document.write( "Question 674921: Use the center, vertices, and asymptotes to graph the hyperbola.\r
\n" ); document.write( "\n" ); document.write( "(x - 1)^2 - 9(y - 2)^2 = 9 \n" ); document.write( "
Algebra.Com's Answer #420505 by lwsshak3(11628)\"\" \"About 
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Use the center, vertices, and asymptotes to graph the hyperbola.
\n" ); document.write( "(x - 1)^2 - 9(y - 2)^2 = 9
\n" ); document.write( "(x-1)^2/9 -(y-2)^2 = 1
\n" ); document.write( "This is a hyperbola with horizontal transverse axis.
\n" ); document.write( "Its standard form of equation:\"+%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\", (h,k)=coordinates of center
\n" ); document.write( "For given hyperbola:
\n" ); document.write( "center: (1,2)
\n" ); document.write( "a^2=9
\n" ); document.write( "a=3
\n" ); document.write( "vertices: (1±a,2)=(1±3,2)=(-2,2) and (4,2)
\n" ); document.write( "b^2=1
\n" ); document.write( "b=1
\n" ); document.write( "asymptotes are straight lines that go thru center and take the form: y=mx+b, m=slope, b=y-intercept
\n" ); document.write( "slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±1/3
\n" ); document.write( "..
\n" ); document.write( "equation of asymptote with negative slope:
\n" ); document.write( "y=-x/3+b
\n" ); document.write( "solve for b using coordinates of center
\n" ); document.write( "2=-1/3+b
\n" ); document.write( "b=7/3
\n" ); document.write( "equation: y=-x/3+7/3
\n" ); document.write( "...
\n" ); document.write( "equation of asymptote with positive slope:
\n" ); document.write( "y=x/3+b
\n" ); document.write( "solve for b using coordinates of center
\n" ); document.write( "2=1/3+b
\n" ); document.write( "b=5/3
\n" ); document.write( "equation: y=x/3+5/3
\n" ); document.write( "see graph below:\r
\n" ); document.write( "\n" ); document.write( "y=±(((x-1)^2-9)/9)^.5+2
\n" ); document.write( " \n" ); document.write( "
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