Chapter 11.4, Problem 11E

To determine

To Identify:

The function with its graph by using horizonal asymptotes as aids.

Answer to Problem 11E

The given function $\frac{1-2x}{x-2}$ is match with the graph $\left(h\right)$ .

Explanation of Solution

Given Information:

  

  $f\left(x\right)=\frac{1-2x}{x-2}$

Calculation:

The function is $f\left(x\right)=\frac{1-2x}{x-2}$

To get the horizontal asymptote and vertical asymptotes follow below steps:

For the horizontal asymptote,

  $y=\underset{x\to \pm \infty }{\lim }f\left(x\right)$

But,

  $f\left(x\right)=\frac{1-2x}{x-2}$

Then,

  $\begin{array}{l}y=\underset{x\to \pm \infty }{\lim }f\left(x\right)\\ =\underset{x\to \pm \infty }{\lim }\frac{1-2x}{x-2}\\ =\underset{x\to \pm \infty }{\lim }\frac{2x\left(\frac{1}{2x}-1\right)}{x\left(1-\frac{2}{x}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{2\left(\frac{1}{2x}-1\right)}{x\left(1-\frac{2}{x}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{2\left(\frac{1}{2x}-1\right)}{\left(1-\frac{2}{x}\right)}\\ =\frac{2\left(\frac{1}{2\left(\pm \infty \right)}-1\right)}{\left(1-\frac{2}{\left(\pm \infty \right)}\right)}\\ =\frac{2\left(\pm 0-1\right)}{\left(1\pm 0\right)}\\ =2\end{array}$

Therefore, the horizontal asymptote is $y=-2$ .

Since the values of $x=2$ , function is unbounded

Then, the vertical asymptote is $x=2$

Then, the vertical asymptotes are $x=2$ and the horizontal asymptotes is $y=-2$ and then the function $f\left(x\right)$ has same behaviour as the graph $\left(h\right)$ .

Hence, the match of the function with its graph is $\left(h\right)$ .