Question

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

Answer 1

Answer:

The average rate of change for the given function in the interval (-5, -1) is 0 (zero)

Step-by-step explanation:

The average rate of change of a function over an interval is the quotient between the difference between the function evaluated at the ends of the interval divided by the length of the interval. That is for our case:

the average rte of change of h(t) in the interval (-5, -1) is:

$\frac{h(-1)-h(-5)}{-1+5}$

so we find:

$h(-1)=(-1+3)^2+5=2^2+5=4+5=9\\and\\h(-5)=(-5+3)^2+5=(-2)^2+5=4+5=9$

then the average rate of change becomes:

$\frac{h(-1)-h(-5)}{-1+5}=\frac{9-9}{4} =\frac{0}{4} =0$