find the domain range and asymptotes of?

Question

f(x)=2/3x-1

Answer 1

Domain: all x not equal to 1/3
Range: all y not equal to 0

Vertical Asymptote: x=1/3
Horizontal Asymptote: y=0

Answer 2

I assume you mean f(x) = 2 / (3x - 1)

Domain: all the values that you can use for x. Because there is a division, you must avoid dividing by 0. This would happen if x = 1/3.
Therefore, the domain is any value except 1/3.

Range: all the values that the function can reach.

Is there a number that you could put for f(x), for which you cannot find an x that could give you that answer?

f(x) = 2 / (3x - 1)
(3x - 1) = 2 / f(x) (this tells us that f(x) cannot be zero)
3x = [2/f(x)] + 1
x = {[2/f(x)] + 1} / 3

Any value of f(x) (except 0) allows us to find an x. Therefore, the range can be anything except 0.

Lim [2 / (3x - 1)] as x goes to + infinity.
Starting from above x = 1/3, and letting x increase towards + infinity, the value of f(x) is positive and diminishes all the time. It can never become negative.

Lim [2 / (3x - 1)] as x goes to - infinity.
Starting from below x = 1/3, and going towards minus infinity, the value of f(x) is negative and increases (becomes less negative) all the time. It can never become positive.

We have shown above that f(x) can have any value (except zero). We have just shown that the line f(x) is moving towards 0 as you move outwards, but it can never reach it.
Therefore, f(x) is asymptotic to 0.

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Look at what happens at x = 67.
f(67) = 2 / (201 - 1) = 2/200 = 0.1
How many values of x are left as we go up? An infinity of them.
How much space is left for the line to go? Only 0.1 (since it is going down and it cannot cross 0). Therefore, the slope must become flatter and flatter, without ever becoming exactly 0. That smells like an asymptote.