The range of the function `sin^2x-5sinx -6` is

Options

A. [-10,0]
B. [-1,1]
C. `[0,pi]`
D. `[-49/4,0]`

Answer

To find the range of the function \( f(x) = \sin^2 x - 5 \sin x - 6 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sin^2 x - 5 \sin x - 6 \]
Let \( y = \sin x \). Then, we can rewrite the function as:
\[ f(y) = y^2 - 5y - 6 \]

### Step 2: Identify the quadratic form
The function \( f(y) = y^2 - 5y - 6 \) is a quadratic function in the standard form \( ay^2 + by + c \), where \( a = 1 \), \( b = -5 \), and \( c = -6 \).

### Step 3: Find the vertex of the quadratic
The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula:
\[ y = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ y = -\frac{-5}{2 \cdot 1} = \frac{5}{2} \]

### Step 4: Determine the maximum and minimum values
Since \( \sin x \) can only take values in the range \([-1, 1]\), we need to evaluate \( f(y) \) at the endpoints of this interval.

1. **Calculate \( f(-1) \)**:
\[ f(-1) = (-1)^2 - 5(-1) - 6 = 1 + 5 - 6 = 0 \]

2. **Calculate \( f(1) \)**:
\[ f(1) = (1)^2 - 5(1) - 6 = 1 - 5 - 6 = -10 \]

### Step 5: Determine the range
From the calculations:
- The maximum value occurs at \( y = -1 \) and is \( 0 \).
- The minimum value occurs at \( y = 1 \) and is \( -10 \).

Thus, the range of the function \( f(x) = \sin^2 x - 5 \sin x - 6 \) is:
\[ [-10, 0] \]

### Final Answer
The range of the function is \([-10, 0]\).

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