Which statement is true about the discontinuities of the function f(x)?
f(x)=(x^2-4)/(x^3-x^2-2x)
a. there is a hole at x = 2.
b. there are asymptotes at x = 0 and x = –1.
c. there are asymptotes at x = 0 and x = –1 and a hole at (2, 2/3)
d. there are holes at x = 0 and x = –1 and an asymptote at x = 2.

Here we need to see which statements are true about the discontinuities of the given function.

The correct option is b: "There are asymptotes at x = 0 and x = –1."

So a rational function has discontinuities when its denominator becomes equal to zero, this is because we can't divide by zero.

Here the function is:

First, we should factorize both numerator and denominator, so we can see if some of the factors appear in both of them and we can remove it.

For the numerator we have:

For the denominator we have:

Now we need to find the factorization of the thing inside the parenthesis.

The roots are given by Bhaskara's formula:

So the roots are:

x = (1 + 3)/2 = 2

x = (1 - 3)/2 = -1

And we can rewrite:

Then the function can be rewritten as:

Then the discontinuities are at the zeros of the denominator, which are at x = 0 and x = -1.

Because the denominator goes to zero at these points, we will have vertical asymptotes, thus the correct option is b

"There are asymptotes at x = 0 and x = –1."

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