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