Chapter 2.6, Problem 43E

Solution

The asymptotes and intercepts, describe the behaviour of vertical asymptotes using limits, analyse and graph the function.

The vertical asymptotes, $x=-2$ , slant asymptotes, $y=x$ with intercepts $\left(1,0\right)$ , $\left(-3,0\right)$ and $\left(0,-1.5\right)$ .

Given:

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

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

To find vertical asymptotes, find the zeros of denominator, so $x+2=0\text{ or }x=-2$

Thus, the zeros of denominator of $h\left(x\right)$ is $x=-2$ , Means the vertical asymptotes are $x=-2$ .

To find the behaviour of vertical asymptotes, compute the limit when x approaches to and $x=-2$ from left and right,

  $\begin{array}{c}\underset{x\to -2^{-}}{\lim }h\left(x\right)=\underset{x\to -2^{-}}{\lim }\frac{x^2+2x-3}{x+2}\\ =\infty \end{array}$

  $\begin{array}{c}\underset{x\to -2^{+}}{\lim }h\left(x\right)=\underset{x\to -2^{+}}{\lim }\frac{x^2+2x-3}{x+2}\\ =-\infty \end{array}$

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $h\left(x\right)=\frac{x^2+2x-3}{x+2}$ 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, $n>m$ end behaviour asymptotes, rewrite $h\left(x\right)$ using division.

  

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

This implies, the end behaviour is slant asymptotes $y=x$ .

Now find the intercept, the x -intercept is given by zeros of numerator that are not zero of denominator.

Hence, to solve $x^2+2x-3=0$ , factorize:

  $\begin{array}{c}{x}^2+2x-3=x^2+3x-x-3\\ =x\left(x+3\right)-1\left(x+3\right)\\ =\left(x+3\right)\left(x-1\right)\end{array}$

That gives, $x=1,-3$ , the x -intercept of $h\left(x\right)$ is $\left(1,0\right)$ and $\left(-3,0\right)$ .

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

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

Thus y -intercept is $\left(0,-\frac{3}{2}\right)\text{ or }\left(0,-1.5\right)$ .

Hence, using vertical asymptotes, $x=-2$ , slant asymptotes, $y=x$ with intercepts $\left(1,0\right)$ , $\left(-3,0\right)$ and $\left(0,-1.5\right)$

the graph obtain is as:

  

Interpretations form graph:

1) Domain of $f\left(x\right)$ is $\left(-\infty ,-2\right)\cup \left(-2,\infty \right)$

2) Range is $\left(-\infty ,\infty \right)$ .

3) Continuous everywhere except $x=-2$ .

4) Increasing in $\left(-\infty ,-2\right)\cup \left(-2,\infty \right)$ .

5) No extremes.

6) Not symmetric.

7) Unbounded.

Conclusion:

The vertical asymptotes, $x=-2$ , slant asymptotes, $y=x$ with intercepts $\left(1,0\right)$ , $\left(-3,0\right)$ and $\left(0,-1.5\right)$ .