Posted by Alfred on Monday, December 5, 2011 at 10:06pm.
How do i graph this hyperbola?
X^2+6x3y^2=0

college algebra (math 90)  Reiny, Monday, December 5, 2011 at 10:59pm
complete the square
x^2 + 6x + 9  3y^2 = 9
(x+3)^2  3y^2 = 9
divide by 9
(x+3)^2 /9  y^2/3 = 1
centre (3,0)
a = 3, b = √3
vertices at (3,0) and (3,0)
faintly sketch a rectangle with corners
at (±3,±√3)
draw in the diagonals and extend them
sketch the hyperbola with the diagonals as asymptotes and the vertices as noted above

college algebra (math 90)  Anonymous, Tuesday, December 6, 2011 at 10:27am
except the whole thing is shifted 3 units to the left  because it's (x+3)^2
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