document.write( "Question 525019: How do you graph this? (x-4)^2+(y+6)=16 \n" ); document.write( "
Algebra.Com's Answer #348011 by lwsshak3(11628)\"\" \"About 
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How do you graph this? (x-4)^2+(y+6)=16
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\n" ); document.write( "(x-4)^2+(y+6)=16
\n" ); document.write( "rewrite:
\n" ); document.write( "(x-4)^2+y+6=16
\n" ); document.write( "y= -(x-4)^2+10
\n" ); document.write( "This is an equation of a parabola of the standard form:y=-A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the center. The negative coefficient A means the parabola opens downward and the parabola has a maximum:
\n" ); document.write( "For given equation:
\n" ); document.write( "vertex:(4,10)
\n" ); document.write( "x-intercepts
\n" ); document.write( "set y=0
\n" ); document.write( "-(x-4)^2+10=0
\n" ); document.write( "(x-4)^2=10
\n" ); document.write( "(x-4)=±√10≈±3.16
\n" ); document.write( "x≈4±3.16=7.16 and 0.84
\n" ); document.write( "you now have 3 points with which you can draw the graph: (4,10), (7.16,0) and (0.84,0) \n" ); document.write( "
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