Question

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

Answer 1

Answer:

C

Step-by-step explanation:

Find the average rate of change in each option:

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

$\dfrac{h(0)-h(-2)}{0-(-2)}\\ \\=\dfrac{((0+3)^2+5)-((-2+3)^2+5)}{2}\\ \\=\dfrac{(9+5)-(1+5)}{2}\\ \\=\dfrac{14-6}{2}\\ \\=4$

Positive

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

$\dfrac{h(4)-h(1)}{4-1}\\ \\=\dfrac{((4+3)^2+5)-((1+3)^2+5)}{3}\\ \\=\dfrac{(49+5)-(16+5)}{3}\\ \\=\dfrac{54-21}{3}\\ \\=11$

Positive

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

$\dfrac{h(-3)-h(-4)}{-3-(-4)}\\ \\=\dfrac{((-3+3)^2+5)-((-4+3)^2+5)}{1}\\ \\=\dfrac{(0+5)-(1+5)}{1}\\ \\=\dfrac{5-6}{1}\\ \\=-1$

Negative

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

$\dfrac{h(4)-h(-3)}{4-(-3)}\\ \\=\dfrac{((4+3)^2+5)-((-3+3)^2+5)}{7}\\ \\=\dfrac{(49+5)-(0+5)}{7}\\ \\=\dfrac{54-5}{7}\\ \\=7$

Positive