question 5 the curve above is the graph of a sinusoidal function. it goes through the points ( - 5, - 2) and…

question 5 the curve above is the graph of a sinusoidal function. it goes through the points ( - 5, - 2) and (1, - 2). find a sinusoidal function that matches the given graph. if needed, you can enter π=3.1416... as pi in your answer, otherwise use at least 3 decimal digits. f(x) =

question 5 the curve above is the graph of a sinusoidal function. it goes through the points ( - 5, - 2) and (1, - 2). find a sinusoidal function that matches the given graph. if needed, you can enter π=3.1416... as pi in your answer, otherwise use at least 3 decimal digits. f(x) =

Answer

Explanation:

Step1: Find the period

The distance between $x = - 5$ and $x = 1$ is half - period. So, $T=2\times(1 - (-5)) = 12$. Then the angular frequency $\omega=\frac{2\pi}{T}=\frac{2\pi}{12}=\frac{\pi}{6}$.

Step2: Determine the amplitude

The maximum value is $y = 2$ and the minimum value is $y=-2$. The amplitude $A=\frac{2 - (-2)}{2}=2$.

Step3: Find the vertical shift

The mid - line is $y = 0$, so $D = 0$.

Step4: Find the phase shift

Let the general form of the sinusoidal function be $y = A\sin(\omega(x - \varphi))+D$. Using the point $(-5,-2)$ and $A = 2$, $\omega=\frac{\pi}{6}$, $D = 0$, we have $-2=2\sin(\frac{\pi}{6}(-5-\varphi))$. Then $\sin(\frac{\pi}{6}(-5 - \varphi))=-1$. So, $\frac{\pi}{6}(-5-\varphi)=-\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$. Solving for $\varphi$ gives $\varphi = - 2$.

Answer:

$f(x)=2\sin\left(\frac{\pi}{6}(x + 2)\right)$