Asked by Rry
Transform 4y²+12y-3x+41=0 into standard form parabola?
Answers
Answered by
oobleck
4y^2+12y-3x+41 = 0
4(y^2+3y) = 3x-41
4(y^2+3y+(3/2)^2 = 3x-41 + 4(3/2)^2
4(y + 3/2)^2 = 3x-32
(y + 3/2)^2 = 3/4 (x - 32/3)
4(y^2+3y) = 3x-41
4(y^2+3y+(3/2)^2 = 3x-41 + 4(3/2)^2
4(y + 3/2)^2 = 3x-32
(y + 3/2)^2 = 3/4 (x - 32/3)
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