Chapter 2.5, Problem 40E

To determine

To find: the horizontal and vertical asymptotes of the given curve. And check the curve by using the graph.

Answer to Problem 40E

  $\frac{1}{2}\text{ and }2,\frac{-1}{2}$

Explanation of Solution

Given:

The givencurve is $y=\frac{x^2+1}{2x^2-3x-2}$ .

Calculation:

The given function is expressed as $y=\frac{x^2+1}{2x^2-3x-2}=\frac{x^2+1}{\left(2x+1\right)\left(x-2\right)}$ .

Now horizontal asymptotes:

  $\underset{x\to \infty }{\lim }\frac{x^2+1}{2x^2-3x-2}=\underset{x\to \infty }{\lim }\frac{1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}{2-\frac{3}{x}-\frac{2}{x^2}}=\frac{1}{2}$

and vertical asymptote:

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

Hence, the horizontal and vertical asymptotes of the given curve is $\frac{1}{2}\text{ and }2,\frac{-1}{2}$ .