Answer:
B
Step-by-step explanation:
Given the function [tex]h(t)=(t+3)^2+5.[/tex]
Find the average rate of change on interval [tex][t_1,t_2]:[/tex]
[tex]\dfrac{h(t_2)-h(t_1)}{t_2-t_1}=\\ \\=\dfrac{((t_2+3)^2+5)-((t_1+3)^2+5)}{t_2-t_1}=\\ \\=\dfrac{(t_2^2+6t_2+9+5)-(t_1^2+6t_1+9+5)}{t_2-t_1}=\\ \\=\dfrac{t_2^2-t_1^2+6t_2-6t_1}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1)+6(t_2-t_1)}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1+6)}{t_2-t_1}=\\ \\=t_2+t_1+6[/tex]
This value is negative only for [tex]t_1=-4,\ t_2=-3,[/tex] so correct option is B
