Transform inputs and outputs in the right order

Try

Suppose a recorded function has a known point f(4)=3. A new display uses g(x)=2f(x−1)−5. Find the input at which the new display uses that known old output. Solve x−1=4 before changing the output: the new point is (5,1).

This small task separates two operations often confused in a graph transformation. The inside expression selects an old input. The outside expression changes the old output after it has been selected.

Predict

Compare f(x−3) and f(x)−3. Which moves a known point right, and which moves it down? Compare f(2x) and 2f(x). Which reduces the horizontal distance needed to reach a known old input? Predict using a particular point rather than memorizing the phrase “opposite inside.”

Experiment

Use f(u)=u² on the restricted domain −2≤u≤2. Construct g(x)=−f(2(x−1))+3. For each old point (u,v), solve 2(x−1)=u, giving x=1+u/2. The new output is 3−v.

Old input u	Old output v	New input x	New output
−2	4	0	−1
−1	1	1/2	2
0	0	1	3
1	1	3/2	2
2	4	2	−1

Plot both curves with labeled, shared units. The transformed domain is [0,2], and the transformed range is [−1,3]. The graph is compressed horizontally, shifted right and reflected vertically before the upward shift. These descriptions must agree with the point table.

Understand

For g(x)=Af(B(x−h))+k, let (u,v) be a point of f. With B nonzero, the matching new input solves B(x−h)=u, so x=h+u/B. The output becomes Av+k. This gives the point mapping (u,v) to (h+u/B, Av+k).

Here h is a horizontal shift, k is a vertical shift, A multiplies outputs, and B multiplies the input presented to f. Horizontal distances scale by 1/|B|; vertical distances scale by |A|. A negative factor reverses the relevant direction. The formula requires B nonzero. If B=0, the inside expression is always zero: the new rule is constant when f(0) exists, or undefined everywhere when it does not. Ordinary stretch language then fails.

Transformations can change order-sensitive results. Shifting upward by 4 and then doubling gives 2f(x)+8. Doubling first and shifting upward by 4 gives 2f(x)+4. The previously added shift is scaled only in the first order.

For a transformed domain, solve an input condition rather than moving interval endpoints by intuition. If f accepts 2≤u≤8 and g(x)=f(−2x+4), require 2≤−2x+4≤8. Reversing order when dividing by −2 yields −2≤x≤1. The negative factor also reverses which old endpoint corresponds to each new endpoint.

Apply

In the written practice, derive a general point mapping and test it with a supplied point. Draw arrows labeled “old input,” “new input” and “new output.” Check both a numerical point and the whole transformed domain; correct-looking vertices alone do not verify a transformation.

The existing quadratic investigation supplies a restricted vertex-form example. It uses positive curvature, so the downward-opening transformations here extend beyond its controls.

Continue

A single formula need not apply throughout a function’s domain. The next lesson uses branch conditions and endpoint markers to keep piecewise rules unambiguous.