Question 474193
Find the vertices of a hyperbola with the equation (x + 3)2 – 4(y – 2)2 = 4.
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(x+3)^2–4(y–2)^2=4
divide by 4
(x+3)^2/4–(y–2)^2/1=1
This is a hyperbola with horizontal transverse axis of the standard form: 
(x-h)^2/a^2-(y-k)^2/b^2=1
..
For given equation:
Center:(-3,2)
a^2=4
a=2
length of transverse axis=2a=4
vertices: (-3ħa,2)=(-3ħ2,2)=(-5,2) and (-1,2)