## Precalculus (6th Edition) Blitzer

$\text{Vertex}= (-1,-1)$ Focus: $f(-1,-2)$ Directrix is: $y=0$ 'Graph d'
General form of a horizontal parabola is given as: $(x-h)^2=4p(y-k)$...(1) Here, $\text{Vertex}=(h,k)$ and focus is: $(h, k+p)$ As we are given $(x-(-1))^2=-4(y-(-1))$ From equation (1), we get: $k=-1, h=-1$ and $4p=-4 \implies p=-1$ $\text{Vertex}=(h,k) \implies (-1,-1)$ Also, focus is: $(h,k+p) \implies f(-1,-1-1)=f(-1,-2)$ Directrix is: $y=k-p \implies y=-1+1=0$ Hence, $\text{Vertex}= (-1,-1)$ Focus: $f(-1,-2)$ Directrix is: $y=0$ Thus, the 'Graph d' matches with all the above conditions.