Find the horizontal for oblique asymptote of f(x)=-3x^2+7x+1/x-2

For greater clarity, please enclose the divisor (x-2) inside parentheses.  Thx.

Divide x-2 into 3x^2 + 7x + 1:
    
             3x    + 13
       
x-2  /   3x^2  +  7x  +   1
           3x^2   - 6x
           
                     13x     +  1
                     13x     - 26
                   
                                -27 (remainder)
                                                 -27
The quotient here is  3x + 13 -
                                                  x-2

The slant asymptote is the line y = 3x + 13.  Note that -27 / (x-2) goes to zero as x grows large in either direction (positive or neg).

Note that you are looking for the "oblique" or "slant" asymptote here; there is no "horizontal" involved.

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4