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

Concept: For a quadratic expression \( f(x) = ax^2 + bx + c \) with \( a > 0 \):

• The extremum point occurs at its vertex where \( x = -\frac{b}{2a} \).

• The minimum value at this vertex is given by \( f\left(-\frac{b}{2a}\right) = -\frac{D}{4a} = \frac{4ac - b^2}{4a} \).

Step 1: Equating the vertex position.
We are given that the minimum value occurs at \( x = -\frac{5}{6} \): \[ -\frac{b}{2a} = -\frac{5}{6} \implies \frac{b}{2a} = \frac{5}{6} \implies 6b = 10a \implies 3b = 5a \implies a = \frac{3b}{5} \quad \cdots (1) \]

Step 2: Using the minimum value expression.
The minimum value is given as \( \frac{49}{12} \): \[ \frac{4ac - b^2}{4a} = \frac{49}{12} \implies c - \frac{b^2}{4a} = \frac{49}{12} \] Substitute \( a = \frac{3b}{5} \) into the denominator of the fraction: \[ c - \frac{b^2}{4\left(\frac{3b}{5}\right)} = \frac{49}{12} \implies c - \frac{5b^2}{12b} = \frac{49}{12} \] \[ c - \frac{5b}{12} = \frac{49}{12} \]

Step 3: Isolating the target expression \( 12c - 5b \).
Multiply the entire equation by 12 to eliminate the fractional denominators: \[ 12 \left(c - \frac{5b}{12}\right) = 12 \left(\frac{49}{12}\right) \] \[ 12c - 5b = 49 \]