Question 975398
For the following hyperbola: 
(x+1)^2/16 – (y–3)^2/9 = 1 
Find: a)Center : (-1, 3)
b)Vertices:  (-3, 3) and (5, 3)
c) Foci: (-4, 3) and (6, 3)
d) Asymptotes: (list separately)
 y=-3x/4+9/4 
 y=3x/4+15/4
***
Given hyperbola has a 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: (-1, 3)
..
a^2=16
a=√16=4
vertices: (-1±a, 3)=(1±4, 3)= (-3, 3) and (5, 3)
..
b^2=9
b=√9=3
..
c^2=a^2+b^2=16+9=25
c=√25=5
..
foci: (-1±c, 3)= (1±5, 3= (-4, 3) and (6, 3)
..
Asymptotes are straight lines that go thru center.
slopes of asymptotes=±b/a=±3/4
..
Equation of asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates of center (-1, 3) which are on the line
b=y+3x/4=3+(3*(-1)/4)=9/4
b=9/4
equation: y=-3x/4+9/4
..
Equation of asymptote with positive slope:
y=3x/4+b
solve for b using coordinates of center (-1, 3) which are on the line
b=y-3x/4=3-(3*(-1)/4)=9/4
b=15/4
equation:  y=3x/4+15/4