Answer
$f(t)=10 sin (\frac{\pi}{12}(t-8))+90$ (mmHg)
Work Step by Step
Step 1. Identify the quantities as $p=24,2|a|=100-80=20, f(2)=80$, where $f(2)=80$ means a minimum of 80mmHg happens at t=2 (2 A.M), and we also get $|a|=10$
Step 2. With $p=\frac{2\pi}{\omega}=24$, we have $\omega=\frac{\pi}{12}$. The function becomes $f(t)=10 sin (\frac{\pi}{12}(t-c))+b$
Step 3. Since $b$ is the average blood pressure, we have $b=90$ and condition $f(2)=80$ lead to
$10 sin (\frac{\pi}{12}(2-c))+90=80$ we have $sin (\frac{\pi}{12}(2-c))=-1$ and thus $2-c=-6,c=8$
Step 4. The final form of the function is $f(t)=10 sin (\frac{\pi}{12}(t-8))+90$ (mmHg)
