Question

Math problem: intervals

Answer 1

Answer:

B

Step-by-step explanation:

Given the function $h(t)=(t+3)^2+5.$

Find the average rate of change on interval $[t_1,t_2]:$

$\dfrac{h(t_2)-h(t_1)}{t_2-t_1}=\\ \\=\dfrac{((t_2+3)^2+5)-((t_1+3)^2+5)}{t_2-t_1}=\\ \\=\dfrac{(t_2^2+6t_2+9+5)-(t_1^2+6t_1+9+5)}{t_2-t_1}=\\ \\=\dfrac{t_2^2-t_1^2+6t_2-6t_1}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1)+6(t_2-t_1)}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1+6)}{t_2-t_1}=\\ \\=t_2+t_1+6$

This value is negative only for $t_1=-4,\ t_2=-3,$ so correct option is B