Chapter 11.4, Problem 12E

To determine

To Identify:

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

Answer to Problem 12E

The given function $\frac{1-4x^2}{x^2-4}$ is match with the graph $\left(e\right)$ .

Explanation of Solution

Given Information:

  

  $f\left(x\right)=\frac{1-4x^2}{x^2-4}$

Calculation:

The function is $f\left(x\right)=\frac{1-4x^2}{x^2-4}$

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,

  $f\left(x\right)=\frac{1-4x^2}{x^2-4}$

Then,

  $\begin{array}{l}y=\underset{x\to \pm \infty }{\lim }\\ =\underset{x\to \pm \infty }{\lim }\frac{1-4x^2}{x^2-4}\\ =\underset{x\to \pm \infty }{\lim }\frac{4x^2\left(\frac{1}{4x^2}-1\right)}{x^2\left(1-\frac{4}{x^2}\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{4\left(\frac{1}{4x^2}-1\right)}{\left(1-\frac{4}{x^2}\right)}\\ =\frac{4\left(\frac{1}{4{\left(\pm \infty \right)}^2}-1\right)}{\left(1-\frac{4}{{\left(\pm \infty \right)}^2}\right)}\\ =\frac{4\left(\pm 0-1\right)}{\left(1\pm 0\right)}\\ =-4\end{array}$

Therefore, the horizontal asymptote is $y=-4$ .

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

Then, the vertical asymptote is $x=\pm 2$

Then, the vertical asymptotes are $x=\pm 2$ and the horizontal asymptotes is $y=-4$ and then the function $f\left(x\right)$ has same behaviour as the graph $\left(e\right)$ .

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