document.write( "Question 189678: Could someone help me with the following problem? \r
\n" ); document.write( "\n" ); document.write( " Problem : A quadratic funtion has its vertex at the point of (0,-5). The function passes throuhg the point (-5,2). When written in vertex form the function is f(x)=a(x-h)^2+k. Where a = ? , h=?, and k=?\r
\n" ); document.write( "\n" ); document.write( "Ok, I figured out what h and k are ( h = 0 , and k = -5) but I dont know how to get a.. could someone help me.
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Algebra.Com's Answer #142316 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"f%28x%29=a%28x-h%29%5E2%2Bk\" Start with the general vertex form.\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=a%28x-0%29%5E2%2B%28-5%29\" Plug in \"h=0\" and \"k=-5\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=ax%5E2-5\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"2=a%28-5%29%5E2-5\" Plug in \"x=-5\" and \"f%28x%29=2\"\r
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\n" ); document.write( "\n" ); document.write( "Note: since the graph passes through the point (-5,2), this means that if \"x=-5\", then \"f%28x%29=2\" (f(x) is really \"y\")\r
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\n" ); document.write( "\n" ); document.write( "\"2=a%2825%29-5\" Square -5 to get 25\r
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\n" ); document.write( "\n" ); document.write( "\"2=25a-5\" Rearrange the terms\r
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\n" ); document.write( "\n" ); document.write( "\"2%2B5=25a\" Add 5 to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"7=25a\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"7%2F25=a\" Divide both sides by 25 to isolate \"a\".\r
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\n" ); document.write( "\n" ); document.write( "\"a=7%2F25\" Rearrange the equation\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=ax%5E2-5\" Go back to the previous function\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=%287%2F25%29x%5E2-5\" Plug in \"a=7%2F25\" \r
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\n" ); document.write( "\n" ); document.write( "So the function with a vertex of (0,-5) and passes through (-5,2) is \"f%28x%29=%287%2F25%29x%5E2-5\"
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