Practice a College Board-style free response question on a rational-function model. Write your response, then reveal the model answer to see exactly what earns each point.
After a patient takes a dose of medication, the concentration of the drug in the bloodstream is modeled by the rational function C(t) = 20t / (t² + 4), where t is the time in hours since the dose (t ≥ 0) and C is the concentration in milligrams per liter (mg/L). Selected values of the model are shown below.
| t (hours) | C(t) (mg/L) |
|---|---|
| 0 | 0 |
| 1 | 4 |
| 2 | 5 |
| 4 | 4 |
| 8 | ≈ 2.35 |
The average rate of change is (C(2) − C(0))/(2 − 0) = (5 − 0)/(2 − 0) = 2.5 mg/L per hour. This means that over the first two hours after the dose, the drug concentration in the bloodstream increased at an average rate of 2.5 milligrams per liter each hour.
C is decreasing on the interval from t = 2 to t = 8 hours, since the output values fall from 5 to 4 to about 2.35 as t increases. On this interval the drug concentration in the bloodstream is declining over time as the body clears the medication. The value at t = 2, where the concentration reaches its highest value of 5 mg/L before it begins to fall, represents a local (relative) maximum — the peak concentration.
In C(t) = 20t/(t² + 4), the degree of the numerator (1) is less than the degree of the denominator (2), so the horizontal asymptote is y = 0. As t → ∞, C(t) → 0, which means that over the long term the drug concentration approaches 0 mg/L — the medication is essentially cleared from the bloodstream.
The revised model D(t) = C(t) + 1 is an additive transformation: a vertical translation of the graph of C upward by 1 unit. Because every output is raised by 1, the horizontal asymptote also shifts up by 1, so the horizontal asymptote of D is y = 1. In context, D approaches a long-term concentration of 1 mg/L rather than 0.