Let's look at the first case: $g(x) = f(x-c)$ .
Suppose that you were making two $xy$-tables; one for the original graph $y = f(x)$ and the second one for the new graph $y = f(x-c)$. How would these tables compare? Look, for example, what happens when $x = 0$ in the first table and $x = c$ in the second table. For the $y$-value in the first we get $f(0)$ and for the $y$-value in the second, we get $f(c-c) = f(0)$, the same thing. This gives the point $(0, f(0))$ on the first graph and $(c, f(0))$ on the second. Note that this second point is c units to the right of the first. Similarly, corresponding to the point $(1, f(1))$ on the first, we have a point $(1+c, f(1))$ on the second, again $c$ units to the right. In general, for each point $(a, f(a))$ on the graph of $y = f(x)$ there is a point $(a+c, f(a))$, c units to the right, on the graph of y = f(xc). (In all of this we have assumed that $c$ is positive.
Now think about what happens when $g(x) = f(x+c)$.
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