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

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

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

Answer

Explanation:

Step1: 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 $f(x)=\frac{(x + 2)(x - 2)}{x(x - 2)(x + 1)}$, $x\neq0,2,- 1$.

Step2: Simplify the function

Cancel out the common - factor $(x - 2)$ (for $x\neq2$), we get $y=\frac{x + 2}{x(x + 1)}$, $x\neq0,-1$.

Step3: Find the holes and asymptotes

A hole occurs when a factor can be canceled out in the numerator and denominator. When $x = 2$, substituting $x = 2$ into the simplified function $y=\frac{x + 2}{x(x + 1)}$, we have $y=\frac{2+2}{2\times(2 + 1)}=\frac{4}{6}=\frac{2}{3}$. So there is a hole at $(2,\frac{2}{3})$. Vertical asymptotes occur when the denominator of the simplified function is zero. Set $x(x + 1)=0$, we get $x=0$ and $x=-1$.

Answer:

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