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:
[tex]f(x) = \frac{x^2 - 4}{x^3 - x^2 - 2x}[/tex]
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:
[tex]x^2 - 4 = x^2 - 2^2 = (x + 2)*(x - 2)[/tex]
For the denominator we have:
[tex]x^3 - x^2 - 2x = x*(x^2 - x - 2)[/tex]
Now we need to find the factorization of the thing inside the parenthesis.
The roots are given by Bhaskara's formula:
[tex]x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*1*(-2)} }{2*1} \\\\x = \frac{1 \pm 3 }{2}[/tex]
So the roots are:
x = (1 + 3)/2 = 2
x = (1 - 3)/2 = -1
And we can rewrite:
[tex]x^3 - x^2 - 2x = x*(x^2 - x - 2) = x*(x + 1)*(x - 2)[/tex]
Then the function can be rewritten as:
[tex]f(x) = \frac{x^2 - 4}{x^3 - x^2 - 2x} = \frac{(x + 2)*(x - 2)}{x*(x + 1)*(x - 2)} = \frac{x + 2}{x*(x + 1)}[/tex]
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."
If you want to learn more, you can read:
https://brainly.com/question/17317969
