91Ó°ÊÓ

Given function is $M\left(t\right)=\frac{40944t^2+454512t-1732032}{t^2+9t-36}.$

Given value oftis$t=0,2,\&4.$

Determine the value of the function at $t=0.$

role="math" localid="1649798710639" $$\begin{gathered} M\left(0\right)=\frac{40944{\left(0\right)}^2+454512\left(0\right)-1732032}{{\left(0\right)}^2+9\left(0\right)-36} \\ M\left(0\right)=48112 \end{gathered}$$

Determine the value of the function at $t=2.$

role="math" localid="1649798704483" $$\begin{gathered} M\left(2\right)=\frac{40944{\left(2\right)}^2+454512\left(2\right)-1732032}{{\left(2\right)}^2+9\left(2\right)-36} \\ M\left(0\right)=47088 \end{gathered}$$

Determine the value of the function at $t=4.$

role="math" localid="1649798697163" $$\begin{gathered} M\left(4\right)=\frac{40944{\left(4\right)}^2+454512\left(4\right)-1732032}{{\left(4\right)}^2+9\left(4\right)-36} \\ M\left(4\right)=46320 \end{gathered}$$

The function has specific values at role="math" localid="1649798785858" $t=0,2,\&4$so function pass through the data points at$t=0,2,\&4.$

Given function is $M\left(t\right)=\frac{40944t^2+454512t-1732032}{t^2+9t-36}.$

The Limit that needs to determine is$\underset{t→∞}{\lim }M\left(t\right).$

Determine the limit $\underset{t→∞}{\lim }M\left(t\right).$

role="math" localid="1649799143891" $$\begin{gathered} \underset{t→∞}{\lim }M\left(t\right)=\underset{t→∞}{\lim }\frac{40944t^2+454512t-1732032}{t^2+9t-36} \\ =\underset{t→∞}{\lim }\frac{40944+{\displaystyle \frac{454512}{t}}-{\displaystyle \frac{1732032}{t^2}}}{1+{\displaystyle \frac{9}{t}}-{\displaystyle \frac{36}{t^2}}} \\ \underset{t→∞}{\lim }M\left(t\right)==40944 \end{gathered}$$

role="math" localid="1649799411648" $\underset{t→∞}{\lim }M\left(t\right)=40944$state that the population will never go less than $40944$with increasing time.