Answer:
(c) r units right
Explanation:
The translation transformations are ...
g(x) = f(x -h) +k . . . . . . . shifts h units right and k units up
Your transformation has h=r and k=0, so the graph is shifted r units right and no units vertically.
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Additional comment
This in not terribly mysterious. Suppose you have f(1) = 6. This means the point (1, 6) is on the graph.
Now, suppose you transform this to ...
g(x) = f(x -2)
When the value of x is 3, the value of g(x) is f(3-2) = f(1) = 6. That means the point (3, 6) is on the graph of g(x).
It is the same as the point (1, 6) on the graph of f(x), but shifted 2 units to the right.
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You can shift a function r units right by replacing x with (x-r), as we just discussed. You can shift a function u units up by replacing y with (y-u). When we write equations in "functional form", the "u" gets "moved."
y-u = f(x) . . . . . graph of f(x) shifted u units up
y = f(x) +u . . . . graph of f(x) shifted u units up. (u added to both sides of the above equation)