Question

What are the points of discontinuity? Are they all removable?

y = (x-3)/x^2 - 12x + 27

A. x=4 x=9 x=1
B. x= -9 c= -3
C. x= -9 x= -3 x= -1
D. x= 9 x= 3

Answer 1

did you mean to write y=(x-3)/(x^2-12x+27)
or is really y=(x-3)/x^2-12x+27

Answer 2

based on the answers I think you meant to write y=(x-3)/(x^2-12x+27)

to find discontinuities's find when the denominator is 0

to determine if they are removable you will have to "simplify" the fraction
and see if the zero's of the bottom are no longer the zero's of the bottom of the "new fraction"

Answer 3

@myininaya
So does that mean it would be D?

Answer 4

the answers seem incomplete but if the equation was y=(x-3)/(x^2-12x+27) then yes the discontinuities are at x=9 , x=3
i say the answers are incomplete because none of the answers answer the question "are they all removable?"