Chapter 2.6, Problem 9E

Solution

To Describe:

The series of transformations that will change the reciprocal function $g\left(x\right)=\frac{1}{x}$ into the given function $f\left(x\right)$ .

To identify and describe behavior using limits at:

The horizontal and vertical asymptotes.

To Graph:

The function $f\left(x\right)$ .

  $f\left(x\right)=13g\left(x+4\right)-2$ . The graph of $f$ is the graph of the reciprocal function shifted $4$ units to the left, then scaled vertically by a factor of $13$ , and finally translated $2$ units down.

The vertical asymptote is $x=-4$ and the horizontal asymptote is $y=-2$ . $\underset{x\to -4^{-}}{\lim }f\left(x\right)\to -\infty$ , $\underset{x\to -4^{+}}{\lim }f\left(x\right)\to \infty$ , and $\underset{x\to \pm \infty }{\lim }f\left(x\right)=-2$ describe the behavior at these asymptotes.

  

Given:

  $f\left(x\right)=\frac{5-2x}{x+4}$ .

Concepts Used:

The reciprocal function $g\left(x\right)=\frac{1}{x}$ has horizontal asymptote at $y=0$ and vertical asymptote at $x=0$ . Also, $\underset{x\to \pm \infty }{\lim }g\left(x\right)=0$ , and $\underset{x\to 0^{\pm }}{\lim }g\left(x\right)\to \pm \infty$ .

If the reciprocal function $g\left(x\right)$ is modified to the form $b+kg\left(x-a\right)$ where $a\text{, }b\text{, and }k$ are constants, then the vertical asymptote shifts to $x=a$ and the horizontal asymptote is shifted to $y=b$ . Also, $\underset{x\to \pm \infty }{\lim }b+kg\left(x-a\right)=b$ , and $\underset{x\to a^{\pm }}{\lim }b+kg\left(x-a\right)\to k\times \pm \infty$ .

The behavior of a function $f$ at a vertical asymptote $x=c$ can be determined using the one sided limits at the point $x=c$ . The behavior at the horizontal asymptote is same as the end behavior of a function. The end behavior of a function $f\left(x\right)$ is indicated by the values of the limits $\underset{x\to -\infty }{\lim }f\left(x\right)$ and $\underset{x\to \infty }{\lim }f\left(x\right)$ .

Calculations:

Rewrite $f\left(x\right)$ in terms of the reciprocal function $g$ .

  $\begin{array}{c}f\left(x\right)=\frac{5-2x}{x+4}\\ =\frac{5-2\left(x+4-4\right)}{x+4}\\ =\frac{5-2\left(x+4\right)-2\left(-4\right)}{x+4}\\ =\frac{5-2\left(x+4\right)+8}{x+4}\\ =\frac{13-2\left(x+4\right)}{x+4}\\ =\frac{13}{\left(x+4\right)}-\frac{2\left(x+4\right)}{\left(x+4\right)}\\ =\frac{13}{\left(x+4\right)}-2\\ =13g\left(x+4\right)-2\end{array}$

Thus, $f\left(x\right)=13g\left(x+4\right)-2$ . The graph of $f$ is the graph of the reciprocal function shifted $4$ units to the left, then scaled vertically by a factor of $13$ , and finally translated $2$ units down.

Compare $f\left(x\right)=13g\left(x+4\right)-2$ with $b+kg\left(x-a\right)$ and obtain the vertical asymptote $x=-4$ and the horizontal asymptote $y=-2$ .

Using limits, describe the behavior at the vertical asymptote $x=-4$ .

  $\begin{array}{c}\underset{x\to -4^{-}}{\lim }f\left(x\right)=\underset{x\to -4^{-}}{\lim }13g\left(x+4\right)-2\\ =13\underset{x\to -4^{-}}{\lim }g\left(x+4\right)-2\\ \to 13\times -\infty -2\\ \to -\infty \end{array}$

  $\begin{array}{c}\underset{x\to -4^{+}}{\lim }f\left(x\right)=\underset{x\to -4^{+}}{\lim }13g\left(x+4\right)-2\\ =13\underset{x\to -4^{+}}{\lim }g\left(x+4\right)-2\\ \to 13\times \infty -2\\ \to \infty \end{array}$

Thus, $\underset{x\to -4^{-}}{\lim }f\left(x\right)\to -\infty$ and $\underset{x\to -4^{+}}{\lim }f\left(x\right)\to \infty$ .

Using limits, describe the behavior at the horizontal asymptote $y=-2$ .

  $\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\left(13g\left(x+4\right)-2\right)\\ =13\underset{x\to \pm \infty }{\lim }g\left(x+4\right)-2\\ =13\times 0-2\\ =-2\end{array}$

Thus, $\underset{x\to \pm \infty }{\lim }f\left(x\right)=-2$ .

Graph:

  

Conclusion:

  $f\left(x\right)=13g\left(x+4\right)-2$ . The graph of $f$ is the graph of the reciprocal function shifted $4$ units to the left, then scaled vertically by a factor of $13$ , and finally translated $2$ units down.

The vertical asymptote is $x=-4$ and the horizontal asymptote is $y=-2$ . $\underset{x\to -4^{-}}{\lim }f\left(x\right)\to -\infty$ , $\underset{x\to -4^{+}}{\lim }f\left(x\right)\to \infty$ , and $\underset{x\to \pm \infty }{\lim }f\left(x\right)=-2$ describe the behavior at these asymptotes.

The graph of $f$ is shown above.