I would start by using completing the square to rewrite the function 'y' in vertex form:
[tex]y+6+\frac{1}{4} = (x- \frac{1}{2})^2 \\ \\ y+\frac{25}{4} = (x- \frac{1}{2})^2[/tex]
Now since you want the inverse, swap the x,y variables
[tex]x+\frac{25}{4} = (y- \frac{1}{2})^2[/tex]
Make a substitution to introduce parameter 't'.
[tex]t = y - \frac{1}{2} [/tex]
Therefore
[tex]x = t^2 - \frac{25}{4}\\ \\ y = t + \frac{1}{2}[/tex]
