Chapter 2, Problem 33RE

(a)

To determine

To find: The power function end behavior modelfor $f\left(x\right)=\frac{2x+1}{x^2-2x+1}$ .

Answer to Problem 33RE

The power function end behavior model for $f\left(x\right)=\frac{2x+1}{x^2-2x+1}$ is $\frac{2}{x}$ .

Explanation of Solution

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

Calculation:

The power function end behavior model of a polynomial function is the highest power term.

Consider the function.

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

The highest power term of the numerator of the function is $2x$ and the highest power term of denominator is $x^2$ . So, the power function end behavior model of the function is:

  $\frac{2x}{x^2}=\frac{2}{x}$

Therefore, the power function end behavior model for $f\left(x\right)=\frac{2x+1}{x^2-2x+1}$ is $\frac{2}{x}$ .

(b)

To determine

To find: The horizontal asymptote of the function $f\left(x\right)=\frac{2x+1}{x^2-2x+1}$ .

Answer to Problem 33RE

The horizontal asymptote is $y=0$ .

Explanation of Solution

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

Calculation:

The horizontal asymptote of the function $y=f\left(x\right)$ is the line $y=b$ if $\underset{x\to \infty }{\lim }f\left(x\right)=b$ or $\underset{x\to -\infty }{\lim }f\left(x\right)=b$ .

The end behavior model for $f\left(x\right)=\frac{2x+1}{x^2-2x+1}$ is $\frac{2}{x}$ . So, apply limit to the end behavior model to find the horizontal asymptote as:

  $\underset{x\to \infty }{\lim }\frac{2}{x}=0$

Therefore, horizontal asymptote is $y=0$ .