What is the domain and range of a hyperbola with vertices (1,-2) (1,6)

`4x^2-y^2-8x-4y+16=0`

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Given the hyperbola:

`4x^2 - y^2 - 8x -4y +16 = 0`

First we will write into the standard form.

==> Divide by 4.

`==gt x^2 - y^2/4 - 2x - y + 4 = 0`

Now we will complete the squares,

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

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

Now we will divide by -2 so the right side is 1

`==gt (y+2)^2/8 - (x-1)^2/2 = 1`

The hyperbola is opening up/down==> Then we have a vertical hyperbola.

1. Domain: we do not have any restrictions on the domain

==> The domain is all real numbers.

2. range: We will find the vertices's.

Vertices's are: `(h, k+-a) = (1, -2+-2sqrt2)`

==> Vertices are:`(1, -2+2sqrt2)`  and `(1, -2-2sqrt2) `

==> Then, the range is all real numbers except the interval `(-2-2sqrt2, -2+2sqrt2)` .

==> Range = `R- (-2-2sqrt2, -2+2sqrt2)`

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