Question 629704
What are the correct vertices, foci, and asymptotes for the following hyperbola?: 
x^2-y^2+2x+6y-9=0
complete the squares
x^2+2x-y^2+6y-9=0
(x^2+2x+1)-(y^2-6y+9)=9+1-9
(x+1)^2-(y-3)^2=1 (This is where you made a fatal error)
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form:{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center
For given equation:
center: (-1,3)
a^2=1
a=1
vertices: (-1±a,-3)=(-1±1,-3)=(-2,-3) and (0,-3)
..
b^2=1
b=1
..
c^2=a^2+b^2=1+1=2
c=√2≈1.4
Foci: (-1±c,-3)=(-1±1.4,-3)=(-2.4,-3) and (0.4,-3)
..
Asymptotes are straight lines that go thru the center(-1,3)
Standard form of equation for straight lines: y=mx+b, m=slope, b=y-intercept
slopes of asymptotes for hyperbolas: ±b/a=±1/1=±1
Equations for asymptotes:
y=x+b
solve for b using coordinates of the center
3=-1+b
b=4
equation: y=x+4
..
y=-x+b
solve for b using coordinates of the center
3=1+b
b=2
equation: y=-x+2