SOLUTION: Given the standard equation (y-2)^2=16(x+4) determine which conic section it represents, and name the vertex.
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Question 40873: Given the standard equation (y-2)^2=16(x+4) determine which conic section it represents, and name the vertex.
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
Since one variable is squared and the other isn't, you know right away that this is a parabola...
(y-2)^2 = 16(x+4) now let's continue and solve for x...
(1/16)(y-2)^2 = x+4
x = (1/16)(y - 2)^2 - 4
vertex at (-4, 2)
It is noticeably wider than usual and concave right...
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