document.write( "Question 112191: Write y=2x^2+12x+14 in vertex form \n" ); document.write( "
Algebra.Com's Answer #81840 by jim_thompson5910(35256)\"\" \"About 
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\"y=2x%5E2%2B12x%2B14\" Start with the given quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"y-14=2x%5E2%2B12x\" Subtract \"14++\" from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"y-14=2%2A%28x%5E2%2B6%2Ax%29\" Factor out the leading coefficient \"2++\". This step is important since we want the \"x%5E2\" coefficient to be 1. \r
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\n" ); document.write( "\n" ); document.write( "Take half of the x coefficient \"6\" to get \"3\" (ie \"6%2F2=3\")\r
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\n" ); document.write( "\n" ); document.write( "Now square \"3\" to get \"9\" (ie \"%283%29%5E2=9\")\r
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\n" ); document.write( "\n" ); document.write( "\"y-14=2%2A%28x%5E2%2B6%2Ax%2B9%29\" Add this number (\"9\") to the expression inside the parenthesis. Now the expression \"x%5E2%2B6%2Ax%2B9\" is a perfect square trinomial.\r
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\n" ); document.write( "\n" ); document.write( "\"y-14%2B%282%29%289%29=2%2A%28x%5E2%2B6%2Ax%2B9%29\" Since we added \"9\" inside the parenthesis, we need to add \"%282%29%289%29\" to the other side (remember, we factored out the leading coefficient \"2\").\r
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\n" ); document.write( "\n" ); document.write( "\"y-14%2B%282%29%289%29=2%2A%28x%2B3%29%5E2\" Factor \"x%5E2%2B6%2Ax%2B9\" into \"%28x%2B3%29%5E2\". Now the right side is a perfect square (note: if you need help with factoring, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"y%2B4=2%2A%28x%2B3%29%5E2\" Multiply \"2\" and \"9\" to get \"18\".\r
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\n" ); document.write( "\n" ); document.write( "\"y=2%2A%28x%2B3%29%5E2-4\" Subtract 4 from both sides\r
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\n" ); document.write( "\n" ); document.write( "Now the equation is in vertex form\r
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