Asked by Ali
inding vertex and focus of parabola y^2 +4y +8x - 12 =0
Answers
Answered by
Damon
y^2 + 4 y = -8 x + 12
y^2 + 4 y + 4 = -8 x + 16
(y+2)^2 = -8 (x-2)
vertex at (2,-2) opens left etc
y^2 + 4 y + 4 = -8 x + 16
(y+2)^2 = -8 (x-2)
vertex at (2,-2) opens left etc
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