Chapter 3.7, Problem 2CFU

To determine

To draw: graph to illustrate the given asymptote.

Answer to Problem 2CFU

  

The vertical and horizontal asymptote in the above graph is $x=0\text{ and }y=0$ .

  

The vertical and horizontal asymptote in the above graph is $x=-1\text{ and }y=3$

  

The vertical and slant asymptote in the above graph is $x=-1\text{ and}$

  $y=3x-3$ .

Explanation of Solution

Given information:

Types of asymptotes to be illustrated are as follows:

Vertical

Horizontal

Slant

Calculation:

The types of asymptotes are discussed below.

Case 1: $f\left(x\right)=\frac{3}{x}$

The graph for $f\left(x\right)$ is as follows:

  

In this case, vertical asymptote is value of $x$ which makes the denominator of the function zero. Example: Vertical asymptote is $x=0$ .

Since the denominator of the function is of higher degree than the numerator, then the horizontal asymptote will be x-axis. Horizontal asymptote is $y=0$ .

Case 2: $f\left(x\right)=\frac{3x}{x+1}$

The graph for $f\left(x\right)$ is as follows:

  

Since the numerator and denominator of the function are of same degree, then the horizontal asymptote will fraction of leading coefficient. For this case, it is $y=\frac{3}{1}$ .

It is clear shown in the graph that the vertical asymptotes for this case is $x=-1$ .

Case 3: $f\left(x\right)=\frac{3x^2}{x+1}$

The graph for $f\left(x\right)$ is as follows:

  

It is clear shown in the graph that the vertical asymptotes for this case is $x=-1$ .

Since the numerator is exactly one degree higher than that of the denominator, then slant asymptote is present. Divide the numerator by the denominator to get the quotient as slant asymptote.

  $x+1\begin{array}{c}\hfill 3x-3\\ \hfill \overline{)\begin{array}{l}3x^2+0x+0\\ \underset{\_}{-\left(3x^2+3x\right)}\downarrow \\ \underset{\_}{\begin{array}{l}-3x+0\\ -\left(-3x-3\right)\end{array}}\\ 3\end{array}}\end{array}$

Thus, the slant asymptote is $y=3x-3$ .