Question

How do you graph `y=log_2(x+1)+3`?

Answer 1

Please see below.

Explanation:

As the equation is `y=log_2(x+1)+3`,

the domain of `y` is the set of all `x` values such that `x+1>0` or `x> -1`

and the range of `y` is the interval `(-oo,+oo)`.

Hence as `x->-1`, `y->-oo` i.e. `x+1=0` is the asymptote.

Now let us select a few values of `x`, choosing so that calculating `log_2(x+1)` is easy. Hence, let `x` take values `{0,1,3,7,15}` and for these values, `y` takes values `{3,4,5,6,7}` and hence points on the graph are

`(0,3)`; `(1,4)`; `(3,5)`; `(7,6)` and `(15,7)`

and graph appears as follows:
graph{2^(y-3)=x+1 [-4.87, 15.13, -1.92, 8.08]}