document.write( "Question 590198: I have to name the center, vertices, foci and slopes of the asymptotes\r
\n" ); document.write( "\n" ); document.write( "x^2 - y^2 -4x +2y = -2 add 4\r
\n" ); document.write( "\n" ); document.write( "(x-2)^2 add 4 to the right side
\n" ); document.write( "now I am stuck because I am not sure what the negative in front of the y^2
\n" ); document.write( "I am not sure how to do the (y- or +2)^2 because of the - in front of the y^2 and do I add a neg or positive 2 on the left side?
\n" ); document.write( "What would my equation be?\r
\n" ); document.write( "\n" ); document.write( "Thank you!
\n" ); document.write( "Connie
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Algebra.Com's Answer #375078 by lwsshak3(11628)\"\" \"About 
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I have to name the center, vertices, foci and slopes of the asymptotes
\n" ); document.write( "x^2 - y^2 -4x +2y = -2 add 4
\n" ); document.write( "**
\n" ); document.write( "I will assume this is the original equation:
\n" ); document.write( "x^2 - y^2 -4x +2y = -2
\n" ); document.write( "To get all the answers, you must first put this equation in standard form by completing the squares:
\n" ); document.write( "(x^2-4x+4) -(y^2-2y+1) = -2+4-1
\n" ); document.write( "(x-2)^2-(y-1)^2=1
\n" ); document.write( "This is an equation of a hyperbola with horizontal transverse axis of the standard form:
\n" ); document.write( "(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center.
\n" ); document.write( "..
\n" ); document.write( "For given equation: (x-2)^2-(y-1)^2=1
\n" ); document.write( "center: (2,1)
\n" ); document.write( "a^2=1
\n" ); document.write( "a=1
\n" ); document.write( "vertices:(2±a,1)=(2±1,1)=(1,1) and (3,1)
\n" ); document.write( "..
\n" ); document.write( "b^2=1
\n" ); document.write( "b=1
\n" ); document.write( "..
\n" ); document.write( "c^2=a^2+b^2=1+1=2
\n" ); document.write( "c=√2
\n" ); document.write( "Foci: (2±c,1)=(2±√2,1)=(2-√2,1) and (2+√2,1)
\n" ); document.write( "..
\n" ); document.write( "slopes of asymptotes=±b/a=±1/1=±1
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