Which statement is true about the discontinuities of the function f(x)?

f(x)= x^2-4/x^3-x^2-2x

A. There is a hole at x = 2.

B. There are asymptotes at x = 0 and x = –1.

C. There are asymptotes at x = 0 and x = –1 and a hole at (2,2/3).

D. There are holes at x = 0 and x = –1 and an asymptote at x = 2.

ANSWER IS C. There are asymptotes at x = 0 and x = –1 and a hole at (2,2/3).

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

if i did my math right then the answer should but 108

-11/5> -11/3