Answer
$${\text{hyperbola}}$$
Work Step by Step
$$\eqalign{
& - 4{x^2} + 8x + {y^2} + 6y = - 6 \cr
& \left( {{y^2} + 6y} \right) - \left( {4{x^2} - 8x} \right) = - 6 \cr
& \left( {{y^2} + 6y} \right) - 4\left( {{x^2} - 2x} \right) = - 6 \cr
& {\text{Complete the square}} \cr
& \left( {{y^2} + 6y + 9} \right) - 4\left( {{x^2} - 2x + 1} \right) = - 6 + 9 - 4\left( 1 \right) \cr
& {\left( {y + 3} \right)^2} - 4{\left( {x - 1} \right)^2} = - 1 \cr
& 4{\left( {x - 1} \right)^2} - {\left( {y + 3} \right)^2} = 1 \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{{1/4}} - {\left( {y + 3} \right)^2} = 1 \cr
& {\text{The equation is in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Therefore, the equation represents a hyperbola}} \cr
& {\text{Graph}} \cr} $$
