What is y-12y+4x+4=0 as a parabola

Question

*Please explain the first few steps of the algebraic process for getting it into vertex form.

Answer 1

Put the x terms on the other side and complete the square using the y's

y² - 12y = -4x - 4. take 1/2 (12) an square it = 6² = 36 add to both sides.

y² -12y + 36 = -4x - 4 + 36

(y - 6)² = -4x + 32. Note this is a parabola that faces to the left

Answer 2

The y term is the squared one, so step one is to get the x term to the other side. Its coefficient will stay with it until the end. That gives us

(y²-12y )+4=-4x

I put in the parentheses because that's the next thing to work on. We want to turn that into (y-h)². To help with that, remember that (y-h)² multiplies out to y²-2hy+h². Looking at the parabola's equation, that means -12=-2h and h=6. We're still missing the h² term of 36. We'll just add that in as well as immediately subtract it so we're really adding zero and not changing the equation. Doing all that gives us this

(y²-12y+36)-36+4=-4x

The parentheses collapse into the squared binomial, and combining terms gives

(y-6)²-32=-4x

Now we divide the -4 to solve for x and get

-0.25(y-6)²-9=x