Chapter 2.6, Problem 54E

To determine

Todescribe:The error in finding vertical and horizontal asymptotes on the graph of given function.

Answer to Problem 54E

The error is that line $y=c$ is considered as a horizontal asymptoteof function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ instead of $y=-c$ .

Explanation of Solution

Given information:

The graph of function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ has a vertical asymptote at $x=c$ and a horizontal asymptote at $y=c$ .

Methodused:

Let $f\left(x\right)=\frac{a_0{x}^m+a_1{x}^{m-1}+\cdots }{b_0{x}^n+b_1{x}^{n-1}+\cdots }$ be a rational function. The line $x=a$ is the vertical asymptote of function $f\left(x\right)$ it $\left(x-a\right)$ is a factor of the denominator. If $m<n$ then x- axis is the horizontal asymptote of $f\left(x\right)$ , and if $m=n$ then $y=\frac{a_0}{b_0}$ is the horizontal asymptote of $f\left(x\right)$ . If $m>n$ there is no horizontal asymptote of $f\left(x\right)$ .

The given function is $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ that may be written as $f\left(x\right)=-\frac{cx}{x-c},c\ne 0$ .

The graph of function $f\left(x\right)=-\frac{cx}{x-c},c\ne 0$ has line $x=c$ as its vertical asymptote because $x=c$ is a factor of denominator.

Thus, the point $x=c$ is a vertical asymptote of the function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ .

The given function is $f\left(x\right)=-\frac{cx}{x-c}=-\frac{c{x}^1}{1\cdot x^1-c}$ , therefore, $y=-\frac{c}{1}=-c$ is horizontal asymptote of the function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ .

Thus, the line $y=-c$ is a horizontal asymptote of the function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ .

The error is that line $y=c$ is given as horizontal asymptote instead of $y=-c$ .

Conclusion:

The graph of function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ has a vertical asymptote at $x=c$ and a horizontal asymptote at $y=-c$ .The error is that line $y=c$ is considered as a horizontal asymptoteof function $f\left(x\right)=\frac{cx}{c-x},c\ne 0$ instead of $y=-c$ .