document.write( "Question 388126: y=1/3x^2-2x+7
\n" ); document.write( "I need to find the vertex. I know I first use the axis of symmetry which is x=-b/2a. I did that and I got 2/2/3. I don't know what to do next. Please help. Thank you.
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Algebra.Com's Answer #274439 by Theo(13342)\"\" \"About 
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y = (1/3)*x^2 - 2*x + 7
\n" ); document.write( "x = -b/2a
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\n" ); document.write( "a = (1/3)
\n" ); document.write( "b = -2
\n" ); document.write( "c = 7
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\n" ); document.write( "-b/2a = -(-2)/(2/3) = 2/(2/3) = 2*3/2 = 6/2 = 3
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\n" ); document.write( "you found x
\n" ); document.write( "now you need to find f(x) to get the y value
\n" ); document.write( "when x = 3, (1/3)*x^2 - 2*x + 7 = (1/3)*3^2 - 2*3 + 7
\n" ); document.write( "this equals 3 - 6 + 7 = 4
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\n" ); document.write( "your vertex should be at the point pair of (x,y) = 3,4)
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\n" ); document.write( "the graph of this equation looks like this:
\n" ); document.write( "\"graph%28600%2C600%2C-5%2C5%2C-2%2C8%2C%281%2F3%29x%5E2+-2x+%2B+7%29\"
\n" ); document.write( "you can see from the graph that the vertex is at the point (x,y) = 3,4)\r
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