Question

If x is real, the minimum value of \(f(x) = x^2 - 8x + 17\) would be

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Vertex form: $x^2 - 8x + 17 = (x-4)^2 + 1$. Since $(x-4)^2 \ge 0$, minimum value is $+1$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The minimum value of a quadratic polynomial \(a x^2 + b x + c\) with \(a > 0\) occurs at its vertex \(x = -b / (2a)\).
Key Formula or Approach:
\[ f(x) = a(x - h)^2 + k \implies f_{\text{min}} = k \quad \text{at } x = h \]

Step 2: Detailed Explanation:
Method 1: Completing the square:
\[ f(x) = x^2 - 8x + 17 = (x^2 - 8x + 16) + 1 = (x - 4)^2 + 1 \]
Since \((x - 4)^2 \ge 0\) for all real \(x\):
\[ f(x) \ge 0 + 1 = 1 \]
The minimum value is 1, achieved when \(x = 4\).

Method 2: Calculus:
\[ f'(x) = 2x - 8 = 0 \implies x = 4 \]
\[ f''(x) = 2 > 0 \text{ (Local Minimum)} \]
\[ f(4) = 4^2 - 8(4) + 17 = 16 - 32 + 17 = 1 \]

Step 3: Final Answer:
Therefore, the minimum value of \(f(x)\) is 1, corresponding to option (A).