Chapter 2.6, Problem 58E

Solution

To find: The intercepts, vertical asymptotes, and end behavior, and graph the function $k\left(x\right)=\frac{2x^5+x^2-x+1}{x^2-1}$ .

  $x-$ intercept is at $\left(-1.1,0\right)$ .

  $y-$ intercept is at $\left(0,-1\right)$ .

Vertical asymptotes: $x=1\text{ and }x=-1$

Given Information:

  $k\left(x\right)=\frac{2x^5+x^2-x+1}{x^2-1}$

Concept Used:

• For $x-$ intercepts, the value of $y$ is 0.
• For $y-$ intercepts, the value of $x$ is 0.
• Vertical Asymptotes: When the denominator becomes 0.

Calculation:

For $x-$ intercepts.

Put $k\left(x\right)=0$ and solve for $x$ ,

  $\begin{array}{c}\frac{2x^5+x^2-x+1}{x^2-1}=0\\ 2x^5+x^2-x+1=0\\ x\approx -1.1\end{array}$

Thus, $x-$ intercept is at $\left(-1.1,0\right)$ .

Put $y-$ intercepts.

Put $x=0$ and solve for $y$ ,

  $\begin{array}{l}y=\frac{2{\left(0\right)}^5+{\left(0\right)}^2-\left(0\right)+1}{{\left(0\right)}^2-1}\\ y=-1\end{array}$

Thus, $y-$ intercept is at $\left(0,-1\right)$ .

Vertical asymptotes: Put denominator to 0.

  $\begin{array}{c}{x}^2-1=0\\ x=-1\\ x=1\end{array}$

The degree of the rational function is 3 and leading coefficient is positive.

As $x$ tends to infinity, $y$ tends to infinity.

As $x$ tends to negative infinity, $y$ tends to negative infinity.

The graph of the function is,