Chapter 14, Problem 13CT

To determine

The values where rational function $R\left(x\right)=\frac{x^3+6x^2-4x-24}{x^2+5x-14}$ is undefined and then find an asymptote or a hole appears at such numbers.

Answer to Problem 13CT

Solution:

1) The function undefined at $x=-7$ and $x=2$.

2) The vertical asymptote is $x=-7$.

3) The graph of $R\left(x\right)$ has hole at point $\left(2,\frac{32}{9}\right)$.

Explanation of Solution

Given information:

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

Explanation:

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

By factoring the numerator,

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

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

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

Now factoring the denominator

$\Rightarrow R\left(x\right)=\frac{\left(x+2\right)\left(x-2\right)\left(x+6\right)}{x^2+7x-2x-14}$

$\Rightarrow R\left(x\right)=\frac{\left(x+2\right)\left(x-2\right)\left(x+6\right)}{x\left(x+7\right)-2\left(x+7\right)}$

$\Rightarrow R\left(x\right)=\frac{\left(x+2\right)\left(x-2\right)\left(x+6\right)}{\left(x-2\right)\left(x+7\right)}$

The given function is rational function.

Rational function is undefined where denominator is zero.

Hence $R\left(x\right)$ is undefined for $\left(x-2\right)\left(x+7\right)=0$.

That is $R\left(x\right)$ is undefined at $x=2$ and, $x=-7$.

The vertical asymptote occurs where denominator of function is zero but numerator is not equal to zero at that point.

$R\left(x\right)=\frac{\left(x+2\right)\left(x-2\right)\left(x+6\right)}{\left(x-2\right)\left(x+7\right)}$

Cancelling common term from numerator and denominator,

$\Rightarrow R\left(x\right)=\frac{\left(x+2\right)\left(x+6\right)}{\left(x+7\right)}$

At $x=-7$ denominator of $R\left(x\right)=\frac{\left(x+2\right)\left(x+6\right)}{\left(x+7\right)}$ is zero but numerator is not zero.

Therefore, vertical asymptote is at $x=-7$.

For $x=2$ the denominator of simplified function $R\left(x\right)=\frac{\left(x+2\right)\left(x+6\right)}{\left(x+7\right)}$ is not equal to zero so to find the hole find the limit as $x$ approaches to $2$

$\underset{x\to 2}{\lim }R\left(x\right)=\underset{x\to 2}{\lim }\frac{\left(x+2\right)\left(x+6\right)}{\left(x+7\right)}$

$=\frac{\underset{x\to 2}{\lim }\left(x+2\right)\left(x+6\right)}{\underset{x\to 2}{\lim }\left(x+7\right)}$

$=\frac{\left(2+2\right)\left(2+6\right)}{\left(2+7\right)}$

$=\frac{\left(4\right)\left(8\right)}{\left(9\right)}$

$=\frac{32}{9}$

Therefore, graph of $R\left(x\right)$ has hole at point $\left(2,\frac{32}{9}\right)$.