document.write( "Question 967790: given that x^2+2x+1>0 , can i say that x^2+2x+1 is the y values of the graph? \n" ); document.write( "
Algebra.Com's Answer #591524 by Theo(13342)\"\" \"About 
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your equation is x^2 + 2x + 1 > 0\r
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\n" ); document.write( "\n" ); document.write( "if you set your equation to y, then the equation becomes:\r
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\n" ); document.write( "\n" ); document.write( "y = x^2 + 2x + 1\r
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\n" ); document.write( "\n" ); document.write( "if x^2 + 2x + 1 > 0, then y > 0 as well because y = x^2 + 2x + 1.\r
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\n" ); document.write( "\n" ); document.write( "you would graph y = x^2 + 2x + 1.\r
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\n" ); document.write( "\n" ); document.write( "you would look for all values of y that are > 0.\r
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\n" ); document.write( "\n" ); document.write( "the graph of your equation looks like this:\r
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\n" ); document.write( "\n" ); document.write( "\"graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Cx%5E2+%2B+2x+%2B+1%29\"\r
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\n" ); document.write( "\n" ); document.write( "your graph touches the x-axis at x = -1.\r
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\n" ); document.write( "\n" ); document.write( "that means that y = 0 when x = -1.\r
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\n" ); document.write( "\n" ); document.write( "the solution to x^2 + 2x + 1 > 0 is all values of x except at x = -1.\r
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\n" ); document.write( "\n" ); document.write( "this means that y > 0 at all values of x except at x = -1, because y = x^2 + 2x + 1.\r
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