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

2 Answers
Dec 10, 2017

Answer:

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

Dec 10, 2017

Answer:

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#