Question 315003
{{{(x+3)^2-4(y-2)^2 =4}}}
{{{(x+3)^2/4-(y-2)^2 =1}}}
The center of the hyperbola is ({{{-3}}},{{{2}}})
The distance from the center to the vertices is {{{a=sqrt(4)=2}}}
So the vertices are ({{{-3+2}}},{{{2}}})=({{{-1}}},{{{2}}}) and ({{{-3-2}}},{{{2}}})=({{{-5}}},{{{2}}})
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 {{{ drawing( 300, 300, -8,2,-4,8, grid(1),circle(-3,2,.2),circle(-5,2,.2),circle(-1,2,.2),graph( 300, 300, -8,2,-4,8, 2+sqrt(((x+3)^2-4)/4),2-sqrt(((x+3)^2-4)/4))
)}}}