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HELP QUICK. PRE-CALC NATURAL LOG?

Please help i can't figure out this information.
2.Use the function f(x)=ln(x-1)+2 to answer the following:

a.Find the domain of f(x). Write your answer using interval notation.

b.Find the range of f(x). Write your answer using interval notation.

c.Find the equation of any asymptote(s) of f(x).

d.Find the coordinates of the x- and y- intercept(s) of f(x), if any.

Please help as much as u can. Im so confused. THanks!

Remember that you can take the natural log of any number that is larger than zero. This means that the function ln(x) is defined for all numbers greater than zero. This set of numbers is the DOMAIN of ln(x).

In this function, you have ln(x-1). You are taking the natural log of the quantity (x-1). This means, than your function will be defined for all numbers where (x-1) is greater than zero. Therefore, since x-1 is greater than zero when x>1, your domain is all real numbers greater than 1.
In interval notation, this is (1, infinity). All numbers between 1 and infinity.

The range is easier.
At the asymptote, the natural log function goes all the way to negative infinity. And as x gets bigger and bigger and bigger, ln(x), (or ln(x-1) for that matter) goes to positive infinity. Therefore, the range of your function extends to all real numbers.
In interval notation, this is (-infinity, infinity). This means all real numbers between negative infinity, and positive infinity.

To find any asymptotes, remember that the function ln(x) has a vertical asymptote when x=0. Here, you have ln(x-1). Since x-1 = 0, when x=1, your asymptote is x=1.

To find the x-intercept, ask yourself, when does f(x) = 0?
Solve the equation f(x) = ln(x-1) + 2 = 0.
ln(x-1) = -2
e^-2 = x-1
x= e^-2 + 1 = about 1.135.
Therefore, you x intercept is (1.135, 0).

To find your y-intercept, plug in x=0. You will find that your function is undefined! This is because ln(0-1) = ln(-1) is undefined.
Therefore, there are no y-intercepts.

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