Answer:
the vertex is:
(2, -1)
Step-by-step explanation:
First solve the equation for the variable y
[tex]x^2-16y-4x-12=0[/tex]
Add 16y on both sides of the equation
[tex]16y=x^2-16y+16y-4x-12[/tex]
[tex]16y=x^2-4x-12[/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=-4\\c=-12[/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{-4}{2}) ^ 2[/tex]
[tex](\frac{b}{2}) ^ 2=(-2) ^ 2[/tex]
[tex](\frac{b}{2}) ^ 2=4[/tex]
[tex]16y=(x^2-4x+4)-4-12[/tex]
Factor the expression that is inside the parentheses
[tex]16y=(x-2)^2-16[/tex]
Divide both sides of the equality between 16
[tex]\frac{16}{16}y=\frac{1}{16}(x-2)^2-\frac{16}{16}[/tex]
[tex]y=\frac{1}{16}(x-2)^2-1[/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=2\\k =-1[/tex]
the vertex is:
(2, -1)
