As $ t \to \infty $, the term $ \frac{8}{t + 2} \to 0 $, so the slant asymptote is:
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- Divide $ t^2 $ by $ t $ to get $ t $. Multiply $ t + 2 $ by $ t $ to get $ t^2 + 2t $. Subtract:
- Next, divide $ -2t $ by $ t $ to get $ -2 $. Multiply $ t + 2 $ by $ -2 $ to get $ -2t - 4 $. Subtract:
- I(t) = t - 2 + \frac{8}{t + 2}.
- \boxed{y = t - 2}
- Question: A herpetologist models the population of a rare frog species in Madagascar with the function $ F(t) = \frac{100t}{t^2 + 4} $, where $ t $ is time in years. What is the maximum population growth rate?
- Solution: To find the maximum of $ F(t) = \frac{100t}{t^2 + 4} $, we compute its derivative using the quotient rule:
