Chapter 4.2, Problem 91E

To determine

Find the asymptotes of the graph of the rational function.

Answer to Problem 91E

  $x=-2$; $y=\frac{1}{2}$

Explanation of Solution

Given: the rational function $f(x)=\frac{2x}{x-3}$

Concept Used:

There are three kinds of asymptotes: horizontal, vertical and oblique.

An asymptote is a line that the graph of a function approaches but never actually touches. There are vertical asymptotes, horizontal asymptotes, and oblique asymptotes. Oblique asymptotes are also called slant asymptotes. Vertical and horizontal asymptotes are vertical and horizontal lines, respectively. An oblique, or slant, asymptote is an asymptote that is neither vertical nor horizontal

  

Calculation:

To find the vertical asymptotes, simply make denominator equal to zero and solve for x.

For Horizontal Asymptotes, if the degree os the numerator is equal to the degree of the denominator, in this case the asymptote is the horizontal line $y=\frac{a}{b}$ , where a is the leading coefficient in the numerator and b is the leading coefficient in the denominator.

  $f(x)=\frac{x^2+3x-10}{2x^2-8}=\frac{(x+5)(x-2)}{2(x^2-4)}=\frac{(x+5)(x-2)}{2(x+2)(x-2)}=\frac{(x+5)}{2(x+2)}$

Vertical Asymptote:   $f(x)=\frac{x^2+3x-10}{2x^2-8}$   $f(x)=\frac{x^2+3x-10}{2x^2-8}=\frac{(x+5)}{2(x+2)}$ Denominator = 0   $x+2=0\Rightarrow x=-2$ Horizontal Asymptotes: For   $f(x)=\frac{x^2+3x-10}{2x^2-8}=\frac{(x+5)}{2(x+2)}$; the leading coefficient in Numerator a = 1 and the leading coefficient in Denominator b = 2 Horizontal Asymptote is the horizontal line   $y=\frac{a}{b}\Rightarrow y=\frac{1}{2}[a=1;b=2]$ Vertical Asymptote:   $x=2$ Horizontal Asymptote:   $y=\frac{1}{2}$

Thus, Vertical Asymptote: $x=-2$ Horizontal Asymptote: $y=\frac{1}{2}$