Answer:
The vertex is: [tex](6, 8)[/tex]
Step-by-step explanation:
First solve the equation for the variable y
[tex]x^2-4y-12x+68=0[/tex]
Add 4y on both sides of the equation
[tex]4y=x^2-4y+4y-12x+68[/tex]
[tex]4y=x^2-12x+68[/tex]
Notice that now the equation has the general form of a parabola
[tex]ax^2 +bx +c[/tex]
In this case
[tex]a=1\\b=-12\\c=68[/tex]
Add [tex](\frac{b}{2}) ^ 2[/tex] and subtract [tex](\frac{b}{2}) ^ 2[/tex] on the right side of the equation
[tex](\frac{b}{2}) ^ 2=(\frac{-12}{2}) ^ 2\\\\(\frac{b}{2}) ^ 2=(-6) ^ 2\\\\(\frac{b}{2}) ^ 2=36[/tex]
[tex]4y=(x^2-12x+36)-36+68[/tex]
Factor the expression that is inside the parentheses
[tex]4y=(x-6)^2+32[/tex]
Divide both sides of the equality between 4
[tex]\frac{4}{4}y=\frac{1}{4}(x-6)^2+\frac{32}{4}[/tex]
[tex]y=\frac{1}{4}(x-6)^2+8[/tex]
For an equation of the form
[tex]y=a(x-h)^2 +k[/tex]
the vertex is: (h, k)
In this case
[tex]h=6\\k =8[/tex]
the vertex is: [tex](6, 8)[/tex]
