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)![]() ![]() You can put this solution on YOUR website! your equation is x^2 + 2x + 1 > 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if you set your equation to y, then the equation becomes:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "y = x^2 + 2x + 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if x^2 + 2x + 1 > 0, then y > 0 as well because y = x^2 + 2x + 1.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you would graph y = x^2 + 2x + 1.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you would look for all values of y that are > 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the graph of your equation looks like this:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your graph touches the x-axis at x = -1.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that means that y = 0 when x = -1.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the solution to x^2 + 2x + 1 > 0 is all values of x except at x = -1.\r \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |