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

(x-2)^2 - 4(y-3)^2 = 36

or

(x-2)^2 /36 - (y-3)^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+4-36
(x^2 -4x+4)/4-4(y^2 -6y+9)/4=4/4
((x-2)^2)/4- (y-3)^2 =1
hyperbola, center at (2,3)

from

(x-2)^2 /36 - (y-3)^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+4-36

((x-2)^2)/4- (y-3)^2 =1

from there, a=2, b = 1, c = √5

adjust from there

Haha ok i totally get it thanks!

Yes, you are correct! In order to find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of the graph of the given hyperbola equation, completing the square is indeed the first step.

Starting with the given equation:
(x^2) - (4y^2) - (4x) + (24y) - (36) = 0

To complete the square for the x-terms, you want to group the x-terms together:
(x^2 - 4x) - (4y^2) + (24y) - 36 = 0

Now, to complete the square for the x-terms, you take half of the coefficient of the x-term (-4 in this case), square it, and add/subtract it to both sides of the equation:
(x^2 - 4x + (2^2)) - (4y^2) + (24y) - 36 + 4 = 4
(x^2 - 4x + 4) - (4y^2) + (24y) - 32 = 4

Similarly, to complete the square for the y-terms, you take half of the coefficient of the y-term (24 in this case), square it and add/subtract it to both sides of the equation:
(x^2 - 4x + 4) - (4y^2 - 24y + (12^2)) - 32 + 144 = 4
(x^2 - 4x + 4) - (4y^2 - 24y + 144) + 112 = 4

Now, let's simplify the equation further:
(x^2 - 4x + 4) - (4y^2 - 24y + 144) = 4 - 112
(x^2 - 4x + 4) - (4y^2 - 24y + 144) = -108

Next, we can factor the squares in both x and y terms:
(x - 2)^2 - 4(y - 6)^2 = -108

To isolate the left side of the equation and make it equal to 1 (as is the standard equation for a hyperbola), we divide both sides of the equation by -108:
[(x - 2)^2 / -108] - [4(y - 6)^2 / -108] = -108 / -108
[(x - 2)^2 / 108] - [(y - 6)^2 / 27] = 1

Now the equation is in the standard form for a hyperbola with a horizontal transverse axis: (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1.

Comparing this to our equation, we have:
Center: (h, k) = (2, 6)
a^2 = 108, so a = √(108)
b^2 = 27, so b = √(27)

The coordinates of the center of the hyperbola are correct: (-2, -3).

Now, for a hyperbola, the distance between the center and the vertices is given by a, so the vertices are located at (2 ± a, 6), which gives us (2 ± √(108), 6).

To find the foci, we use the relationship c^2 = a^2 + b^2, where c is the distance between the center and the foci. Therefore, c = √(a^2 + b^2), which gives us the foci at (2 ± c, 6).

Lastly, the equations of the asymptotes of the hyperbola can be found using the equation y - k = ±(b/a)(x - h), which gives us:
y - 6 = ±(√(27)/√(108))(x - 2)

So, to summarize:
Center: (-2, -3)
Vertices: (2 + √(108), 6) and (2 - √(108), 6)
Foci: (2 + c, 6) and (2 - c, 6), where c = √(a^2 + b^2)
Equations of asymptotes: y - 6 = ±(√(27)/√(108))(x - 2)