Question 625961
Graph the equation. Identify the center, vertices and Foci
{{{(x+3)^2/16 - (y-1)^2/25=1}}}
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation:{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
For given equation:
center: (-3,1)
a^2=16
a=√16=4
vertices: (-3±a,1)=(-3±4,1)=(-7,1) and (1,1)
..
b^2=25
b=√25=5
..
c^2=a^2+b^2=16+25=41
c=√41≈6.4
Foci:  (-3±c,1)=(-3±6.4,1)=(-9.4,1) and (3.4,1)
see graph below:
y=±(25(x+3)^2/16-25)^.5+1

 {{{ graph( 300, 300, -10, 10, -10, 10,(25(x+3)^2/16-25)^.5+1,-(25(x+3)^2/16-25)^.5+1) }}}