Calculus I: Returning to starting position?

Question

The third question in a problem I'm given asks if and when the object returns to its starting position. What would the starting position on a graph be? If the domain is [0, infinity), the starting position is (0,9). Is there none, considering the equation is x(t)=t + 9/(t+1)? Or would it be when x(t)=9 again, which in this case, would be t=8?
Thanks a bunch

Answer 1

you are absolutely right about the second part

for x(t) = t+9/(t+1)

x(0) = 0 + 9/(0+1) = 9

it wold be when x(t) = t+9/(t+1) = 9
or t(t+1) + 9 = 9(t+1)
Or t^2 + t +9 = 9t + 9
or t^2 -8t =0
t(t-8) = 0
t=0 or 8
so at 8 it is same position

Answer 2

By definition, your starting position is t=0, x=9, not
at t=8, x=9

Answer 3

x(t)=9