find the dicontinuities of the function.\n$f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$.\nthere is a removable…

find the dicontinuities of the function.\n$f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$.\nthere is a removable discontinuity at (-3, -3).\ncomplete\nwhere is the vertical asymptote(s)?\n$x=-3,x=-1$\n$x=-1$\n$x=-9$\n$x=-9,x=-3$\ndone

find the dicontinuities of the function.\n$f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$.\nthere is a removable discontinuity at (-3, -3).\ncomplete\nwhere is the vertical asymptote(s)?\n$x=-3,x=-1$\n$x=-1$\n$x=-9$\n$x=-9,x=-3$\ndone

Answer

Explanation:

Step1: Factor the numerator and denominator

Factor $x^{2}+12x + 27=(x + 3)(x+9)$ and $x^{2}+4x + 3=(x + 3)(x + 1)$. So $f(x)=\frac{(x + 3)(x + 9)}{(x + 3)(x + 1)}$.

Step2: Simplify the function

Cancel out the common factor $(x + 3)$ (for $x\neq - 3$), we get $f(x)=\frac{x + 9}{x + 1},x\neq - 3$.

Step3: Find the vertical - asymptote

A vertical asymptote occurs where the denominator of the simplified function is zero. Set $x+1 = 0$, then $x=-1$.

Answer:

B. $x=-1$