Question

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

Over which interval does h have a negative average rate of change?


Choose 1 answer:


(Choice A)


−2≤t≤0


(Choice B)


1≤t≤4


(Choice C)


−4≤t≤−3


(Choice D)


−3≤t≤4

Answer 1

Choice C

`-4\leq t\leq-3`

Explanation:

`h(t)=(t+3)^2+5`

Rate of change
`=h'(t)`

`h'(t)=(d)/(dt)h(t)\\ h'(t)=2(t+3)`

For negative rate of change
`h'(t)\leq0`

`2(t+3)\leq0\\(t+3)\leq0\\t\leq-3`

Choice C satisfies this condition.