## Calculus: Early Transcendentals 9th Edition

a) $f(x)=8-e^{x}$ b) $f(x)=e^{4-x}$
a) Because we have to reflect $e^{x}$ across the line y=4, we must think about how far our horizontal asymptote is from the line of reflection. In this case, the function $e^{x}$ has an asymptote and the line y=0 (4 units below y=4). Therefore, when we reflect this function the asymptote will be 4 units above the line of reflection at y=8. In order to make sure that the function will exponentially decrease, it must be 8 (the max asymptote value) minus $e^{x}$. Hence, $f(x)=8-e^{x}$. b) Yet again we can see that the y-intercept of the original function is at the point (0,1) and that this point is two units away from the reflection line of x=2. Therefore, a new point in this graph has to be (0,1) reflected over the line x=2, so, the point will be (4,1). Next, we need to reflect the line over the a vertical line, which will reverse the direction of the opening of the function. The function $e^{-x}$ will do this. Now all we need to add in is the 4 units that the graph must be moved to the right, hence, $f(x)=e^{4-x}$.