question 5 the curve above is the graph of a sinusoidal function. it goes through the points ( - 5, - 1) and…

Explanation:

Step1: Determine the period

The function passes through $(-5, - 1)$ and $(1,-1)$. The distance between these two - points is half of the period. So, $T = 2\times(1-( - 5))=12$. Using the formula $T=\frac{2\pi}{|B|}$, and since $T = 12$, we have $|B|=\frac{2\pi}{T}=\frac{\pi}{6}$. Let's assume $B=\frac{\pi}{6}$ for a standard - form sinusoidal function $y = A\sin(Bx - C)+D$.

Step2: Determine the amplitude

The maximum value of the function is $y = 1$ and the minimum value is $y=-1$. The amplitude $A=\frac{\text{max}-\text{min}}{2}=\frac{1-( - 1)}{2}=1$.

Step3: Determine the vertical shift

The mid - line of the function is $y=\frac{1+( - 1)}{2}=0$, so $D = 0$.

Step4: Determine the phase shift

We can use the point $(-5,-1)$ and the form $y = A\sin(Bx - C)+D$. Substituting $A = 1$, $B=\frac{\pi}{6}$, $D = 0$ and $x=-5,y=-1$ into $y=\sin(\frac{\pi}{6}x - C)$ gives $-1=\sin(\frac{\pi}{6}\times(-5)-C)$. We know that $\sin\theta=-1$ when $\theta=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$. So, $\frac{\pi}{6}\times(-5)-C=\frac{3\pi}{2}+2k\pi$. Solving for $C$: [ \begin{align*} -\frac{5\pi}{6}-C&=\frac{3\pi}{2}+2k\pi\ -C&=\frac{3\pi}{2}+\frac{5\pi}{6}+2k\pi\ -C&=\frac{9\pi + 5\pi}{6}+2k\pi\ -C&=\frac{14\pi}{6}+2k\pi\ C&=-\frac{7\pi}{3}-2k\pi \end{align*} ] Let $k = 0$, then $C=-\frac{7\pi}{3}$. The sinusoidal function is $f(x)=\sin(\frac{\pi}{6}x+\frac{7\pi}{3})$. We can simplify $\frac{7\pi}{3}=2\pi+\frac{\pi}{3}$, and since $\sin(x + 2\pi)=\sin x$, the function can be written as $f(x)=\sin(\frac{\pi}{6}x+\frac{\pi}{3})$.

Answer:

$f(x)=\sin(\frac{\pi}{6}x+\frac{\pi}{3})$