Ans: A
Note that the coordinates of the vertex of the graph of the function $y=-(x+h)^2 + k$ are $(-h,k)$. According to the given figure, the vertex lies on the quadrant II. Hence, we have $-h<0$ and $k>0$. Therefore, we have $h>0$ and $k>0$.
Note that the coordinates of the vertex of the graph of the function $y=-(x+h)^2 + k$ are $(-h,k)$. According to the given figure, the vertex lies on the quadrant II. Hence, we have $-h<0$ and $k>0$. Therefore, we have $h>0$ and $k>0$.
