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

C

Explanation:

Find the average rate of change in each option:

A. For
`-2\le t\le 0:`

`(h(0)-h(-2))/(0-(-2))\\ \\=(((0+3)^2+5)-((-2+3)^2+5))/(2)\\ \\=((9+5)-(1+5))/(2)\\ \\=(14-6)/(2)\\ \\=4`

Positive

B. For
`1\le t\le 4:`

`(h(4)-h(1))/(4-1)\\ \\=(((4+3)^2+5)-((1+3)^2+5))/(3)\\ \\=((49+5)-(16+5))/(3)\\ \\=(54-21)/(3)\\ \\=11`

Positive

C. For
`-4\le t\le -3:`

`(h(-3)-h(-4))/(-3-(-4))\\ \\=(((-3+3)^2+5)-((-4+3)^2+5))/(1)\\ \\=((0+5)-(1+5))/(1)\\ \\=(5-6)/(1)\\ \\=-1`

Negative

D. For
`-3\le t\le 4:`

`(h(4)-h(-3))/(4-(-3))\\ \\=(((4+3)^2+5)-((-3+3)^2+5))/(7)\\ \\=((49+5)-(0+5))/(7)\\ \\=(54-5)/(7)\\ \\=7`

Positive