Question

Need help with this maybe question

Answer 1

Answer:

the vertex is:

(2, -1)

Step-by-step 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 $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-4}{2}) ^ 2$

$(\frac{b}{2}) ^ 2=(-2) ^ 2$

$(\frac{b}{2}) ^ 2=4$

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

Factor the expression that is inside the parentheses

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

Divide both sides of the equality between 16

$\frac{16}{16}y=\frac{1}{16}(x-2)^2-\frac{16}{16}$

$y=\frac{1}{16}(x-2)^2-1$

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)