Answer
$y^2=\frac{1}{2}(x+2)$
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 x-axis and the vertex is in $(-2,0)$.
Thus the equation becomes $(y-0)^2=4k(x-(2))\\y^2=4k(x+2)$.
We also know that $(0,1)$ is on the graph, hence plugging in $x=0,y=1$ gives us: $1^2=4k(0+2)\\1=8k\\k=\frac{1}{8}$.
Hence our equation: $y^2=4\frac{1}{8}(x+2)\\y^2=\frac{1}{2}(x+2)$
