Precalc
posted by Maya .
Heres the equation:
(x^2)(4y^2)(4x)+(24y)(36)=0
Your supposed to find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of its graph.
So first you complete the square right?
(x^2 4x+4)4(y^2 +6y+9)=36+4+36
[((x+2)^2)/76] [(y+3)^2)/19]=1
So its a hyperbola, and is the center at (2, 3)??

Picking it up from your
(x^2 4x+4)4(y^2 +6y+9)=36+4+36
should have been
(x^2 4x+4)4(y^2 +6y+9)=36+4+36
notice on the left you had
... 4(..... +9) , so you subtracted 36, thus you must subtract 36 from the right side as well
(you had added 36)
final version
(x2)^2  4(y3)^2 = 36
or
(x2)^2 /36  (y3)^2 /9 = 1
centre would be (2,3)
a = 1 , b = 3 , c = √10
vertices: (1,3) and (3,3)
foci :( 2√10 , 3) and (2+√10 , 3)
asymptotes:
first: y = 3x + b , with (2,3) on it
3 = 6+b
b = 3 > y = 3x  3
second: y = 3x + b
3 = 6+b
b = 9 > y = 3x + 9 
Watch your signs!
(x^2 4x+4)4(y^2 +6y+9)=36+4+36
should be (x^2 4x+4)4(y^2 6y+9)=36+436
(x^2 4x+4)/44(y^2 6y+9)/4=4/4
((x2)^2)/4 (y3)^2 =1
hyperbola, center at (2,3) 
from
(x2)^2 /36  (y3)^2 /9 = 1 , the rest should say
centre is (2,3)
a = 6, b = 3, c = √45
vertices: (4,3) and (8,3)
foci :( 2√45 , 3) and (2+√45 , 3)
asymptotes:
first: y = (1/2)x + b , with (2,3) on it
3 = 1+b
b = 2 > y = (1/2)x + 2
second: y = (1/2)x + b
3 = 1+b
b = 2 > y = (1/2)x + 2
check my arithmetic 
Thank you guys so much :)

I am sorry, don't know where I got that 4 on the RS from
Go with Nadine's final equation
(x^2 4x+4)4(y^2 6y+9)=36+436
((x2)^2)/4 (y3)^2 =1
from there, a=2, b = 1, c = √5
adjust from there 
Haha ok i totally get it thanks!