The graph of the function `f(x)=-(1)/(2)(x+4)(x+8)` in the xy-plane is a parabola. Which of the following is an equivalent form of function f in which the maximum value of the function appears as a constant?

Options

A. `f(x)=(x+4)(4-(1)/(2)x)`
B. `f(x)=-(1)/(2)(x+6)^(2)+2`
C. `f(x)=-(1)/(2)x(x+12)+16`
D. `f(x)=(1)/(2)(x+6)^(2)+20`

Answer

To find an equivalent form of the function \( f(x) = -\frac{1}{2}(x + 4)(x + 8) \) where the maximum value appears as a constant, we will follow these steps:

### Step 1: Expand the function
We start by expanding the function \( f(x) \).

\[ f(x) = -\frac{1}{2}(x + 4)(x + 8) \]

Using the distributive property (FOIL method):

\[ = -\frac{1}{2}(x^2 + 8x + 4x + 32) \]
\[ = -\frac{1}{2}(x^2 + 12x + 32) \]

### Step 2: Factor out the negative half
Next, we factor out the \(-\frac{1}{2}\):

\[ f(x) = -\frac{1}{2}x^2 - 6x - 16 \]

### Step 3: Complete the square
Now, we will complete the square for the quadratic expression. The general form for completing the square is:

\[ a^2 + 2ab + b^2 = (a + b)^2 \]

In our case, we have:

\[ -\frac{1}{2}(x^2 + 12x + 32) \]

We need to complete the square for \(x^2 + 12x\). First, take half of the coefficient of \(x\) (which is 12), square it, and add/subtract that value:

\[ \left(\frac{12}{2}\right)^2 = 36 \]

Now, we rewrite the function:

\[ f(x) = -\frac{1}{2}(x^2 + 12x + 36 - 36 + 32) \]
\[ = -\frac{1}{2}((x + 6)^2 - 4) \]

### Step 4: Simplify the expression
Now we simplify the expression:

\[ f(x) = -\frac{1}{2}(x + 6)^2 + 2 \]

### Conclusion
The equivalent form of the function \( f(x) \) where the maximum value appears as a constant is:

\[ f(x) = -\frac{1}{2}(x + 6)^2 + 2 \]