Answer
(a) 25 insects
(b) About 596 insects
(c) The largest population is determined by the horizontal asymptote since the more time passes, the closer the population will get to the value of the asymptote, which is 2500 insects.
Work Step by Step
(a) Solve P(0):
$P(0)=\dfrac{50(1+0.5(0))}{2+0.01(0)}$
$P(0)=\dfrac{50(1)}{2}$
$P(0)=\dfrac{50}{2}=25$ insects
(b) Solve P(60) (since 5 years equals 60 months):
$P(60)=\dfrac{50(1+0.5(60))}{2+0.01(60)}$
$P(60)=\dfrac{50(1+30)}{2+0.6}$
$P(60)=\dfrac{50(31)}{2.6}$
$P(60)=\dfrac{1550}{2.6}\approx596$ insects
(c) The degrees of both the numerator and denominator are the same, so the horizontal asymptote is the ratio of the leading coefficients:
$\dfrac{50(1+0.5t)}{2+0.01t}\rightarrow\dfrac{50+25t}{2+0.01t}\rightarrow\dfrac{25}{0.01}\rightarrow y=2500$
The largest population is determined by the horizontal asymptote since the more time passes, the closer the population will get to the value of the asymptote, which is 2500 insects.
