Question

For the quadratic function \(f(x) = x^2 - kx + 12\), the minimum value of \(f(x)\) is 3. Which of the following is the value of \(k\)?

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

A quick way to find the minimum/maximum value of a quadratic function \(ax^2 + bx + c\) is to use the formula \(\frac{4ac - b^2}{4a}\). For this problem, setting \(\frac{4(1)(12) - (-k)^2}{4(1)} = 3\) gives \(48 - k^2 = 12\), which quickly leads to \(k^2 = 36\).

Answer 1

The Correct Option is A

Step 1: Since 3 is the minimum value of \( f \), the line \( y=3 \) touches the parabola at exactly one point (its vertex), it can never cross it.
Step 2: Setting \( f(x)=3 \) gives \( x^{2}-kx+12=3 \), i.e. \( x^{2}-kx+9=0 \). Touching at one point means this equation must have equal roots, so its discriminant is zero.
Step 3:
\[ k^{2}-4(1)(9) = 0 \implies k^{2}=36 \implies k=\pm6. \]
Step 4: Among the given choices only 6 appears.

Final Answer: \( k=6 \). (Option 1)