document.write( "Question 90683: Determine whether each relation is a function.\r
\n" ); document.write( "\n" ); document.write( "x^2=1+y^2
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Algebra.Com's Answer #65822 by vertciel(183)\"\" \"About 
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This relation is not a function.\r
\n" ); document.write( "\n" ); document.write( "Proof:\r
\n" ); document.write( "\n" ); document.write( "x^2 = 1 + y^2\r
\n" ); document.write( "\n" ); document.write( "x = sqrt(1 + y^2)\r
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\n" ); document.write( "\n" ); document.write( "Substitute y for 1:\r
\n" ); document.write( "\n" ); document.write( "x = sqrt(1 + 1^2)\r
\n" ); document.write( "\n" ); document.write( " = sqrt(2)\r
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\n" ); document.write( "\n" ); document.write( "Substitute y for -1:\r
\n" ); document.write( "\n" ); document.write( "x = sqrt(1 + (-1)^2)\r
\n" ); document.write( "\n" ); document.write( " = sqrt(2)\r
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\n" ); document.write( "\n" ); document.write( "Therefore, sqrt(2) has two y values: (√2, 1) and (√2, -1)
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