Question 674921
Use the center, vertices, and asymptotes to graph the hyperbola. 
(x - 1)^2 - 9(y - 2)^2 = 9
(x-1)^2/9 -(y-2)^2 = 1
This is 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
For given hyperbola:
center: (1,2)
a^2=9
a=3
vertices: (1±a,2)=(1±3,2)=(-2,2) and (4,2)
b^2=1
b=1
asymptotes are straight lines that go thru center and take the form: y=mx+b, m=slope, b=y-intercept
slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±1/3
..
equation of asymptote with negative slope:
y=-x/3+b
solve for b using coordinates of center
2=-1/3+b
b=7/3
equation: y=-x/3+7/3
...
equation of asymptote with positive slope:
y=x/3+b
solve for b using coordinates of center
2=1/3+b
b=5/3
equation: y=x/3+5/3
see graph below:

y=±(((x-1)^2-9)/9)^.5+2
{{{ graph( 300, 300, -10, 10, -10, 10,(((x-1)^2-9)/9)^.5+2,-(((x-1)^2-9)/9)^.5+2,-x/3+7/3,x/3+5/3) }}}