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

Approach: For an upward parabola, the minimum is the value at the vertex. The slick shortcut is the formula: minimum \(= c - \dfrac{b^2}{4a}\). Set it equal to 3 and solve.

Step 1: Identify the coefficients.
For \(f(x) = x^2 - kx + 12\): \(a = 1,\ b = -k,\ c = 12\). Since \(a > 0\), the parabola opens up and does have a minimum.

Step 2: Apply the minimum-value formula.
\[ f_{\min} = c - \frac{b^2}{4a} = 12 - \frac{(-k)^2}{4(1)} = 12 - \frac{k^2}{4}. \]

Step 3: Set the minimum equal to 3 and solve.
\[ 12 - \frac{k^2}{4} = 3 \implies \frac{k^2}{4} = 9 \implies k^2 = 36 \implies k = \pm 6. \]

Step 4: Match the option.
Both \(k = 6\) and \(k = -6\) give the same minimum, but only \(6\) appears in the choices.

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

Answer 2

Approach (calculus): For an upward parabola, the minimum occurs where the derivative is zero.

Step 1: \( f(x) = x^{2}-kx+12 \), so \( f'(x) = 2x-k \). Setting \( f'(x)=0 \) gives the vertex at \( x=\frac{k}{2} \).
Step 2: Substitute back:
\[ f\left(\frac{k}{2}\right) = \frac{k^{2}}{4} - k\cdot\frac{k}{2} + 12 = 12-\frac{k^{2}}{4}. \]
Step 3: Set this equal to the given minimum, 3:
\[ 12-\frac{k^{2}}{4} = 3 \implies k^{2} = 36 \implies k=\pm6. \]
Step 4: Only \( k=6 \) is among the choices.

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