Chapter 2.6, Problem 45E

In Exercises 45−50, find the end behavior asymptote of the given rational function f and graph it together with f in two windows:

a. (a) One showing the details around the vertical asymptote(s) of f.

b. (b) One showing a graph of f that resembles its end behavior asymptote.

45. $f\left(x\right)=\frac{x^2-2x+3}{x-5}$

Solution

The end behaviour asymptotes of function f and graph it in two window a) One showing around vertical asymptote(s). b) One showing a graph of f that resembles the end behaviour asymptote.

The end behaviour asymptotes of $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ is $y=x+3$ , with a vertical asymptote is $x=5$ .

Given:

The function is $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ .

Concept Used:

The x -intercept is given by zeros of numerator that are not zero of denominator. And y -intercept is given by $f\left(0\right)$ .

And,

If a polynomial function in the form $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ , then, vertical asymptotes is given by real zeros of denominator provided it should not be zero of numerator.

And,

The end behaviour asymptote given by $\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\text{ or }\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is horizontal asymptotes.

The condition can be concluded as,

1) If $n<m$ , the end behaviour is horizontal asymptotes that is $y=0$ ,

2) If $n=m$ the end behaviour is horizontal asymptotes is horizontal asymptotes that is $y=\frac{a_n}{b_n}$

3) If $n>m$ , the quotient $q\left(x\right)$ of $f\left(x\right)/g\left(x\right)$ is the end behaviour asymptotes, where $q\left(x\right)$ is given by $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ with $r\left(x\right)$ is remainder of $f\left(x\right)/g\left(x\right)$ .

Calculation:

Consider the function, $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ .

Find vertical asymptotes, so find the zeros of denominator,

  $\begin{array}{c}x-5=0\\ x=5\end{array}$

Since, the zero of denominator is vertical asymptotes, so vertical asymptote is $x=5$ .

Therefore, $\underset{x\to 5^{-}}{\lim }f\left(x\right)=\underset{x\to 5^{+}}{\lim }f\left(x\right)=\infty \text{ or }-\infty$

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ with $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ .

Since, $n>m$ end behaviour asymptotes is given by quotient of $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ . So rewrite the function using long division as follows,

  

Thus, $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ can be rewritten as,

  $\begin{array}{c}f\left(x\right)=\frac{x^2-2x+3}{x-5}\\ =\frac{\left(x+3\right)\left(x-5\right)+18}{x-5}\\ =\left(x+3\right)+\frac{18}{x-5}\end{array}$

Or, $f\left(x\right)=\left(x+3\right)+\frac{18}{x-5}$ .

Hence, the end behaviour asymptotes is, $y=x+3$ .

Now, graph the end behaviour asymptote together with f in two window, a) One showing around vertical asymptote and b) One showing a graph of f that resembles the end behaviour asymptote.

a) The graph of f with end behaviour asymptote, $y=x+3$ around vertical asymptotes, $x=5$ is as follows,

  

And, b) The graph of f that resembles the end behaviour asymptote is as follows,

  

Clearly, the graph of f resembles the end behaviour asymptote.

Conclusion:

The end behaviour asymptotes of $f\left(x\right)=\frac{x^2-2x+3}{x-5}$ is $y=x+3$ , with a vertical asymptote is $x=5$ .