y = 3sin(4x + 12)+3\nthe given function has an amplitude of \ncheck

Explanation:

Step1: Recall amplitude formula

For $y = A\sin(Bx - C)+D$, amplitude is $|A|$. Here $A = 3$, so amplitude is $|3|=3$.

Step2: Recall period formula

The period of $y = A\sin(Bx - C)+D$ is $T=\frac{2\pi}{|B|}$. Given $B = 4$, so $T=\frac{2\pi}{4}=\frac{\pi}{2}$.

Step3: Recall phase - shift formula

For $y = A\sin(Bx - C)+D$, phase - shift is $\frac{C}{B}$. Rewrite $y = 3\sin(4x + 12)+3$ as $y=3\sin(4(x + 3))+3$, so phase - shift is $- 3$ (3 units to the left).

Step4: Recall vertical - shift formula

For $y = A\sin(Bx - C)+D$, vertical - shift is $D$. Here $D = 3$.

Answer:

Amplitude: 3, Period length: $\frac{\pi}{2}$, Phase shift: 3, Direction: left, Vertical shift: 3