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

which statement is true about the discontinuities of the function f(x)?\nf(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)?\nf(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 $f(x)=\frac{(x + 2)(x - 2)}{x(x - 2)(x + 1)}$.

Step2: Simplify the function

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

Step3: Analyze holes and asymptotes

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

Answer:

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