Question 927550
Given the hyperbola with the equation x^2−y^2−6x−2y+7=0, find the vertices, the foci, and the equations of the asymptotes. 
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x^2−6x−y^2−2y+7=0
complete the square:
(x^2−6x+9)−(y^2+2y+1)=-7+9-1
(x-3)^2-(y+1)^2=1
(x-3)^2/2-(y+1)^2/2=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=coordinates of center.
center:(3,-1)
a^2=1
a=1
vertices:(3±a,-1)=(3±1,-1)=(4,-1) and (2,-1)
b^2=1
b=1
c^2=a^2+b^2=1+1=2
c=√2
foci:(3±c,-1)=(3±√2,-1)=(3+√2,-1) and (3-√2,-1)
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Asymptotes:Straight line equations of the form: y=mx+b, m=slope, b=y-intercept
Asymptotes go thru center
slope of asymptotes: ±b/a=±1
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equation of asymptote with positive slope:
y=x+b
solve for b with coordinates of center
-1=3+b
b=-4
equation: y=x-4
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equation of asymptote with negative slope:
y=-x+b
solve for b with coordinates of center
-1=-3+b
b=2
equation: y=-x+2