The vertex form of a parabola is given by:
[tex]x=a(y-k)^2+h[/tex]
Where the vertex is:
[tex]\begin{gathered} V(h,k)=(-2,4) \\ so\colon \\ x=a(y-4)^2-2 \\ x=a(y-4)^2-2 \end{gathered}[/tex]
for (1,-2):
[tex]\begin{gathered} 1=a(-2-4)^2-2 \\ 1=a(-6)^2-2 \\ 1=36a-2 \\ solve_{\text{ }}for_{\text{ }}a\colon \\ 36a=1+2 \\ 36a=3 \\ a=\frac{3}{36} \\ a=\frac{1}{12} \\ \end{gathered}[/tex]
therefore:
[tex]x=\frac{1}{12}(y-4)^2-2[/tex]
