Question

In Problems 7-44, follow Steps 1 through 7 on page 228 to analyze the graph of each function.

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

Short answer

The graph is

Step-by-step solution

Given information

Given function$H\left(x\right)=\frac{x^2+4}{x^4-1}$

Factoring the numerator and denominator

Factoring, we get

$$\begin{gathered} H\left(x\right)=\frac{x^2+4}{x^4-1} \\ H\left(x\right)=\frac{x^2+4}{(x^2-1)(x^2+1)} \\ H\left(x\right)=\frac{x^2+4}{(x-1)(x+1)(x^2+1)} \end{gathered}$$

Calculate the domain

Setting denominator to zero, we get

$$\begin{gathered} (x-1)(x+1)(x^2=1)=0 \\ x=1,-1 \end{gathered}$$

The domain is all real numbers except$1,-1$

Calculate the x intercept 

Setting numerator to zero, we get

$$\begin{gathered} x^2+4=0 \\ {x}^2=-4 \end{gathered}$$

There is no $x$intercept

The$y$intercept is$H\left(0\right)=-4$

Calculate the vertical asymptote

Setting denominator to zero, we get

$$\begin{gathered} x^2-1=0 \\ {x}^2=1 \\ x=1,-1 \end{gathered}$$is the vertical asymptote

Calculate the horizontal asymptote

The degree of numerator is less than denominator so the horizontal asymptote is $y=0$

The intersection point is

$$\begin{gathered} H\left(x\right)=\frac{x^2+4}{x^4-1} \\ =0 \\ {x}^2+4=0 \\ {x}^2=-4 \end{gathered}$$

It doesn't intersects$H$

Sketching the graph

The graph is