Identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither. 5x/ x3 + 5x2 + 6x

Function is discontinuous at

x=0   (Hole)

x=-3  (Vertical asymptote)

x=-2  (Vertical asymptote)

Step-by-step explanation:

Given:

We need to identity the discontinuity of the function. As we know function is discontinuous where it is not defined.

So, The function is discontinuous at hole, asymptote and break point.

For hole, we will cancel like factor from numerator and denominator.

At x=0 we get hole.

For vertical asymptote, we set denominator to 0

x+3=0  and   x+2=0

Vertical asymptote:

x=-3 and x=-2

Function is discontinuous at

x=0   (Hole)

x=-3  (Vertical asymptote)

x=-2  (Vertical asymptote)

The vertical asymptote of a function are the zeroes of the denominator of a rational function

The function is discontinuous at:

The function is given as:

Factor out x, from the denominator

Expand

Factorize

Factor out x + 2

Cancel out like factors

There is no coefficient of x in the numerator.

So, we have:

--- hole

Equate the denominator of to 0

Solve for x

So, the vertical asymptote is:

Read more about vertical asymptote and holes at:

link

idk

step-by-step explanation:

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: