Chapter 2.2, Problem 30E

(a)

To determine

To find: The vertical asymptotes of the graph of $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ .

Answer to Problem 30E

The vertical asymptotes of the graph of $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ are $x=-\frac{1}{2}$ and $x=3$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ .

Calculation:

To graph a function $y=\frac{1-x}{2x^2-5x-3}$ , 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 $y=\frac{1-x}{2x^2-5x-3}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{1}\boxed{-}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{2}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{5}\boxed{\text{X}}\boxed{-}\boxed{3}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(1-\text{X}\right)/\left(\text{2X}^2-5\text{X}-3\right)$

Press the window key and adjust the window to $\left[-2,4\right]$ by $\left[-2,2\right]$ .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 $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ , the vertical asymptotes to the graph are $x=-\frac{1}{2}$ and $x=3$ .

Therefore, the vertical asymptotes of the graph of $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ are $x=-\frac{1}{2}$ and $x=3$ .

(b)

To determine

To find:The behavior of the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ to the right and left of the vertical asymptotes.

Answer to Problem 30E

The value of $\underset{x\to -{\frac{1}{2}}^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -{\frac{1}{2}}^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively and the value of $\underset{x\to 3^{-}}{\lim }f\left(x\right)$ , $\underset{x\to 3^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively.

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ .

Calculation:

First take the vertical asymptote $x=-\frac{1}{2}$ of function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ .

As observed from the graph of function drawn in part (a), the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ approaches to $\infty$ as $x$ approaches to $-\frac{1}{2}$ from the left side. So, the value of $\underset{x\to -{\frac{1}{2}}^{-}}{\lim }f\left(x\right)$ is $\infty$ .

As observed from the graph of functiondrawn in part (a), the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ approaches to $-\infty$ as $x$ approaches to $-\frac{1}{2}$ from the right side. So, the value of $\underset{x\to -{\frac{1}{2}}^{+}}{\lim }f\left(x\right)$ is $-\infty$ .

Now take the vertical asymptote $x=3$ of the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ .

As observed from the graph of functiondrawn in part (a), the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ approaches to $\infty$ as $x$ approaches to $3$ from the left side. So, the value of $\underset{x\to 3^{-}}{\lim }f\left(x\right)$ is $\infty$ .

As observed from the graph of functiondrawn in part (a), the function $f\left(x\right)=\frac{1-x}{2x^2-5x-3}$ approaches to $-\infty$ as $x$ approaches to $3$ from the right side. So, the value of $\underset{x\to 3^{+}}{\lim }f\left(x\right)$ is $-\infty$ .

Therefore, the value of $\underset{x\to -{\frac{1}{2}}^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -{\frac{1}{2}}^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively and the value of $\underset{x\to 3^{-}}{\lim }f\left(x\right)$ , $\underset{x\to 3^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively.