I have some confusion about shifts concerning exponential functions. I can best describe my question with an example.
Take $y = e^{-(x-3)}$. This graph has a reflection over the $y$-axis and is shifted right $3$ units.
Why right instead of left, though. Considering that this is equivalent to $y=e^{(-x+3)}$ I thought that the shift would be opposite of the sign being that it is in parentheses.
Take $y = e^{x+3}$ for example. The graph of this function is shifted left $3$ because of the parentheses.
How am I supposed to figure out the shifts in the graphs? Am I missing a detail here?
Best Answer
All you have written is correct. You only have to take care on the order of the transformations. For this, ask: 'What happens to $x$?' and reverse the order and the operations.
In the case of $e^{-(x-3)}$, $x$ is first decreased by $3$, then multiplied by $-1$. If we reverse these operations, we see that first we have to reflect the graph of $e^x$ along the $y$-axis and then shift it to the right by $3$ (shift it to the left by $-3$).
For the same $e^{-x+3}$, we find that $x$ is first multiplied by $-1$ then the gotten expression is increased by $3$, so, reversing these, we first shift, indeed to the left, and then reflect.
Update:
The transformation for $e^{-(x-3)}$ corresponds to the substitions: let $u:=x-3$. First, from $u\mapsto e^u$ we go to $u\mapsto e^{-u}$ by reflecting the original graph on the $y$ axis. Then making the substition $x\mapsto x-3$ i.e. $x\mapsto u$ in the variable will give us the second step. You will be convinced if you plug in (enough) concrete values of $x$: e.g. if $x=3$ then $u=0$ and then $e^{-(x-3)}=e^{-u}=1$. If $x=4$ then $u=1$, and so on..
In general, the graph of $g(x)=f(x-3)$ is shifted to the right (to the left by $-3$) compared to the graph of $f(x)$, because $$g(x+3)=f(x)\,.$$