Chapter 11.4, Problem 9E

To determine

To Identify:

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

Answer to Problem 9E

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

Explanation of Solution

Given Information:

  

  $f\left(x\right)=\frac{x^2}{x^2-1}$

Calculation:

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

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,

  $f\left(x\right)=\frac{x^2}{x^2-1}$

Then,

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

Therefore, the horizontal asymptote is $y=1$ .

Since the both values of $x=\pm 1$ , function is unbounded

Then, the vertical asymptote is $x=\pm 1$

Then, the vertical asymptotes are $x=\pm 1$ and the horizontal asymptotes is $y=1$ and then the function $f\left(x\right)$ has same behaviour as the graph $\left(f\right)$ .

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