The minimum value `(x-6)^(2) + (x + 3)^(2) + (x-8)^(2) + (x + 4)^(2) + (x-3)^(2)` is

Options

A. 114
B. 141
C. 104
D. 2

Answer

To find the minimum value of the expression \((x-6)^{2} + (x + 3)^{2} + (x-8)^{2} + (x + 4)^{2} + (x-3)^{2}\), we will follow these steps:

### Step 1: Write down the expression
We start with the expression:
\[ f(x) = (x-6)^{2} + (x + 3)^{2} + (x-8)^{2} + (x + 4)^{2} + (x-3)^{2} \]

### Step 2: Expand each term
Next, we expand each squared term:
1. \((x-6)^{2} = x^{2} - 12x + 36\)
2. \((x + 3)^{2} = x^{2} + 6x + 9\)
3. \((x-8)^{2} = x^{2} - 16x + 64\)
4. \((x + 4)^{2} = x^{2} + 8x + 16\)
5. \((x-3)^{2} = x^{2} - 6x + 9\)

### Step 3: Combine all the expanded terms
Now, we combine all the expanded terms:
\[ f(x) = (x^{2} - 12x + 36) + (x^{2} + 6x + 9) + (x^{2} - 16x + 64) + (x^{2} + 8x + 16) + (x^{2} - 6x + 9) \]

Combining like terms:
- The \(x^{2}\) terms: \(5x^{2}\)
- The \(x\) terms: \(-12x + 6x - 16x + 8x - 6x = -20x\)
- The constant terms: \(36 + 9 + 64 + 16 + 9 = 134\)

So, we have:
\[ f(x) = 5x^{2} - 20x + 134 \]

### Step 4: Find the vertex of the quadratic
The expression \(f(x) = 5x^{2} - 20x + 134\) is a quadratic function. The minimum value occurs at the vertex, which can be found using the formula:
\[ x = -\frac{b}{2a} \]
where \(a = 5\) and \(b = -20\).

Calculating:
\[ x = -\frac{-20}{2 \cdot 5} = \frac{20}{10} = 2 \]

### Step 5: Substitute back to find the minimum value
Now we substitute \(x = 2\) back into the function \(f(x)\):
\[ f(2) = 5(2)^{2} - 20(2) + 134 \]
Calculating:
\[ f(2) = 5(4) - 40 + 134 = 20 - 40 + 134 = 114 \]

### Conclusion
Thus, the minimum value of the expression is:
\[ \boxed{114} \]