Answer
$x^2=4(y+1)$
Work Step by Step
If the major axis of the parabola is parallel to the y-axis then it is in the form of $4k(y-a)=(x-b)^2$, where $(b,a)$ is the vertex of the parabola and $k$ is a constant (positive if open up, negative if open down).
If the major axis of the parabola is parallel to the x-axis then it is in the form of $4k(x-a)=(y-b)^2$, where $(a,b)$ is the vertex of the parabola and $k$ is a constant. (positive if open right, negative if open left).
Here the major axis is parallel to the y-axis and the vertex is in $(0,-1)$.
Thus the equation becomes $(x-0)^2=4k(y-(-1))\\x^2=4k(y+1)$.
We also know that $(2,0)$ is on the graph, hence plugging in $x=12y=0$ gives us: $2^2=4k(0+1)\\4=4k\\k=1$.
Hence our equation: $x^2=4(1)(y+1)\\x^2=4(y+1)$
