document.write( "Question 151404: How would I go about writing the quadratic function, in (standard) form, which has a vertex of (-6, 2) and passes through (5, 12)?\r
\n" ); document.write( "\n" ); document.write( "The standard form is y= a(x-h)^2+k , so i would put -6 as h and 2 as k but from there I am stuck.
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Algebra.Com's Answer #111294 by jim_thompson5910(35256)\"\" \"About 
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\"y=+a%28x-h%29%5E2%2Bk\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"y=+a%28x-%28-6%29%29%5E2%2B2\" Plug in \"h=-6\" and \"k=2\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=+a%28x%2B6%29%5E2%2B2\" Rewrite \"x-%28-6%29\" as \"x%2B6\"\r
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\n" ); document.write( "\n" ); document.write( "Since the quadratic passes through (5, 12), this means that if you plug in \"x=5\", you will get \"y=12\". So we can plug in \"x=5\" and \"y=12\" to solve for \"a\"\r
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\n" ); document.write( "\n" ); document.write( "\"12=+a%285%2B6%29%5E2%2B2\" Plug in \"x=5\" and \"y=12\"\r
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\n" ); document.write( "\n" ); document.write( "\"12=+a%2811%29%5E2%2B2\" Add\r
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\n" ); document.write( "\n" ); document.write( "\"12=+a%28121%29%2B2\" Square 11 to get 121\r
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\n" ); document.write( "\n" ); document.write( "\"12=+121a%2B2\" Rearrange the terms\r
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\n" ); document.write( "\n" ); document.write( "\"10=+121a\" Subtract 2 from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"10%2F121=+a\" Divide both sides by 121.\r
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\n" ); document.write( "\n" ); document.write( "So the value of \"a\" is \"a=10%2F121\"\r
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\n" ); document.write( "\n" ); document.write( "This means that the equation is \r
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\n" ); document.write( "\n" ); document.write( "\"y=+%2810%2F121%29%28x%2B6%29%5E2%2B2\"
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