function**:
We have the quadratic function:
\[ y = x^2 - 4x - 9 \]

2. **Find the Vertex**:
The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -4 \).
\[ x_{vertex} = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2 \]

3. **Evaluate the Function at the Vertex**:
Now, we will find the value of \( y \) at \( x = 2 \):
\[ y(2) = 2^2 - 4 \cdot 2 - 9 = 4 - 8 - 9 = -13 \]

4. **Evaluate the Function at the Endpoints**:
Next, we evaluate \( y \) at the endpoints of the interval \( x = -2 \) and \( x = 3 \):
- For \( x = -2 \):
\[ y(-2) = (-2)^2 - 4 \cdot (-2) - 9 = 4 + 8 - 9 = 3 \]
- For \( x = 3 \):
\[ y(3) = 3^2 - 4 \cdot 3 - 9 = 9 - 12 - 9 = -12 \]

5. **Determine the Range**:
Now we have the values of \( y \) at the critical point and the endpoints:
- \( y(-2) = 3 \)
- \( y(2) = -13 \)
- \( y(3) = -12 \)

The minimum value of \( y \) is \( -13 \) and the maximum value is \( 3 \). Thus, the range of \( y \) is:
\[ [-13, 3] \]
Here, \( a = -13 \) and \( b = 3 \).

6. **Calculate \( a + 5b \)**:
Now we calculate \( a + 5b \):
\[ a + 5b = -13 + 5 \cdot 3 = -13 + 15 = 2 \]

### Final Answer:
\[ a + 5b = 2 \]