Question

Need help with this math question

Answer 1

Answer:

The vertex is: $(6, 8)$

Step-by-step explanation:

First solve the equation for the variable y

$x^2-4y-12x+68=0$

Add 4y on both sides of the equation

$4y=x^2-4y+4y-12x+68$

$4y=x^2-12x+68$

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

$ax^2 +bx +c$

In this case

$a=1\\b=-12\\c=68$

Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-12}{2}) ^ 2\\\\(\frac{b}{2}) ^ 2=(-6) ^ 2\\\\(\frac{b}{2}) ^ 2=36$

$4y=(x^2-12x+36)-36+68$

Factor the expression that is inside the parentheses

$4y=(x-6)^2+32$

Divide both sides of the equality between 4

$\frac{4}{4}y=\frac{1}{4}(x-6)^2+\frac{32}{4}$

$y=\frac{1}{4}(x-6)^2+8$

For an equation of the form

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

the vertex is: (h, k)

In this case

$h=6\\k =8$

the vertex is: $(6, 8)$