Chapter 2, Problem 28RE

(a)

To determine

To find: The vertical asymptotes of the given function.

Answer to Problem 28RE

The vertical asymptote is $x=-2$ and $x=0$ .

Explanation of Solution

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

Calculation:

To graph a function $y=\frac{x-1}{x^2\left(x+2\right)}$ , 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 $y=\frac{x-1}{x^2\left(x+2\right)}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{x^2}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{2}\boxed{)}\boxed{)}$ .

The display will show the equation,

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

Press the window key and adjust the window to $\left[-4,4\right]$ by $\left[-3,3\right]$ . 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)

From the graph it can be observed that the vertical asymptote of the graph of the functions are $x=-2$ and $x=0$ .

Therefore, the vertical asymptote is $x=-2$ and $x=0$ .

(b)

To determine

To describe: The behavior of $f\left(x\right)$ to the left and right of any vertical asymptote.

Answer to Problem 28RE

The behavior of $f\left(x\right)$ for $x=-2$ to the left hand limit is $\infty$ and right hand limit is $-\infty$ , for $x=0$ , both the limits left as well as right hand limit is $-\infty$ .

Explanation of Solution

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

Calculation:

For vertical asymptote $x=-2$ .

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

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

For vertical asymptote $x=0$ .

As observed from the graph of function from part (a), the function $f\left(x\right)=\frac{x-1}{x^2\left(x+2\right)}$ approaches to $\infty$ as $x$ approaches to 0 from the left hand side. So, the value of $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ is $-\infty$ .

As observed from the graph of function from part (a), the function $f\left(x\right)=\frac{x-1}{x^2\left(x+2\right)}$ approaches to $\infty$ as $x$ approaches to 0 from the right hand side. So, the value of $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ is $-\infty$ .

Therefore, the behavior of $f\left(x\right)$ for $x=-2$ to the left hand limit is $\infty$ and right hand limit is $-\infty$ , for $x=0$ , both the limits left as well as right hand limit is $-\infty$ .