# Graphing $2^{4-x}$ (very basic question!)

I approached graphing $2^{4-x}$ by simplifying the function to $2^{-x}$ (i.e. the reflection in the y-axis of $2^{x}$), and then applying what I've learned that adding a constant to $x$ translates the graph to the left by $x$ units. My answer was wrong. The correct answer is $2^{-x}$ shifted to the right by 4 units.

I hope someone can provide a clear rule, or show how I should've thought about this properly. I figure it's because the coefficient on $x$ is negative that reverses the rule, but there's probably a cleaner way to think about it.

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To get $2^{4-x}$ from $2^{-\color{maroon}x}$ you subtract $4$: $2^{-(\color{maroon}{x-4})}=2^{-x+4}=2^{4-x}$ (note the parentheses here). So essentially, what you said about the negative reversing the rules is correct. –  David Mitra Dec 11 '12 at 12:46
How about looking at it as $2^{4} / 2^x$? Does that help you? What you did was also not wrong in theory, you were just using $-x$ (subtracting $4$ from $x$). –  Amzoti Dec 11 '12 at 12:53

Had you been adding $4$ to the $x$, you'd indeed have been shifting to the left. However, in this case, you had $-x$. Thus, $$4-x=-x+4=-(x-4),$$ so in effect, you were subtracting $4$ from the $x$, when all was said and done. The negative did, indeed, reverse the rule.
Put another way, if we have two functions $f(x)$ and $g(x)$ and a positive constant $a$, then $g(x)$ is obtained from shifting $f(x)$ left by $a$ if $g(x)=f(x+a)$, and right by $a$ if $g(x)=f(x-a)$. In this case, $f(x)=2^{-x}$, $g(x)=2^{4-x}$ and $a=4$. $$f(x+4)=2^{-(x+4)}=2^{-4-x}\neq g(x),$$ but $$f(x-4)=2^{-(x-4)}=2^{4-x}=g(x).$$
I appreciate the thorough answer. I see now that if I had considered $2^{-x}$ as $2^{-(x)}$ then I would've subtracted the 4 from $x$ and got $2^{4-x}$ –  PeteUK Dec 11 '12 at 15:02