consider the function f(x) = 1/x - 1 for x not=0 and f(x) = 0 for x = 0.

decide whether f(x) has

a)one symptote and two discontinuities
b)two asymptotes and two discontinuities
c)two asymptotes and one discontinuity

I'm not sure if you mean 1/(x-1) or (1/x) - 1 here, but I'll assume you mean the second one.
You should be able to see that 1/x has one discontinuity and two asymptote, the two axis. When we subtract 1 from it to get 1/x -1, all we've done is shift the function down 1 unit, but it still has similar behavior.
Defining the function to be 0 at x-0 does not make f(x) continuous there; 0 is not a removable discontinuity for 1/x.

To determine the number of asymptotes and discontinuities of the function f(x) = 1/x - 1, we need to analyze its behavior.

First, let's consider the function 1/x. This function has one discontinuity at x = 0, as division by zero is undefined. It also has two asymptotes: a vertical asymptote at x = 0 (since the function approaches positive or negative infinity as x approaches 0 from the left or right, respectively) and a horizontal asymptote at y = 0 (since the function approaches 0 as x approaches positive or negative infinity).

Now, when we subtract 1 from 1/x, we are simply shifting the graph down by 1 unit. So the resulting function f(x) = 1/x - 1 retains the same number of asymptotes and discontinuities as 1/x.

Therefore, the correct answer is b) two asymptotes and two discontinuities.