Question

How do you find the center, vertices, foci and asymptotes of `4x^2-y^2-24x-4y+28=0`?

Answer 1

Center `(h, k)=(3, -2)`
Vertex `(h+a, k)=(4, -2)` and `(h-a, k)=(2, -2)`
Foci `(h+c, k)=(5.23, -2)` and `(h-c, k)=(0.77, -2)`
Asymptotes `y=2x-8` and `y=-2x+4`

Explanation:

From the given equation
`4x^2-y^2-24x-4y+28=0`

rearrange first so that the variables are together

`4x^2-24x-y^2-4y+28=0`

Perform completing the square

`4(x^2-6x)-(y^2+4y)+28=0`

`4(x^2-6x+9-9)-(y^2+4y+4-4)+28=0`

`4(x-3)^2-36-(y+2)^2+4+28=0`

`4(x-3)^2-(y+2)^2-4=0`

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

From the form of hyperbola

`(x-h)^2/a^2-(y-k)^2/b^2=1`

`a=1` and `b=2` and `c=sqrt5`

Center `(h, k)=(3, -2)`

Vertex
`(h+a, k)=(3+1, -2)=(4, -2)`
`(h-a, k)=(3-1, -2)=(2, -2)`

Foci
`(h+c, k)=(3+sqrt5, -2)=(5.23, -2)`
`(h-c, k)=(3-sqrt5, -2)=(0.77, -2)`

Asymptotes

`y=+-b/a(x-h)+k`

`y=+2/1(x-3)+(-2)`
`y=2x-6-2`
`y=2x-8`
and

`y=-2/1(x-3)+(-2)`

`y=-2x+6-2`

`y=-2x+4`

kindly see the graph

hyperbola-colored red
asymptotes-colored blue
center-colored violet
vertices-colored orange
foci-colored black

God bless...I hope the explanation is useful.