document.write( "Question 72735: I keep finding examples that have the problem squared but none without. I would appreciate any and all help with this.
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document.write( "Find the vertex of y=-2(x -3) + 4 and then graph.
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document.write( "I think the vertex is (6,4) and the -2 makes it open downward if im correct. I came up with (5,0) and (-1,0) as the x intercepts and (0,10) as the y intercept. When I plot them on a graph with the vertex, the points are all over the graph so Im doing something wrong. Please please help! \n" );
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Algebra.Com's Answer #52005 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Find the vertex of y=-2(x -3)^2 + 4 and then graph.\r \n" ); document.write( "\n" ); document.write( "------------ \n" ); document.write( "There are several way to find the vertex. \n" ); document.write( "You can rewrite the equation as y-4 = -2(x-3)^2 \n" ); document.write( "This follows the form y-k = a(x-h)^2 \n" ); document.write( "where (h,k) is the vertex. \n" ); document.write( "So your vertex is at (3,4). \n" ); document.write( "-------- \n" ); document.write( "To graph you could plot some points: \n" ); document.write( "You know the x=3 is at the vertex, or on the axis of symmetry. \n" ); document.write( "If x=4 you get y=-2*1+4 = 2; Point at (4,2) and by symmetry at (2,2) \n" ); document.write( "If x=5 you get y=-2*5+4 = -6; Point at (5,-6) and by symmetry at (1,-6) \n" ); document.write( "etc. \n" ); document.write( " \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |