document.write( "Question 824771: Please help me solve this: Why does the \"c\" parameter in standard form, f(x) = ax^2 + bx + c, not equal the \"k\" parameter in vertex form, y= a (x-h)^2+k? \r
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Algebra.Com's Answer #496808 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
Why does the \"c\" parameter in standard form, f(x) = ax^2+bx+c, not equal the \"k\" parameter in vertex form, y = a(x-h)^2+k?
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document.write( "        f(x) = ax˛+bx+c\r\n" );
document.write( "           y = a(x-h)˛+k\r\n" );
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document.write( "Set f(x) identically equal to y\r\n" );
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document.write( "Then for every value of x,\r\n" );
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document.write( "    ax˛+bx+c = a(x-h)˛+k   \r\n" );
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document.write( "In particular it is true when x=0, so\r\n" );
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document.write( "a(0)˛+b(0)+c = a(0-h)˛+k\r\n" );
document.write( "           c = ah˛+k\r\n" );
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document.write( "So c is not equal to k except when a or h is 0.\r\n" );
document.write( "If h=0 then the vertex is (0,k) which is a point\r\n" );
document.write( "on the y-axis and if a=0 the equation is not a \r\n" );
document.write( "quadratic, but only f(x) = c and y = k, a\r\n" );
document.write( "horizontal line. \r\n" );
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document.write( "Edwin
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