Question

lesson 18: more on polynomial functions and modeling
score: 57.07/100 answered: 9/15
question 11
population growth
the graph of ( p(t) ) below shows the population (in thousands of people) of a city ( t ) years after 2008.
note that 8 years after 2008, the population is approximately 5.15 thousand people.
and 16 years after 2008, the population is approximately 8.86 thousand people.
approximate the average rates of change over each of the following time intervals. round to two decimal places.
0 years to 8 years:
8 years to 16 years:
0 years to 16 years:
complete the statement to explain the meaning of the rate of change in this situation.
the citys population is at rate.
what do the values above tell you about the rate of change of ( v(t) )?
the rate of change for ( d(t) ) is

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).

Step2: Calculate the average rate of change from \(0\) to \(8\) years

We know that \(a = 0\), \(b=8\), \(P(0)=3\) (from the \(y\) - intercept of the graph, when \(t = 0\), \(P(t)=3\)) and \(P(8)=5.15\).
Using the formula \(\frac{P(8)-P(0)}{8 - 0}=\frac{5.15 - 3}{8}=\frac{2.15}{8}=0.27\)

Step3: Calculate the average rate of change from \(8\) to \(16\) years

Here \(a = 8\), \(b = 16\), \(P(8)=5.15\) and \(P(16)=8.86\)
Using the formula \(\frac{P(16)-P(8)}{16 - 8}=\frac{8.86-5.15}{8}=\frac{3.71}{8}=0.46\)

Step4: Calculate the average rate of change from \(0\) to \(16\) years

Here \(a = 0\), \(b = 16\), \(P(0)=3\) and \(P(16)=8.86\)
Using the formula \(\frac{P(16)-P(0)}{16 - 0}=\frac{8.86 - 3}{16}=\frac{5.86}{16}=0.37\)

Answer:

0 years to 8 years: \(0.27\)
8 years to 16 years: \(0.46\)
0 years to 16 years: \(0.37\)
The city's population is increasing at an increasing rate.
The rate of change for \(P(t)\) is positive and the function is non - linear (since the average rate of change over different intervals is not constant)