Chapter 3, Problem 12T

(a)

To determine

To find: The function that has a horizontal asymptote out of the given functions.

Answer to Problem 12T

The function that has a horizontal asymptote are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Explanation of Solution

Given information:

The given functions are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ , $s\left(x\right)=\frac{x^3+27}{x^2+4}$ , $t\left(x\right)=\frac{x^3-9x}{x+2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Calculation:

Check the functions that has a finite limit as $x$ tends to $\pm \infty$ .

Check the limit.

  $\begin{array}{c}\underset{x\to \pm \infty }{\lim }r\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{2x-1}{x^2-x-2}\\ =\underset{x\to \pm \infty }{\lim }\frac{x\left(2-\frac{1}{x}\right)}{x^2\left(1-\frac{1}{x}-\frac{2}{x^2}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{\left(2-\frac{1}{x}\right)}{x\left(1-\frac{1}{x}-\frac{2}{x^2}\right)}\\ =0\end{array}$

So, the horizontal asymptote of the function $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ is equal to $y=0$ i.e. $x$ -axis.

Check the limit.

  $\begin{array}{c}\underset{x\to \pm \infty }{\lim }s\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{x^3+27}{x^2+4}\\ =\underset{x\to \pm \infty }{\lim }\frac{x^3\left(1+\frac{27}{x^3}\right)}{x^2\left(1+\frac{4}{x^2}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{x\left(1+\frac{27}{x^3}\right)}{\left(1+\frac{4}{x^2}\right)}\\ =\pm \infty \end{array}$

So, there is no horizontal asymptote of the function $s\left(x\right)=\frac{x^3+27}{x^2+4}$ .

Check the limit.

  $\begin{array}{c}\underset{x\to \pm \infty }{\lim }t\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{x^3-9x}{x+2}\\ =\underset{x\to \pm \infty }{\lim }\frac{x^3\left(1-\frac{9}{x^2}\right)}{x\left(1+\frac{2}{x}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{x^2\left(1-\frac{9}{x^2}\right)}{\left(1+\frac{2}{x}\right)}\\ =\infty \end{array}$

So, there is no horizontal asymptote of the function $t\left(x\right)=\frac{x^3-9x}{x+2}$ .

Check the limit.

  $\begin{array}{c}\underset{x\to \pm \infty }{\lim }u\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{x^2+x-6}{x^2-25}\\ =\underset{x\to \pm \infty }{\lim }\frac{x^2\left(1+\frac{1}{x}-\frac{6}{x^2}\right)}{x^2\left(1-\frac{25}{x^2}\right)}\\ =\frac{\left(1+0-0\right)}{\left(1-0\right)}\\ =1\end{array}$

So, the horizontal asymptote of the function $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ is equal to $y=1$ .

Therefore, the function that has a horizontal asymptote are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

(b)

To determine

To find: The function that has a slant asymptote.

Answer to Problem 12T

The function that has a slant asymptote is $s\left(x\right)=\frac{x^3+27}{x^2+4}$ .

Explanation of Solution

Given information:

The given functions are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ , $s\left(x\right)=\frac{x^3+27}{x^2+4}$ , $t\left(x\right)=\frac{x^3-9x}{x+2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Calculation:

For the function to have a slant asymptote the degree of the numerator must one greater than the degree of the denominator.

Out of the given functions only $s\left(x\right)=\frac{x^3+27}{x^2+4}$ satisfies this property.

Therefore, the function that has a slant asymptote is $s\left(x\right)=\frac{x^3+27}{x^2+4}$ .

(c)

To determine

To find: The function that has no vertical asymptote.

Answer to Problem 12T

The function that has no vertical asymptote is $s\left(x\right)=\frac{x^3+27}{x^2+4}$ .

Explanation of Solution

Given information:

The given functions are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ , $s\left(x\right)=\frac{x^3+27}{x^2+4}$ , $t\left(x\right)=\frac{x^3-9x}{x+2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Calculation:

For the function to have a vertical asymptote the denominator of the function must not be equal to $0$ for any value of $x$ .

Out of the given functions only $s\left(x\right)=\frac{x^3+27}{x^2+4}$ satisfies this property as $x^2+4$ is always positive rest all the function have denominator equal to zero for some value of $x$ .

Therefore, the function that has no vertical asymptote is $s\left(x\right)=\frac{x^3+27}{x^2+4}$ .

(d)

To determine

To graph: The function $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ with horizontal and vertical asymptotes.

Explanation of Solution

Given information:

The given functions are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ , $s\left(x\right)=\frac{x^3+27}{x^2+4}$ , $t\left(x\right)=\frac{x^3-9x}{x+2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Graph:

As the function $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ has two vertical asymptote equal to $x=5$ and $x=-5$ . Also it has one horizontal asymptote $y=1$ .

Sketch the graph of the function with horizontal and vertical asymptotes.

  

Figure(1)

(e)

To determine

To find: The polynomial $P$ that has a same end behavior as $t\left(x\right)=\frac{x^3-9x}{x+2}$ and also graph both in same screen.

Answer to Problem 12T

The polynomial $P$ that has a same end behavior as $t\left(x\right)=\frac{x^3-9x}{x+2}$ is equal to $x^2-2x-5$ . The graph of both the function and polynomial is shown in Figure(2).

Explanation of Solution

Given information:

The given functions are $r\left(x\right)=\frac{2x-1}{x^2-x-2}$ , $s\left(x\right)=\frac{x^3+27}{x^2+4}$ , $t\left(x\right)=\frac{x^3-9x}{x+2}$ and $u\left(x\right)=\frac{x^2+x-6}{x^2-25}$ .

Calculation:

Simplify the function with division algorithm.

  $x+2\begin{array}{c}\hfill x^2-2x-5\\ \hfill \overline{)\begin{array}{c}{x}^3-9x\\ \underset{\_}{x^3+2x^2}\\ -2x^2-9x\\ \underset{\_}{-2x^2-4x}\\ -5x\\ \underset{\_}{-5x-10}\\ 10\end{array}}\end{array}$

So, by the division the function $t\left(x\right)=\frac{x^3-9x}{x+2}$ is written as $x^2-2x-5+\frac{10}{x+2}$ .

Therefore, the polynomial $P$ that has a same end behavior as $t\left(x\right)=\frac{x^3-9x}{x+2}$ is equal to $x^2-2x-5$ .

Sketch the graph of the function $t\left(x\right)=\frac{x^3-9x}{x+2}$ and polynomial $y=x^2-2x-5$ in the same window:

  

Figure(2)