Chapter 1.2, Problem 62E

Solution

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

Vertical asymptote: $x=-2$ .

Horizontal asymptote: $y=0$ .

Given information: 

The function $h\left(x\right)=\frac{2x-4}{x^2-4}$ .

Calculation:

Graph of

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

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

Putting the value of $x=0$ then $y=1$ .

And $x=2$ , $y=0$

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

And $x=3$ , $y=0.4$

Plot the graph:

  

By inspection of the graph, a vertical asymptote is at $x=-2$ and horizontal asymptote is at $y=0$ .

Algebraically, the vertical asymptote corresponds to the fact that $x=-2$ is

excluded from the domain, as that would cause a division by zero.

Even though $x=2$ is also excluded from the domain of $h$ , it does not produce a vertical asymptote, factor that out of the function (creating an equivalent function without that discontinuity):

  $\begin{array}{c}\frac{2x-4}{x^2-4}=\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\frac{2}{x+2}\end{array}$

To demonstrate the horizontal asymptote. Right now, it's sufficient to say that, as $x$

approaches infinity, the denominator becomes increasingly larger than the numerator, and the quotient of the division approaches zero.

Therefore, horizontal asymptote: $y=0$ and vertical asymptote: $x=-2$ .