which statement is true about the discontinuities of the function (f(x))? (f(x)=\frac{x^{2}-4}{x^{3}-x^{2}-2x…

Explanation:

Step1: Factor the function

First, factor the numerator and denominator. The numerator $x^{2}-4=(x + 2)(x - 2)$. The denominator $x^{3}-x^{2}-2x=x(x^{2}-x - 2)=x(x - 2)(x+1)$. So, $f(x)=\frac{(x + 2)(x - 2)}{x(x - 2)(x + 1)}$.

Step2: Identify holes and asymptotes

A hole occurs when a factor can be canceled from both the numerator and denominator. Here, the factor $(x - 2)$ can be canceled for $x\neq2$. When $x = 2$, we have a hole. The vertical asymptotes occur at the values of $x$ that make the denominator equal to zero after canceling common - factors. The denominator of the simplified function (after canceling $(x - 2)$) is $x(x + 1)$. Setting $x(x + 1)=0$, we get $x=0$ and $x=-1$ as the vertical asymptotes. To find the $y$ - coordinate of the hole, substitute $x = 2$ into the simplified function $y=\frac{x + 2}{x(x + 1)}$. When $x = 2$, $y=\frac{2+2}{2\times(2 + 1)}=\frac{4}{6}=\frac{2}{3}$.

Answer:

There are asymptotes at $x = 0$ and $x=-1$ and a hole at $(2,\frac{2}{3})$.