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

which statement is true about the discontinuities of the function f(x)? f(x) = (x^2 - 4)/(x^3 - x^2 - 2x) there is a hole 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, 2/3). there 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)? f(x) = (x^2 - 4)/(x^3 - x^2 - 2x) there is a hole 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, 2/3). there 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 $f(x)=\frac{x + 2}{x(x + 1)}$, $x\neq0,2,-1$.

Step3: Find the holes and asymptotes

A hole occurs when a factor can be canceled out. Since we canceled out $(x - 2)$, to find the $y$ - coordinate of the hole, substitute $x = 2$ into the simplified function $y=\frac{2+2}{2(2 + 1)}=\frac{4}{6}=\frac{2}{3}$. So there is a hole at $(2,\frac{2}{3})$. Vertical asymptotes occur at the values of $x$ that make the denominator of the simplified function equal to 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 $(2,\frac{2}{3})$.