Question

For a quadratic expression \( ax^{2}+bx+c \), if the minimum value \( \frac{49}{12} \) exists at \( x=\frac{-5}{6} \), then \( 12c-5b = \)

• A. \( 35 \)
• B. \( 61 \)
• C. \( 49 \)
• D. \( 37 \)

Hint:

Always remember that a quadratic expression can be cleanly written in vertex form as \( f(x) = a(x-h)^2 + k \), where \( (h,k) = \left(-\frac{b}{2a}, \frac{4ac-b^2}{4a}\right) \) represents the coordinates of the turning vertex point.

Answer 1

The Correct Option is C

Step 1: Recall the vertex facts.
For $f(x)=ax^2+bx+c$ with $a>0$, the lowest point sits at $x = -\dfrac{b}{2a}$, and the smallest value there is $\dfrac{4ac-b^2}{4a}$.
Step 2: Match the vertex position.
The minimum is at $x=-\tfrac{5}{6}$, so $-\dfrac{b}{2a} = -\dfrac{5}{6}$. This gives $6b = 10a$, so $a = \dfrac{3b}{5}$.
Step 3: Use the minimum value.
The minimum value is $\tfrac{49}{12}$, so \[ \frac{4ac-b^2}{4a} = \frac{49}{12} \;\Rightarrow\; c - \frac{b^2}{4a} = \frac{49}{12} \]
Step 4: Plug in a.
Substitute $a=\tfrac{3b}{5}$ into the denominator: $\dfrac{b^2}{4a} = \dfrac{b^2}{4(3b/5)} = \dfrac{5b}{12}$. So $c - \dfrac{5b}{12} = \dfrac{49}{12}$.
Step 5: Clear the fractions.
Multiply every term by 12: \[ 12c - 5b = 49 \]
Step 6: State the answer.
The needed value appears directly. \[ \boxed{12c - 5b = 49} \]