document.write( "Question 177947: Starting with the graph of y = e^x, find the equation of the graph that results from reflecting about the line y=5. \n" ); document.write( "
Algebra.Com's Answer #132902 by jim_thompson5910(35256)\"\" \"About 
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In order to reflect about the line y=5, we first need to reflect about the line y=0 (which is much easier). To do that, simply shift the graph \"y=e%5Ex\" 5 units down (to effectively move the line \"y=5\" to \"y=0\"). Algebraically, you just subtract 5 from \"e%5Ex\" to get\r
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\n" ); document.write( "\n" ); document.write( "\"y=e%5Ex-5\"\r
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\n" ); document.write( "\n" ); document.write( "Now to reflect over the line \"y=0\" (the x-axis), simply replace \"y\" with \"-y\" to get\r
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\n" ); document.write( "\n" ); document.write( "\"-y=e%5Ex-5\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=-e%5Ex%2B5\" Multiply both sides by -1 to make y positive.\r
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\n" ); document.write( "\n" ); document.write( "Now since we shifted \"y=e%5Ex\" down 5 units (ie subtracted 5), we need to shift \"y=-e%5Ex%2B5\" back up (by adding 5). So add 5 to \"y=-e%5Ex%2B5\" to get\r
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\n" ); document.write( "\n" ); document.write( "\"y=-e%5Ex%2B5%2B5\" \r
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\n" ); document.write( "\n" ); document.write( "\"y=-e%5Ex%2B10\" Add\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So after reflecting \"y=e%5Ex\" over the line \"y=5\", we get the equation \"y=-e%5Ex%2B10\"\r
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\n" ); document.write( "\n" ); document.write( "Here's a graph to visually verify the answer\r
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\n" ); document.write( "\n" ); document.write( "\"+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+exp%28x%29%2C+-exp%28x%29%2B10%2C5%29+\" Graph of the original equation \"y=e%5Ex\" (red) and the equation \"y=-e%5Ex%2B10\" (green) which is a reflection over the line \"y=5\" (blue)
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