Since (2,1) is the turning point, let's write the function in its vertex form:
[tex]\begin{gathered} y=a(x-h)^2+k \\ so\colon \\ y=a(x-2)^2+1 \end{gathered}[/tex]
Let's use one of the points:
[tex]\begin{gathered} (3,2) \\ 2=a(3-2)^2+1 \\ 2=a(1)^2+1 \\ so\colon \\ a=2-1 \\ a=1 \end{gathered}[/tex][tex]y=(x-2)^2+1[/tex]
Therefore:
[tex]\begin{gathered} x=0 \\ y=(0-2)^2+1 \\ y=5 \end{gathered}[/tex]
