For finding range we sometimes use the quadratic equation for example `f(x)=x^(2)+2x+2` will have range `[1,oo)` as, `f(x)=x^(2)+2x+2=(x+1)^(2)+1 ge 1`
Range of `-x^(2)+4x+6` is

Options

A. `[4,oo)`
B. `(-oo,10]`
C. `(0,5)`
D. `(0,oo)`

Answer

To find the range of the quadratic function \( f(x) = -x^2 + 4x + 6 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = -x^2 + 4x + 6 \]
To make it easier to analyze, we will factor out the negative sign:
\[ f(x) = - (x^2 - 4x - 6) \]

### Step 2: Complete the square
Next, we will complete the square for the expression inside the parentheses:
1. Take the coefficient of \( x \) (which is -4), halve it to get -2, and then square it to get 4.
2. We will add and subtract this square inside the parentheses:
\[ f(x) = - \left( (x^2 - 4x + 4) - 4 - 6 \right) \]
This simplifies to:
\[ f(x) = - \left( (x - 2)^2 - 10 \right) \]
Now, distribute the negative sign:
\[ f(x) = 10 - (x - 2)^2 \]

### Step 3: Identify the maximum value
The expression \( (x - 2)^2 \) is a perfect square and is always non-negative. Its minimum value is 0, which occurs when \( x = 2 \). Thus:
\[ f(x) \text{ is maximized when } (x - 2)^2 = 0 \]
Substituting \( x = 2 \) into \( f(x) \):
\[ f(2) = 10 - 0 = 10 \]

### Step 4: Determine the range
Since \( (x - 2)^2 \) can take any non-negative value, \( f(x) \) will decrease as \( (x - 2)^2 \) increases. Therefore, the maximum value of \( f(x) \) is 10, and as \( (x - 2)^2 \) increases, \( f(x) \) will approach negative infinity. Thus, the range of \( f(x) \) is:
\[ (-\infty, 10] \]

### Conclusion
The range of the function \( f(x) = -x^2 + 4x + 6 \) is:
\[ (-\infty, 10] \]