the graph of a sinusoidal function has a minimum point at (0, - 10) and then has a maximum point at (2…

Explanation:

Step1: Find the amplitude $A$

The amplitude is half the distance between the maximum and minimum values. The maximum value is $y = - 4$ and the minimum value is $y=-10$. So $A=\frac{-4 - (- 10)}{2}=\frac{-4 + 10}{2}=3$.

Step2: Find the vertical shift $D$

The vertical - shift is the average of the maximum and minimum values. So $D=\frac{-4+( - 10)}{2}=\frac{-4 - 10}{2}=-7$.

Step3: Find the period $T$

The distance between a minimum and the next - maximum is half of a period. The $x$ - values of the minimum point $(0,-10)$ and the maximum point $(2,-4)$ give that $\frac{T}{2}=2$, so $T = 4$. Then, using the formula $\omega=\frac{2\pi}{T}$, we get $\omega=\frac{2\pi}{4}=\frac{\pi}{2}$.

Step4: Determine the phase shift $C$

Since the function has a minimum at $x = 0$, for a cosine - type function $y = A\cos(\omega(x - C))+D$, when $x = 0$, we have $y=-10$. Substituting $A = 3$, $\omega=\frac{\pi}{2}$, and $D=-7$ into $y = A\cos(\omega(x - C))+D$, we get $-10=3\cos(\frac{\pi}{2}(0 - C))-7$. Then $- 3=3\cos(-\frac{\pi}{2}C)$, so $\cos(-\frac{\pi}{2}C)=-1$. Since $\cos\theta=-1$ when $\theta=(2k + 1)\pi,k\in\mathbb{Z}$, we have $-\frac{\pi}{2}C=(2k + 1)\pi$. Let $k = 0$, then $C=-2$.

Step5: Write the function formula

The general form of a sinusoidal function is $y = A\cos(\omega(x - C))+D$. Substituting $A = 3$, $\omega=\frac{\pi}{2}$, $C=-2$, and $D=-7$ into it, we get $f(x)=3\cos(\frac{\pi}{2}(x + 2))-7$. Using the identity $\cos(a + b)=\cos a\cos b-\sin a\sin b$, $\cos(\frac{\pi}{2}(x + 2))=\cos(\frac{\pi}{2}x+\pi)=-\cos(\frac{\pi}{2}x)$. So $f(x)=-3\cos(\frac{\pi}{2}x)-7$.

Answer:

$f(x)=-3\cos(\frac{\pi}{2}x)-7$