Question

find the discontinuities of the function.

$f(x)=\frac{x^{2}+12 x+27}{x^{2}+4 x+3}$

there is a removable discontinuity at

Explanation:

Step1: Factor numerator and denominator

$$\begin{align*} x^{2}+12x + 27&=(x + 3)(x+9)\\ x^{2}+4x + 3&=(x + 1)(x + 3) \end{align*}$$

So, \(f(x)=\frac{(x + 3)(x + 9)}{(x + 1)(x + 3)}\)

Step2: Simplify the function (for \(x

eq - 3\))
Cancel out the common factor \((x + 3)\) (when \(x
eq-3\)), we get \(y=\frac{x + 9}{x + 1}\)

Step3: Find the removable - discontinuity

A removable discontinuity occurs when a factor cancels out. Set the canceled - out factor equal to zero. \(x+3 = 0\) gives \(x=-3\)
Substitute \(x =-3\) into the simplified function \(y=\frac{x + 9}{x + 1}\) (we use the simplified function because the original function is undefined at \(x=-3\) in its original form, but we can find the limit value which represents the \(y\) - coordinate of the removable discontinuity).
\(y=\frac{-3 + 9}{-3+1}=\frac{6}{-2}=-3\)

Answer:

\((-3,-3)\)