Chapter 14, Problem 13CT

To determine

Whether an asymptote or a hole appears or not.

Answer to Problem 13CT

The function has a hole in $x=-7\text{ and }x=2$ .

Explanation of Solution

Given:

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

Calculation:

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

The domain of the function is,

  $\left(-\infty ,-7\right)\cup \left(-7,2\right)\cup \left(2,\infty \right)$

Thus, the function is not defined in,

  $x=-7\text{ and }x=2$ .

Therefore, factorise the numerator also and simplify the function if possible,

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

Therefore,

The function has a hole in $x=-7\text{ and }x=2$ .

Hence,the function has a hole in $x=-7\text{ and }x=2$ .