Question

The maximum value of $f\left(x\right)=(x+3)(4-x)+3$ is

• A. $\frac{61}{2}$
• B. $\frac{61}{4}$
• C. $3$
• D. $15$

Answer 1

The correct option is B $\frac{61}{4}$

$f\left(x\right)=-x^2+x+15$,
$a=-1(<0)$
Maximum value $=\frac{4ac-b^2}{4a}$
$=\frac{4\times (-1)\times 15-1}{-4}=\frac{61}{4}$