the graph of a sinusoidal function intersects its midline at (0, -2) and then has a minimum point at…

Explanation:

Step1: Determine the vertical shift

The mid - line of a sinusoidal function is given by $y = k$. Since the function intersects its mid - line at $(0, - 2)$, then $k=-2$.

Step2: Determine the amplitude

The amplitude $A$ is the distance from the mid - line to a maximum or minimum point. The mid - line is $y = - 2$ and the minimum point is $(\frac{3\pi}{2},-7)$. So $A=\vert-2-(-7)\vert = 5$. Since it is a minimum point after the mid - line intersection, $A=-5$ (for a cosine - type function starting at the mid - line and going down first).

Step3: Determine the period

The distance from the mid - line intersection point $(0, - 2)$ to the minimum point $(\frac{3\pi}{2},-7)$ is $\frac{3\pi}{2}$ units. This is $\frac{3}{4}$ of a period for a cosine - type function. Let the period be $T$. Then $\frac{3}{4}T=\frac{3\pi}{2}$, so $T = 2\pi$. The formula for the angular frequency $\omega=\frac{2\pi}{T}$, and since $T = 2\pi$, $\omega = 1$.

Step4: Determine the phase shift

The general form of a cosine function is $y = A\cos(\omega(x - h))+k$. Since the function passes through $(0, - 2)$ and we are using a cosine function, when $x = 0$, $y=-2$. Substituting $A=-5$, $\omega = 1$, $k=-2$ into $y = A\cos(\omega(x - h))+k$ gives $-2=-5\cos(-h)-2$. This implies $\cos(-h)=0$. A simple choice for $h$ is $\frac{\pi}{2}$. The function is $f(x)=-5\cos(x-\frac{\pi}{2})-2$. Using the identity $\cos(x - \frac{\pi}{2})=\sin(x)$, we can rewrite it as $f(x)=-5\sin(x)-2$.

Answer:

$-5\sin(x)-2$