Track five labelled points through eight transformations

The Teacher asks that composite transformations and replacements be emphasised. Here replace x by $x-2$: $(p,q)\mapsto(p+2,q)$. The source gives the entire first transformed graph. All later coordinate calculations use the printed rounded B/D data consistently; they do not assert more precision for the underlying curve.

Replace x by $-x$ to get $y=f(-x)$, then y by $y-1$ to get $y=f(-x)+1$. Intermediate points are $A\prime(1,0)$, $B\prime(0.549,-0.631)$, $C\prime(0,0)$, $D\prime(-1.215,2.113)$, $E\prime(-2,0)$. The overall map is $(p,q)\mapsto(-p,q+1)$. Both originals print $B\prime\prime(0.5,0.369)$; the x-coordinate must be 0.549 from the given B. Preserve that source typo in native/review evidence; the answer and reconstructed graph use 0.549.

Translate left 1, obtaining $y=f(x+1)$; then replace x by $2x$ to obtain $y=f(2x+1)$. The intermediate points are $A\prime(-2,0)$, $B\prime(-1.549,-0.631)$, $C\prime(-1,0)$, $D\prime(0.215,2.113)$, $E\prime(1,0)$. The overall map is $(p,q)\mapsto((p-1)/2,q)$. The Teacher says this repeats Example 8 and need not be treated in detail again; reuse that replacement method while preserving both source diagrams and every coordinate here.

First retain original inputs $p\ge0$ and reflect them to form $y=f(|x|)$. The intermediate D images are $(\pm1.215,2.113)$ and E images $(\pm2,0)$, with C unchanged. Then replace y by $y/2$: $y/2=f(|x|)$, hence $y=2f(|x|)$. The source final plot has doubled heights 4.226 but is mislabelled $y=f(|x|)$; preserve that typo in native/review text and label the actual final graph with its factor 2. Negative-input A/B are not carried into this construction.

First reflect in the y-axis to obtain $y=f(-x)$, with intermediate A′(1,0), B′(0.549,−0.631), C′(0,0), D′(−1.215,2.113), E′(−2,0). Retain the right-hand half of this reflected graph and mirror it to obtain $y=f(-|x|)$. Equivalently, retain original inputs $p\le0$, then duplicate them symmetrically. The displayed intermediate and final source sketches are both retained.

Translate right 2 to obtain $y=f(x-2)$; the intermediate coordinates are exactly those in part (a). Then reflect only the negative-output arcs in the x-axis, giving $y=|f(x-2)|$. The map is $(p,q)\mapsto(p+2,|q|)$. B changes output sign; A/C/E stay on the axis and D stays above it. Both source stages and all coordinates are retained.

First retain the original $p\ge0$ side and mirror it: $y=f(|x|)$ has C(0,0), D-images $(\pm1.215,2.113)$ and E-images $(\pm2,0)$. Then replace x by $x-2$ to obtain $y=f(|x-2|)$, symmetric about $x=2$. For each retained input p, the two images have x-coordinates $2\pm p$. At p0 the two images coincide, so C is counted once. This differs from taking output modulus in part (f).

A, C and E have zero outputs, so they have no finite reciprocal images and their x-coordinates become poles in the supplied continuous sketch. From the printed rounded data, $1/(-0.631)\approx-1.585$ and $1/2.113\approx0.473$. B becomes a local maximum and D a local minimum. Preserve all four reciprocal branches and the original unbounded tail behaviour, giving horizontal limit 0. The reciprocal is undefined at 0, so it has no y-intercept; it has no x-intercept either. The source repeats the original given diagram beside this reciprocal, and both are retained.