document.write( "Question 590198: I have to name the center, vertices, foci and slopes of the asymptotes\r
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document.write( "x^2 - y^2 -4x +2y = -2 add 4\r
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document.write( "(x-2)^2 add 4 to the right side
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document.write( "now I am stuck because I am not sure what the negative in front of the y^2
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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?
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document.write( "What would my equation be?\r
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document.write( "Thank you!
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document.write( "Connie \n" );
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Algebra.Com's Answer #375078 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " \n" ); document.write( " |