We are given the cooling model of `T(t)=100-70(1-(1/2)^(t/3))`
We can rewrite this:
`100-70(1-(1/2)^(t/3))`
`=100-70+70(1/2)^(t/3)`
`=70(1/2)^(t/3)+30`
So the simplified model is `T(t)=70(1/2)^(t/3)+30`
The domain is `t>=0` as the physical model only makes sense for positive time.
The range is `30<T(t)<=100` . The function starts at 100, and approaches 30 as t increases without bound. Note that `70(1/2)^(t/3)>0` for all t.
** As stated, your problem probably has a typo. If the room is `20^@` C then the liquid will approach 20 degrees, not 30 degrees as t increases. **
The T intercept is where t=0 so it is 100.
There is no t-intercept as the function is always greater than 30.
There is a horizontal asymptote at T=30.
Using `(1/2)^t` as the base function the transformations are as follows:
A vertical stretch of factor 70.
A horizontal stretch of factor 1/3.
A vertical translation (shift) up 30 units.
The graph: