Question 177947
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^x}}} 5 units down (to effectively move the line {{{y=5}}} to {{{y=0}}}). Algebraically, you just subtract 5 from {{{e^x}}} to get



{{{y=e^x-5}}}



Now to reflect over the line {{{y=0}}} (the x-axis), simply replace "y" with "-y" to get



{{{-y=e^x-5}}}



{{{y=-e^x+5}}} Multiply both sides by -1 to make y positive.



Now since we shifted {{{y=e^x}}} down 5 units (ie subtracted 5), we need to shift {{{y=-e^x+5}}} back up (by adding 5). So add 5 to {{{y=-e^x+5}}} to get


{{{y=-e^x+5+5}}} 



{{{y=-e^x+10}}} Add



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Answer:


So after reflecting {{{y=e^x}}} over the line {{{y=5}}}, we get the equation {{{y=-e^x+10}}}



Here's a graph to visually verify the answer


{{{ graph( 500, 500, -10, 10, -10, 10, exp(x), -exp(x)+10,5) }}} Graph of the original equation {{{y=e^x}}} (red) and the equation {{{y=-e^x+10}}} (green) which is a reflection over the line {{{y=5}}} (blue)