SOLUTION: x2+6x+4y+5=0 conic name?
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Question 855001: x2+6x+4y+5=0 conic name?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
x2+6x+4y+5=0 conic name?
complete the square:
(x^2+6x+9)=-4y-5+9
(x+3)^2=-4y+4
(x+3)^2=-4(y-1)
This is an equation of a parabola that opens down with vertex at (-3,1)
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