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)\$.