[tex]\bf y=a(x-{{ h}})^2+{{ k}}\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})[/tex]
those are the vertex form of a parabola... so hmmm
the vertex of this one is at 0,1 and intercepts or "solutions" are at -1 and 1, so is opening downwards, notice the picture below
that means, the squared variable is the "x", thus the form is [tex]\bf y=a(x-{{ h}})^2+{{ k}}[/tex]
now, we know the vertex is at 0,1, and two x-intercepts of [tex]\pm 1,0[/tex]
thus [tex]\bf y=a(x-{{ h}})^2+{{ k}}\qquad
\begin{cases}
h=0\\
k=1\\
when
\\
x=\pm 1\\
y=0
\end{cases}
\\\\\\
\textit{so.. hmmm let us use the point ... say hmmm 1,0}
\\\\
y=a(x-{{ h}})^2+{{ k}}\implies 0=a(1-0)^2+1[/tex]
solve for "a", to see what that coefficient is, then plug it back in the vertex form equation
