posted by Lucy .
Find the coordinates of the center, the foci, the vertices, and the equations of the asymptotes of the graph of the equation: (x+1)^2/2-y^2/8=1.
vertices: (-1+/-sqrt2, 0)
equations of the asymptotes: y-0=sqrt8/sqrt2(x+1)
Is this correct and if not, how do I figure this out??
hyperbola due to - sign
yes, center at -1,0 because of(x+1)^2 and (y)^2
a =sqrt 2
b = sqrt 8 = 2 sqrt 2
so vertices at y =0 x = -1+sqrt 2 and -1 -sqrt 2
center to focus = sqrt(2+8) = sqrt 10
so at y = 0 and x = -1 +/- sqrt 10
slope of asymptotes = +/- b/a = sqrt2
y = (+/- sqrt 2) - 1
so yes, correct