Chapter 1.2, Problem 59E

Solution

To find: all horizontal and vertical asymptotes of the function.

Vertical asymptote: $x=1,x=-1$

Horizontal asymptote: $y=1$ .

Given information: 

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

Calculation:

Graph of

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

Let $y=f\left(x\right)$ .

Putting the value of $x=0$ then $y=-2$ .

And $x=1$ , $y=\infty$

And $x=-1$ , $y=\infty$

And $x=-2$ , $y=2$

And $x=2$ , $y=2$

Plot the graph:

  

By inspection of the graph a horizontal asymptote is at $y=1$ and vertical asymptotes is at $x=-1$ and $x=1$ .

Algebraically, the vertical asymptotes correspond to the fact that $x=1$ and $x=-1$ are excluded from the domain, as that would cause a division by zero.

To demonstrate the horizontal asymptote. Right now, it's sufficient to say that, $x$ approaches infinity, the $+2$ and $-1$ in the function become irrelevant and the function approaches $\frac{x^2}{x^2}=1$ .

Therefore, vertical asymptote: $x=1,x=-1$ and horizontal asymptote: $y=1$ .