Outside the function moves it up/down: y = f(x) + k slides the whole graph up by k (down if k is negative). Changes outside the function act on the y-values, exactly as they look.
IB-style question — vertical shift
y = f(x) passes through (2, 5). Where does y = f(x) + 4 pass through (at x = 2)?
Step by step
- Add 4 to the y-coordinate.
Final answer
(2, 9).
Outside = as expected: +k outside means up k; −k means down k. No surprises here.
Inside the function moves it the OPPOSITE way: y = f(x − a) slides the graph right by a — the opposite of the sign you see. f(x + a) moves it left. Changes inside f act on the x-values, and they're counterintuitive.
IB-style question — horizontal shift
Describe the transformation from y = f(x) to y = f(x − 3).
Step by step
- Inside is x − 3; move the opposite of the sign.
Final answer
Translation 3 units to the right.
Don't be fooled by the minus: f(x − 3) goes right (not left). Think: to get the same output, x must be 3 bigger, so the graph sits 3 to the right.
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Combine into one vector: A horizontal shift a and vertical shift b together form the translation vector with top a (right) and bottom b (up). So y = f(x − a) + b is a translation by that vector.
IB-style question — read the vector
Write the translation taking y = f(x) to y = f(x + 1) − 4 as a vector.
Step by step
- Inside x + 1 → left 1 (a = −1); outside − 4 → down 4 (b = −4).
Final answer
Translation by the vector (−1, −4) (1 left, 4 down).
Move every point by the vector: A translation slides every point (and asymptote, intercept, vertex) by the same vector: add a to the x, b to the y.
IB-style question — image of a point
y = f(x) passes through (3, 5). Find its image under y = f(x − 2) + 1.
Step by step
- Right 2 (add to x), up 1 (add to y).
Final answer
(5, 6).
Asymptotes move too: If f has a vertical asymptote x = 0, then f(x − 2) + 1 has it at x = 2 (and a horizontal asymptote rises by 1).