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April 16, 2014

April 16, 2014

Posted by **Alfred** on Monday, December 5, 2011 at 10:06pm.

X^2+6x-3y^2=0

- college algebra (math 90) -
**Reiny**, Monday, December 5, 2011 at 10:59pmcomplete 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 10:27amexcept the whole thing is shifted 3 units to the left - because it's (x+3)^2

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