the graph of a sinusoidal function has a minimum point at (0, 3) and then intersects its midline at (5π, 5)…

Explanation:

Step1: Determine the amplitude $A$

The mid - line value $D$ is the average of the maximum and minimum values. The minimum value is $y = 3$ and a point on the mid - line is $(5\pi,5)$. The amplitude $A$ is the distance from the mid - line to the maximum or minimum. So $A=5 - 3=2$. Since the function starts at a minimum, $A=- 2$.

Step2: Determine the vertical shift $D$

The mid - line value $D$ is the $y$ - coordinate of the mid - line. Since a point on the mid - line is $(5\pi,5)$, $D = 5$.

Step3: Determine the period $T$

The distance from a minimum point to the mid - line is a quarter of the period. The $x$ - value changes from $x = 0$ to $x=5\pi$ from the minimum to the mid - line. So $\frac{T}{4}=5\pi$, then $T = 20\pi$. The formula for the angular frequency $\omega=\frac{2\pi}{T}$, so $\omega=\frac{2\pi}{20\pi}=\frac{1}{10}$.

Step4: Determine the phase shift $C$

The general form of a sinusoidal function is $y = A\sin(\omega(x - C))+D$. Since the function has a minimum at $x = 0$, for a sine function $y=-A\sin(\omega x)+D$ (equivalent to a phase - shift of $C = 0$ in the general form). Substituting $A=-2$, $\omega=\frac{1}{10}$, $D = 5$ into the formula $y = A\sin(\omega(x - C))+D$, we get $y=-2\sin(\frac{1}{10}x)+5$.

Answer:

$f(x)=-2\sin(\frac{1}{10}x)+5$