Question

Need help with this maybe question

Answer 1

the vertex is:

(2, -1)

Explanation:

First solve the equation for the variable y

`x^2-16y-4x-12=0`

Add 16y on both sides of the equation

`16y=x^2-16y+16y-4x-12`

`16y=x^2-4x-12`

Notice that now the equation has the general form of a parabola

`ax^2 +bx +c`

In this case

`a=1\\b=-4\\c=-12`

Add
`((b)/(2)) ^ 2` and subtract
`((b)/(2)) ^ 2` on the right side of the equation

`((b)/(2)) ^ 2=((-4)/(2)) ^ 2`

`((b)/(2)) ^ 2=(-2) ^ 2`

`((b)/(2)) ^ 2=4`

`16y=(x^2-4x+4)-4-12`

Factor the expression that is inside the parentheses

`16y=(x-2)^2-16`

Divide both sides of the equality between 16

`(16)/(16)y=(1)/(16)(x-2)^2-(16)/(16)`

`y=(1)/(16)(x-2)^2-1`

For an equation of the form

`y=a(x-h)^2 +k`

the vertex is: (h, k)

In this case

`h=2\\k =-1`

the vertex is:

(2, -1)