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 function \( f(x) = -x^2 + 4x + 6 \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting the function in a form that makes it easier to identify the range. We can factor out the negative sign from the quadratic expression:

\[ f(x) = - (x^2 - 4x - 6) \]

### Step 2: Complete the square
Next, we complete the square for the expression inside the parentheses. To do this, we take the coefficient of \( x \) (which is -4), halve it, and square it:

\[ \text{Half of } -4 = -2 \quad \text{and} \quad (-2)^2 = 4 \]

Now, we can rewrite the expression:

\[ x^2 - 4x = (x - 2)^2 - 4 \]

Substituting this back into our function gives:

\[ f(x) = -((x - 2)^2 - 4 - 6) = -((x - 2)^2 - 10) \]

### Step 3: Simplify the function
Now we simplify the expression:

\[ f(x) = - (x - 2)^2 + 10 \]

### Step 4: Analyze the function
The term \( (x - 2)^2 \) is always non-negative (i.e., \( (x - 2)^2 \geq 0 \)). Therefore, the maximum value of \( - (x - 2)^2 \) occurs when \( (x - 2)^2 = 0 \):

\[ f(x) \text{ is maximized when } (x - 2)^2 = 0 \implies f(x) = 10 \]

### Step 5: Determine the range
Since \( - (x - 2)^2 \) can take any non-positive value, the function \( f(x) \) will take values from \( -\infty \) up to 10. Therefore, the range of \( f(x) \) is:

\[ (-\infty, 10] \]

### Final Answer
The range of the function \( f(x) = -x^2 + 4x + 6 \) is \( (-\infty, 10] \).

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