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

7/4 -25/4 3/4 -25/8

-25/8

$\frac{dy}{dx} = 4x+7 , \frac{d^2y}{dx^2} = 4 >0$ $\Rightarrow \frac{dy}{dx} = 0 \text{ at x = }-\frac{7}{4}$ At at this point y will attain its minimum value $y_{min} = \frac{-25}{8}$