Chapter 2.6, Problem 37E

Solution

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

The vertical asymptote is $x=\frac{3}{2}$ and $x=-1$ , horizontal asymptotes is $y=0$ with intercept $\left(0,-0.66\right)$ .

Given:

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

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

To find vertical asymptotes, find the zeros of denominator, so factorize the denominator $2x^2-x-3$ :

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

Thus, the zeros of denominator of $f\left(x\right)$ is $x=\frac{3}{2}$ and $x=-1$ , Means the vertical asymptotes are $x=\frac{3}{2}$ and $x=-1$ .

To find the behaviour of vertical asymptotes, compute the limit when x approaches to $\frac{3}{2}$ and $-1$ from left and right,

  $\begin{array}{c}\underset{x\to {\frac{3}{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\frac{3}{2}}^{-}}{\lim }\frac{2}{2x^2-x-3}\\ =-\infty \text{ since, }2x^2-x-3<0\text{ as }x\to \frac{3}{2}\text{ from left}\end{array}$

  $\begin{array}{c}\underset{x\to {\frac{3}{2}}^{+}}{\lim }f\left(x\right)=\underset{x\to {\frac{3}{2}}^{+}}{\lim }\frac{2}{2x^2-x-3}\\ =\infty \text{ since, }2x^2-x-3>0\text{ as }x\to \frac{3}{2}\text{ from right}\end{array}$

  $\begin{array}{c}\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{-}}{\lim }\frac{2}{2x^2-x-3}\\ =\infty \text{ since, }2x^2-x-3>0\text{ as }x\to -1\text{ from left}\end{array}$

  $\begin{array}{c}\underset{x\to -1^{+}}{\lim }f\left(x\right)=\underset{x\to -1^{+}}{\lim }\frac{2}{2x^2-x-3}\\ =-\infty \text{ since, }2x^2-x-3<0\text{ as }x\to -1\text{ from right}\end{array}$

To find end behaviour asymptotes, find $m\text{ and }n$ compare, $f\left(x\right)=\frac{2}{2x^2-x-3}$ 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 horizontal asymptotes, $y=0$ .

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

Since, the numerator is constant, no x -intercept is present.

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

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

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

Hence, using vertical asymptotes, $x=\frac{3}{2}$ and $x=-1$ , horizontal asymptotes, $y=0$ with intercept is $\left(0,-\frac{2}{3}\right)$ or $\left(0,-0.66\right)$ the graph obtain is as:

  

Interpretations form graph:

1) Domain of $f\left(x\right)$ is $\left(-\infty ,-1\right)\cup \left(-1,3/2\right)\cup \left(3/2,\infty \right)$

2) Range is $\left(-\infty ,-0.6\right]\cup \left(0,\infty \right)$ .

3) Continuous everywhere except $x=-1,3/2$ .

4) Increasing $\left(-\infty ,-1\right)\cup \left(-1,0.3\right)$ and decreasing in $\left(0.3,1.5\right)\cup \left(1.5,\infty \right)$ .

5) Local maxima $\left(0.3,-0.6\right)$ .

6) Not symmetric.

7) Unbounded.

Conclusion:

The vertical asymptote is $x=\frac{3}{2}$ and $x=-1$ , horizontal asymptotes is $y=0$ with intercept $\left(0,-0.66\right)$ .