Question

question consider the graph of the function $f(x)=\frac{x^{3}-6x^{2}+8x}{x^{2}-2x - 8}$. which is a removable discontinuity for the graph? select all that apply. select all that apply: $x=-4$ $x=-2$ $x = 0$ $x = 2$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{3}-6x^{2}+8x=x(x^{2}-6x + 8)=x(x - 2)(x - 4)$. The denominator $x^{2}-2x - 8=(x - 4)(x+2)$. So $f(x)=\frac{x(x - 2)(x - 4)}{(x - 4)(x + 2)}$, $x
eq4,x
eq - 2$.

Step2: Identify removable discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x - 4)$ cancels out. When $x = 4$, the original function is undefined but $\lim_{x
ightarrow4}f(x)$ exists.

Answer:

No correct options provided in the given list. A removable - discontinuity is at $x = 4$ which is not in the list of $x=-4,x=-2,x = 0,x = 2$.