Question

H(t)=(t+3)^2+5

What is the average rate of change of h over the interval −5≤t≤−1?

Answer 1

Answer:0

Explanation:

Answer 2

0

Explanation:

The average rate of change of the function H(t) over the interval
`-5\le t\le -1` can be calculated as

`(H(-1)-H(-5))/((-1)-(-5))`

Find H(-5) and H(-1):

`H(-5)-(-5+3)^2+5=(-2)^2+5=4+5=9\\ \\H(-1)=(-1+3)^2+5=2^2+5=9`

Then

`H(-1)-H(-5)=9-9=0`

and the average rate of change is 0.