The function f(x) = 2(x+3)² - 8 has a minimum value of what?

The function is in vertex form, $a(x-h)^2+k$, with $a=2$, $h=-3$, and $k=-8$. Since $a>0$, the parabola opens upward, so its minimum value is the $y$-coordinate of the vertex, $-8$, choice F. The squared term $2(x+3)^2$ is never negative, so the smallest it can be is $0$, giving $f=-8$. Choice G ($8$) has the wrong sign. Choice H ($-3$) is the $x$-coordinate of the vertex, where the minimum occurs, not the minimum value. Choice J ($3$) uses the wrong sign for that $x$-coordinate and also confuses it with the minimum value.