Question

Determine where the rational function

$R\left(x\right)=\frac{x^3+6x^2-4x-24}{x^2+5x-14}$

is undefined. Determine whether an asymptote or a hole appears at such number

Short answer

The function is undefined at$c=-7$and has a vertical asymptote at this point and a hole at$x=-2$

Step-by-step solution

Step 1. Values at which function is undefined

We can write the given function as ,

$$\begin{gathered} R\left(x\right)=\frac{x^3+6x^2-4x-24}{x^2+5x-14}=\frac{(x+6)(x+2)(x-2)}{(x-2)(x+7)} \\ x-2=0\Rightarrow x=2 \\ x+7=0\Rightarrow x=-7 \end{gathered}$$

Step 2. Solve the limit atc=2

Now, apply the limits,

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^3+6x^2-4x-24}{x^2+5x-14}=\underset{x\to 2}{\lim }\frac{(x+6)(x+2)(x-2)}{(x-2)(x+7)} \\ =\underset{x\to 2}{\lim }\frac{x^2+8x+12}{x+7} \\ =\frac{{\lim }_{x\to 2}\left(x^2+8x+12\right)}{{\lim }_{x\to 2}(x+7)} \\ =\frac{{\lim }_{x\to 2}{x}^2+{\lim }_{x\to 2}8x+{\lim }_{x\to 2}12}{{\lim }_{x\to 2}x+{\lim }_{x\to 2}7} \\ =\frac{2^2+8\left(2\right)+12}{2+7} \\ =-\frac{32}{8} \\ =-4 \end{gathered}$$

Therefore, at the point$(2,-4)$, there will be a hole.

Step 3. Solve the limit at c=-7

On applying the limits,

$$\begin{gathered} \underset{x\to -7}{\lim }\frac{x^3+6x^2-4x-24}{x^2+5x-14}=\underset{x\to -7}{\lim }\frac{(x+6)(x+2)(x-2)}{(x-2)(x+7)} \\ =\underset{x\to -7}{\lim }\frac{x^2+8x+12}{x+7} \\ =\frac{{\lim }_{x\to -7}\left(x^2+8x+12\right)}{{\lim }_{x\to -7}(x+7)} \\ =\frac{{\lim }_{x\to -7}{x}^2+{\lim }_{x\to -7}8x+{\lim }_{x\to -7}12}{{\lim }_{x\to -7}x+{\lim }_{x\to -7}7} \\ =\frac{(-7{)}^2+8(-7)+12}{(-7)+7} \\ =\frac{5}{0} \end{gathered}$$

Since the limit comes out to be infinity, therefore there will be an asymptote at this value.