Chapter 2.2, Problem 29E

(a)

To determine

To find: The vertical asymptotes of the graph of $f\left(x\right)=\frac{x^2-2x}{x+1}$ .

Answer to Problem 29E

The vertical asymptote of the graph of $f\left(x\right)=\frac{x^2-2x}{x+1}$ is $x=-1$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{x^2-2x}{x+1}$ .

Calculation:

To graph a function $y=\frac{x^2-2x}{x+1}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{x^2-2x}{x+1}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{2}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\text{X}^2-2\text{X}\right)/\left(\text{X}+1\right)$

Press the window key and adjust the window to $\left[-6,6\right]$ by $\left[-12,6\right]$ .Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from graph of function $f\left(x\right)=\frac{x^2-2x}{x+1}$ , the vertical asymptote to the graph is $x=-1$ .

Therefore, the vertical asymptote of the graph of $f\left(x\right)=\frac{x^2-2x}{x+1}$ is $x=-1$ .

(b)

To determine

To find:The behavior of the function $f\left(x\right)=\frac{x^2-2x}{x+1}$ to the right and left of the vertical asymptotes.

Answer to Problem 29E

The value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively.

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{x^2-2x}{x+1}$ .

Calculation:

As observed from the graph of function in part (a), the function $f\left(x\right)=\frac{x^2-2x}{x+1}$ approaches to $-\infty$ as $x$ approaches to $-1$ from the left side.So, the value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ is $-\infty$ .

As observed from the graph of function, the function $f\left(x\right)=\frac{x^2-2x}{x+1}$ approaches to $\infty$ as $x$ approaches to $-1$ from the right side.So, the value of $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ is $\infty$ .

Therefore, the value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively.