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3x^2-12y^2+6x+48y-93=0

Given the equation: 3x^2-12y^2+6x+48y-93=0 determine: 
a: The center C
b: the two vertices
c: the slopes of the asymptotes

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Nader O. | Experienced and Patient Math, Business, and Economics TutorExperienced and Patient Math, Business, ...
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This is equation of hyperbola in general form which you need to change it to standard form to see the information asked. You need to complete the square for both x and y.
 
3(x^2+2x) - 12(y^2-4y) = 93
3(x^2+2x+1) - 12(y^2-4y+4) = 93 + 3 -48
3(x+1)^2 - 12(y-2)^2 = 48
 
divide both side by 48
 
(x+1)^2/16 - (y-2)^2/4 = 1
 
now center is at (-1, 2)
vertices (-5,2) and (3,2)
Slopes of the asymptotes - 1/2 and 1/2