A parabola is said to be written in vertex form if it is express as:
[tex]y-k=a(x-h)^2[/tex]
in this form the vertex is the point (h,k)
Let's complete the squares to write the parabola in the form shown above:
[tex]\begin{gathered} x^2-4y-12x+68=0 \\ 4y-68=x^2-12x \\ 4y-68+(-\frac{12}{2})^2=x^2-12x+(-\frac{12}{2})^2 \\ 4y-68+36=(x-6)^2 \\ 4y-32=(x-6)^2 \\ 4(y-8)=(x-6)^2 \\ y-8=\frac{1}{4}(x-6)^2 \end{gathered}[/tex]
Hence, the parabola in vertex form is:
[tex]y-8=\frac{1}{4}(x-6)^2[/tex]
Comparing with the vertex form we conclude that the vertex is (6,8)
