Question

H(t)=(t+3)^2+5 Over which interval does h have a negative average rate of change?

Answer 1

Answer: -4<t<-3

Step-by-step explanation:

Answer 2

Answer:

( -∞, -3 ) or any subinterval of  ( -∞, -3 )

Step-by-step explanation:

h(t)=(t+3)^2+5 has the form

h(t) =(t-h)^2 + 5, indicating that h(t)=(t+3)^2+5 represents a parabola with vertex at (-3,5) that opens up.  This (-3,5) is the minimum point of the graph.

To the left of t = -3, h(t) is decreasing; to the right of t = -3, h(t) is increasing.

Thus, it is on any interval to the left of t = -3 that the rate of change of this function is negative (the function is decreasing  on ( -∞, -3 ).