Question 573785
vertices, foci, and asymptotes for 
x^2-4y^2-4x+24y-36=0
complete the square
(x^2-4x+4)-4(y^2-6y+9)=36+4-36
(x-2)^2-4(y-3)^2=4
(x-2)^2/4-(y-3)^2=1
This is an equation for a hyperbola with horizontal transverse axis of the standard form: (x-h)^2/a^2-(y-k)^2/b^2=1
For given hyperbola:
center: (2,3)
a^2=4
a=2
vertices: (2±a,3)=(2±2,3)=(0,3) and (4,3)
b^2=1
c^2=a^2+b^2=4+1=5
c=√5≈2.24
foci: (2±c,3)=(2±√5,3)=(.24,3) and (4.24,3)
asymptotes:
slope of asymptotes=±b/a=±1/2=±.5
Equation of asymptotes:
y=mx+b(straight lines which go thru center)
solving for b
3=-.5*2+b
b=4
Equation for one of the asymptotes with the negative slope
y=-.5x+4
..
solving for b
3=.5*2+b
b=2
Equation for other asymptote with the positive slope
y=.5x+2