college algebra (math 90)
posted by Alfred on .
How do i graph this hyperbola?
X^2+6x3y^2=0

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 
except the whole thing is shifted 3 units to the left  because it's (x+3)^2