Answer
$\text{Vertex}= (-1,-1)$
Focus: $f(-1,0)$
Directrix is: $y=-2$
'Graph a'
Work Step by Step
General form of a horizontal parabola is given as: $(x-h)^2=4p(y-h)$...(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: $(h,k+p) \implies f(-1,-1+1)=f(-1,0)$
Directrix is: $y=k-p \implies y=-1-1=-2$
Hence, $\text{Vertex}= (-1,-1)$
Focus: $f(-1,0)$
Directrix is: $y=-2$
Thus, the 'Graph a' matches with all the above conditions.
