Question

1. the percent of voters between the ages of 18 and 29 that participated in each united states presidential election between the years 1988 to 2016 are shown in the table.

year	1988	1992	1996	2000	2004	2008	2012	2016
percentage of voters ages 18 - 29	35.7	42.7	33.1	34.5	45.0	48.4	40.9	43.4

the function p gives the percent of voters between 18 and 29 years old that participated in the election in year t.
a. determine the average rate of change for p between 1992 and 2000.
b. pick two different values of t so that the function has a negative average rate of change between the two values. determine the average rate of change.
c. pick two values of t so that the function has a positive average rate of change between the two values. determine the average rate of change.

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, $t$ is the independent variable (year) and $P(t)$ is the percentage of voters.

Step2: Solve part a

For the years 1992 and 2000, $t_1 = 1992$, $P(t_1)=42.7$, $t_2 = 2000$, $P(t_2)=34.5$.
The average rate of change is $\frac{P(2000)-P(1992)}{2000 - 1992}=\frac{34.5 - 42.7}{8}=\frac{-8.2}{8}=-1.025$.

Step3: Solve part b

Let's choose $t_1 = 1996$, $P(t_1)=33.1$ and $t_2 = 2008$, $P(t_2)=48.4$.
The average rate of change is $\frac{P(2008)-P(1996)}{2008 - 1996}=\frac{48.4 - 33.1}{12}=\frac{15.3}{12}=1.275$.

Step4: Solve part c

Let's choose $t_1 = 2012$, $P(t_1)=40.9$ and $t_2 = 2016$, $P(t_2)=43.4$.
The average rate of change is $\frac{P(2016)-P(2012)}{2016 - 2012}=\frac{43.4 - 40.9}{4}=\frac{2.5}{4}=0.625$.

Answer:

a. - 1.025
b. For $t_1 = 1996$ and $t_2 = 2008$, the average rate of change is 1.275
c. For $t_1 = 2012$ and $t_2 = 2016$, the average rate of change is 0.625