SAT Question a7711fe8: Nonlinear Functions answer and explanation with Desmos solution

Choice A is correct. The given quadratic function $f$ is in vertex form, $f(x)=(x-h{)}^2+k$, where $(h,k)$ is the vertex of the graph of $y=f(x)$ in the $xy$-plane. Therefore, the vertex of the graph of $y=f(x)$ is $(2,-4)$. In addition, the $y$-coordinate of the vertex represents either the minimum or maximum value of a quadratic function, depending on whether the graph of the function opens upward or downward. Since the leading coefficient of $f$ (the coefficient of the term $(x-2{)}^2$) is $1$, which is positive, the graph of $y=f(x)$ opens upward. It follows that at $x=2$, the minimum value of the function $f$ is $-4$.