Function Transformations: Shifts

Add to x and the graph slides, add to the output and it lifts, multiply and it stretches. Augmentation, normalisation and positional encodings are all transformations worn as tools.

The bowl x squared slides right 3 and lifts 2 to become (x minus 3) squared plus 2, carrying its vertex from the origin to (3, 2)

Why Does This Exist?

Graphs rarely arrive in position. A parabola centres at the origin; the data sits three units right and two up. A signal spans millivolts; the model wants unit range. Transformations move, stretch and flip graphs into place without recomputing their essence: shifting, scaling and reflecting are cheaper than refitting.

They are also the vocabulary of preprocessing. Standardisation is a shift plus a stretch. Cropping translates coordinates. Augmentation shifts and flips deliberately. The slopes these moves preserve or rescale are derivatives and partial derivatives; following transformed landscapes downhill is gradients.

The one idea you need first. A graph plots $(x, f(x))$ . Changing what goes into $f$ moves points sideways; changing what comes out moves them vertically. Inside versus outside is the whole compass.

Think of It Like This

Sliding a stencil on graph paper

Cut a parabola from a stencil and lay it on graph paper: vertex at the origin. Slide it 3 right and 2 up without rotating: same cutout, new position, vertex now at $(3, 2)$ . Stretch it taller: same centre, steeper sides. Flip it over: a hill instead of a bowl.

Every transformation is stencil handling, never redrawing. And the stencil explains the backwards-feeling horizontal shift: to keep the same cutout point over a new $x$ , the formula must subtract, feeding the old input to the unchanged shape.

Where it stops: stencils slide rigidly, while scaling stretches the cutout itself. The analogy shows position; stretch needs the multiplier below.

How It Actually Works

Four moves. Vertical shift: $f(x) + k$ lifts every point by $k$ ; $(x - 3)^2 + 2$ sits 2 above $(x-3)^2$ . Horizontal shift: $f(x - h)$ slides right by $h$ , backwards-feeling but strict, because $x - h$ feeds the old input $h$ units late. Vertical stretch: $a \cdot f(x)$ multiplies heights. Horizontal stretch: $f(bx)$ divides widths by $b$ . Reflection: $-f(x)$ flips over the $x$ -axis.

Combine with care: order matters. Shift-then-scale differs from scale-then-shift, since scaling moves already-shifted positions.

Worked example: carry the vertex to (3, 2)

Start $f(x) = x^2$ , vertex $(0, 0)$ . Slide right 3: replace $x$ with $x - 3$ , giving $(x - 3)^2$ , vertex $(3, 0)$ . Lift 2: add outside, $g(x) = (x - 3)^2 + 2$ , vertex $(3, 2)$ .

Verify. $g(3) = 0 + 2 = 2$ . At the old centre: $g(0) = 9 + 2 = 11$ , the original $0$ lifted by $2$ and pushed right off the axis. The diagram above shows the original bowl in place and the carried bowl arriving at $(3, 2)$ .

Code

1def f(x: float) -> float:2    return x**23
4
5def g(x: float) -> float:6    # Right 3 (inside, backwards), up 2 (outside, as written).7    return f(x - 3.0) + 2.08
9
10print(g(3.0))  # -> 2.011print(g(0.0))  # -> 11.012print(f(0.0), f(3.0))  # -> 0.0 9.0

The last line proves the shift is not a value change: $f$ still returns $9.0$ at $3$ , while $g$ returns $2.0$ there. Same shape, different position, and the code says so in one line each.

Watch Out For

Shifting left with (x minus 3)

$(x - 3)^2$ moves the bowl RIGHT by 3, and most beginners slide it left. The test is instant: the vertex solves $x - 3 = 0$ , giving $x = 3$ , on the right. Inside-the-brackets moves run opposite to intuition because the formula compensates to reuse old outputs.

Find the new zero of the inside expression to locate the shift. If a shifted activation behaves mirrored, this sign is the cause.

Stacking scale and shift in the wrong order

"Stretch by 2, then shift right 3" is $2f(x - 3)$ ... applied as $2(x-3)^2$ shifts first in reading order but the stretch scales the shifted shape. Meanwhile $2x$ shifted gives $(2x - 3)^2 = 4(x - 1.5)^2$ : the shift halved. Same words, different order, vertex at 3 versus 1.5.

Parenthesise each stage and verify with the vertex. If a preprocessing chain lands off-centre, unpeel the stages one at a time.

The Quick Version

Outside changes move vertically: $f(x) + 2$ lifts by 2, as written.
Inside changes move horizontally and backwards: $f(x - 3)$ slides right 3.
Multipliers stretch: $a$ outside scales heights, $b$ inside divides widths.
$(x-3)^2 + 2$ carries $(0,0)$ to $(3,2)$ ; $g(3) = 2$ , $g(0) = 11$ .
Locate shifts by zeroing the inside; parenthesise stacked stages.