Question

Help check work with changing equations of conics to standard vertex form.

Answer 1

y2 − 12y − 8x + 20 = 0

This would be a horizontal parabola

Answer 2

because we are missing x^2

Answer 3

yes thats correct

Answer 4

(y^2-12y)-8x=-20

we would square now

(y-6y)^2-8x=-20

Answer 5

(y-6)^2=8x-20

Answer 6

\[y^2-12y-8x+20=0\]\[(y^2-12y)-8x=-20\]\[(y^2-12y\color{red}{+36})-8x=-20\color{red}{+36}\]\[(y-6)^2-8x=16\]

@perl is this more like it? Not 100% sure.

Answer 7

where did 36 come from?

Answer 8

If this is in terms of completing the square, I believe that is what you are trying to accomplish.

Answer 9

looks good so far

Answer 10

@DarkBlueChocobo
we are using the fact that
$$ \Large x^2 + bx + (b/2)^2 = (x+b/2)^2

$$

Answer 11

Yes, or what is called `completing the square` :)

Answer 12

and 6^2 is 36 alright

Answer 13

y^2 -12y = y^2 + (-12)*y

y^2 + (-12)*y + (-12/2)^2 = ( y + (-12/2) ) ^2

Answer 14

\[(y-6)^2\] but how would you finish the right side?