document.write( "Question 23682: how would you graph the following fuctions?\r
\n" ); document.write( "\n" ); document.write( "1) y=1/2(x-3)2 +1\r
\n" ); document.write( "\n" ); document.write( "and y=-(x+4)2 -5 ?
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Algebra.Com's Answer #12492 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
You probably already know that the graph of \"y+=+x%5E2\" looks like this:\r
\n" ); document.write( "\n" ); document.write( "\"+graph%28300%2C300%2C+-10%2C+10%2C-10%2C+10%2C+x%5E2+%29\"\r
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\n" ); document.write( "\n" ); document.write( "Well, these graphs will be similar, but because of the (x-3) you will be moving the graph of y =x^2 to the RIGHT 3 units. Because of the +1 added to the end of the equation, the graph will be moved UP 1 unit. And because of the 1/2 coefficient, it opens up the graph and makes it wider. \r
\n" ); document.write( "\n" ); document.write( "\"graph+%28400%2C400%2C+-10%2C10%2C-10%2C10%2C+%281%2F2%29%2A%28x-3%29%5E2+%2B1%29+\"\r
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\n" ); document.write( "\n" ); document.write( "For your second graph, it says \"y=-%28x%2B4%29%5E2+-5\", the (x+4) moves the graph to the LEFT 4, the -5 at the end of the equation moves the graph DOWN 5 units, and the negative coefficient in front of the squared term INVERTS the graph! It should look like this:
\n" ); document.write( "\"graph+%28400%2C400%2C+-10%2C10%2C-10%2C10%2C-%28x%2B4%29%5E2+-5%29\"\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC \r
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