document.write( "Question 202962: Explain how to get the graph of the function f(x)=e^(x-2) +3 from the graph of the function g(x)=e^x \n" ); document.write( "

Algebra.Com's Answer #153116 by jsmallt9(3758) $\"\"$ $\"About$

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In general, if you take a function and replace the x's with (x - h) you move the graph to the right h units. And if you take a function and replace it's y's with (y - k) then you move the graph up k units.

\n" ); document.write( "In your problem $\"g%28x%29+=+e%5Ex\"$ . If we replace the x with (x-2) we get the function e^(x-2) and the graph this function is the same as g(x) except it is 2 units to the right. If we take g(x) and replace it's y (or g(x)) with (y - 3) (or (g(x) - 3) we get $\"g%28x%29+-+3+=+e%5Ex\"$ . Then if we add 3 to both sides we get $\"g%28x%29+=+e%5Ex+%2B+3\"$ and the graph of this new function is 3 units up from the original g(x).

\n" ); document.write( "Combining both we see that f(x) = e^(x-2) + 3 will have a graph which the same as g(x) except it has been moved 2 units to the right and 3 units up. The graph below is close to the being your problem. (Algebra.com's graphing software will not graph e^x so I used 3^x.)
\n" ); document.write( " $\"graph%28800%2C+800%2C+-10%2C+10%2C+-10%2C+10%2C+%283%5Ex%29+%2C+%283%5E%28x-2%29+%2B+3%29%29\"$ \n" ); document.write( "

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