Left and Right

Now put the change inside the brackets. Take $y = f\left(x-4\right)$ with $f\left(x\right) = {x}^{2}$. The parent reaches its lowest output, 0, when its input is 0. Here the input that f receives is $x-4$, and that is 0 when x is 4. So the vertex is at (4, 0): the graph has moved 4 to the right.

Every other point follows. The parent gives 1 at input 1. On the new graph, f has to receive 1, so $x-4 = 1$ and x is 5: the point (1, 1) becomes (5, 1). Each output now arrives 4 steps later, which means 4 places further right.

So $y = f\left(x-h\right)$ moves the graph right h. The heights are the same as before; only the inputs where they happen have changed.