Answer:
Step-by-step explanation:
I'm assuming you meant to type in
[tex]f(x)=\frac{x^2+12x+27}{x^2+4x+3}[/tex] because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to
[tex]f(x)=\frac{(x+3)(x+9)}{(x+1)(x+3)}[/tex] and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.
The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!
