Question

which statement is true about the discontinuities of the function $f(x)$?
$f(x) = \frac{x^2 - 4}{x^3 - x^2 - 2x}$
there is a hole at $x = 2$.
there are holes at $x = 0$ and $x = -1$ and an asymptote at $x = 2$.
there are asymptotes at $x = 0$ and $x = -1$.
there are asymptotes at $x = 0$ and $x = -1$ and a hole at $(2, \frac{2}{3})$.

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^2 - 4\) as a difference of squares: \(x^2 - 4=(x - 2)(x + 2)\).
Then, factor the denominator \(x^3 - x^2 - 2x\):
Factor out an \(x\) first: \(x(x^2 - x - 2)\).
Then factor the quadratic: \(x^2 - x - 2=(x - 2)(x + 1)\).
So the denominator becomes \(x(x - 2)(x + 1)\).

Now, the function is \(f(x)=\frac{(x - 2)(x + 2)}{x(x - 2)(x + 1)}\).

Step2: Identify holes and asymptotes

A hole occurs when a factor cancels out from numerator and denominator. Here, \((x - 2)\) cancels out (for \(x
eq2\), since \(x = 2\) makes the original denominator zero). To find the \(y\)-coordinate of the hole, substitute \(x = 2\) into the simplified function (after canceling \((x - 2)\)):
Simplified function: \(f(x)=\frac{x + 2}{x(x + 1)}\) (for \(x
eq2,0,-1\)).
Substitute \(x = 2\): \(\frac{2 + 2}{2(2 + 1)}=\frac{4}{6}=\frac{2}{3}\). So there is a hole at \((2,\frac{2}{3})\).

Vertical asymptotes occur where the denominator is zero and the factor does not cancel. The remaining denominator factors are \(x\) and \(x + 1\), so vertical asymptotes at \(x = 0\) and \(x=-1\).

Answer:

There are asymptotes at \(x = 0\) and \(x=-1\) and a hole at \((2,\frac{2}{3})\) (the last option in the bottom left: "There are asymptotes at \(x = 0\) and \(x=-1\) and a hole at \((2,\frac{2}{3})\)").