we have the function
[tex]P(t)=\frac{60(1+0.4t)}{0.01t+3}[/tex]
Part A
For t=0
substitute
[tex]\begin{gathered} P(t)=\frac{60(1+0.4(0))}{0.01(0)+3} \\ \\ P(t)=\frac{60(1)}{3} \\ \\ P(t)=20 \end{gathered}[/tex]
The initial population was 20 insects
Part B
For t=5 years -------> Convert to months
t=60 months
substitute
[tex]\begin{gathered} P(t)=\frac{60(1+0.4(60))}{0.01(60)+3} \\ \\ P(t)=\frac{60(1+24)}{0.6+3} \\ \\ P(t)=\frac{1500}{3.6} \\ \\ P(t)=416.67 \end{gathered}[/tex]
The answer is 417 insects
Part C
Determine horizontal asymptote
[tex]\begin{gathered} P(t)=\frac{60\left(1+0.4t\right)}{0.01t+3} \\ \\ rewrite \\ P(t)=\frac{60+24t}{0.01t+3} \end{gathered}[/tex]
The horizontal asymptote is given by the ratio of
[tex]\frac{24}{0.01}=2,400[/tex]
P(t)=2,400 ---------> horizontal asymptote
that means
as the value of time t increases ------> the value of the population tends to 2,400 insects
the population cannot be greater than 2400 insects
The range for the function P(t) is the interval [20, 2400)
All real numbers greater than or equal to 20 insects and less than 2400 insects
Part D
Using a graphing tool
