# Graphing Basic Exponential Functions

How can we use the graph of $y=2^x$ to sketch the graph of $y=2^{x-1}$?

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Note that for any $x,y$ graph: $y = f(x-1)$ is just the graph $y = f(x)$ shifted over one unit to the right.

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Why does it shift over one unit to the right instead of left? – jaykirby Jul 3 '13 at 11:19
I'll explain it like this: where you once had $f(0)$, you now have $f(0-1)=f(-1)$. That is, $f(-1)$ was shifted to the right to cover $x=0$. – Omnomnomnom Jul 3 '13 at 11:21
Could you try explaining it to me without the f's..I get confused easily by that.. – jaykirby Jul 3 '13 at 11:24
@omnomnomnon: What happens to the y value, it is always going to be one less isn't it? – jaykirby Jul 3 '13 at 11:44
So to explain it without all the $f$'s, let's look at $2^x$ and $2^{x-1}$ over the numbers $\{1,2,3\}$. We have: $$2^1=2;2^{1-1}=1\\ 2^2=4; 2^{2-1}=2\\ 2^3=8; 2^{3-1}=4$$ At each $x$, the second graph takes the value that was the $y$ at one to its left. The result, if you graphed this, is that $2^{x-1}$ is shifted to the right by one. Does that make sense? – Omnomnomnom Jul 3 '13 at 14:01