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

Here $A=5$, $B=1$, $C=0$, $D=2$.

• Amplitude $=|A|=5$.
• Period $=\frac{2\pi }{|1|}=2\pi$.
• Phase shift $=-\frac{0}{1}=0$ (no slide).
• Midline $y=2$; max $=2+5=7$; min $=2-5=-3$.

$A=\frac{1}{2}$, $B=4$, $C=0$, $D=0$.

• Amplitude $=\left|\frac{1}{2}\right|=\frac{1}{2}$.
• Period $=\frac{2\pi }{4}=\frac{\pi }{2}$.
• Phase shift $=-\frac{0}{4}=0$.
• Midline $y=0$; max $=\frac{1}{2}$; min $=-\frac{1}{2}$.

$A=3$, $B=2$, $C=\frac{\pi }{3}$, $D=-4$.

• Period $=\frac{2\pi }{|2|}=\pi$.
• Phase shift $=-\frac{C}{B}=-\frac{\pi /3}{2}=-\frac{\pi }{6}$. The minus means the wave slides left by $\frac{\pi }{6}$. Check by factoring: $2x+\frac{\pi }{3}=2\left(x+\frac{\pi }{6}\right)$. The $\left(x+\frac{\pi }{6}\right)$ form means "replace $x$ by $x+\frac{\pi }{6}$", which drags the graph left — consistent. ✔
• Midline $y=-4$; max $=-1$; min $=-7$.

$A=1$, $B=\frac{1}{2}$, $C=-\frac{\pi }{4}$, $D=0$.

• Period $=\frac{2\pi }{|1/2|}=2\pi \cdot 2=4\pi$. Why longer? $B=\frac{1}{2}<1$: the input crawls, so it needs a huge $x$-range to reach $2\pi$.
• Phase shift $=-\frac{C}{B}=-\frac{-\pi /4}{1/2}=\frac{\pi /4}{1/2}=\frac{\pi }{2}$ (right). Check: $\frac{x}{2}-\frac{\pi }{4}=\frac{1}{2}\left(x-\frac{\pi }{2}\right)$ → shift right by $\frac{\pi }{2}$. ✔

• Midline $D=\frac{\max +\min }{2}=\frac{9+1}{2}=5$.
• Amplitude $A=\frac{\max -\min }{2}=\frac{9-1}{2}=4$.
• $B=\frac{2\pi }{T}=\frac{2\pi }{6}=\frac{\pi }{3}$.
• A plain $\sin$ rises through its midline at input $=0$. We want that at $x=2$, i.e. a slide right by $2$. From $\text{shift}=-\frac{C}{B}=2$ we get $C=-2B=-2\cdot \frac{\pi }{3}=-\frac{2\pi }{3}$.

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

Verify the start point: at $x=2$, input $=\frac{\pi }{3}(2)-\frac{2\pi }{3}=0$, so $y=4\sin 0+5=5$ (on the midline) and $\sin$ is increasing there. ✔

• $T_f=\frac{2\pi }{3}$, $T_g=\frac{2\pi }{6}=\frac{\pi }{3}$. So $g$'s period is half of $f$'s — $g$ repeats twice as fast (because its $B$ is twice as large).
• "Return to $0$ going upward again" means completing one full cycle from the start. That happens after exactly one period. So $f$ first does this at $x=T_f=\frac{2\pi }{3}$, and $g$ at $x=T_g=\frac{\pi }{3}$.

• $\max =40$, $\min =2\Rightarrow D=\frac{40+2}{2}=21$, amplitude $=\frac{40-2}{2}=19$.
• Period $60\Rightarrow B=\frac{2\pi }{60}=\frac{\pi }{30}$.
• Starting at the bottom at $t=0$: a plain $\sin$ starts at the midline going up, but we need it to start at the minimum. The cleanest fit is a flipped cosine: recall $-\cos$ starts at its minimum. So use amplitude $\times (-\cos )$: $h(t)=-19\cos \left(\frac{\pi }{30}t\right)+21.$ Check: at $t=0$, $h=-19(1)+21=2$ ✔ (lowest). At $t=30$ (half a turn), $h=-19(-1)+21=40$ ✔ (highest). The red curve in the figure shows exactly this: starts low, peaks at the top, returns.

(If you prefer $\sin$ form: $h(t)=19\sin \left(\frac{\pi }{30}t-\frac{\pi }{2}\right)+21$, since $\sin (\theta -\frac{\pi }{2})=-\cos \theta$ — see cos as shifted sin.)

• Symmetric about $0\Rightarrow D=0$; amplitude $=0.3$.
• Frequency $256$ per second means $256$ cycles fit in $1$ second, so period $T=\frac{1}{256}$ s, giving $B=\frac{2\pi }{T}=2\pi \cdot 256=512\pi$.
• Starts at $0$ moving positive → a plain $\sin$ (no shift), since $\sin$ leaves the origin heading up. This is exactly the Simple harmonic motion pattern. $s(t)=0.3\sin \left(512\pi t\right).$ Check: $s(0)=0$, and $\sin$ is increasing at $0$, so it moves positive. ✔

(a) $A=-2$, $B=\pi$, $C=-\frac{\pi }{2}$, $D=3$.

• Amplitude $=|-2|=2$ (the minus reflects the wave, height unchanged).
• Period $=\frac{2\pi }{\pi }=2$.
• Phase shift $=-\frac{C}{B}=-\frac{-\pi /2}{\pi }=\frac{1}{2}$ (right by $\frac{1}{2}$).
• Midline $y=3$; max $=5$, min $=1$.

(b) $y=3$ means $-2\sin \left(\pi x-\frac{\pi }{2}\right)+3=3\Rightarrow \sin \left(\pi x-\frac{\pi }{2}\right)=0.$ Sine is zero when its input is any integer multiple of $\pi$ (see Solving trig equations): $\pi x-\frac{\pi }{2}=k\pi ,k\in \mathbb{Z}\Rightarrow \pi x=\frac{\pi }{2}+k\pi \Rightarrow x=\frac{1}{2}+k.$ For $x\in [0,4]$: $k=0,1,2,3\Rightarrow x=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}.$ (Note $k=-1$ gives $x=-\frac{1}{2}<0$, excluded; $k=4$ gives $x=\frac{9}{2}>4$, excluded — we covered both ends.)

$A=4$, $B=\frac{\pi }{2}$, $D=1$. Max value $=1+4=5$; period $=\frac{2\pi }{\pi /2}=4$.

(a) Max occurs when $\sin \left(\frac{\pi }{2}x\right)=1$, i.e. input $=\frac{\pi }{2}+2k\pi$: $\frac{\pi }{2}x=\frac{\pi }{2}+2k\pi \Rightarrow x=1+4k.$ First two positive: $x=1$ and $x=5$ (one period apart, as expected).

(b) $y=3\Rightarrow 4\sin \left(\frac{\pi }{2}x\right)+1=3\Rightarrow \sin \left(\frac{\pi }{2}x\right)=\frac{1}{2}.$ The base angle with sine $\frac{1}{2}$ is $\frac{\pi }{6}$. Rising means we want the branch where sine increases through $\frac{1}{2}$, i.e. input $=\frac{\pi }{6}$ (not the falling $\frac{5\pi }{6}$): $\frac{\pi }{2}x=\frac{\pi }{6}\Rightarrow x=\frac{\pi /6}{\pi /2}=\frac{1}{3}.$ So the first rising crossing of $y=3$ is at $x=\frac{1}{3}$.