question 1 (4 points)\nsketch a sinusoidal function that passes through (0, -3) and has a period of 9, an…

Explanation:

Step1: Recall the general form of a sinusoidal function

The general form of a sinusoidal function is $y = A\sin(B(x - C))+D$ (we could also use cosine - here we use sine for illustration). The amplitude is $|A|$, the period is $T=\frac{2\pi}{|B|}$, the phase - shift is $C$, and the equation of the axis is $y = D$. Given that the amplitude $|A| = 3$, the period $T = 9$, and the equation of the axis $y=-1$, so $D=-1$. Since $T=\frac{2\pi}{|B|}=9$, we can solve for $B$: $|B|=\frac{2\pi}{9}$. Let's assume $B=\frac{2\pi}{9}$ (we can choose the positive value for simplicity, the negative value would just result in a reflection). So far, the function is $y = 3\sin(\frac{2\pi}{9}(x - C))-1$.

Step2: Use the given point to find the phase - shift $C$

The function passes through the point $(0, - 3)$. Substitute $x = 0$ and $y=-3$ into the function $y = 3\sin(\frac{2\pi}{9}(x - C))-1$: $-3=3\sin(-\frac{2\pi}{9}C)-1$. First, add 1 to both sides of the equation: $-3 + 1=3\sin(-\frac{2\pi}{9}C)$, so $-2 = 3\sin(-\frac{2\pi}{9}C)$. Then, $\sin(-\frac{2\pi}{9}C)=-\frac{2}{3}$. We know that $\sin(-\alpha)=-\sin(\alpha)$, so $\sin(\frac{2\pi}{9}C)=\frac{2}{3}$. $\frac{2\pi}{9}C=\arcsin(\frac{2}{3})+2k\pi$ or $\frac{2\pi}{9}C=\pi-\arcsin(\frac{2}{3})+2k\pi,k\in\mathbb{Z}$. Let $k = 0$ and choose the principal value. $\frac{2\pi}{9}C=\arcsin(\frac{2}{3})$, then $C=\frac{9}{2\pi}\arcsin(\frac{2}{3})$. The function is $y = 3\sin(\frac{2\pi}{9}(x-\frac{9}{2\pi}\arcsin(\frac{2}{3})))-1$. To sketch the function:

• The axis of the function is $y=-1$.
• The amplitude is 3, so the maximum value of the function is $y=-1 + 3=2$ and the minimum value is $y=-1-3=-4$.
• The period is 9. Mark the key - points:
  • Start with the point $(0,-3)$.
  • The mid - line is $y = - 1$.
  • Since the period $T = 9$, we can find other key - points by moving along the $x$ - axis in increments of $\frac{9}{4}=2.25$.
  • The zero - crossing points of the sine function can be found by setting $\frac{2\pi}{9}(x-\frac{9}{2\pi}\arcsin(\frac{2}{3}))=k\pi,k\in\mathbb{Z}$, and then solving for $x$.

Answer:

The sinusoidal function is $y = 3\sin(\frac{2\pi}{9}(x-\frac{9}{2\pi}\arcsin(\frac{2}{3})))-1$ and its key features for sketching are: axis $y=-1$, amplitude 3, period 9, passing through $(0, - 3)$.