Chapter 2.6, Problem 29E

Solution

The asymptotes and intercepts of function and graph the function.

The vertical asymptotes is $x=-2$ ,end behaviour asymptotes is $y=x-4$ and no horizontal asymptotes with only intercept $\left(0,\frac{\text{3}}{2}\right)$ .

Given:

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

Concept Used:

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 $f\left(x\right)/g\left(x\right)$ . Hence, no horizontal asymptotes.

And,

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

From using asymptotes and intercepts the graph can be drawn.

Calculation:

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

To find vertical asymptotes, find the zeros of denominator, $x+2$ :

  $x+2=0\text{ or }x=-2$

Thus, the vertical asymptotes are $x=-2$ .

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{x^2-2x+3}{x+2}$ 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 $n>m$ ,the quotient of $f\left(x\right)=\frac{x^2-2x+3}{x+2}$ is the end behaviour asymptotes,

  $\begin{array}{l}\stackrel{x-4}{x+2\overline{)\begin{array}{l}{x}^2-2x+3\\ -\underset{¯}{\left(x^2+2x\right)}\\ -4x+3\\ -\underset{¯}{\left(-4x-8\right)}\end{array}}}\\ 11\end{array}$

Thus, rewrite, $f\left(x\right)=\frac{x^2-2x+3}{x+2}$ .

  $\begin{array}{c}f\left(x\right)=\frac{x^2-2x+3}{x+2}\\ =\frac{\left(x-4\right)\left(x+2\right)+11}{x+2}\\ =\left(x-4\right)+\frac{11}{x+2}\end{array}$

Hence, the end behaviour asymptote is $y=x-4$ .

Now find the intercept, the x -intercept is given by zeros of numerator that are not zero of denominator. Hence solve $x^2-2x+3$ for x .

  $x=1\pm i\sqrt{2}$

Since the roots of $x^2-2x+3$ is not real, there is no x -intercept.

And y -intercept is given by $f\left(0\right)$ , so from $f\left(x\right)=\frac{x^2-2x+3}{x+2}$ .

  $\begin{array}{c}f\left(0\right)=\frac{0^2-2\left(0\right)+3}{0+2}\\ =\frac{3}{2}\end{array}$

Thus y -intercept is $\left(0,\frac{\text{3}}{2}\right)$ .

Hence, using vertical asymptotes $x=-2$ ,end behaviour asymptotes is $y=x-4$ and no horizontal asymptotes with only intercept $\left(0,\frac{\text{3}}{2}\right)$ the graph obtain is as:

  

Conclusion:

The vertical asymptotes is $x=-2$ ,end behaviour asymptotes is $y=x-4$ and no horizontal asymptotes with only intercept $\left(0,\frac{\text{3}}{2}\right)$ .