What is the minimum value of y = (2x+1)(x+3) ?

$\frac{dy}{dx} = 4x+7 , \frac{d^2y}{dx^2} = 4 >0$

$\Rightarrow \frac{dy}{dx} = 0 \text{ at x = }-\frac{7}{4}$

At at this point y will attain its minimum value

$y_{min} = \frac{-25}{8}$