suppose that an insect population in millions is modeled by f(x) = (10x + 9)/(x + 3), where x≥0 is in…

Explanation:

Step1: Find the horizontal - asymptote

For a rational function $y = \frac{f(x)}{g(x)}=\frac{10x + 9}{x + 3}$, where $f(x)=10x + 9$ and $g(x)=x + 3$ are polynomials. Since the degree of the numerator and the denominator are the same (both degree 1), the horizontal - asymptote is given by $y=\frac{a}{b}$, where $a$ is the leading coefficient of the numerator and $b$ is the leading coefficient of the denominator. Here, $a = 10$ and $b = 1$, so the horizontal - asymptote is $y = 10$.

Step2: Analyze the function behavior

As $x\to+\infty$, $f(x)=\frac{10x+9}{x + 3}=\frac{10x+30-21}{x + 3}=\frac{10(x + 3)-21}{x + 3}=10-\frac{21}{x + 3}$. As $x$ increases (after several months), $\frac{21}{x + 3}\to0$, and $f(x)\to10$.

Step3: Find the initial population

To find the initial population, we substitute $x = 0$ into the function $f(x)$. So $f(0)=\frac{10\times0+9}{0 + 3}=\frac{9}{3}=3$ (in millions).

Answer:

(a) To graph the function $y=\frac{10x + 9}{x + 3}$ in the window $[0,16]$ by $[0,16]$, we can use a graphing utility. The horizontal - asymptote is $y = 10$. (b) The initial insect population is $3$ million. The equation of the horizontal asymptote is $y = 10$. (c) After several months, the insect population approaches $10$ million. (d) The horizontal asymptote $y = 10$ represents the long - term limit of the insect population. As time (in months) goes on, the population of insects will get closer and closer to $10$ million. For the multiple - choice part about the graph, without seeing the actual graphs, we know that the function starts at $y = 3$ (when $x = 0$) and approaches $y = 10$ as $x$ increases. The graph should start at the point $(0,3)$ and have a horizontal asymptote at $y = 10$.