Both at Once

The two kinds of move combine. Take $y = f\left(x+1\right)-2$ with $f\left(x\right) = {x}^{2}$. Inside, $x+1$ moves the graph 1 left. Outside, the minus 2 moves it 2 down. So the vertex goes from (0, 0) to (-1, -2).

The pattern is worth naming: for $f\left(x\right) = {x}^{2}$, $y = f\left(x-h\right)+k$ has its vertex at (h, k). Read h from inside the brackets, where the sign looks flipped ($x+1$ is $x-\left(-1\right)$, so h is -1), and read k from outside, exactly as written. With $f\left(x\right) = {x}^{2}$ this is $y = {\left(x-h\right)}^{2}+k$, called the vertex form of a parabola.

The order of the two moves does not matter: slide then lift, or lift then slide, you land in the same place. After the vertex, the shape does the rest. One step right of it the curve is 1 higher, so (0, -1) is on this graph too.