Work Step by Step
RECALL: (1) The vertex form of a quadratic function is $f(x)=a(x-h)^2+k$ where $(h, k)$ are the coordinates of its vertex. (2) The graph of the quadratic function $f(x) = a(x-h)^2+k$ opens (a) upward when $a \gt 0$, (b) downward when $a \lt0$. The equation $y=(x+2)^2+4$ can be written as $y=[x-(-2)]^2+4$. With $a=1$, then the graph of the function is a parabola that opens downward. This equation is in vertex form with its vertex at $(-2, 4)$. Therefore, the answer is Option A.