Which of the following correctly shows the range of the function f(x)=`sqrt(-x^(2)+4x+12)` ?

Options

A. ` 0lt y lt4`
B. ` -1 le y le 4`
C. `-1 lt y lt 5`
D. `0 le y le 4`

Answer

To find the range of the function \( f(x) = \sqrt{-x^2 + 4x + 12} \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sqrt{-x^2 + 4x + 12} \]

### Step 2: Set up the equation
Let \( y = f(x) \). Then we have:
\[ y = \sqrt{-x^2 + 4x + 12} \]
Squaring both sides gives:
\[ y^2 = -x^2 + 4x + 12 \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ -x^2 + 4x + 12 - y^2 = 0 \]
This can be rewritten as:
\[ x^2 - 4x + (y^2 - 12) = 0 \]

### Step 4: Analyze the quadratic equation
For \( x \) to have real solutions, the discriminant of this quadratic equation must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4(1)(y^2 - 12) \]
Calculating this gives:
\[ D = 16 - 4(y^2 - 12) = 16 - 4y^2 + 48 = 64 - 4y^2 \]

### Step 5: Set the discriminant greater than or equal to zero
To ensure real solutions for \( x \), we need:
\[ 64 - 4y^2 \geq 0 \]
This simplifies to:
\[ 4y^2 \leq 64 \]
Dividing both sides by 4 gives:
\[ y^2 \leq 16 \]

### Step 6: Solve for \( y \)
Taking the square root of both sides, we find:
\[ |y| \leq 4 \]
This implies:
\[ -4 \leq y \leq 4 \]

### Step 7: Consider the range of the square root function
Since \( f(x) = \sqrt{-x^2 + 4x + 12} \) represents a square root, \( y \) must be non-negative. Therefore, we restrict \( y \) to:
\[ 0 \leq y \leq 4 \]

### Conclusion
Thus, the range of the function \( f(x) = \sqrt{-x^2 + 4x + 12} \) is:
\[ [0, 4] \]