Using 4x^2 - y^2 -8x -4y +16=0 I got -(x^2-1)^2/4 - (y^2 - 2/16=0 for standard form. I need to locate centre, vertices and state domain and range of curve. I'm not sure how could you please explain.

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    The center is (1, -2) from the numerators of standard form. The denominator of the first term is a^2 so a = 4. To find the vertices, add and subtract "a" to the coordinate of the center that is listed in the positive fraction- in this case , add and subtract 4 to the y coordinate. The vertices are (1, -6) and (1, 2) Since this hyperbola is vertical the domain is all reals. The range is (-∞, -6] unioned with [2, ∞)