From the graph, we notice that the parabola has to be a horizontal parabola with vertex at:
[tex](-2,3),[/tex]
that opens to the right.
Recall that the standard form of a horizontal parabola is:
[tex]\mleft(y-k\mright)^2=4p\mleft(x-h\mright)\text{.}[/tex]
Where, (h,k) are the coordinates of the vertex, and p is the distance to the vertex to focus.
Substituting the vertex in the above equation, we get:
[tex](y-3)^2=4p(x-(-2))\text{.}[/tex]
From the diagram, we get that:
[tex]p=3.[/tex]
Therefore:
[tex](y-3)^2=12(x+2)\text{.}[/tex]
Solving the above equation for x, we get:
[tex]\begin{gathered} \frac{(y-3)^2}{12}=x+2, \\ x=\frac{(y-3)^2}{12}-2. \end{gathered}[/tex]
Answer:
[tex]x=\frac{(y-3)^2}{12}-2.[/tex]
