## Algebra 2 (1st Edition)

The function will have a minimum, and the value of the minimum will be $y=-2$.
The value of $A$ is positive, so the function will open upwards. Therefore, the function will have a minimum. The $x$-coordinate of the minimum will be $-\frac{B}{2A}$, if the function is in the form $Ax^{2}+Bx+C$. $A=\frac{3}{2}$, $B=6$, and $C=4$, so the $x$ coordinate will be: $=-6\div((2)\times(\frac{3}{2}))=-6\div(\frac{6}{2})=-\frac{6\times2}{6}=-\frac{12}{6}=-2$. Therefore, the value of the minimum is $y=-2$.