Question

How do you graph `y=log_2 (x-2)`?

Answer 1

Answer Below:

Explanation:

We know how to sketch `log_2 x` rather simply:

graph{lnx [-4.04, 15.96, -4.88, 5.12]}

But `log_2 (x-2)` is just `log_2 x` shifted `2` to the right:

`log_2 (x-2 )`: graph{ln(x-2) [-4.04, 15.96, -4.88, 5.12]}

The asymptote for `log_2 x` is `x=0`
The `x` intercept is `1`

So the asymptote for `log_2 (x-2)` is `x=2`
The new `x` intercept is hence `3`

As indicated on the second graph

Answer 2

See below

Explanation:

`y=log_2(x-2)`

First, Let's find the domain:
`x-2>0quad=>quadx>2`

Now we know that this logaritmic function will be approching x=2 but will never get there.The base equals 2 which is greater than 1 so that means It's increasing its value on the whole domain.

Intercept x axis: [3,0]
We can find it by equation but it's much easier to simply think like this: `ifquadlogxquad"must equal 0 then x must equal 1"quad`which means `x-2=1quad=>quad x=3`

Doesn't intercept y axis

Next point could be when y=1. (Let's do an equation this time)
`y=log_axquad=>quad1=log_2(x-2)quad`
`a^y=xquad=>quad2^1=x-2`

`y=1quad=>quadx=4`