Transformations of trig graphs — A·sin(Bx + C) + D (amplitude, period, phase, vertical shift): Worked examples

The scenario matrix

Every sinusoid question is built from choices of the four numbers $A, B, C, D$ . Here is the full grid of "cases" — each row is a class of question, and the last column names the worked example that lands in that cell.

Cell	What makes it special	Danger it hides	Covered by
C1	All four numbers "friendly" positive	none — the warm-up	Example 1
C2	$A < 0$ (negative amplitude)	flip vs. shrink confusion	Example 2
C3	$B < 0$ (negative inside $x$ )	period uses $∣ B ∣$ ; direction flips	Example 3
C4	$C \neq = 0$ and $B \neq = 1$	must factor $B$ before reading the slide	Example 4
C5	$B \to$ large / $B \to 0^{+}$ (limiting)	period races to $0$ / blows up to $\infty$	Example 5
C6	Degenerate $A = 0$	"wave" collapses to a flat line	Example 6
C7	Reverse-engineer from max/min/start	inverting all four formulas at once	Example 7
C8	Real-world word problem (tides)	translating words → $A, B, C, D$ , units	Example 8
C9	Exam twist: a cosine disguised as sine	using cos as shifted sin	Example 9

Before each example, cover the answer and forecast — guessing first is what makes the rule stick.

Example 1 — the friendly case (Cell C1)

Worked example

$y = 2 sin (3 x + \frac{π}{2}) + 1$ . Find amplitude, period, phase shift, midline, max, min. Forecast: All four numbers are simple. Guess each of the five outputs before reading on.

Step 1 — Amplitude $= ∣ A ∣ = ∣2∣ = 2$ . Why this step? $sin$ only ever outputs values in $[- 1, 1]$ ; multiplying by $A$ stretches that to $[- 2, 2]$ . The height above the midline is $∣ A ∣$ .

Step 2 — Period $= \frac{2 π}{∣ B ∣} = \frac{2 π}{3}$ . Why this step? One full cycle needs the input to grow by $2 π$ . Input is $3 x$ , so $3 Δ x = 2 π \Rightarrow Δ x = \frac{2 π}{3}$ .

Step 3 — Phase shift. Factor: $3 x + \frac{π}{2} = 3 (x + \frac{π}{6})$ , so shift $= - \frac{π}{6}$ (left by $\frac{π}{6}$ ). Why this step? The slide lives in $x$ , not in $3 x$ ; factoring $B = 3$ out exposes the true $x$ -shift.

Step 4 — Midline / max / min. $D = 1 \Rightarrow$ midline $y = 1$ ; max $= 1 + 2 = 3$ ; min $= 1 - 2 = - 1$ . Why this step? Adding $D$ lifts every point; peaks sit $∣ A ∣$ above the midline, troughs $∣ A ∣$ below.

Verify: At $x = - \frac{π}{6}$ (the start of a cycle), $3 x + \frac{π}{2} = 0$ , so $y = 2 sin 0 + 1 = 1$ — exactly the midline, rising. ✔ Everything consistent.

Figure — Transformations of trig graphs — A·sin(Bx + C) + D (amplitude, period, phase, vertical shift)

Example 2 — negative amplitude (Cell C2)

Worked example

$y = - 4 sin (x) + 2$ . Amplitude? Where does it start at $x = 0$ — peak or trough? Forecast: Does the minus make the wave shorter, or upside-down? Guess before reading.

Step 1 — Amplitude $= ∣ - 4∣ = 4$ . Why this step? Amplitude is a distance (midline to peak) so it is never negative. The minus is not a size, it is a direction.

Step 2 — What the minus really does. Write $- 4 sin x = 4 \cdot (- sin x)$ . The factor $- sin x$ is the ordinary sine flipped across the midline. Why this step? Reflecting swaps "rising through midline" for "falling through midline", so the shape starts by going down, not up.

Step 3 — Value at $x = 0$ . $y = - 4 sin 0 + 2 = - 4 (0) + 2 = 2$ . Why this step? $x = 0$ is where an un-flipped sine crosses its midline going up; here it crosses going down, but it is still on the midline, value $2$ .

Step 4 — Max/min. Midline $y = 2$ , max $= 2 + 4 = 6$ , min $= 2 - 4 = - 2$ .

Verify: Just after $x = 0$ , $sin x > 0$ , so $- 4 sin x < 0$ , so $y < 2$ : the graph dips below the midline immediately — confirming the flip (a normal sine would rise above). ✔

Example 3 — negative inside the $x$ (Cell C3)

Worked example

$y = sin (- 2 x)$ . Find the period and describe the shape versus $sin (2 x)$ . Forecast: Does the minus change the period? Does it flip the wave? Guess both.

Step 1 — Period uses $∣ B ∣$ . Here $B = - 2$ , so $T = \frac{2 π}{∣ - 2∣} = π$ . Why this step? Period is a positive length; the sign of $B$ cannot make a length negative, so we always take $∣ B ∣$ .

Step 2 — What the minus does to the shape. Use the identity $sin (- θ) = - sin θ$ (sine is an odd function, from the Unit circle and sine definition). So $sin (- 2 x) = - sin (2 x)$ . Why this step? It converts a "negative- $B$ " wave into a "negative- $A$ " wave, which we already know is a vertical flip of $sin (2 x)$ .

Step 3 — Conclusion. $sin (- 2 x)$ has the same period $π$ as $sin (2 x)$ but is reflected across the midline $y = 0$ (it starts going down).

Verify: At $x = \frac{π}{4}$ : $sin (- 2 \cdot \frac{π}{4}) = sin (- \frac{π}{2}) = - 1$ , while $sin (2 \cdot \frac{π}{4}) = sin (\frac{π}{2}) = + 1$ . Opposite signs, same size → confirmed flip. ✔

Example 4 — factoring $B$ when $B \neq = 1$ (Cell C4)

Worked example

$y = 5 sin (4 x - π) - 3$ . Find the exact phase shift (with direction) and the midline. Forecast: Is the shift $π$ ? Or $π /4$ ? Left or right? Commit to a guess.

Step 1 — Never read the shift off $C$ alone. The number $- π$ is added to $4 x$ , not to $x$ . Why this step? This is the single most common error (see the parent's mistake box). The slide must be measured in the same units as $x$ .

Step 2 — Factor $B = 4$ . $4 x - π = 4 (x - \frac{π}{4})$ . Why this step? Now the input has the clean form $B (x - h)$ with $h = \frac{π}{4}$ , and $h$ is the true horizontal shift, rightward-positive.

Step 3 — Read it. Shift $= - \frac{C}{B} = - \frac{- π}{4} = + \frac{π}{4}$ → right by $\frac{π}{4}$ . Why this step? $f (x - h)$ moves the graph toward positive $x$ by $h$ ; here $h = + \frac{π}{4}$ .

Step 4 — Midline. $D = - 3$ , so midline $y = - 3$ ; max $= - 3 + 5 = 2$ , min $= - 3 - 5 = - 8$ .

Verify: A cycle should start (rising through midline) where the input is $0$ : $4 x - π = 0 \Rightarrow x = \frac{π}{4}$ . That matches the "right by $\frac{π}{4}$ " claim. ✔ Contrast the wrong answer: if the shift were $π$ , the input at $x = π$ would be $4 π - π = 3 π \neq = 0$ — not a cycle start, so $π$ is definitely wrong.

Example 5 — limiting behaviour of $B$ (Cell C5)

Worked example What happens to the graph of

$sin (B x)$ as (a) $B \to \infty$ and (b) $B \to 0^{+}$ ? Forecast: One case makes the waves cram together; the other stretches them flat. Which is which?

Step 1 — Write the period as a function of $B$ . $T (B) = \frac{2 π}{∣ B ∣}$ . Why this step? Every limiting question about wave spacing is really a limit of this formula.

Step 2 — Case (a) $B \to \infty$ . $T = \frac{2 π}{B} \to 0$ . Why this step? A huge $B$ shoves the input past $2 π$ almost instantly, so cycles pile up: the graph becomes an infinitely fast wiggle — a "blur" of oscillation.

Step 3 — Case (b) $B \to 0^{+}$ . $T = \frac{2 π}{B} \to \infty$ . Why this step? A tiny $B$ means the input crawls; over any screen you'd see only the very start of one enormous cycle, which looks almost like a straight line through the origin. Indeed for small $B x$ , $sin (B x) \approx B x$ (a near-straight slope).

Verify (numeric anchor): $B = 100 \Rightarrow T = \frac{2 π}{100} \approx 0.0628$ (tiny). $B = 0.01 \Rightarrow T = \frac{2 π}{0.01} \approx 628.3$ (huge). The two limits point in opposite directions, as claimed. ✔

Example 6 — the degenerate case $A = 0$ (Cell C6)

Worked example Describe

$y = 0 \cdot sin (7 x + 2) + 4$ . Is it even a wave? Forecast: With $A = 0$ , what shape survives? Guess before reading.

Step 1 — Substitute $A = 0$ . $y = 0 \cdot sin (\dots) + 4 = 0 + 4 = 4$ for every $x$ . Why this step? Zero times anything (including the whole wiggling sine) is zero, so nothing is left but the constant $D$ .

Step 2 — Interpret geometrically. The graph is the flat horizontal line $y = 4$ — the midline itself, with no oscillation. Why this step? Amplitude $= ∣ A ∣ = 0$ means "height above midline is zero", i.e. the peaks and troughs have collapsed onto the midline.

Step 3 — Period/shift are meaningless here. Although $B = 7$ and $C = 2$ appear, they act on a term that has been multiplied to $0$ , so they change nothing observable. Why this step? A transformation of an invisible wave produces an invisible change — worth stating so you don't waste time computing $T = \frac{2 π}{7}$ for a flat line.

Verify: Pick any two $x$ values, say $x = 0$ and $x = 1$ : $y = 0 \cdot sin (2) + 4 = 4$ and $y = 0 \cdot sin (9) + 4 = 4$ . Same output → truly constant. ✔

Example 7 — reverse-engineer from data (Cell C7)

Worked example A sinusoid has

maximum $9$ , minimum $1$ , period $6$ , and rises through its midline (a cycle start) at $x = 2$ . Write it as $A sin (B x + C) + D$ . Forecast: Four unknowns, four clues. Guess $D$ and $A$ first (they're the easy pair).

Step 1 — Midline $D = \frac{max + min}{2} = \frac{9 + 1}{2} = 5$ . Why this step? The midline is the average level the wave wobbles around — halfway between top and bottom.

Step 2 — Amplitude $A = \frac{max - min}{2} = \frac{9 - 1}{2} = 4$ . Why this step? The full swing is $max - min = 8$ ; half of that (midline to peak) is the amplitude.

Step 3 — $B$ from the period. $B = \frac{2 π}{T} = \frac{2 π}{6} = \frac{π}{3}$ . Why this step? Invert $T = \frac{2 π}{∣ B ∣}$ . We take the positive root because we want a plain, un-flipped sine.

Step 4 — $C$ from the start point. The wave rises through the midline at $x = 2$ , i.e. it is shifted right by $2$ . Shift $= - \frac{C}{B} = 2 \Rightarrow C = - 2 B = - 2 \cdot \frac{π}{3} = - \frac{2 π}{3}$ . Why this step? "Rises through midline" is exactly where a plain sine starts a cycle; matching that instant to $x = 2$ fixes the horizontal slide, and the shift formula converts slide → $C$ .

Answer: $y = 4 sin (\frac{π}{3} x - \frac{2 π}{3}) + 5.$

Verify: At $x = 2$ : input $= \frac{π}{3} (2) - \frac{2 π}{3} = 0$ , so $y = 4 sin 0 + 5 = 5 =$ midline. ✔ Max reached when input $= \frac{π}{2}$ : $y = 4 (1) + 5 = 9$ ; min when input $= - \frac{π}{2}$ : $y = - 4 + 5 = 1$ . ✔

Example 8 — real-world word problem: tides (Cell C8)

Worked example The water depth in a harbour is

$8 m$ at high tide and $2 m$ at low tide. High tides are $12 hours$ apart. At $t = 0$ (midnight) the depth is exactly at the average level and rising. Model depth $h (t)$ in metres, $t$ in hours. Forecast: Which of $A, B, C, D$ does each sentence pin down? Match them before reading.

Step 1 — $D$ (average level). $D = \frac{8 + 2}{2} = 5 m$ . Why this step? "Average level" is the midline; average of high and low.

Step 2 — $A$ (tidal range). $A = \frac{8 - 2}{2} = 3 m$ . Why this step? Half the total rise-and-fall; peak sits $3$ m above the $5$ m midline.

Step 3 — $B$ (period). Consecutive high tides are $12$ h apart, so $T = 12$ h and $B = \frac{2 π}{12} = \frac{π}{6} per hour$ . Why this step? The gap between identical points of the wave is the period; invert to get $B$ .

Step 4 — $C$ (start condition). At $t = 0$ the depth is at the midline rising — exactly the start of a plain sine cycle — so no shift: $C = 0$ . Why this step? $sin (B t)$ already starts at its midline going up, which is precisely the stated midnight condition.

Answer: $h (t) = 3 sin (\frac{π}{6} t) + 5 (metres) .$

Verify (units + values): $[B] = \frac{rad}{h}$ , so $B t$ is dimensionless — good, sine needs a pure number. First high tide when $\frac{π}{6} t = \frac{π}{2} \Rightarrow t = 3$ h: $h = 3 (1) + 5 = 8$ m ✔. First low tide at $t = 9$ h: $h = 3 sin \frac{3 π}{2} + 5 = 3 (- 1) + 5 = 2$ m ✔. Related model: this is a Simple harmonic motion oscillation.

Example 9 — exam twist: a cosine in disguise (Cell C9)

Worked example Rewrite

$y = 6 cos (2 x) - 1$ in the form $A sin (B x + C) + D$ , then state its amplitude, period and phase shift. Forecast: How far must you slide a sine to make it look like a cosine? Guess the shift.

Step 1 — Recall the sine↔cosine relation. $cos θ = sin (θ + \frac{π}{2})$ . Why this step? Cosine is just sine shifted; see cos as shifted sin. This lets us reuse all the sine machinery instead of learning a separate cosine ruleset.

Step 2 — Substitute $θ = 2 x$ . $6 cos (2 x) - 1 = 6 sin (2 x + \frac{π}{2}) - 1$ . Why this step? Direct application of the identity with the actual input $2 x$ — nothing else changes.

Step 3 — Read off the standard-form numbers. $A = 6$ , $B = 2$ , $C = \frac{π}{2}$ , $D = - 1$ .

Amplitude $= ∣6∣ = 6$ .
Period $= \frac{2 π}{2} = π$ .
Phase shift: factor $2 x + \frac{π}{2} = 2 (x + \frac{π}{4})$ → left by $\frac{π}{4}$ . Why this step? Once in $A sin (B x + C) + D$ form, every earlier formula applies unchanged.

Verify: Check the two forms agree at $x = 0$ : original $= 6 cos 0 - 1 = 6 - 1 = 5$ ; rewritten $= 6 sin \frac{π}{2} - 1 = 6 (1) - 1 = 5$ . ✔ And at $x = \frac{π}{4}$ : original $= 6 cos \frac{π}{2} - 1 = - 1$ ; rewritten $= 6 sin π - 1 = - 1$ . ✔

Recap — which cell did each example hit?

Recall Self-test (click to reveal)

Amplitude of $y = - 4 sin x + 2$ ? ::: $∣ - 4∣ = 4$ (the minus is a flip, not a size). Period of $sin (- 2 x)$ ? ::: $\frac{2 π}{∣ - 2∣} = π$ (use $∣ B ∣$ ). Phase shift of $5 sin (4 x - π) - 3$ ? ::: $\frac{π}{4}$ to the right. As $B \to 0^{+}$ , the period of $sin (B x)$ tends to…? ::: $\infty$ (the wave flattens toward a line). Model form for max $9$ , min $1$ , period $6$ , rising through midline at $x = 2$ ? ::: $4 sin (\frac{π}{3} x - \frac{2 π}{3}) + 5$ . Write $6 cos (2 x) - 1$ as a shifted sine. ::: $6 sin (2 x + \frac{π}{2}) - 1$ .

Connections

Parent topic — the rules these examples exercise
Unit circle and sine definition — source of "sine is odd" ( $sin (- θ) = - sin θ$ ) used in Example 3.
cos as shifted sin — the identity behind Example 9's disguise.
Function transformations — the general "inside acts opposite" rule seen in Examples 1 & 4.
Simple harmonic motion — the physics home of the tide model (Example 8).
Solving trig equations — the natural next step: find when a modelled wave hits a target value.