Question 638489: y^2-4x+6y+29=0
Identify the conic section. It is a parabola, give the vertex.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y^2-4x+6y+29=0
Identify the conic section. It is a parabola, give the vertex.
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y^2-4x+6y+29=0
y^2+6y-4x+29=0
complete the square
(y^2+6y+9)-4x=-29+9
(y+3)^2-4x+20=0
(y+3)^2-4(x-5)=0
conic section is that of a parabola that opens rightwards
Its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
vertex: (5,-3)
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