Question

1. the graph of a quadratic function, f, is shown.

a. find the average rate of change of f on the interval -3, -2.
b. find the average rate of change of f on the interval -2, -1.
c. find the average rate of change of f on the interval -1, 0.
d. find the average rate of change of f on the interval 0, 1.
e. show that the rate of change of the average rates of change of f is constant.

Explanation:

Step1: Recall average - rate - of - change formula

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

Step2: Find values from the graph for part a

From the graph, when $x=-3$, $f(-3)=4$ and when $x = - 2$, $f(-2)=-3$. Then the average rate of change on $[-3,-2]$ is $\frac{f(-2)-f(-3)}{-2-(-3)}=\frac{-3 - 4}{-2 + 3}=\frac{-7}{1}=-7$.

Step3: Find values from the graph for part b

When $x=-2$, $f(-2)=-3$ and when $x=-1$, $f(-1)=-6$. Then the average rate of change on $[-2,-1]$ is $\frac{f(-1)-f(-2)}{-1-(-2)}=\frac{-6+3}{-1 + 2}=\frac{-3}{1}=-3$.

Step4: Find values from the graph for part c

When $x=-1$, $f(-1)=-6$ and when $x = 0$, $f(0)=-5$. Then the average rate of change on $[-1,0]$ is $\frac{f(0)-f(-1)}{0-(-1)}=\frac{-5 + 6}{0+1}=\frac{1}{1}=1$.

Step5: Find values from the graph for part d

When $x = 0$, $f(0)=-5$ and when $x = 1$, $f(1)=0$. Then the average rate of change on $[0,1]$ is $\frac{f(1)-f(0)}{1-0}=\frac{0 + 5}{1}=5$.

Step6: Calculate the rate of change of the average rates of change

Let $A_1=-7$, $A_2=-3$, $A_3 = 1$, $A_4=5$.
The rate of change from $A_1$ to $A_2$ is $\frac{A_2 - A_1}{-2-(-3)}=\frac{-3+7}{-2 + 3}=4$.
The rate of change from $A_2$ to $A_3$ is $\frac{A_3 - A_2}{-1-(-2)}=\frac{1 + 3}{-1+2}=4$.
The rate of change from $A_3$ to $A_4$ is $\frac{A_4 - A_3}{0-(-1)}=\frac{5 - 1}{0 + 1}=4$. So the rate of change of the average rates of change is constant.

Answer:

a. -7
b. -3
c. 1
d. 5
e. Shown above (rate of change of average rates of change is 4).