Worked examples — Transformations — vertical - horizontal shifts, reflections, stretches - compressions

This page is the "battlefield" for transformations. The parent note built the rules. Here we hunt down every kind of case the topic can throw at you — every sign, every degenerate input, every limiting value — and grind each one to a finished answer.

Before we start, one promise: every symbol you see is one we already earned in the parent note. A quick refresher of the master machine:

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The scenario matrix

Every problem in this topic lives in exactly one of these cells. Our examples below tick off all of them.

Cell	What makes it tricky	Covered by
C1 · Pure vertical, $k>0$ and $a>1$	output-side, "normal" direction	Ex 1
C2 · Pure horizontal, $h>0$ (the "minus means right" trap)	input-side, backwards	Ex 2
C3 · Horizontal compression $b>1$ (bigger = squeezed)	counter-intuitive scaling	Ex 3
C4 · Negative $a$ AND negative $b$ (double reflection)	two flips at once	Ex 4
C5 · Full combined form, correct order of operations	composition order	Ex 5
C6 · Zero / degenerate input ($a=1,b=1$ etc.)	"identity" edge case	Ex 6
C7 · Fractional scale $0<b<1$ (stretch) + fractional $a$ (compress)	small-number scaling	Ex 7
C8 · Real-world word problem (Ferris wheel)	translate words → transform	Ex 8
C9 · Exam twist: read the transform backwards from a given graph	reverse engineering	Ex 9
C10 · Limiting behaviour / domain-range under transform	how domain & range move	Ex 10

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Example 1 — Cell C1: pure vertical, normal direction

Forecast: Guess now — does the vertex go up or down, and does the parabola get skinnier or wider?

1. Identify the parts. Here $a=2$, $k=3$, and $b=1$, $h=0$. Why this step? Matching to $af(x)+k$ first tells us we only touch the output — nothing horizontal moves.
2. Vertex. The vertex of $x^2$ is $(0,0)$. Multiply the height by $2$: still $2\cdot 0=0$. Then add $3$: height becomes $3$. New vertex $(0,3)$. Why this step? $a$ multiplies height, so a height of $0$ stays $0$; $k$ then lifts everything.
3. Move the point $(2,4)$. Height $4\to a\cdot 4=8\to 8+k=11$. New point $(2,11)$. Why this step? Vertical transforms keep $x$ fixed and apply "$\times a$ then $+k$" to $y$.

Verify: Plug $x=2$ into $g$: $2(2{)}^2+3=2\cdot 4+3=11$. ✓ And $g(0)=3$. ✓ Parabola is skinnier ($a>1$ stretch) and lifted 3.

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Example 2 — Cell C2: the "minus means right" trap

Forecast: Left or right? By how much?

1. Read the inside. $g(x)=f(x-h)$ with $h=4$. Why this step? $h$ lives inside the bracket, so it's a horizontal move — and horizontal is backwards.
2. Solve for where the "special thing" happens. The inflection happens when the inside equals $0$: $x-4=0\Rightarrow x=4$. Why this step? The cube behaves at its inside-value $0$; we must feed $x=4$ to make the inside $0$ again.
3. State the shift. New inflection $(4,0)$ — a move right by 4, even though we subtracted.

Verify: $g(4)=(4-4{)}^3=0$. ✓ Test the old point $(-1,-1)$: it should move to $(-1+4,-1)=(3,-1)$, and $g(3)=(3-4{)}^3=(-1{)}^3=-1$. ✓

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Example 3 — Cell C3: horizontal compression (bigger $b$ = squeezed)

Forecast: Is $\cos (3x)$ stretched wide or squeezed narrow compared to $\cos (x)$?

1. Read $b$. $g(x)=f(bx)$ with $b=3$. Why this step? $b$ multiplies the input; $|b|>1$ makes the graph run faster, so it compresses.
2. New period. Period $=\frac{2\pi }{|b|}=\frac{2\pi }{3}$. Why this step? One full cycle needs the inside to travel $2\pi$: $3x=2\pi \Rightarrow x=\frac{2\pi }{3}$.
3. First minimum. $\cos$ hits $-1$ when its inside $=\pi$: $3x=\pi \Rightarrow x=\frac{\pi }{3}$. Why this step? We move the event to $\frac{a}{b}$; the event at $x=\pi$ moves to $\frac{\pi }{3}$.

Look at the figure: the coral $\cos (3x)$ fits three humps inside one lavender hump of $\cos (x)$ — that's the squeeze.

Verify: $g(\pi /3)=\cos (3\cdot \pi /3)=\cos (\pi )=-1$. ✓ Period check: $g(0)=\cos 0=1$ and $g(2\pi /3)=\cos (2\pi )=1$. ✓

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Example 4 — Cell C4: negative $a$ AND negative $b$ (double reflection)

Forecast: Which quadrant does the curve end up in — I, II, III, or IV?

1. Inside flip ($b=-1$). $\sqrt{-x}$ reflects across the $y$-axis. The domain flips: $\sqrt{x}$ needs $x\ge 0$, so $\sqrt{-x}$ needs $-x\ge 0\Rightarrow x\le 0$. Why this step? Negating the input mirrors left↔right, and the "allowed inputs" mirror too.
2. Outside flip ($a=-1$). The leading minus reflects across the $x$-axis, pushing outputs negative. Why this step? $a<0$ negates height; positive ​ values become negative.
3. Combine. Input $x\le 0$ (left half) and output $\le 0$ (below axis) → the curve sits in Quadrant III, sweeping down-left.

The figure shows the mint original in Quadrant I and the coral double-reflection mirrored into Quadrant III.

Verify: $g(-4)=-\sqrt{-(-4)}=-\sqrt{4}=-2$. Original $f(4)=2$; a double flip sends $(4,2)\to (-4,-2)$. ✓ Domain $x\le 0$: $g(-9)=-\sqrt{9}=-3$ defined, $g(9)=-\sqrt{-9}$ undefined. ✓

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Example 5 — Cell C5: full combined form, correct order

Forecast: Where does the corner point land after all four moves?

Use the mnemonic Shift → Squeeze → Stretch → Shift (see Function Composition for why inside-out).

1. Horizontal shift ($h=-1$): $\sqrt{x+1}$. Inside $=0$ at $x=-1$. Point $(0,0)\to (-1,0)$. Why this step? $x+1=(x-(-1))$, so $h=-1$ → left by 1.
2. Horizontal compress ($b=3$): $\sqrt{3(x+1)}$. Divide horizontal distance by 3. The corner is where inside $=0$: still $x=-1$ (dividing $0$ by anything is $0$). Why this step? $|b|>1$ squeezes toward the corner; a corner at "distance 0" doesn't move.
3. Vertical flip + stretch ($a=-2$): $-2\sqrt{3(x+1)}$. Heights multiply by $-2$: corner height $0\to 0$. Why this step? $a=-2$ flips below the axis and doubles the drop.
4. Vertical shift ($k=4$): add 4. Corner $(-1,0)\to (-1,4)$. Why this step? $+4$ lifts everything, corner included.

Verify: $g(-1)=-2\sqrt{3\cdot 0}+4=4$. ✓ Test $x=2$: $g(2)=-2\sqrt{3\cdot 3}+4=-2\cdot 3+4=-2$. ✓ Domain (from Domain and Range): need $3(x+1)\ge 0\Rightarrow x\ge -1$. ✓

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Example 6 — Cell C6: the degenerate "identity" case

Forecast: Does anything move?

1. Read the constants. $a=1$, $b=1$, $h=0$, $k=0$. Why this step? We must check for the degenerate case before "transforming" — sometimes the answer is "nothing changes."
2. Apply each. Multiply height by 1 → same. Multiply input by 1 → same. Shift by 0 → same. Why this step? $1$ is the multiplicative identity and $0$ the additive identity; they leave the V-graph untouched.
3. Conclusion. $g(x)=|x|$ exactly — the identity transformation.

Verify: $g(-3)=|-3|=3=f(-3)$. ✓ $g(5)=5=f(5)$. ✓ The graphs coincide everywhere.

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Example 7 — Cell C7: fractional scales (stretch + compress at once)

Forecast: Wider or skinnier? Taller or flatter?

1. Horizontal, $b=\frac{1}{2}$. Since $0<b<1$, this is a horizontal stretch by factor $\frac{1}{b}=2$. A point at $x=a$ moves to $x=\frac{a}{b}=2a$. Why this step? Small $b$ makes the function run slower, so it spreads out.
2. Vertical, $a=\frac{1}{2}$. Since $0<a<1$, this is a vertical compression: heights halved. Why this step? Multiplying output by less than 1 pushes it toward the $x$-axis.
3. Move $(2,4)$. Horizontal: $x=2\to \frac{2}{1/2}=4$. Vertical: $y=4\to \frac{1}{2}\cdot 4=2$. New point $(4,2)$. Why this step? Apply $x\to \frac{x}{b}$ and $y\to ay$ separately.

Verify: $g(4)=\frac{1}{2}{\left(\frac{1}{2}\cdot 4\right)}^2=\frac{1}{2}\cdot 4=2$. ✓ New point $(4,2)$ ✓ — wider and flatter than $x^2$.

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Example 8 — Cell C8: real-world word problem

Forecast: At the bottom the height is smallest — which cosine transform starts at a minimum?

1. Pick the parent. Start from $\cos t$, which starts at its maximum. We want to start at the minimum (bottom of wheel), so use $-\cos$. Why this step? Reflecting $\cos$ over the $x$-axis makes it start low — matching a rider at the bottom.
2. Vertical scale $a=10$. Radius 10 → amplitude 10. So far: $-10\cos (\cdots )$. Why this step? Amplitude is how far the height swings from centre; that's the radius.
3. Vertical shift $k=12$. Centre is 12 m up → add 12: $-10\cos (\cdots )+12$. Why this step? $k$ raises the "midline" to the axle height.
4. Horizontal scale for the period. Period must be $40$ s: need $b=\frac{2\pi }{40}=\frac{\pi }{20}$. Final model: $H(t)=-10\cos \left(\frac{\pi }{20}t\right)+12.$ Why this step? Period $=\frac{2\pi }{b}=40\Rightarrow b=\frac{\pi }{20}$.
5. Evaluate at $t=10$. $H(10)=-10\cos \left(\frac{\pi }{20}\cdot 10\right)+12=-10\cos \left(\frac{\pi }{2}\right)+12=-10\cdot 0+12=12$. Why this step? At quarter-turn the rider is level with the axle, so height = midline = 12 m.

Verify: $H(0)=-10\cos 0+12=-10+12=2$ m (bottom, radius below axle ✓). $H(20)=-10\cos \pi +12=10+12=22$ m (top ✓). $H(10)=12$ m. ✓ Uses Trigonometric Functions.

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Example 9 — Cell C9: exam twist, reverse-engineer from a graph

Forecast: Which sign of $a$ makes sine start by going down?

1. Amplitude → $a$. Amplitude 4 gives $|a|=4$. "Starts downward" (goes negative first) means a flip: $a=-4$. Why this step? $\sin$ normally rises first; a negative $a$ reflects it so it falls first.
2. Period → $b$. Period $\pi$ means $\frac{2\pi }{|b|}=\pi \Rightarrow |b|=2$, take $b=2$. Why this step? We read the horizontal squeeze straight from the period formula.
3. Vertical shift → $k$. Up 1 gives $k=1$. No horizontal shift, so $h=0$. Why this step? Midline moved up 1 → $k=1$.
4. Assemble. $g(x)=-4\sin (2x)+1$.

Verify: Amplitude $|-4|=4$ ✓. Period $\frac{2\pi }{2}=\pi$ ✓. $g(0)=-4\sin 0+1=1$ (midline ✓). Just after $0$, $\sin$ increasing but $\times (-4)$ makes $g$ decrease → starts downward ✓. Max value $=4+1=5$, min $=-4+1=-3$.

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Example 10 — Cell C10: how domain & range move (limiting behaviour)

Forecast: Does the asymptote move with $h$ or with $k$?

1. Domain. The inside must stay positive: $x-5>0\Rightarrow x>5$. Domain $(5,\infty )$. Why this step? Horizontal shift $h=5$ drags the whole "allowed input" region right by 5. See Domain and Range.
2. Vertical asymptote. It sat at $x=0$; the inside $=0$ now at $x=5$. Asymptote: $x=5$. Why this step? The asymptote is a horizontal-position feature, so it moves with $h$, not $k$.
3. Range. $\ln x$ already covers all reals; multiplying by 3 and subtracting 2 still covers all reals. Range $(-\infty ,\infty )$. Why this step? Vertical scaling/shifting a full-real range leaves it full-real — limiting behaviour is unchanged.

Verify: $g(6)=3\ln (6-5)-2=3\ln 1-2=-2$. ✓ As $x\to 5^{+}$, $\ln (x-5)\to -\infty$, so $g\to -\infty$: asymptote confirmed at $x=5$. ✓ $g(5+e)=3\ln (e)-2=3-2=1$. ✓

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Recall Quick self-test

Which way does $f(x+3)$ shift? ::: Left by 3 (inside, backwards). For $f(4x)$, stretch or compress? ::: Compress (bigger $b$ squeezes). In $g=af(b(x-h))+k$, which letter moves the vertical asymptote? ::: $h$ (horizontal shift). Ferris-wheel height at the top with $H(t)=-10\cos (\frac{\pi }{20}t)+12$? ::: $22$ m. Reverse-engineer sine that starts downward with amplitude 4? ::: $a=-4$.