## Precalculus (10th Edition)

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